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UNIT 1
1
E U C L I D I A N S P A C E Թܖ
Unit Structure
1.1 ObMectives
1.2 Introduction
1.3 Inner product and norm on Թ
1.4 Cauchy –SchZar] ineTuality
1.5 Open and Close subset of Թ
1.6 Interior ܣ and boundary ܣ߲ of subset ܣ of Թ
1.7 Operator norm ȁȁܶȁȁ of linear transformation and it’s properties
1.8 Compactness on Թ
1.9 Heine-%orel theorem
1.10 %ol]ano-Weierstrass theorem on Թ
1.11 Continuous function on Թ
1.12 Continuous function and compactness
1.13 Connected subset of Թ
1.14 Lets sum up
1.15 Unit End e[ercise
1.16 Reference
1.1 Objectives
After going through this chapter students Zill be able to under stand:
x Inner product and norm on Թ 1munotes.in
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2A1ALof Թ
x Operator norm ȁȁܶȁȁ of linear transformation
x Compactness on Թ
x Connected subset of Թ
x Continuous function on Թ
1.2 Introduction
This are the generali]e concept Zhich Ze have already study at our graduate level.
So before going to this chapter one should go through the defin ition and basic
e[ample of Vector spaces, Inner product spaces, 1orm linear spa ces and Metric
spaces and also all the above concept over ԹǤ
1.3 Inner product and norm on Թܖ
Euclidian n-space Թܖ is defined as the set of all n-tuples ሺݔଵǡݔଶǡǥǥǡݔ ሻ Zhere
each ݔᇲ௦ሺͳ ݅ ݊ሻ is Must a real number. So, ԹൌԹൈԹൈǥǥǥൈ
Թሺ௧௦ሻൌሼሺ࢞ǡ࢞ǡǥǥǡ࢞ ሻǣ࢞אԹ ǡͳ݅݊ ሽ . An element of Թ is often
called point of Թ, and ԹଵǡԹଶǡԹଷare often called the line, the plane, and the space
respectively.
If ݔdenotes the element of Թ then it looN liNe ݔ ൌ ሺݔ ଵǡݔଶǡǥǥǥǡݔ ሻ Zhere each
ݔis a real number.
The point of Թ is also called a vector in Թ, as Թis a vector space over the real
number, of dimension n Zith operations, ݔ ݕ ൌ ሺݔ ଵݕଵǡݔଶݕଶǡǥǥǤǤǡݔ
ݕሻ and ܽǤݔ ൌ ሺݔܽ ଵǡݔܽଶǡǥǤݔܽ ሻ Zhere ݔǡݕ are element of Թ and ܽis an element
of ԹǤ
In this vector space there is the notation of the length of a v ector ݔǡusually called
the norm ȁȁݔȁȁ of ݔand defined by หȁݔȁหൌඥݔଵଶݔଶଶڮǤݔ ଶǤ I f ݊ൌͳ ǡ then
หȁݔȁหൌȁ ݔȁ mod of ݔ , is the usual absolute value of ݔǤ
1.3.1 Properties of norm on Թǣ-
If ݔǡݕ are element of Թ and ܽis an element of Թ then the folloZing properties of
ȁȁǤȁȁholds,
݅ሻหȁݔȁหͲ ݀݊ܽ ห ȁݔȁหൌͲ ݂݀݊ܽ݅ ൌ Ͳ ሺ
ሻǤ munotes.in
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3Chapter 1: Euclidian Space Rn݅݅ሻหȁݔݕȁหหȁݔȁหȁȁݕȁȁ
݅݅݅ሻหȁܽǤݔȁหൌȁܽȁǤȁȁݔȁȁ
1.4 Cauchy-Schwar] ineTuality on Թ
Statement:- For all ݔǡݕ in Թ ȁσݔݕ
ୀଵȁหȁݔȁหǤหȁݕȁห eTuality holds if and
only if ݔ and ݕ are linearly dependent.
Proof:- If ݔ and ݕ are linearly dependent, then the eTuality clearly holds.
ChecN
If not, then ݕߣെݔ ് Ͳ ሺݎݐܿ݁ݒሻfor all ߣ in Թǡso Ͳ൏หȁݕߣെݔȁหଶൌ
σሺݕߣെݔሻଶൌ
ୀଵ ߣଶσݕଶ
ୀଵെʹߣσݔݕ
ୀଵσݔଶ
ୀଵǤ
Therefore the right side is a Tuadratic eTuation in ߣ Zith no real solution, so its
discriminant must be negative.
Thus Ͷሺσݔݕ
ିଵሻଶെͶσݔ
ୀଵଶǤσݕ
ୀଵଶ൏Ͳ Ǥ
Note:- The Tuantity σݔݕ
ୀଵis called inner product of ݔand ݕ and denoted as ൏
ݔǡݕ Ǥ
1. Open and Closed subsets of Թ
The close interval ሾܽǡܾሿ has a natural analogue in ԹǤ The close rectangle
ሾܽǡܾሿൈሾܿǡ݀ሿǡ defined as the collection of all pairs ሺݔǡݕሻ Zith אݔሾܽǡܾሿאݕ݀݊ܽ
ሾܿǡ݀ሿ. More generally, if ؿܣԹ and ؿܤԹǡ then ܣൈؿܤԹା is defined as
set of all ሺݔǡݕሻאԹା Zith ܤאݕ݀݊ܽܣאݔ Ǥ In particular, ԹାൌԹൈԹǤ
The set ሾܽଵǡܾଵሿൈሾܽଶǡܾଶሿൈǥǥǥǤǤൈ ሾܽǡܾሿؿԹ i s c a l l e d a closed
rectangle i n ԹǢ Z h i l e t h e s e t ሺܽଵǡܾଵሻൈሺܽଶǡܾଶሻൈǥǥǤൈ ሺܽǡܾሻؿԹ i s
called an open rectangle.
1..1 Definition:- A set ؿܷԹ is called open set if for each ܷאݔ there is an
open rectangle ܣ such that ܷؿܣאݔ Ǥ
1..2 Definition:- A subset ܥ of Թis called closed set if Թ̳ܥ is open set.
E[ample: If ܥ contains finitely many points, then ܥ is closed set.
1. Interior and Boundary of subset of Թ
If ؿܣԹand אݔԹǡthen of three possibilities must holds munotes.in
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4A1ALcalled the interior points of ܣ and denoted by ܣ .
݅݅ሻ There is an open rectangle ܤsuch that ؿܤאݔԹ̳ܣǤ Set of all such points
are called the E[terior points of A and denoted by ܣ .
݅݅݅ሻ I f ܤ is any open rectangle Zith ܤאݔ ǡ t h e n ܤ c o n t a i n s p o i n t s o f b o t h ܣ
andԹ̳ܣ .Set of all such points are called the boundary points of ܣ and
denoted by ܣ߲ .
1.7 Operator norm หȁࢀȁห of linear transformation and its
properties
1.7.1 Definition:- Let ܸ and ܹ be a vector space over real field Թǡ and ܶǣܸ ՜ ܹ .
We say that ࢀ is linear if ܶሺݒݏ ݓݐሻൌݏ ǤܶሺݒሻݐǤܶሺݓሻ f o r a l l ݏǡאݐ Թ and
ǡ א .
E[ample:- Let ܸൌԹǡܹൌԹǤ Define ܶǣԹ՜Թ by ܶሺݔሻൌݔ. Then T is
linear transformation.
Note:- Let ࣦሺܸǡܹሻ denote the set of all linea r transformation from vector space ܸ
to vector space ܹǤ One can see each element of ࣦሺԹǡԹሻ as a point of Euclidian
space Թǡ and thus Ze can speaN of open set in ࣦሺԹǡԹሻ, continuous function
of linear transformation etc.
1.7.2 Definition:- Let ܶǣԹ՜Թ be a linear transformation. The norm หȁܶȁห of
ܶ is defined as หȁܶȁหൌ ൛ ห ȁܶሺݒሻȁหǣאݒ Թ݀݊ܽหȁݒȁหͳ ൟ Ǥ
1.7.3 Properties of norm หȁࢀȁห ofࢀ :For all linear maps ܵǡܶǣԹ՜Թ and
RǣԹ՜Թ Ze have the folloZing properties of norm
݅ሻ หȁܵܶȁหหȁܵȁหȁȁܶȁȁ
݅݅ሻ หȁܴܵȁหหȁܴȁหǤหȁܵȁห And
݅݅݅ሻ หȁܶܿȁหൌȁܿȁหȁܶȁหሺאܿԹሻǤ
1. Compactness on Թ
A collection ࣩ of open sets is called open cover of ܣ or briefly cover ܣ if every
points ܣאݔ is in some open sets in the collection ࣩǤ
E[ample 1: If ࣩ is collection of all open interval ሺܽǡ ܽͳሻfor אܽԹ, then ࣩ is
a cover of ԹǤ Clearly no finite number of open sets in ࣩ Zill cover ԹǤ munotes.in
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5Chapter 1: Euclidian Space RnE[ample 2: If ࣩ is collection of all open interval ଵ
ǡͳെଵ
for all integer ݊ͳ ǡ
then ࣩ is open cover of ሺͲǡͳሻǡ but again no finite collection of sets in ࣩ Zill cover
ሺͲǡͳሻǤ
1..1 Definition:- A subset ؿܣԹ is said to be a compact if every open cover ࣩ
contains a finite sub collection of open sets Zhich also covers ܣǤ
E[ample 1: A set Zith finitely many points is compact.
E[ample 2: Let ܣ is infinite set Zhich contains 0 and the numbers ଵ
for all integers
݊ .Reason: If ࣩ is open cover, Ͳܷא for some open set ܷ in ࣩ there are only
finitely many pints of ܣ Zhich are not in ܷ ,each reTuire at most one more open
set .
Note:- One may also define the compactness as a subset ݂ܣԹ is seTuentially
compact if every seTuence ሺܽሻ in ܣ has subseTuence ሺܽሻ that converges to limit
in ܣǤ
1. +eine-Borel theorem
The closed and bounded rectangle of Թ is compact.
1. .1 Corollary: - If ؿܣԹ and ؿܤԹ are compact, then ܣൈܤ is compact.
1. .2 Theorem: - A closed and bounded subset of Թ is compact.
Proof: - If ؿܣԹis closed and bounded, then ܤؿܣ for some closed rectangle
ܤǤ If ࣩ is an open cover of ܣǡ then ࣩ together Zith Թ̳ܣ is an open cover of ܤ .
Hence a finite number ܷଵǡܷଶǡǥǥǤǤǡܷ of sets in ࣩ ,together Zith Թ̳ܣ Zill cover
ܤǤ Then ܷଵǡܷଶǡǥǥǤǤǡܷ cover ܣǤ
1.10 Bol]ano-Weierstrass theorem on Թ
Every bounded seTuence of Թ has convergent subseTuence.
Proof: - A bounded seTuence is contained in a closed and bounded rectang le, Zhich
is compact. Therefore the seTuence has a subseTuence that conve rges to a limit in
the rectangle.
1.11 Continuous function on Թ
A function from Թ୬ to Թ୫ sometimes called a vector valued function of n
variable is the rule Zhich associates to each points of Թ to some points of Թ, munotes.in
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6A1ALtaNes Թinto Թ”) to indicate that ݂ሺݔሻאԹis defined for אݔԹǤ
The notation ݂ǣܣ ՜ Թ indicates that ݂ሺݔሻ is defined only for [ in the set ܣ ,Zhich
is called the domain of ݂Ǥ If ܣؿܤ ǡ Ze define ݂ሺܤሻ as set of all ݂ሺݔሻfor ܤאݔ ,
and if ؿܥԹZe define ݂ିଵሺܥሻൌሼܣאݔ ǣ݂ሺݔሻܥאሽǤ
Let ؿܣԹ, a function ݂ǣܣ ՜ Թ determine ݉ c o m p o n e n t f u n c t i o n
݂ଵǡ݂ଶǡǥǤǤ݂ǣܣ՜Թ by ݂ሺݔሻൌ ሺ݂ଵሺݔሻǡ݂ଶሺݔሻǡǥǥǤǤǡ݂ ሺݔሻሻ.
The notation
௫՜݂ሺݔሻൌܾ means that Ze can get ݂ሺݔሻ as closed to ܾ as desired,
by choosing ݔ sufficiently closed to, but not eTual to ܽǤ
In mathematical terms this means that for every number ߝͲ there is the number
ߜͲ s u c h t h a t หȁ݂ሺݔሻെܾȁห൏ߝ f o r a l l ݔ i n t h e d o m a i n o f ݂ Z h i c h s a t i s f y
Ͳ൏หȁݔെܽȁห൏ߜǤ
A function ݂ǣܣ ՜Թ is called continuous at ܣאܽ if
௫՜݂ሺݔሻൌ ݂ሺܽሻ and simply
called continuous if it is continuous at each ܣאܽ Ǥ
Note: - We may also define continuity as, a function ݂ǣԹ՜Թ is continuous if
and only if ݂ିଵሺܷሻ is open Zhenever אܷԹ is open.
1.11.1 Theorem: - If ؿܣԹǡ a function ݂ǣܣ ՜Թ is continuous if and only if
for every open set ؿԹ୬ there e[ist open set ܷ in Թ such that ݂ିଵሺܷሻൌܣתܸ Ǥ
Proof: - Suppose ݂ i s c o n t i n u o u s . I f ݂אܽିଵሺܷሻǡthen ݂ሺܽሻܷא Ǥ S c i e n c e ܷ i s
open, ther is an open rectangle ܤ Zith ݂ሺܽሻܷؿܤא Ǥ Science f is continuous at ܽ ,
Ze can ensure that ݂ሺݔሻܤא ǡ p r o v i d e d Z e c h o o s e ݔ i n s o m e s u f f i c i e n t l y s m a l l
rectangle ܥ containing ܽǤ We can do this for each ݂אܽିଵሺܷሻǤ Let ܸ be union of
all such ܥ i.e. ܸൌڂ ܥ אషభሺሻǤ Clearly ݂ିଵሺܷሻൌܣתܸ Ǥ The converse can be
prove similarly.
1.11.2 Theorem: - If ݂ǣܭ ՜Թ is continuous Zhere ؿܭԹ and ܭ is compact,
then ݂ሺܭሻؿԹ is compact.
Proof: - Let ࣩ be an open cover of ݂ሺܭሻǤ For each open set ܷ in ࣩ there is an open
set ܸ such that ݂ିଵሺܷሻൌܸܭתǤ The collection of all ܸ is an open cover of ܭǤ
Science ܭis compact, a finite number of ܸభǡܸమǡǥǥǤǡܸ cover ܭǤThen
ܷଵǡܷଶǡǥǥǤǡܷ covers ݂ሺܭሻǤ
munotes.in
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7Chapter 1: Euclidian Space Rn1.13 Connected subset of Թ
A subset ؿܧԹ is said to be disconnected if there e[ist disMoint non-empty open
sets or closed sets ܷ and ܸ of ܧ Zith ܧ ൌ ܸܷ .
A subset ؿܧԹ is said to be connected if it is not disconnected.
Note: - Let ؿܧԹǡ a subset ܧؿܣ is said to be open or closed in E if there e[ist
open or closed subset ܤof Թ such that ܣൌܤתܧ Ǥ
1.13.1 Proposition: - A subset ؿܧԹ is connected if and only if it is not the union
of tZo non-empty disMoint closed sets, or eTuivalently, if and only if there e[ist no
subset of ܧ Zhich is simultaneously open and closed, other than ܧ and Ǥ
1.13.2 Proposition: - A subset ؿܧԹ is connected if and only if for every pair
ܷǡܸ of open subset of Թ such that ܸܷؿܧ and ܣתܸתܷൌǡ then Ze have
either ܷؿܣ or ܸؿܣ Ǥ
1.13.3 Theorem: - A subset ܣof Թ is connected if and only i f it is an interval.
Proof: - Direct part suppose ܣ is not an interval, so there e[ist ܽǡܣאܾ ǡבܿ݀݊ܽ
ܣǡ Z i t h ܽ൏ܿ൏ܾ Ǥ L e t ܷ ൌ ሺെλǡ ܿሻ and ܸൌሺܿǡλሻǤ Then ܷandܸ are open
subset of Թ, Zith ܸܷؿܣ and ܸתܷൌǤ Science ܣתܷאܽ and ܣתܸאܾ ǡ
Ze see from the proposition 1.13.2 that ܣ is disconnected.
Converse part 1oZ suppose that ܣ is an interval, and there e[ists open subset
ܷandܸ of Թ such that ؿ ǡתתൌ Ǥ and neither ܣתܷ nor ܣתܸ is
empty. Let ܣתܷאܽ and ܣתܸאܾ .We may assume that ܽ൏ܾ Ǥ Science ܣ is an
interval, for each ݐ Zith ܽݐܾ ǡ Ze have ܣאݐ and hence either ܷאݐ orܸאݐ Ǥ
Let ܧൌሼאݐሾܽǡܾሿǣܷאݐሽand let ܿൌ ܧ .Then ܾܽܿ ǡ so ܣאܿ Ǥ If ܸאܿ ,
then ܿܽ ǡ and ሺܿെ߳ǡܿ߳ ሻܸؿ for some ߳Ͳ ǡܸ is open. %ut science ܿ is least
upper bound of ܧǡ there e[ist אݐ ሺܿ െ߳ǡܿሿ Zith ܷאݐ Ǥ This contradiction shoZs
that ܸבܿ Ǥ IF ܷאܿ ,then ܿ൏ܾ and ሺܿെ߳ǡܿ߳ ሻܷؿ for some ߳Ͳ Ǥ %ut then
there e[ist ݐܿ Zith אݐሾܽǡܾሿܷתǡ contadicting the fact that ܿ is an upper bound
of ܧǤ Thus our assumption that both ܷתܣ and ܸתܣ are non-empty is invalid.
Thus ܣ is connected.
1.13.4 Theorem: - Let ؿܧԹ is connected subset and ݂ǣܧ ՜Թ is continuous,
then ݂ሺܧሻؿԹ is connected.
Proof: - Let ݂ሺܧሻ is disconnected, then there e[ist open subset ܷ and ܸ of Թ such
that ݂ሺܧሻܸܷؿǡ݂תܸתܷ ሺܧሻൌ Ǣ and ݂תܷሺܧሻ് ǡ݂תܸ ሺܧሻ് Ǥ
%ut then ݂ିଵሺܷሻ a n d ݂ିଵሺܸሻ are open science f is continuous, munotes.in
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8A1ALscience ݂ሺܧሻܸתܷתൌǢ and neither ݂ିଵሺܷሻ nor ݂ିଵሺܸሻ is empty.
Thus ܧ is disconnected Zhenever ݂ሺܧሻ is disconnected.
1.14 Let us sum up
In this chapter Ze have learnt the folloZing:
x Inner product and norm on Թ, open and closed subset of ԹǤ
x Operator norm ȁȁܶȁȁ of linear transformation and its properties.
x Connectedness and compactness on ԹǤ
x Continuous function on Թ a n d i t s b e h a v i o r Z i t h c o n n e c t e d a n d c o m p a c t
subset of ԹǤ
1.1 Unit end E[ercise
1. Define the folloZing terms and find 2 to 3 e[amples for each.
i Vector spaces ii Inner product spaces
iii 1orm Linear spaces iv Metric Spaces
2. Prove the properties of 1.3.1
3. For ݔൌሺݔଵǡݔଶሻאԹଶǤ
We define ሻหȁȁหଵൌȁଵȁȁଶȁ
ሻหȁȁหଶൌඥଵଶଶଶ
ሻหȁݔȁหஶൌ ሼȁݔଵȁǡȁ ݔଶȁሽ
Prove that ȁȁǤȁȁଵǡȁȁǤȁȁଶandȁȁǤȁȁஶ define the norms on ԹଶǤ
1ote - ȁȁǤȁȁଶ is called usual or Euclidean norm.
4. State and prove the Cauchy- SchZar] ineTuality on ԹZhere Թis vector
space over the field ԧǤ
5. Prove the folloZing properties of inner product. Consider ݔǡאݕ Թ and ܽ in
ԹǤ
݅ሻ ൏ݔǡݕ ൌ൏ ݕǡݔ
݅݅ሻ ൏ ݔܽǡݕ ൌ ܽǤ൏ ݔǡݕ
݅݅݅ሻݒሻ ൏ ݔǡݕ ൌหȁݔݕȁหଶെหȁݔെݕȁหଶ
Ͷሺݕݐ݅ݐ݊݁݀݅݊݅ݐܽݖ݅ݎ݈ܽܲሻ
6. Prove that หȁȁݔȁหെหȁݕȁȁห หȁݔെݕȁหǤ
7. Prove that ݂ǣԹ՜Թ defined by ݂ሺݔሻൌȁȁݔȁȁ is continuous function. munotes.in
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9Chapter 1: Euclidian Space Rn8. Let ݂and ݃ be integrable on ሾܽǡܾሿ t h e n p r o v e t h a t ,
ቚ݂Ǥ݃
ቚቀ݂ଶ
ቁభ
మǤቀ݃ଶ
ቁభ
మ This is Cauchy-SchZar] ineTuality in
integrable space Hint:- Consider separately the case Ͳൌሺ݂െ݃ߣሻଶ
for
some אߣԹ and Ͳ൏ሺ݂െ݃ߣሻଶ
for all ߣin ԹǤ
9. Prove that closed rectangle in Թ is indeed a closed set.
10. Prove that interior of any set ܣ and e[terior of any set ܣ i.e. interior of Թ̳ܣ
are open.
11. Prove that finite union of open set is open. Also, prove arbitr ary union of
open set is open.
12. Prove that finite intersection of open set is open. Does the a rbitrary
intersection of open set is open"
13. Prove that finite intersection of closed set is closed. Also, p rove arbitrary
intersection of closed set is closed.
14. Prove that finite unions of closed sets are closed. Does the ar bitrary unions
of closed sets are closed"
15. Prove that boundary ܣ߲ of set ܣ is closed.
16. Prove that ሼאݔ Թǣหȁݔെܽȁห൏ݎ ሽ is open Zhere ݎ is some positive real
number.
17. Find the interior, e[terior and boundary of the folloZing sets.
݅ሻሼאݔ Թǣหȁݔെܽȁห൏ͳ ሽ
݅݅ሻሼאݔ Թǣหȁݔെܽȁหൌͳ ሽ
݅݅݅ሻሼאݔ Թǣ Each ݔrational ሽ
18. ChecN that ܶǣԹ՜Թǡdefined as folloZs are linear transformation or not
݅ሻܶሺݔሻൌݔͳ Թ
݅݅ሻܶሺݔሻൌͲԹ
݅݅݅ሻܶሺݔሻൌʹ ݔ
19. Let ܵǡܶǣԹ՜Թ and ܴǣԹ՜Թ be the linear map then prove that
ܵܶǡܴܵǡܶܿ݀݊ܽሺאܿԹሻ are also a linear map.
20. Prove the properties of ȁȁܶȁȁ of 1.7.3
21. Prove that continuous real values function defined on a compact s e t i s
bounded Ref. C.C. Pugh munotes.in
Page 10
10A1AL ݂ǣؿܣ Թ՜Թ
i The ሺ߳ǡߜሻ- condition
ii The seTuential convergen ce preservation condition
iii The closed set condition i.e. inverse image of closed set in Թ is closed
in ؿܣԹǤ
iv The open set condition i.e. inverse image of open set in Թ is open in
ؿܣԹǤ Ref. C.C. Pugh
23. Prove that every compact set is closed and bounded. Ref. C.C. Pugh
24. The closed interval ሾܽǡܾሿؿԹ is compact. Ref. C.C. Pugh
1.1 References
1. C. C. Pugh, Mathematical A nalysis, Springer UTM.
2. A. %roZder, Mathematical An alysis an Introduction, Springer .
3. W. Rudin, Principals of Math ematical Analysis, Mc*raZ-Hill India.
4. M. SpivaN, Calculus on Manifolds, Harper-Collins Publishers
munotes.in
Page 11
UNIT 2
2
DI))ERENTIATION O) )UNCTIONS
O) SE9ERAL 9ARIABLES
Unit Structure
2.0 ObMectives
2.1 Introduction
2.2 Total Derivative
2.3 Partial Derivatives
2.4 Directional Derivatives
2.5 Summary
2.0 Objectives
After reading this unit you should be able to
x define a differentiable function of several variables
x define and calculate the partial and directional derivatives i f they e[ist of a
function of several variables
x establish the connection betZeen the total, partial and directi onal derivatives
of a differentiable function at a point
2.1 Introduction
betZeen metric spaces. Another important concept is differentia tion. If Ze try to
apply this to functions betZeen metric spaces, Ze encounter a p roblem. We realise
that apart from the distance notion, the domain and codomain al so need to have an
algebraic structure. So, let us consider Euclidean spaces liNen
. Which have Zhich 11munotes.in
Page 12
12A1ALare Zhat Ze call functions of several variables.
In this chapter Ze shall introduce the concept of differentiabi lity of a function of
several variables. The e[tension of this concept from one to se veral variables Zas
not easy. Many different approaches Zere tried before this fina l one Zas accepted.
The definition may seem a little difficult in the beginning, bu t as you Zill see, it
alloZs us to e[tend all our NnoZledge of derivatives of functio ns one variable to
the several variables case. of graduation. So, here Ze shall try to go a little deeper int o these concepts, and
deal Zith vector functions of several variables.
2.2 Total Derivative
To arrive at a suitable definition of differentiability of func tions of several
variables, mathematicians had to closely e[amine the concept of derivative of a
function of a single variable. To decide on the approach to e[t ension of the concept,
it Zas important to NnoZ Zhat Zas the essence and role of a der ivative. So, let us
recall the definition of the derivative of a function :f o
We say that f is differentiable at aR, if the limit, 0lim
oh haf haf e[ists.
In that case, Ze say that the derivative of f at a,
fa 0lim
oh haf haf .......(2.1
So, Ze taNe the limit of the ratio of the increment in f x to the increment in x. 1oZ,
Zhen our function is defined on Rn, the increment in the independent variable Zill
be a vector. Since division by a vector is not defined, Ze cann ot Zrite a ratio similar
to the one in 2.1 . %ut 2.1 can be reZritten as
0lim
oh>haf haf −
fa @ 0, or 0lim
oh>
.fa h fa f ah
h @ 0, or munotes.in
Page 13
13Chapter 2: Diff erentiation of Functions of Several Variables
0 lim
hrh
ho 0, Zhere r h f ah – f a –
fa.h.
So, Ze can Zrite f ah f a
fa.h r h , ........................ 2.2
where the “remainder” r h is so small, that rh
h tends to ]ero as h tends to ]ero.
For a fi[ed a, f a , and
faare fi[ed real numbers. This means, e[cept for the
remainder, r h , 2.2 e[presses f a h as a linear function of h. This also helps us
in “linearizing” f. We say that for points close to a, the graph of the function f can
be appro[imated by a line.
Thus,
fa gives rise to a linear function L from
to
.
L:
o
, h ՜
fa.h, Zhich helps us in lineari]ing the given function f near
the given point a. 2.2 then transforms to
f a h f a L h r h ........................... 2.3
It is this idea of lineari]ation that Ze are noZ going to e[ten d to a function of
several variables.
Definition 2.1 Suppose E is an open set in n
, f : E mo
, and aE. We say
that f is differentiable at a, if there e[ists a linear transformation:nmT o
,
such that 0lim
ohࢌצሺࢇାࢎሻିࢌሺࢇሻିࢀሺࢎሻצ
צࢎצ 0 ......................... 2.4
and Ze Zrite
fa T.
If f is differentiable at every point in E, Ze say that f is differentiable in E.
Remark 2.1
i %old letters indicate vectors.
ii Since E is open, ݎ Ͳ, such that % a, r ؿ E . W e c h o o s e h, such that
צhצ < r, so that a hא E. munotes.in
Page 14
14A1ALdenominator is the norm in Rn.
iv The linear transformation T depends on the point a. So, Zhen Ze have to deal
Zith more than one point, Ze use the notation, Ta, Tb, and so on.
We have seen that in the one variable case, the derivative defi nes a linear function,
h ՜ f1 a .h from R to R. Similarly, here the derivative is a linear transformation
from Rn to Rm. With every such transformation, Ze have an associated m n
matri[. The Mth column of this matri[ is T(ej , Zhere ej is a basis vector in the
standard basis of Rn.
For a given point a, the linear transformation Ta is called the total derivative of f at
a, and is denoted by f1 a or D f a . We can then Zrite
f(a + h ) = f(a) + Ta(h) + r(h),where ࢘ሺࢎሻ
צࢎצ ՜, as h՜. .. (2.)
We noZ give a feZ e[amples.
E[ample 2.1 : Consider f: Rn ՜ Rn, f(x) = a + x, Zhere a is a fi[ed vector in Rn .
Find the total derivative of f at a point p א Rn, if it e[ists.
Solution : 1oZ, f(p + h) – f(p) = h. So, if Ze taNe T to be the identity
transformation from Rn to Rn, then Ze get
f(p + h) – f(p) – T(h) 0, and hence 0lim
ohࢌצሺାࢎሻିࢌሺሻିࢀሺࢎሻצ
צࢎצൌ 0.
Comparing this Zith 2.5, Ze conclude that the identity transfor mation is the total
derivative of f at the point p.
E[ample 2.2 : Find the total derivative, if it e[ists, for f : R2՜ R2, f(x, y) = (x2, y2),
at a point a a1, a2 .
Solution : If f is differentiable, Ze e[pect Ta to be a 22
matri[.
munotes.in
Page 15
15Chapter 2: Diff erentiation of Functions of Several Variables Let h = (h 1, h2). 1oZ,
f(a + h) – f(a) ሺܽଵ݄ଵሻଶǡሺܽଶ݄ଶሻଶሻെሺܽଵଶǡܽଶଶሻ
ʹܽଵ݄ଵ݄ଵଶǡʹ ܽଶ݄ଶ݄ଶଶ
( 2ܽଵ݄ଵǡʹܽଶ݄ଶ ݄ଵଶǡ݄ଶଶ
൬ʹܽଵͲ
Ͳʹ ܽ ଶ൰൬݄ଵ
݄ଶ൰ ݄ଵଶǡ݄ଶଶ
We taNe Ta ൬ʹܽଵͲ
Ͳʹ ܽ ଶ൰ , and r(h) ݄ଵଶǡ݄ଶଶ , and Zrite
f(a+h) = f(a) + Ta(h) + r(h), Zhere ࢘ሺࢎሻ
צࢎצ ൌሺభమǡమమሻ
ξሺభమାమమሻ՜ 0, as h՜.
Thus Ta is the total derivative of f at a.
1oZ that Ze have defined the total derivative, let us see hoZ m any of the results
that Ze NnoZ about derivatives of functions of a single variabl e, hold for these total
derivatives.
Theorem 2.1: If f : Rn ՜ Rm is differentiable at a א Rn, then its total derivative is
uniTue.
Proof : Suppose f has tZo derivatives, T1 and T2 at a, and let T T1 – T2. Let h
א Rn,
h ് 0, and t א R, such that t ՜ 0.
Then th ՜ 0 as t ՜ 0.
Since T1 is a total derivative of f at a, 0lim
ot࢘צሺ௧ࢎሻצ
צ௧צࢎൌ0lim
otࢌצሺࢇା௧ࢎሻିࢌሺࢇሻିࢀሺ௧ࢎሻצ
צ௧צࢎൌ 0 ...................... 2.6
Since T2 is also a total derivative of f at a, 0lim
ot࢘צሺ௧ࢎሻצ
צ௧צࢎൌ0lim
otࢌצሺࢇା௧ࢎሻିࢌሺࢇሻିࢀሺ௧ࢎሻצ
צ௧צࢎൌ 0 ...................... 2.7 munotes.in
Page 16
16A1AL = ࢌצሺࢇࢎݐሻെࢌሺࢇሻെࢀሺࢎݐሻെሾࢌሺࢇࢎݐሻെࢌሺࢇሻെࢀሺࢎݐሻሿצ
ࢌצ ሺࢇࢎݐሻെࢌሺࢇሻെࢀሺࢎݐሻצ ࢌצ ሺࢇࢎݐሻെࢌሺࢇሻെࢀሺࢎݐሻצ
Therefore , ࢀצሺ௧ࢎሻצ
צ௧צࢎ ࢌצሺࢇା௧ࢎሻିࢌሺࢇሻିࢀሺ௧ࢎሻצ
צ௧צࢎ + ࢌצሺࢇା௧ࢎሻିࢌሺࢇሻିࢀሺ௧ࢎሻצ
צ௧צࢎ
Since T is a linear transformation , T(th) = tT( h). Therefore,
ȁ௧ȁࢀצሺࢎሻצ
ȁ௧ȁצࢎצ ࢌצሺࢇା௧ࢎሻିࢌሺࢇሻିࢀሺ௧ࢎሻצ
צ௧צࢎ + ࢌצሺࢇା௧ࢎሻିࢌሺࢇሻିࢀሺ௧ࢎሻצ
צ௧צࢎ .
So, using 2.6 and 2.7 , Ze get
00lim
otࢀצሺࢎሻצ
צࢎצ0lim
otࢌצሺࢇା௧ࢎሻିࢌሺࢇሻିࢀሺ௧ࢎሻצ
צ௧צࢎ0lim
otࢌצሺࢇା௧ࢎሻିࢌሺࢇሻିࢀሺ௧ࢎሻצ
צ௧צࢎൌ0
Since ࢀצሺࢎሻצ
צࢎצ is independent of t, this means ࢀצሺࢎሻצ
צࢎצ = 0,z
Zhich means that ࢀצ ሺࢎሻ צ 0.
1oZ , h Zas any non-]ero vector in Rn. Further, T 0 0. Hence Ze conclude that
T h 0 for all h א Rn. Thus T T1 – T2 is the ]ero linear transformation. Thus,
T1 T2. That is, the derivative is uniTue.
In the ne[t e[ample Ze find the derivatives of some standard fu nctions.
E[ample 2.3 :
i Find the total derivative f1(a), if f : Rn ՜Rm , f(x) = c,
Z h e r e c is a fi[ed vector in Rm and a אRn.
ii If f : Rn ՜Rm is a linear transformation,
s h o Z t h a t Df(a) = f for every a אRn .
munotes.in
Page 17
17Chapter 2: Diff erentiation of Functions of Several Variables Solution :
i Since f is a constant function, Ze e[pect its derivative to be the ]ero
transformation.
H e r e f(a + h) – f(a) = c – c = 0.
If Ze taNe T to be the ]ero transformation,
0lim
ohצሺሻצ
צצൌ0lim
ohצሺା௧ሻିሺሻି்ሺሻצ
צצൌ 0.
H e n c e f1 a e[ists and is eTual to 0 for every a א Rn.
ii Since f is a linear transformation, f(a + h) = f(a) + f(h).
If Ze taNe T f,
r(h) = f (a + h) f(a) f(h) 0 ฺ צሺሻצ
צצൌ Ͳ Ǥ
We have defined the total derivative of a function as a linear transformation. 1oZ
Ze prove a result about linear transformations Zhich Ze may use later.
Proposition 2.1 : Every linear transformation T from Rn to Rm is continuous on Rn.
Proof: If T is the ]ero linear transfo rmation, it is clearly continuou s.
If T ് 0, let p א Rn,
p = (p 1, p2, ..., p n , and ߝͲ. Suppose ^ e1, e2, ..., en` is the standard basis for Rn.
Choose ߜൌߝ ȀM, Zhere M ܶצሺ݁ଵሻצ ܶצሺ݁ଶሻצ....... ܶצሺ݁ሻצ .
If x = (x 1, x2, ..., x n) is such that ࢞צ െ צ ߜ ,then _ xi – pi_ ߜ for i 1, 2, ..., n.
Also, ࢞צ െ צ ߜ ฺܶצ ሺ࢞ሻെܶሺܘሻצ ܶצ ሺ ܠെܘ ሻצ ܶצ ൫ሺݔଵെଵሻࢋ
ሺݔଶെଶሻࢋڮሺݔെሻࢋ൯צ _ሺݔଵെଵሻ| ܶצ ሺࢋሻצ+ |ሺݔଶെଶሻ| צ
ܶሺࢋሻצ+ ....... | ሺݔെሻ| ܶצ ሺࢋሻצ ߜ(ܶצሺࢋሻצܶצ ሺࢋ ሻצ+.......+ צ
ܶሺࢋሻצ = )ߝ
Thus, T is continuous at p. Since p Zas an arbitrary point of Rn, Ze conclude that
T is continuous on Rn.
In fact, since ߜ did not depend on p, Ze can conclude that T is uniformly continuous
on Rn. munotes.in
Page 18
18A1ALThe ne[t theorem shoZs that thi s holds for functions of several variables too.
Theorem 2.2 : If f : Rn ՜Rm is differentiable at p, then f is continuous at p.
Proof : Since f is differentiable at p, there e[ists a linear transformation Tp such
that 0lim
ohࢌצሺାࢎሻିࢌሺሻିࢀሺࢎሻצ
צࢎצൌ 0.
Thus, ߝ Ͳǡߜ ଵͲ ǡ such that
צ݄צ ൏ߜ ଵฺ ࢌצሺࢇାࢎሻିࢌሺࢇሻିࢀሺࢎሻצ
צࢎצ ߝȀʹ
Choose ߜଶൌ ሺͳǡߜଵሻǤ Then
צ݄צ ൏ߜ ଶ ฺ݂צ ሺ݄ሻെ݂ሺሻെܶሺ݄ሻצ ߝȀʹ צ݄צ ߝȀ ʹ
%y Proposition 2.1, Tp is continuous at 0, and Tp 0 0. So, there e[ists ߜଷͲ ǡ
such that
צࢎצ ൏ߜ ଷ ฺ ܂צܘሺࢎሻצ൏ ߝȀʹ.
1oZ choose ߜൌ ሺߜଶǡߜଷሻǤ Then
צࢎצ ൏ߜ ฺࢌצ ሺࢎሻെࢌሺሻצࢌצ ሺࢎሻെࢌሺሻെࢀሺࢎሻצࢀ ሺࢎሻ צ
ఌ
ଶఌ
ଶൌ ߝ .
Thus,
ࢎ՜ࢌሺࢎሻൌࢌ ሺ ሻ, and f is continuous at p.
With your NnoZledge of functions of one variable, you Zould e[p ect that the
converse of Theorem 2.2 does not hold. That is, continuity does n o t i m p l y
differentiability. The folloZing e[ample shoZs that it is indee d so.
E[ample 2.4 : Consider the function f : R ՜R2, f(x) = (|x|, |x|). We shall shoZ that
f is continuous at 0, but is not differentiable there.
*iven ߝͲ ǡ choose ߜൌ ߝ Ȁξʹ. Then
_[_ ߜ ฺࢌצ ሺݔሻצൌצሺȁݔȁǡȁݔȁሻצ൏ξߜଶߜଶൌξʹ ߜൌ ߝ .
Hence, f is continuous at x 0. munotes.in
Page 19
19Chapter 2: Diff erentiation of Functions of Several Variables 1oZ suppose f is differentiable at x 0. Then there e[ists a linear transformation
T : R ՜ R2, such that 0lim
ohࢌሺሻିࢌሺሻିࢀሺሻ
0 ฺ0lim
ohሺȁȁǡȁȁሻିࢀሺሻ
0
ฺ
՜ሺȁȁ
ሺͳǡͳሻെࢀሺͳሻሻ ൌ
Now, (1, 1) and ( −1, −1) are two distinct points in R2,
and % 1, 1 , 1 תB((−1, −1), 1) = .For ߝ 1, ߜ ! 0, such that
צ݄צ ൏ߜ ฺצȁȁ
ሺͳǡͳሻെࢀሺͳሻሻ צ൏ ߝ ..……… ...............2.8
Putting h ߜȀʹ in 2.8 , Ze get צȁȁ
ሺͳǡͳሻെࢀሺͳሻሻ צൌ ԡሺͳǡͳሻെܶሺͳሻԡ൏ͳ.
This means T 1 א % 1, 1 , 1 .
Similarly, taking h = − ߜȀʹ, Ze get that T 1 א B((−1, −1), 1). But this contradicts
the fact that B((1, 1), 1) and B(( −1, − 1), 1) are disjoint.
Thus, f is not differentiable at x 0.
If f : Rn ՜Rm , then, as you NnoZ, Ze can Zrite f = (f 1,f2, ...,f m),
Zhere each fi : Rn ՜R,
i 1, 2,...,m. These fis are called coordinate functions of f. Similarly, a linear
transformation
T : Rn ՜Rm can be Zritten as T T 1,T2, ...,T m , Zhere each T i i s a l i n e a r
transformation from Rn to R.
Theorem 2.3 : Let f = (f 1,f2, ...,f m) : Rn ՜Rm, and p א Rn. f is differentiable at p,
if and only if each fi, 1݅ m is differentiable at p.
Proof : f i s d i f f e r e n t i a b l e a t p if and if there e[ists a linear transformation
Tp : Rn ՜Rm, such
that 0lim
ohࢌצሺାࢎሻିࢌሺሻିࢀሺࢎሻצ
צࢎצ 0, that is, if only if munotes.in
Page 20
20A1ALohצσሾభሺାࢎሻିሺሻି்ሺࢎሻሿࢋצ
צࢎצ 0, Zhere ^ e1, e2, ..., e m` is the standard basis of Rm,
if and only if, 0lim
ohȁሺାࢎሻିሺሻି்ሺࢎሻȁ
צࢎצ 0, i, 1݅ m.
That is, if and only if each fi is differentiable and Dfi = Ti, ݅ ,1݅ m.
Thus, Df(p) = Tp = (Df 1(p), Df 2(p), ....., Df m(p)).
Theorem 2.4 : Let f : Rn ՜Rm and g : Rn ՜Rm be tZo functions differentiable at
p א Rn. If k א R, then f g and kf are also differentiable at p. Moreover,
D f + g p D f p D g p , and D kf p kDf p .
Proof : Let D f p T1, and D g p T2. Then T1 T2 is also a linear transformation
from Rn to Rm, and
0 0lim
ohצሺࢌାࢍሻሺାࢎሻିሺࢌାࢍሻሺሻିሺࢀାࢀሻሺࢎሻצ
צࢎצ
0lim
ohצሾࢌሺାࢎሻିࢌሺሻିࢀሺࢎሻሿାሾࢍሺାࢎሻିࢍሺሻିࢀሺࢎሻሿצ
צࢎצ
0lim
ohࢌצሺାࢎሻିࢌሺሻିࢀሺࢎሻצ
צࢎצ 0lim
ohࢍצሺାࢎሻିࢍሺሻିࢀሺࢎሻצ
צࢎצ 0.
Therefore, f + g is differentiable at p, and D f + g p T1 T2 Df p D g p .
1oZ, 0lim
ohצࢌሺାࢎሻିࢌሺሻିࢀሺࢎሻצ
צࢎצൌ_N_ 0lim
ohࢌצሺାࢎሻିࢌሺሻିࢀሺࢎሻצ
צࢎצ 0.
Therefore, kf is also differentiable and D kf p kT1 kDf p .
2.3 Partial Derivatives
We NnoZ that the derivative of a function of one variable denot es the rate at Zhich
the function value changes Zith change in the domain variable. In the case of
functions of several variables, change in the domain vector var iable means a change
in any or all of its components. %ut if Ze consider change in o nly one component
and study the rate at Zhich the function value changes, Ze get Zhat is NnoZn as
the partial derivative of the function. Corresponding to each c omponent of the
variable, there Zill be a partial derivative. Here is the forma l definition. munotes.in
Page 21
21Chapter 2: Diff erentiation of Functions of Several Variables Definition 2.2 Let f : E՜Rm, Zhere E كRn. Let x = (x 1, x2, ..., x n be an interior
point of E. Then for every i, i 1, 2, ..., n, the limit 0lim
ohࢌሺ୶భǡ୶మǡǥǡ୶ା୦ǡ୶శభǡǥǡ୶ሻିࢌሺ୶భǡ୶మǡǥǡ୶ሻ
, if it e[ists, is called the ith partial derivative
of f Zith respect to [ i at [. It is denoted by డ
డ௫ǡ݂௫ǡܦݎ݂ .We Zrite డ
డ௫ሺ࢞ሻ to
indicate the point at Zhich the pa rtial derivative is calculate d.
Remark 2.2 :
i If a function f has partial derivatives at every point of the set E,
Ze say that f has partial derivatives on E.
ii It is clear from the definiti on that a partial derivative can be defined at an
interior point of E, and not on its boundary.
iii If a function has a partial derivative at a point, its va lue depends on the values
of the function in a neighbourhood of that point. So, if the fu nction values
outside this neighbourhood are changed, it does not affect the value of the
partial derivative.
The folloZing e[amples Zill maNe the concept clear.
E[ample 2. : Find the partial derivative of the function, f x, y, z) = xyz + x2z.
Solution : This is a real-valued fun ction. differentiation of such a function.
డ
డ௫ 0lim
ohሺ௫ାሻ௬௭ାሺ௫ାሻమ௭ି௫௬௭ି௫మ௭
= y z + 2 x z . S i m i l a r l y , y o u c a n c h e c N t h a t
fy xz, and f] xy + x2.
Let us taNe a vector-valued function in the ne[t e[ample.
E[ample 2. : Find the partial derivatives of the function,
f : R3 ՜R2, f(x, y, z) = (xy, z2), if they e[ist.
Solution : 0lim
ohቀሺ௫ାሻ௬ǡ௭మቁିሺ௫௬ǡ௭మሻ
0lim
ohሺሺ௫ାሻ௬ି௫௬ǡሻ
munotes.in
Page 22
22A1ALohሺ௫ାሻ௬ି୶୷
ǡ0lim
oh
ሻ y, 0 .
Therefore, డ
డ௫ y, 0 .
Proceeding similarly, Ze find that డ
డ௬ x, 0 , and డ
డ௭ 0, 2 z .
by taNing the partial derivatives of its coordinate functions. In fact Ze have the
folloZing theorem, Zhich establishes the connection betZeen dif ferentiability of a
vector-valued function and the e[istence of partial derivatives o f i t s c o o r d i n a t e
functions
Theorem 2. : Let E be an open subset of Rn, and f : E ՜Rm.
Suppose f f1,f2, ...,f m is differentiable at p א E. Then the partial derivatives
డ
డ௫ೕ e[ist for i 1, 2, ..., m, M 1, 2, ..., n.
Proof : Since f i s d i f f e r e n t i a b l e a t p, there e[ists a linear transformation T,
such that 0lim
ohࢌצሺାࢎሻିࢌሺሻିࢀሺࢎሻצ
צࢎצൌͲ. L e t h teM, Zhere ^ e1, e2, ...,en` is the
standard basis of Rn. Then, h ՜ 0 if and only if t ՜ 0. Thus,
0lim
otࢌצ൫ା௧ࢋ൯ିࢌሺሻିࢀሺ௧ࢋೕሻצ
ȁ࢚ȁൌͲ. Therefore, 0lim
otࢌ൫ା௧ࢋ൯ିࢌሺሻ
࢚ൌ T ݁ .
That is,
0lim
otࢌ൫ା௧ࢋ൯ିࢌሺሻ
࢚ǡ0lim
otࢌ൫ା௧ࢋ൯ିࢌሺሻ
࢚ǡǥǡ0lim
otࢌ൫ା௧ࢋ൯ିࢌሺሻ
࢚
T ݁ .
Hence the limits e[ist, and డࢌ
డ௫ೕ p e[ists for all i 1, 2, ..., m.
Since M Zas arbitrary, Ze conclude that డࢌ
డ௫ೕ p e[ists for all i 1, 2, .., m, M 1, 2, .., n.
If f : E ՜Rm, Zhere E is an open subset of Rn, and if f is differentiable at p א E ,
then using Theorem 2.5, the matri[ of the linear transformation T can be Zritten as munotes.in
Page 23
23Chapter 2: Diff erentiation of Functions of Several Variables
ۉۈۈۈۈۈۇμ݂ଵ
μݔଵሺሻμ݂ଵ
μݔଶሺሻ Ǥ Ǥ ڄμ݂ଵ
μݔሺሻ
μ݂ଶ
μݔଵሺሻμ݂ଶ
μݔଶሺሻ Ǥ Ǥ Ǥμ݂ଶ
μݔሺሻ
Ǥ
ڄ
μ݂
μݔଵሺሻμ݂
μݔଶሺሻ Ǥ ڄڄ μ݂
μݔሺሻیۋۋۋۋۋۊ
This m [ n matri[ is called the -acobian matri[ of f at p, and is denoted by > f’(p @
or >D f(p @.
If m n, the determinant of the -acobian matri[ is called the -acobian of f at p, and
is denoted by డሺభǡమǡǥǡሻሺሻ
డሺ௫భǡ௫మǡǥǡ௫ሻ .
Thus, if f is differentiable at p, then the total derivative of f at p, T : Rn ՜Rm is
given by the -acobian matri[. For x x1, x2, ..., x n א Rn,
T x >f’(p @ቌ௫భ
௫మ
ǤǤ
Ǥ
௫ቍ.
When m 1, f is a real-valued function, and T ej డ
డ௫ೕሺሻ. Hence, the -acobian
matri[ of T is the roZ matri[, >డ
డ௫భሺሻడ
డ௫మሺሻǥడ
డ௫ሺሻ@.
The vector form, డ
డ௫భሺሻǡడ
డ௫మሺሻǡǥǡడ
డ௫ሺሻ is called the gradient of f at p , and
is denoted by f(p), or grad f(p).
If h = (h 1, h2, ..., h n א Rn,
Tp h >డ
డ௫భሺሻడ
డ௫మሺሻǥப
ப୶ሺሻ@ ൮భ
మ
ǤǤ
Ǥ
൲.
Thus, T h డ
డ௫భሺሻ݄ଵడ
డ௫మሺሻ݄ଶǥడ
డ௫ሺሻ݄, or Tp h f(p)x h.
So, Ze can say that the total derivative Tp of a real-valued fun ction is given by
Tp h f(p)xh. munotes.in
Page 24
24A1ALi f [, y x2y, exy
ii f x, y, z xsinz, yez at 1, 2, 1 .
Solution : i f1 x, y) = x2y, and f2 x, y) = exy. Therefore, డభ
డ௫ 2xy, డభ
డ௬ x2,
డమ
డ௫ yexy, and డమ
డ௬ xexy.
Hence, > fi x, y @ ʹݔݕݔ ଶ
݁ݕ௫௬݁ݔ௫௬൨
డభ
డ௫ sinyz, and డభ
డ௫ 1, 2, 1 sin2
డభ
డ௬ 1, 2, 1 cos2, డభ
డ௭ 1, 2, 1 2 cos2,
డమ
డ௫ 1, 2, 1 0, డమ
డ௬ 1, 2, 1 e-1, డమ
డ௭ 1, 2, 1 2e-1.
Thus, > fi 1, 2, 1 @ ቀെ݊݅ݏʹ െʹݏܿʹ ʹݏܿʹ
Ͳെ ݁ିଵെʹ݁ିଵቁ
In the ne[t section Ze shall consider yet another type of deriv ative.
2.4 Directional Derivatives
Partial derivatives measure the rate of change of a function in the directions of the
standard basis vectors. Directional derivatives measure the rat e of change in any
given direction.
Definition 2.3 : Let f : E ՜ R, Zhere E is an open subset of Rn. Let u be a unit
vector in Rn, and p א E. If 0lim
otሺା௧࢛ሻିሺሻ
௧ e[ists, then it is called the directional
derivative of f at p in the direction u. It is denoted by డ
డ௨ሺሻ or fu(p).
E[ample 2. : Find the directional derivati ves of the folloZing functions:
i f x, y) 2xy + 3y2 at p 1, 1 , in the direction of v 1, 1 . munotes.in
Page 25
25Chapter 2: Diff erentiation of Functions of Several Variables ii f(x, y x2y at p 3, 4 , in the direction of v 1, 1 .
Solution : i The unit vector u in the given direction is ଵ
ξଶǡଵ
ξଶ . Hence the
reTuired directional derivative is 0lim
ot൬ሺଵǡଵሻା௧ሺభ
ξమǡభ
ξమሻ൰ିሺଵǡଵሻ
௧.
0lim
ot൬ቀଵା
ξమǡଵା
ξమቁ൰ିሺଵǡଵሻ
௧
0lim
otଶሺଵା
ξమሻమାଷሺଵା
ξమሻమିହ
௧ 0lim
otହξଶ௧ାହ௧మȀଶ
௧ 5ξʹ.
ii We have the same unit vector u here. Therefore,
Duf(p) 0lim
ot൬ቀଷା
ξమǡସା
ξమቁ൰ିሺଷǡସሻ
௧ 0lim
otቀଷା
ξమቁమ
ሺସା௧Ȁξଶሻିଷ
௧ ଷଷξଶ
ଶ.
E[ample 2. : Find the directional derivatives, if they e[ist, in the follo Zing
cases:
i f x, y ൜ݔݕǡݕݔ݂݅ ൌͲ
ͳǡ ݁ݏ݅ݓݎ݄݁ݐൠ, at 0, 0 , u u1, u2 , __u__ 1
ii f x, y ൝௫௬మ
௫మା௬రǡ݂݅ሺݔǡݕሻ് ሺͲǡͲሻ
Ͳǡ݂݅ ሺݔǡݕሻൌ ሺͲǡͲሻൡ at 0,0 , u 1 ξʹ, 1ξʹ .
Solution: i if u1 ് 0, u2 ് 0, 0lim
ot൫ሺା௧௨భǡା௧௨ మሻ൯ିሺǡሻ
௧ 0lim
otଵି
௧ , Zhich
does not e[ist. If either u1 or u2 is ]ero, Ze get the standard basis vectors, 1, 0
and 0, 1 .
If u 1, 0 , 0lim
ot൫ሺା௧ǡሻ൯ିሺǡሻ
௧
௧՜௧ି
௧ 1.
Similarly, if u 0, 1 , 0lim
ot൫ሺǡା௧ሻ൯ିሺǡሻ
௧ 1. munotes.in
Page 26
26A1ALIn any other direction, the derivative does not e[ist. 1ote tha t the directional
derivative in the direction 1, 0 is fx, and that in the direction 0, 1 is fy. Thus, this
function has both the partial derivatives at 0, 0 .
ii
௧՜൬ቀା
ξమǡା
ξమቁ൰ିሺǡሻ
௧
௧՜యȀమξమ
మ
మశర
రି
௧
௧՜ξଶ
ሺଶା௧మሻ 1 ξʹ.
Thus, D uf 0, 0 1 ξʹ.
In fact, if Ze taNe u cosߠ , sinߠ , then Ze can shoZ that f has directional
derivative at 0, 0 in the direction of u, Zhatever be ߠ .That is, the directional
derivatives of f at 0, 0 e[ist in all directions. %ut you can easily shoZ that this
function is not continuous at 0, 0 by using the tZo-path test . Recall, that you need
to shoZ that the limits of f, at 0, 0 along tZo different paths are different. Then
by Theorem 2.2 Ze can conclude that f is not differentiable at 0, 0 .
This e[ample shoZs that the e[istence of all directional deriva tives at a point does
not guarantee differentiability there. %ut Ze have the folloZin g theorem:
Theorem 2.7 : Let f : E ՜ R, Zhere E is an open subset of Rn. If f is differentiable
at p א Rn, then the directional derivatives of f at p e[ist in all directions.
Proof : Since f i s d i f f e r e n t i a b l e a t p, there e[ists a linear transformation,
T: Rn ՜ R, such that
0lim
ohȁሺାࢎሻିሺሻି்ሺࢎሻȁ
צࢎצൌͲ.
Let u be any unit vector in Rn, and taNe h tu. Then h ՜ 0, as t ՜ 0. Therefore, 0lim
otȁሺା௧࢛ሻିሺሻି்ሺ௧࢛ሻȁ
ȁ௧ȁൌͲ. This means,
0lim
otȁሺା௧࢛ሻିሺሻି௧்ሺ࢛ሻ
ȁ௧ȁȁൌͲ. That is, 0lim
otሺା௧࢛ሻିሺሻ
௧ൌ T u , or, D uf(p T u . ...................... 2.5 munotes.in
Page 27
27Chapter 2: Diff erentiation of Functions of Several Variables Since u Zas an arbitrary unit vector, Ze conclude that the directional derivatives of
f at p e[ist in all directions.
1oZ, if u = (u 1, u2, ..., u n , T u T u1e1 + u2e2 + ... + u nen , Zhere ^ e1, e2, ..., e n`
is the standard basis of Rn. Therefore, by 2.5 ,
T u u1T e1 u2T(e2) + ... + u nT(en)
u1 ܦభf(p) + u2 ܦమf(p) + ... + un ܦf(p)
u1డሺሻ
డ௫భ u2డሺሻ
డ௫మ ... unడሺሻ
డ௫
f(p)x u
Thus, D uf(p f(p)x u ........................ 2.6
2.6 gives an easy Zay to find a directional derivative of a d ifferentiable function,
if its partial derivatives are NnoZn. For e[ample, if f [, y x2 + y2, then fx and fy
at 1, 2 are 2 and 4, respectively. So, the directional deriva tive of f at 1, 2 in the
direction 2i – 3M is given by 2i 4M לሺଶିଷ
ξଵଷሻ ି଼
ξଵଷ.
This concept of directional derivatives can be e[tended to vect or-valued functions.
The directional derivative of a vector-valued function is a vec tor formed by the
directional derivatives of its coordinate functions. Thus, to f ind the directional
derivative of f(x, y) = (x + y, x2), at 1, 2 in the direction of 3, 4 , Ze first find
the directional derivatives of f1 x, y) = x + y, and f2(x, y) = x2 . these are 75 and 65, respectively. Therefore, the reTuired di rectional derivative of
f is 75, 65 .
We have seen in Theorems 2.6 and 2.7, that differentiability of f at a point
guarantees the e[istence of partial and directional derivatives there. We have also
noted that the converse statements are not true. Our ne[t theor em gives us a
sufficient condition Zhich guarantees the differentiability of a function at a point.
Theorem 2. : Let E be an open subset of Rn, and f : E ՜Rm, f = (f 1,f2, ...,f m . If
all the partial derivatives, D jfi(x) of all the coordinate functions of f e[ist in an open munotes.in
Page 28
28A1ALat a.
Proof : In the light of Th eorem 2.3, it is enough to prove this theo rem for the case
m 1. So, Ze consider a scalar function f from Rn to R, all Zhose partial derivatives
DMf are continuous at a. Since E is open, for a given ߝ ! 0, Ze can find r ! 0, such
that the open ball,
% a, r ܧؿ ,and __ x – a _ _ r ฺ_ D jf(x) Djf(a _ ߝn, for M 1, 2, ... , n.
. ................. 2 .7
1oZ, suppose h = (h 1, h2, ... , h n), __h__ r. Let v0 = 0, v1 = h 1e1, v2 = v 1 + h 2e2, .. ,
vn = vn – 1 + h nen. Then f(a + h) – f (a) σሾ݂൫ࢇ࢜ ൯െ݂ሺࢇ࢜ ିሻሿ
ୀଵ .
. .............. 2. 8
Since __ vj|_ r, vj א % a, r , and since % a, r is conve[, the line segment Moining the
points,
a+ vj – 1 and a + vj lies in it, for all M 1, 2, ... , n. Therefore, Ze can appl y the Mean
Value Theorem to the Mth term in the sum 2.8 , and get
f(a + vj) – f(a + vj − 1) = h jDjf(a + vj – 1 + ߠhjej) , for some ߠא 0, 1 . Then, using
2.7 , Ze can Zrite
_f(a + h) – f(a) σ݄ሺ୨݂ሻ
ୀଵ (a)| _σ݄ሺ୨݂ሻ
ୀଵ (a + vj − 1 + ߠhjej )- σ݄ሺ݂ሻ
ୀଵ (a)|
ଵ
σȁ݄ȁߝ
ୀଵ ȁȁࢎȁȁߝ ,for all h, such that __ h__ r.
This means that 0lim
ohࢌצሺࢇାࢎሻିࢌሺࢇሻିࢌ′ሺࢎሻצ
צࢎצൌ 0, Zhere ݂′ is the lin ear transformation, Zhose matri[
>݂′ሺࢇሻ@ consists of the roZ, D 1f(a), D2f(a), ...., Dnf(a)).
Thus, f is differentiable at a.
Definition 2.4 : A function f : E ՜Rm, f = (f 1,f2, ...,f m), Zhere E is an open subset
of Rn, i s s a i d t o b e continuously differentiable , or, a C1 f u n c t i o n , i f D Mfi i s
continuous on E for all M, M 1, 2, ..., n, and for all i, i 1, 2, ..., m. munotes.in
Page 29
29Chapter 2: Diff erentiation of Functions of Several Variables The continuity of partial derivatives assumed in Theorem 2.8, i s only a sufficient
condition, and not a necessary one. That is, there may be funct ions Zhich are
differentiable at a point, but do not have continuous partial d erivatives there. We
noZ give you an e[ample, and asN you to ZorN out the details S ee E[ercise 3.
E[ample 2.10 : Consider the function f : R2→ R given by
f(x, y) ەۖ۔ۖۓݔଶ݊݅ݏଵ
௫ݕଶ݊݅ݏଵ
௬ǡݕݔ݂݅ ് Ͳ
ݔଶ݊݅ݏଵ
௫ǡݔ݂݅ ് Ͳǡݕ ൌ Ͳ
ݕଶ݊݅ݏଵ
௬ǡݔ݂݅ ൌ Ͳǡݕ ് Ͳ
Ͳǡݔ݂݅ ൌ Ͳ ൌ ݕ
This function is differentiable at 0, 0 , but neither
݂௫ൌቊʹ݊݅ݏݔଵ
௫െݏܿଵ
௫ǡݔ݂݅ ് Ͳ
Ͳǡ ݔ݂݅ ൌ Ͳ, nor ݂௬ൌቊʹ݊݅ݏݕଵ
௬െݏܿଵ
௬ǡݕ݂݅ ് Ͳ
Ͳǡ ݕ݂݅ ൌ Ͳis
continuous at 0, 0 .
Here are some e[ercises that you should try.
E[ercises :
1 ShoZ that the folloZing function is differentiable at all x in Rn.
f : Rn ՜ ࡾ ,f(x) = xx T x , Zhere T : Rn ՜Rn is a linear transformation.
2 Let f(x, y x3 + x, x2 – y2, 2x + 3y3 , p 2, 1 , v 4, 5 . Compute the
partial derivatives of f, and the directional derivative of f in the direction v,
at p.
3 Prove the assertions in E[ample 2.10. Hint: To shoZ that f i s d i f f e r e n t i a b l e ,
checN that f(h, k) - f(0, 0) – h(hsinଵ
) + k(ksinଵ
) 0, and so, D f hsinଵ
, ksinଵ
) .
munotes.in
Page 30
30A1ALIn this unit Ze have e[tended the concept of differentiation fr om functions of one
variable to functions of several variables. Apart from the tota l derivatives, Ze have
also defined partial derivatives, and directional derivatives. We have proved that
differentiability implies the e[istence of all partial and dire ctional derivatives at a
point, but the converse is not true. As in the case of function s of one variable, Ze
prove that differentiable functions are continuous, but not vic e versa. We have also
derived a sufficient condition for differentiability in terms o f the partial derivatives.
munotes.in
Page 31
UNIT 2
3
DERI9ATI9ES O) +I*+ER ORDER
Unit Structure
3.0 ObMectives
3.1 Introduction
3.2 -acobian Matri[ and Chain Rule
3.3 Higher order partial derivatives
3.4 Mean Value Theorem
3.5 Summary
3.0 Objectives
After reading this chapter, you should be able to
x differentiate a composite of tZo vector-valued functions
x define and calculate derivatives of higher order
x derive the conditions for the e Tuality of mi[ed partial derivat ives
x state and prove the Mean Value Theorem
3.1 Introduction
In the last chapter you have seen hoZ functions of several vari ables are
differentiated. 1oZ Ze shall start by discussing hoZ a composit e function of tZo
differentiable functions can be differentiated. The -acobian ma tri[ introduced in
the last chapter proves useful in this.
One of the important applications of derivatives is the locatio n of e[treme points
of a function. In the ne[t chapter Ze are going to see hoZ this c o n c e p t c a n b e
e[tended to scalar functions of several variables. %ut Ze shall do the necessary
spade-ZorN in this chapter. So, Ze shall introduce higher order derivatives. We 31munotes.in
Page 32
32A1ALmay recall that the Mean Value Theorem Zas one of the most impo rtant theorems
that you studied in Calculus in F. <. %. Sc. We shall see Zheth er this theorem can
be applied to functions of several variables.
3.2 Jacobian Matri[ and Chain Rule
We have seen in Theorem 3.5, that if f: Rn Rm, is differentiable at p, then all
partial derivatives of all coordinate functions of f e[ist at p. That is,
if f = (f 1, f2, ... , f m , then D jfi(p) e[ists for all i 1, 2, ..., m and all M 1, 2, ..., n.
We have also seen that if ^ e1, e2, ..., en` is the standard basis for Rn, then
(p)(ej) = (Djf1(p), Djf2(p), ..., Djfm(p)).
If h is a vector in Rn, then (p)(h) (p),
Zhich is a linear transformation from Rn to Rm, thus has the matri[,
As Ze have already mentioned in Chapter 3, this m [ n matri[, c alled the -acobian
matri[, is denoted by >D f(p @. The Nth roZ of this matri[ is the gradient vector,
fk(p , and the Mth column is the image of ej under the linear transformation D jf(p).
Thus, the -acobian matri[ of f is formed by all first order partial derivatives of f.
This means, Ze can Zrite the -acobian matri[ of any function, a ll of Zhose partial
derivatives e[ist. As Ze have noted earlier, the e[istence of p artial derivatives does
not guarantee differentiability. So, even Zhen a function is no t differentiable Ze
Zould be able to Zrite its -acobian matri[, provided all its pa rtial derivatives e[ist.
If f : Rn R, then its -acobian matri[, if it e[ists, Zill be a 1 [ n matri[, or a matri[
vector.
If f : Rn Rm is differentiable at p Rn, and if h is any vector in Rn, then
(p)(h) = >Df(p)@h is obtained by multiplying the m [ n matri[ >D f(p @ Zith the
n [ 1 column matri[ h. Thus,
f
R
munotes.in
Page 33
33Chapter 3: Derivatives of Higher Orderצצࢌᇱሺሻሺࢎሻצצൌצצσୀଵԝԝ൫݂ሺሻήࢎ൯ࢋצצσୀଵԝצצ൫݂ሺሻήࢎ൯ࢋצצൌσୀଵԝห൫݂ሺሻήࢎ൯ห,
Since צצࢋצצൌͳ ǡ ͳ݆
Cauchy-SchZart] ineTuality for inner products says that ȁ࢛ή࢜ȁ צ࢜צצ࢛צ Ǥ
Using this Ze get צצࢌᇱሺሻሺࢎሻצצσୀଵԝצצ݂ሺሻצצצࢎצ ൌ צࢎצσୀଵԝצצ݂ሺሻצצ .
If Ze taNe ൌσ୨ୀଵ୫ԝצצ୨ሺܘሻצצ ,then צצࢌᇱሺሻሺࢎሻצצצࢎצ ....... ..................3.1
We have seen in Theorem 3.4 hoZ to get the derivative of the su m of tZo
differentiable functions, and also that of a scalar multiple of a d i f f e r e n t i a b l e
function. The ne[t theorem, Zhich is NnoZn as the chain rule, t ells us hoZ to get
the total derivative of a composite of tZo functions.
Theorem 3.1 Chain Rule : Let f and g be tZo differentiable functions, such that
the composite function ݂ή݃ is defined in a neighbourhood of a point a Rn.
Suppose g is differentiable at a, g(a) = p, and f is differentiable at
p. Then ݂ή݃ i s d i f f e r e n t i a b l e a t a, and ሺࢍלࢌሻᇱሺࢇሻ ൌ ࢌᇱሺሻࢍלᇱሺࢇሻ ൌ
ሾࢌሺሻሿሾࢍሺࢇሻሿ
Proof : If ࢎ is such that צࢎצ i s s m a l l , t h e n ࢇࢎ Z i l l b e l o n g t o t h e a b o v e
neighbourhood of ܽ ,in Zhich ݂ή is defined. 1oZ, since ݃ is differentiable at ࢇ ,
ൌࢍ ሺ ࢇࢎ ሻെࢍ ሺ ࢇ ሻൌࢍᇱሺࢇሻሺࢎሻࡱצࢎצ ࢇሺࢎሻ, ....... 3.2
Zhere Ea(h) 0, as h 0.
f is differentiable at p = g(a , and therefore, f(g(a + h)) – f(g(a)) = f(p + k) – f(p)
= ( p)(k) + || k || Ep(k), Zhere Ep(k) 0, as k 0.
= ( g(a))[ g(a + h) – g(a)] + || k || Ep(k)
= ( g(a))[ ( a)(h) + || h || Ea(h)] + || k || Ep(k), using 3.2 .
= ( g(a)) ( a)(h) + ( g(a)) [|| h || Ea(h)] + || k || Ep(k), since g a is a
linear transformation. Thus, Ze can Zrite f(g(a + h)) – f(g(a)) = ( g(a)) ( a)(h)
+ || h ||> (g(a)) Ea(h) + Ep(k)@, if h 0. ……. 3.3
R
[
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34A1AL 3.3 tends to ]ero, as h tends to ]ero.
We NnoZ that Ea(h) 0, as h 0. ..............
__ k || = || g(a + h) – g(a) || || ( a)(h) || + || h || || E a(h) __, using 3.2 .
If M , then using 3.1 , Ze can Zrite __ a h __ M __ h __. Thus,
__ k __ M __ h | | + | | h || || E a(h) _ _ _ _ h __ M __ Ea(h) _ _ . T h e r e f o r e ,
M __ Ea(h __. This means that is bounded. Thus,
0, as h 0, since h 0 . ....
Using
and
, Ze can say that the term in the sTuare brac Nets in 3.3 tends to
]ero as h
0. Therefore,
ሺሺାሻሻିሺሺሻሻିᇲሺሺሻሻᇲሺሻሺሻ
צצ՜Ͳ as ݄՜Ͳ
This shoZs that ݂ή݃ is differentiable at a, and ሺࢌήࢍሻᇱሺࢇሻ ൌ ࢌᇱሺࢍሺࢇሻሻࢍלᇱሺࢇሻ
The Chain Rule can be Zritten in terms of -acobian matrices as folloZs:
ሺࢌήࢍሻሺࢇሻൌቂ ቀ ࢌ ൫ ࢍ ሺࢇሻ൯ቁቃൣ൫ࢍሺࢇሻ൯൧.
Here the product on the right hand side is matri[ multiplicatio n. If y = g(x), and z
= f(y), comparing the entries in the matrices in 3.3 , Ze get
ப௭
ப௫ೖൌσୀଵԝப௭
ப௬ೕப௬ೕ
ப௫ೖǡ Zhere ப௭
ப௫ೖൌ୩ሺ݂ ή݃ሻ ୧ǡப௭
ப௬ೕൌሺ݂ሻ୧, and ப௬ೕ
ப௫ೖൌ୩ሺࢍሻ୨.
E[ample 3.1 : Write the matrices for ࢌᇱǡࢍᇱ and ሺࢌήࢍሻᇱ for the folloZing functions,
and evaluate them at the point ሺʹǡͷሻ݂ڄሺݔǡݕሻ ൌ ሺݔݕǡݔଶݕଶǡʹݔ
͵ݕሻǡ݃ሺݑǡݒሻ ൌ ሺݔǡݕሻ ൌ ሺݑଶǡݒଷሻ
Solution : Here f1(x, y) = x + y, f 2(x, y) = x2 + y2, f3(x, y) = 2x + 3y,
݃ଵሺݑǡݒሻ ൌ ݑଶ and ݃ଶሺݑǡݒሻ ൌ ݒଷǤ T h i s m e a n s , ሺ݂ሻ ൌ൭ͳͳ
ʹݔ ʹݕ
ʹ͵൱,
and ሺ݃ሻ ൌ ቀʹݑ Ͳ
Ͳ͵ ݒଶቁ.
ሺࢌήࢍሻሺݑǡݒሻ ൌ ሺݑଶݒଷǡݑସݒǡʹݑଶ͵ݒଷሻǤ
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35Chapter 3: Derivatives of Higher OrderHence,
ሺࢌήࢍሻ ൌ ൭ʹݑ ͵ݒଶ
Ͷݑଷݒହ
Ͷݑ ͻݒଶ൱
At u, v) 2, 5 , x, y 4, 125 . Therefore,
ሺ݂ሻሺͶǡͳʹͷሻ ൌ ൭ͳͳ
ͺʹ ͷ Ͳ
ʹ͵൱ǡሺ݃ሻሺʹǡͷሻൌቀͶͲ
Ͳ ͷቁ, and
ሺࢌήࢍሻሺʹǡͷሻ ൌ ൭Ͷ ͷ
͵ʹ ͳͺͷͲ
ͺʹ ʹ ͷ൱
3.3 +igher Order Partial Derivatives
calculated the partial derivatives of some functions of n varia bles. If you taNe a looN
at those e[amples, you Zill realise that the partial derivative s are themselves
functions of n variables. So, Ze can talN about their partial d erivatives. These, if
they e[ist, Zill be the second order partial derivatives of the original function. If
Ze differentiate these again, Ze Zill get the third order parti al derivatives of the
original function, and so on. We taNe a simple e[ample to illus trate.
E[ample 3.2 : Find partial derivatives of all possible orders for the func tion,
f(x, y, z) = (x2y2, 3xy3z, xz3).
Solution : Since f is a polynomial function, Ze do not have to Zorry about the
e[istence of partial derivatives. We get fx = (2xy2, 3y3z, z3), fy = (2x2y, 9xy2z, 0),
fz = (0, 3xy3, 3xz2).
Then, ݂௫௫ൌபమ
ப௫మൌሺʹݕଶǡͲǡͲሻǡ݂௫௬ൌப
ப௬ቀப
ப௫ቁൌபమ
ப௬ப௫ൌሺͶݕݔǡͻݕଶݖǡͲሻǡ
݂௫௭ൌሺͲǡ͵ݕଷǡ͵ݖଶሻ
Differentiating fy, Ze get fyx = (4xy, 9y2, 0 , fyy = (2x2, 18xyz, 0 , and fyz = (0, 9xy2, 0 .
Then differentiating fz Ze get fzx = ( 0 , 3 y3, 3z2 , fzy = ( 0 , 9 x y2, 0 , and
fzz = (0, 0, 6xz). munotes.in
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36A1ALalso get fxyz = (0, 9y2, 0), fyxz = (0, 0, 0), fzzz = (0, 0, 6x), and so on. There Zill be
27 third order partial derivatives of f. See if you can get the remaining.
to the variables [ and y. These tZo derivatives have come out t o be eTual in
E[ample 3.2. %ut you may have seen e[amples of scalar functions o f s e v e r a l
variables, for Zhich the tZo may not be the same. Here is an e[ ample, to Mog your
memory.
E[ample 3.3 : C o n s i d e r t h i s f u n c t i o n from ܴଶ to ܴǡ݂ሺݔǡݕሻ ൌ௫௬൫௫మି௬మ൯
௫మା௬మ for
ሺݔǡݕሻ ് ሺͲǡͲሻ , and ݂ሺͲǡͲሻ ൌ Ͳ . ݂௫ሺͲǡͲሻ ൌ Ͳǡ݂ ௬ሺͲǡͲሻ ൌ Ͳǡ݂ ௫ሺͲǡ݇ሻ ൌ ՜ԝ݂ሺ݄ǡ݇ሻെ݂ሺͲǡ݇ሻ
݄ൌെ ݇
݂௬ሺ݄ǡͲሻ ൌ ՜ԝ݂ሺ݄ǡ݇ሻെ݂ሺ݄ǡͲሻ
݇ൌ݄
Then, ݂௫௬ሺͲǡͲሻ ൌ ՜ԝೣሺǡሻି ೣሺǡሻ
ൌ ՜ԝିି
ൌെ ͳ, and similarly,
݂௬௫ሺͲǡͲሻ ൌ ͳǤ
Thus, the mi[ed partial derivatives of this function both e[ist , but are not eTual.
Remark 3.1 : If f is a function from Rn to R, the partial derivative of f Zith respect
to the ith variable, xi, is denoted by D if, and the partial derivative of D if Zith respect
to xj , that is, D M Dif is denoted by D Mif.
The folloZing theorem gives a sufficient condition for the tZo mi[ed partial
derivatives of a function to be eTual. Since the behaviour of a vector-valued
function is decided by the behaviour of its coordinate function s, it is enough to
derive this sufficient condition for a scalar function. Without loss of generality, Ze
state the theorem for a function of tZo variables.
Theorem 3.2 : Let f : R2 R, such that the partial derivatives, D 1f, D2f, D12f and
D21f e [ i s t o n a n o p e n s e t S i n R2. If a, b S , a n d D 12f and D 21f a r e b o t h
continuous at a, b , then D 12f(a, b) D 21f(a, b).
Proof : We choose positive real numbers , h and k, Zhich are small enough so that
the rectangle Zith vertices a, b), (a + h, b), (a, b + k), (a + h, b + k lies Zithin S.
1oZ Ze consider a function
h, N >f a h, b N – f a h, b @ – >f a, b N – f a, b @.
munotes.in
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37Chapter 3: Derivatives of Higher OrderWe also define a function * on > a, a+h @, * [ f [, b N – f [, b .
1oZ Ze can Zrite h, k * a + h – * a . Since * is defined in terms of f, and
since f h a s a l l t h e n e c e s s a r y p r o p e r t i e s , * i s c o n t i n u o u s o n > a, a + h @, and is
differentiable in a, a + h . So, Ze apply the Mean Value Theorem for functions of
a single variable to *, and get
* a + h – * a) h c , for some c (a, a + h). 1oZ x D 1f(x, b + k) –
D1f x, b). So, Ze Zrite h, k) = * a + h – * a = h>D1f(c, b + k) – D1f c, b)@.
1oZ D 1f (c, y) is a differentiable function of one variable Zith derivative e Tual to
D21f. So applying MVT to D 1f(c, y on the interval > b, b + k @, Ze get
h,k) = h >D1f(c, b + k) – D1f(c, b) @ = hk D21f(c, d), ............................... 3.4
for some d (b, b + k).
We noZ Zrite h, k) = [f(a + h, b + k) – f(a, b + k)] – [f(a + h, b) – f(a, b)], and
define
H y) = f(a + h, y) – f(a, y), so that h, k) = H(b + k) – H(b . Using the same
arguments that Ze used for *, Ze apply MVT to H, and then to D 2f(x, p , Ze get
h, k)= k >D2f(a + h, p) – D2f(a, p) @ = kh D12f(q, p), …………………… 3.5
for some p (b, b + k , and q (a, a + h).
From 3.4 and 3.5 Ze get D 21f(c, d) = D12f(q, p). Since D 12f and D 21f a r e
continuous, taNing the limit as h, k 0,0 , Ze get D 12f(a, b) D 21f(a, b).
As Ze have mentioned earlier, the conditions of this theorem ar e sufficient, and not
necessary. In fact, the continuity of Must one of the mi[ed par tial derivatives is also
sufficient to guarantee eTuality. Functions Zhose partial deriv atives are continuous
play an important role in Calcul us. We classify these functions as folloZs:
Definition 3.1 : A f u n c t i o n f from Rn to Rm i s s a i d t o b e continuously
differentiable , or belong to class C1, if all its partial derivatives D if are continuous.
It is said to belong to class C’’, if all its second order partial derivatives are
continuous, and so on. If all its partial derivatives of all or ders are continuous, then
it is said to belong to class .
We have proved that a function in class C1 is differentiable in Theorem 3.8. In
Theorem 3.2 Ze have seen that the mi[ed partial derivatives of a function belonging
to class C’’ are eTual.
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38A1ALpartial derivatives of order up to N are continuous, can be app ro[imated by means
of a polynomial of order N. We shall also discuss the techniTue to find the ma[imum
and minimum values of a function belonging to class C’’.
3.4 Mean 9alue Theorem
The Mean Value Theorem MVT is an important theorem in Calculu s. It is used
as a tool to derive many other results. In the last section Ze have used it in the proof
of Theorem 3.2. In this section Ze shall see if it also holds g ood for functions of
several variables. %ut first, let us recall the one-variable ca se.
M9T single variable : If f : >a, b@ R is continuous on > a, b@, and differentiable
on a, b , then there e[ists c (a, b), such that f b – f a b - a c .
If Ze Zrite b = a + h , then there e[ists , such that
f a h – f a = h .
Unfortunately, it is not possible to e[tend this theorem to a f unction f : Rn Rm,
Zhen m ! 1. This Zill be Tuite clear from the folloZing e[ample .
E[ample 3.4 : Consider f : >0, 2 @ R2, f t = (cost, sint . This function is
continuous on >0, 2 @ and differentiable on 0, 2 .
1oZ, f (2 ) – f(0) 1, 0 – 1, 0 0, 0 .
(t) = ( − sint, cost). For the extension of MVT to hold, we mus t have
f(2 ) – f(0) = 2 (c for some c in 0, 2 . So, Ze should have 0, 0 2
sinc, cosc). %ut this is impossible, since sinc and cosc both cannot be ]ero.
So, the e[tension of MVT in its stated form does not hold. %ut there is a Zay around
this difficulty. A slightly modified version of MVT does hold t rue for all functions
of several variables. We noZ state and prove this modified theo rem for functions
from Rn to Rm. As a special case of this theorem you Zill reali]e that MVT h olds
for real-valued functions of several variables.
Theorem 3.3 : Mean Value Theorem Let ݂ǣܵ ՜ Թ, Zhere S is an open subset
of Թ. Suppose f is differentiable on S. Let x and y be tZo points in S, such that the
line segment Moining x and y, L x, y ^ tx + 1 t y _ Ͳݐͳ `, also lies in S.
Then for every a Rm, there is a point z S,
such that a{f(y) – f(x)} = a { ( z)(y x ` ........................................ 3.6
)
)
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39Chapter 3: Derivatives of Higher Order%efore Ze start the proof, let us understand the geometry invol ved. Let u = y – x.
Then x + tu gives us a point on the line segment L x, y), if 0 1. Since S is
open, Ze can find a ! 0, such that S, and S. See Fig.
3.1, in Zhich Ze shoZ the sit uation Zhen n 2. The point p is on the e[tension of
L x, y) and is eTual to x + (1 + ) u. Similarly the point q is also on the e[tension
of L x, y , and is eTual to x – u for some ! 0.
p
)igure 3.1
Thus Ze get a ! 0, such that x + tu S for every t . 1 o Z Z e
start the formal proof.
Proof : Let a Rn. We define a function F : R, F(t) = a f(x + tu .
This F is a differentiable function on , and (t) = a, using chain rule.
Recall, that is a linear transformation.
Thus, Ze can apply MVT for functions of a single variable, and get
ܨሺͳሻെܨሺͲሻ ൌ ܨᇱሺߠሻ, for some ߠǡͲ൏ߠ ൏ͳ ........................................... 3.7
ǡܨሺͳሻ ൌ ࢇήࢌሺ࢛࢞ሻ ൌ ࢇήࢌሺ࢟ሻǡܨሺͲሻ ൌ ࢇήࢌሺ࢞ሻ , and
ܨᇱሺߠሻ ൌ ࢇή ሼࢌᇱሺ࢛࢞ߠሻሺ࢛ሻ ሽൌࢇήሼࢌᇱሺࢠሻሺ࢟െ࢞ሻ ሽ, Zhere ࢠൌ࢞ܮא࢛ߠ ሺ ࢞ ǡ࢟ ሻ
Therefore, from 6.7 Ze get ࢇήሼࢌሺ࢟ሻെࢌሺ࢞ሻሽ ൌ ࢇή ሼࢌᇱሺࢠሻሺ࢟െ࢞ሻ ሽ
for some אࢠ .
)
p
q Y X S munotes.in
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40A1ALi 3.6 is true for all x, y in S, such that the line segment Moining x and y is also
in S. This means, if S is a conve[ open set in Rn, then 3.6 Zill be true for
all x, y in S.
ii If f is a real-valued function, then m 1, and a R. Then for a 1 Ze
have 1. ሼ݂ሺ࢟ሻെ݂ሺ࢞ሻሽ ൌ ͳǤ ሼ݂ᇱሺܢሻሺ࢟െ࢞ሻ ሽൌ સ݂ሺࢠሻήሺ࢟െ࢞ሻ ,
for some אࢠ.
So, the MVT for functions of a single variable e[tends directly t o r e a l - v a l u e d
functions of several variables. We can also directly prove MVT for scalar functions.
The proof runs e[actly similar to that of Theorem 3.3, if Ze pu t a 1.
The MVT has a Zell-NnoZn conseTuence, Zhich Ze noZ state:
Theorem 3.4 : Let f : S Rm, Zhere S is an open connected subset of Rn.
Suppose f is differentiable on S, and 0 for every p S. Then f is a
constant function on S.
Proof : The set S is polygonally connected, since it is open and con nected. Let x
and y be tZo points in S. Then x and y are Moined by line segments L 1, L2, L3, ... ,
Lr, lying entirely in S. Suppose L i is a line segment Moining pi and pi+1, 1
r, p1 = x, and pr+1 = y.
Let ܴאܽ. Then using Theorem ͵Ǥ͵, Ze have
ܽήሼ݂ሺାଵሻെ݂ሺሻሽൌ ܽήሼࢌᇱሺܢሻሺାଵെሻሽǡݖא
ൌ Ͳǡ since ࢌᇱሺܢሻൌ
This means,
ܽήሼ݂ሺݕሻെ݂ሺݔሻሽ ൌ ܽή ሼ݂ሺାଵሻെ݂ሺଵሻሽൌσԝ
ୀଵܽήሼ݂ሺାଵሻെ݂ሺሻሽൌ
Ͳ……ሺ͵Ǥͺሻ
ሺ͵Ǥͺሻ is true for every ܽ in ܴ. So, in particular, it is true for ݂ሺݕሻെ݂ሺݔሻ .
Thus,
ሼ݂ሺݕሻെ݂ሺݔሻሽήሼ݂ሺݕሻെ݂ሺݔሻሽ ൌ݂צ ሺݕሻെ݂ሺݔሻ צଶൌͲ
So, ݂ሺݕሻെ݂ሺݔሻ ൌ Ͳǡ or ݂ሺݕሻ ൌ ݂ሺݔሻ .
Since x and y Zere any arbitrary points in S, Ze have thus proved that f is a constant
function on S.
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41Chapter 3: Derivatives of Higher OrderTry a feZ e[ercises noZ.
E[ercises :
1 Find the partial derivatives, D 1f, D2f, D12f and D 21f at 0, 0 , if they e[ist, for
the folloZing function f from R2 to R.
f (x, y) = y , if (x, y) (0, 0), and f(0, 0) = 0.
2 If u(x, y) = x +y2, x(t) = 3t2 + 4, and y(t) = sin2t, find t and t .
3 If u(x, y) = x – 2y + 3, x = r + s + t, y = rs + t2, find ur, us and ut at
1, 2, 4 .
4 Let f : R2 R2, and g : R3 R2 be tZo vector functions, defined as:
f(x, y) = (sin(2x + y), cos(x + 2y)), g(r, s, t) = (2r – s – 3t, r2 – 3st).
i Write the -acobian matrices for ݂ and ݃Ǥ If ݄ is the composite function,
݂ή݃ ,compute the -acobian matri[ of ݄ at the point ሺͳǡͲǡെʹሻ .
5 If f i s a f u n c t i o n f r o m R2 to R, and D 1f 0 at all points, shoZ that f i s
independent of the first variable. If D 1f D 2f 0 at all points, shoZ that f is
a constant function.
3. Summary
In this chapter Ze have derived the chain rule for differentiat ion of composite of
tZo functions. We have also seen that the -acobian matri[ for t he composite
function is the product of the -acobian matrices of the tZo giv en functions. We
have defined higher order partial derivatives of functions of s everal variables. We
have seen functions, Zhose second order mi[ed partial derivativ es depend on the
order of the variables Zith respect to Zhich the function is di fferentiated. On the
other hand, Ze have derived sufficient conditions for such mi[e d partial derivatives
to be eTual. Finally, through an e[ample Ze have seen that the Mean Value
Theorem cannot be e[tended to all vector functions. We have pro ved a restricted
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42A1ALvalued functions of several variables. As a result of MVT Ze ha ve proved that a
function defined on an open connected set is constant, if its d erivative is uniformly
]ero over its domain.
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UNIT 2
4
APPLICATIONS O) DERI9ATI9ES
Unit Structure
4.0 ObMectives
4.1 Introduction
4.2 Taylor’s Theorem
4.3 Ma[ima and Minima
4.4 Lagrange’s Multipliers
4.5 Summary
4.0 Objectives
After reading this chapter, you should be able to
x state Taylor’s theorem for real -valued functions of several variables
x obtain Taylor’s expansions for some simple functions
x define, locate and classify e[treme points of a function of se veral variables
x obtain the e[treme values of a function of n variables, subMect t o s o m e
constraints
4.1 Introduction
In the tZo previous chapters Ze have discussed differentiation of scalar and vector
functions of several variables. 1oZ Ze shall tell you about som e applications of
derivatives. In your study of functions of one variable you hav e seen that a maMor
application of the concept of derivatives is the location of ma [ima and minima of
a function. This NnoZledge is very crucial for curve tracing. H ere Ze shall see hoZ
the derivatives help us in locating the e[treme values of a rea l-valued function of
several v ariables. But before we do that, we are going to discuss Taylor ’s theorem
and Taylor’s expansions, which help us approximate a function w ith the help of
polynomials. This NnoZledge Zill help us derive some tests for locating and
classifying the e[treme points of a function. 43munotes.in
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44A1ALIt will be useful to recall Taylor’s theorem for functions of o ne variable, Zhich you
have studied in F. <. %. Sc. Here Ze shall also give you the pr oof of this theorem.
Our method of proof involves the use of Rolle’s theorem. You ha ve studied this
theorem too in F. Y. We now state Rolle’s theorem, and then mov e on to Taylor’s
theorem.
Theorem 4.1 (Rolle’s Theorem): If f: >a, b@ ՜
ࡾ
ሾܽǡܾሿǡሺܽǡܾሻǡ݂ሺܽሻൌ
݂ሺܾሻǡאܿ ሺܽǡܾሻǡ
݂ᇱ(c) 0.
Theorem 4.2 (Taylor’s theorem for real functions of one variable): Let f be a real-
valued function defined on the open interval p, q . Suppose f has derivatives of all
orders up to and including n 1 in p, q . Let a be any point in p, q . Then for any
[ א p, T , f(x) = f(a) +1ax݂ᇱ(a) +2 2ax݂ᇱᇱሺܽሻǤǤǤ
naxn݂ሺሻሺܽሻ 1 1
naxn݂ሺାଵሻ(c),………………………………………………………. .. 4.1
Zhere c א a, b .
Proof: We noZ define a neZ function g on >a, x@, or > x, a@, according as a x, or x
a, by g(y) = f(y) +1 yx݂ᇱሺݕሻ2 2yx݂ᇱᇱሺݕሻǤǤǤ
nyxn݂ሺሻሺݕሻ
ሺݔ െݕሻାଵA, ………………………………………………………….. .. 4.2
Zhere A is a constant, chosen so as to satisfy g(x) = g(a). We can easily Zrite the
e[pression for A by using this condition. We leave this to you as an e[ercise. See
E[ercise 1 .
Using the properties of f, Ze can see that g satisfies all the conditions of Rolle’s
theorem on its domain. Thus, Ze can conclude that there e[ists a point c א
ሺܽǡݔሻǡሺሺݔǡܽሻሻ
݃ᇱሺܿሻൌ
ͲǤǡሺͶǤʹሻǡ݃ᇱሺݕሻൌ݂ᇱሺݕሻെ݂ᇱሺݕሻሺݔെ
ݕሻ݂ᇱᇱሺݕሻെሺݔെݕሻ݂ᇱᇱሺݕሻ2 2yx݂ᇱᇱᇱሺݕሻെǤǤǤെ 1 1
nyxn݂ሺሻሺݕሻ
nyxn݂ሺାଵሻሺݕሻȂሺ݊ͳሻሺݔ െݕሻA.
ሺݔ െݕሻሾ 1
ny fnെ (n + 1)A]. munotes.in
Page 45
45Chapter 4: Applications of Derivatives Hence, ݃ᇱሺܿሻൌሺݔ െܿሻሾ 1
nc fn − (n + 1)A] = 0.
This means that A = 1
nc fn
Substituting this value of A in 4.2 , Ze get
f(x) = g(x) =
g(a) = f(a)+1ax݂ᇱ(a) +2 2ax݂ᇱᇱሺܽሻǤǤǤ
naxn݂ሺሻሺܽሻ 1 1
naxn݂ሺାଵሻ(c), thus proving the theorem.
Remark 4.1 : If the function in Theorem 4.2 has derivatives of all orders in p, q ,
then Ze can Zrite a Taylor e[pansion as in 7.1 for any n א
ࡺǤ ǡ orders are bounded by a positive number ܯ ,
that is, if ௗ
ௗ௫פ൏ ܯ f o r a l l ݊ , and at all points in ሺǡݍሻ,
then ቚሺ௫ିሻశభ
ሺାଵሻǨ݂ሺାଵሻሺܿሻቚ ቚሺ௫ିሻశభ
ሺାଵሻǨቚ՜Ͳ as ݊՜λ for every ݔ in some interval
ሼݔǣȁݔ െܽȁ ൏ ܴሽǤ Therefore, in this case Ze can Zrite
݂ሺݔሻൌ݂ሺܽሻ௫ି
ଵǨ݂ᇱሺܽሻሺ௫ିሻమ
ଶǨ݂ᇱᇱሺܽሻǥሺ௫ିሻ
Ǩ݂ሺሻሺܽሻሺ௫ିሻశభ
ሺାଵሻǨ݂ሺାଵሻሺܿሻǡǥ,...(4.3)
The infinite series in 4.3 is convergent under the given cond itions, and is called
the Taylor series of f about a.
1oZ, 7.1 can be Zritten as f(x) = P n(x) + R n(x , Zhere Pn(x) = f a 1ax݂ᇱ a 2 2ax݂ᇱᇱሺሻǥ
naxn݂ሺ୬ሻሺሻ
is called the nth ܉ܜܝܗ܊܉ܗܔ܉ܑܕܗܖܡܔܗܘܚܗܔܡ܉܂ǡ
ܴ݊ሺݔሻൌ 1 1
naxn݂ሺାଵሻ(c , is called the remainder.
We now state Taylor’s theorem for functions of two variables, a nd then find Taylor
e[pansions of some functions.
Theorem Ǥ͵ (Taylor’s theorem for ݂ǣܴଶ՜ܴሻǣ Let ݂ be a real-valued ୬ାଵ
function on an open conve[ set ࡾكଶ. Let ሺܽǡܾሻ א .Then for any ሺݔǡݕሻ א ,munotes.in
Page 46
46A1AL݂ሺݔǡݕሻൌ ݂ሺܽǡܾሻቀ݄డ
డ௫݇డ
డ௬ቁ݂ሺܽǡܾሻଵ
ଶǨቀ݄డ
డ௫݇డ
డ௬ቁଶ
݂ሺܽǡܾሻǥ
ଵ
Ǩቀ݄డ
డ௫݇డ
డ௬ቁ
݂ሺܽǡܾሻଵ
ሺାଵሻǨቀ݄డ
డ௫݇డ
డ௬ቁାଵ
f(c, d), ........... 4.4
Zhere h = x – a, k = y – b, and c, d is some point on the line segment Moining a,
b and x, y .
We are not going to prove this th eorem. %ut, note the folloZing points:
1. Recall that f is Cn1 means f has continuous partial derivatives of all orders
n 1. This ensures that all the relevant mi[ed partial derivat ives are eTual.
2. E is conve[. This guarantees that the line segment Moining any tZo points of E,
lies in E, the domain of f.
Pn(x, y) = f(a, b)+ (hడ
డ௫݇డ
డ௬ሻ݂ሺܽǡܾሻଵ
ଶǨቀ݄డ
డ௫݇డ
డ௬ቁଶ
݂ሺܽǡܾሻǥ
ଵ
Ǩቀ݄డ
డ௫݇డ
డ௬ቁ
݂ሺܽǡܾሻ݄ ൌ ݔȂܽǡ݇ ൌ ݕȂܾǡ
nth Taylor polynomial, and
ܴ݊ሺݔǡݕሻൌଵ
ሺାଵሻǨቀ݄డ
డ௫݇డ
డ௬ቁାଵ
f(c, d) is called the remainder of order n.
Let us use this theorem to get the e[pansions of some functions .
E[ample 4.1 : Find the Taylor e[pansions of the folloZing functions about t he
given points up to the third order.
i f(x, y) = x3 + 2xy2 – 3xy + 4x + 5, (a, b) 1, 2
ii f(x, y) = sin(2x + 3y) (a, b) 0, 0 .
Solution: i Since f(x,y) = x3 + 2xy2 – 3xy + 4x + 5 is a polynomial, it has partial
derivatives of all orders. Furt her, its partial derivatives of order ! 3 are all ]ero. In
fact, fx = 3x2 + 2y2 – 3y + 4, f y = 4xy – 3x, f xx = 6x, f xy = 4y – 3, f yy = 4x, f xxx = 6,
fxxy = 0, f xyy = 4, f yyy = 0, and all higher partial derivatives are ]ero. Calculating all
these partial derivatives at 1, 2 , Ze Zrite f(1 + h, 2 + k) = 12 + 9h + 5k +
ଵ
ଶǨሺ݄ʹͳͲ݄݇Ͷ݇ʹሻଵ
ଷǨ(6h3 + 12hk2) + R 3 .
1oZ, R3 involves all fourth order derivatives, and therefore is ]ero. Hence,
f(1 + h, 2 + k) = 12 + 9h + 5k + ଵ
ଶǨሺ݄ʹͳͲ݄݇Ͷ݇ʹሻଵ
ଷǨ(6h3 + 12hk2) .
munotes.in
Page 47
47Chapter 4: Applications of Derivatives ii f(x, y) = sin(2x + 3y also has derivatives of all orders.
fx = 2cos(2x + 3y) 2 at 0, 0 , fy = 3cos(2x + 3y) 3 at 0, 0 ,
fxx = 4sin(2x + 3y), f xy = 6sin(2x + 3y), f yy = 9sin(2x + 3y . These
second order derivatives are all ]ero at 0, 0 .
fxxx = 8cos(2x + 3y), f xxy = 12cos(2x + 3y),
fxyy = 18cos(2x + 3y), f yyy = 27cos(2x + 3y).
These are, respectively, 8, 12, 18, and – 27 at 0, 0 . Thus, f(h, k) = 0 + (2h
+ 3k) + ଵ
ଶǨǤͲଵ
ଷǨሺͺ݄͵Ȃ͵Ǥͳʹ݄ʹ݇Ȃ͵Ǥͳͺ݄݇ʹȂʹ݄͵ሻܴ͵ǡ ͵ ൌ
ଵ
ସǨሺడ
డ௫݇డ
డ௬ 4sin 2c 3d , Zhere c, d is some point on the line segment Mo ining
0, 0 and h, N .
We are now going to state Taylor’s theorem for real -valued functions of n variables.
For this, let us first taNe a close looN at the Taylor e[pansio n of a function of tZo
variables.
If Ze Zrite x, y as a + h, b + k , Ze get f(a + h, b + k) = f(a, b) + (hడ
డ௫
݇డ
డ௬ሻ݂ሺܽǡܾሻଵ
ଶǨቀ݄డ
డ௫݇డ
డ௬ቁଶ
݂ሺܽǡܾሻǤǤǤଵ
Ǩቀ݄డ
డ௫݇డ
డ௬ቁ
݂ሺܽǡܾሻ
ଵ
ሺାଵሻǨቀ݄డ
డ௫݇డ
డ௬ቁାଵ
f(c, d),
If Ze taNe the variables to x1, x2, instead of x and y, taNe a, b to be a1, a2 , and
h, k to be
݂ሺܽଵ݄ǡܽଶ݄ଶሻൌ݂ሺܽଵǡܽଶሻ൬݄ଵμ
μݔଵ݄ଶμ
μݔଶ൰݂ሺܽଵǡܽଶሻ
ͳ
ʹǨ൬݄ଵμ
μݔଵ݄ଶμ
μݔଶ൰ଶ
݂ሺܽଵǡܽଶሻڮ
ͳ
݊Ǩ൬݄ଵμ
μݔଵ݄ଶμ
μݔଶ൰
݂ሺܽଵǡܽଶሻͳ
ሺ݊ͳሻǨ൬݄ଵμ
μݔଵ݄ଶμ
μݔଶ൰ଶ
݂ሺܿǡ݀ሻ
ൌԝ
ୀԝͳ
݇Ǩ൬݄ଵμ
μݔଵ݄ଶμ
μݔଶ൰
݂ሺܽଵǡܽଶሻܴሺܿǡ݀ሻ
ൌԝ
ୀԝͳ
݇Ǩܦభమǥೖ݂ሺܽଵǡܽଶሻ݄భ݄మǥ݄ೖܴሺܿǡ݀ሻ munotes.in
Page 48
48A1ALப௫భப௫మǥಢೣೖ, and ݅ଵǡ݅ଶǡǥǡ݅ൌͳ or 2, and the sum is taken
over all
ordered k-tuples ൫݅ଵǡ݅ଶǡǥǡ݅൯Ǥ For example,
σܦభమሺଵǡଶሻ݄భ݄మൌܦଵ՝݂ሺܽଵǡܽଶሻ݄ଵଶܦଵଶ݂ሺܽଵǡܽଶሻ݄ଵ݄ଶ
ܦଶ՝݂ሺܽଵǡܽଶሻ݄ଶ݄ଵܦଶଶ݂ሺܽǡ a ଶ ) ݄ଶଶ
ൌቆ ݄ଵଶμଶ
μݔଵଶʹ݄ଵ݄ଶμଶ
μݔଵμݔଶ݄ଶଶμଶ
μݔଶଶቇ݂ሺܽଵǡܽଶሻǤ
Similarly,
σܦభమయ݂ሺܽଵǡܽଶሻ݄భ݄మ݄య ൌ ܦͳͳͳ݂ሺܽͳǡܽʹሻ݄ͳ͵
ܦͳͳʹ݂ሺܽͳǡܽʹሻ݄ͳʹ݄ʹܦͳʹͳ݂ሺܽͳǡܽʹሻ݄ͳ݄ʹ݄ͳ
ܦʹͳͳ݂ሺܽͳǡܽʹሻ݄ʹ݄ͳʹܦͳʹʹ݂ሺܽͳǡܽʹሻ݄ͳ݄ʹʹ
ܦʹͳʹ݂ሺܽͳǡܽʹሻ݄ʹ݄ͳ݄ʹܦʹʹͳ݂ሺܽͳǡܽʹሻ݄ʹʹ݄ͳܦʹʹʹ݂ሺܽͳǡܽʹሻ݄ʹ͵ ൌ
ሺ݄ଵଷడయ
డ௫భయ͵݄ଵଶ݄ଶడయ
డ௫భమడ௫మ͵݄ଵ݄ଶଶడయ
డ௫భడ௫మమ݄ଶଷడయ
డ௫మయ )f(a 1, a2) .
e[ample, D12f and D 21f, or D112f , D 121f, and D211f. We could do this, since f ܥאஶ
ensures that that these partial derivatives are equal. Now we s tate Taylor’s theorem
for real-valued functions of several variables.
Theorem 4.4 :
Let f : E ࡾǡ
ࡾnǤ ǡ
ࢇ ൌሺܽͳǡܽʹǡǥǡ݊ܽ ሻא ǡࢎ ൌ ሺ݄ͳǡ݄ʹǡǥǡ݄݊ ሻࡾא nǡ
ࢇאࢎǤ
݉ܥא݂ ǡ݂ሺࢇࢎሻൌ σଵ
Ǩσܦభమǥೖିଵ
ୀ ݂ሺࢇሻ݄భ݄మǤǤǤ݄ೖ
+ R m-1(c), ..................................................... ................................... 4.5
Zhere ݅ଵǡ݅ଶǡǤǤǤǡ݅ taNe values from the set ^1, 2, ..., n`, and the inner summat ion
in 4.5 is taNen over all possible such N-tuples.
Further, the remainder R m-1(c) = ଵ
Ǩσܦభమǥ݂ሺࢉሻ݄భ݄మǥ݄. This sum is
taNen over all possible m-tuples i1, i2, ..., i m , Zhere i1, i2, ..., i m taNe values from
^1, 2, ..., n`,and c is some point on the line segment Moining a and
a + h. munotes.in
Page 49
49Chapter 4: Applications of Derivatives This theorem is used to appro[i mate a given function by a polyn omial. In the ne[t
section Ze shall use it to derive conditions for locating and c lassifying e[treme
points of a function.
E[ercises : 1 Write the e[pression for A appearing in Theorem 4.2.
4.3 Ma[ima And Minima
One of the most interesting and Zell-NnoZn applications of Calc ulus is the location
and classification of e[treme points of a function. problems involving functions of one or tZo variables. We shall noZ e[tend the
definitions of ma[ima and minima to functions of n variables, a nd derive suitable
tests for their location.
Definition 4.1 : Let f : Rn ՜ ࡾǤאࢇ
ࡾn
ሺ ሻ
ࢇǡ
݂ሺ࢞ሻ ݂ሺࢇሻא࢞ N.
f(a) is then called the local or relative ma[imum value.
A local minimum or relative minimum is defined in a similar m anner.
ݔସଶݔହଶ, clearly has a local minimum at ሺͲǡͲǡͲǡͲǡͲሻ . Can you find an e[ample of a
function Zith a local ma[imum" Definition Ǥʹǣ A point ܴאܽ is called a saddle
point of a function ݂ǣܴ՜ܴ ,if every ball ሺࢇǡݎሻǡݎ Ͳ , contains points ࢞ ,such
that ݂ሺ࢞ሻ ݂ሺࢇሻ , and also other points ࢟ ,such that ݂ሺ࢟ሻ ݂ሺࢇሻ .
In general, it is not easy to spot the local ma[imum or local m inimum merely by
observation. For differentiable functions Ze can derive tests t o locate these values.
derivative vanishes at an e[treme point. We have a very similar test for the location
of e[treme points of a function of n variables, as you can see in the ne[t theorem.
Theorem 4. : If f : Rn ՜ ࡾ
ࡾאࢇ nǡ݅ ൌ ͳǡʹǡǤǤǤǡǡడ
డ௫ a , if it e[ists, is eTual to ]ero.
Proof: Since has a local ma[imum at ࢇǡݎ Ͳ ,
such that א࢞ ሺࢇǡݎሻ ฺ ݂ሺ࢞ሻ ݂ሺࢇሻ .
For ݅ൌͳ ǡ ʹ ǡǥǡ , consider a function ݃ǣሺܽെݎǡܽݎሻ՜ࡾ ,such that
݃ሺݔሻ ൌ ݂ሺܽଵǡܽଶǡǥǡܽିଵǡݔǡܽାଵǡǥǡܽሻǤ since ݂ሺܽሻ is the local ma[imum value
of ݂ǡ݃ሺܽሻ is the ma[imum value of ݃. If ப
ப௫ሺܽሻ e[ists, then ݃ᇱሺܽሻ also e[ists,
and the tZo are eTual. %y applying the first derivative test fo r functions of one
variable to ݃,
Ze get
μ݂
μݔሺࢇሻ ൌ ݃ᇱሺܽሻൌͲ munotes.in
Page 50
50A1ALడ௫ a , if it e[ists, is eTual to
]ero, even Zhen a is a local minimum of f.
Thus, if f has a local e[tremum at a, and all the partial derivatives e[ist at a, then
ࢺf(a) 0.
As in the case of functions of one variable, the condition in t heorem 4.5 is a
necessary one, and is not sufficient. That is, if all the parti al derivatives of a
function at a point a are ]ero, Ze cannot say that a is a local ma[imum or local
minimum point. It may be neither.
An e[ample is the function f : R2 ՜ R, f(x, y) = 1 – x2 + y2. Here fx = - 2x , and
fy = 2 y . So, fx 0, 0 0 and fy 0, 0 0. %ut you can see clearly, that f has a
ma[imum in the direction of the [-a[is, and a minimum in the di rection of the y-
a[is at 0, 0 . This means, f has neither a minimum, nor a ma[imum at 0, 0 . In
fact 0, 0 is a saddle point for this function.
Definition 4.3 : Let f : Rn ՜ ࡾ ǡ ࡾאࢇ nǤ డ
డ௫ a is eTual
to ]ero for i 1, 2, ..., n, then a is called a critical point, or a stationary point
of f.
Theorem 7.5, tells us to looN for e[treme points among the crit ical points of a
function. We shall noZ see hoZ to classify these points as loca l ma[ima, local
minima, or saddle points. This involves second order partial de rivatives. This is to
be e[pected, since in one variable functions too, Ze have a sec ond derivative test
to classify stationary points. The proof of the test for severa l variables involves
Tuadratic forms. definition and recall the relevant results.
Definition 4.4 : If A = (a ij is a real symmetric n [ n matri[, and x = (x 1, x2, ..., x n)
ࡾא nǡܳሺ࢞ሻൌ σσ ܽݔݔ
ୀଵ
ୀଵ i s c a l l e d a Tuadratic form associated
with A.
We can Zrite Q(x) = xAxt. I f A i s a d i a g o n a l m a t r i [ , t h e n Q(x)
σܽݔଶ
ୀଵ
Ǥ
ܣ
ǡ
Ͳ࢞ǡܳሺ࢞ሻൌ Ͳ ฺ ࢞ ൌ
Ǥ
Ǥ
Ͳ࢞ǡܳሺ࢞ሻൌ Ͳ ฺ x = 0. Such a Tuadratic form is called negative
definite.
It may not be very easy to get the eigen values. %ut Ze have an easier Zay to decide.
A principal minor of a sTuare matri[, A, is the determinant of the matri[ obtained
by taNing the first k roZs, and the first k columns of A, 1 ݇ n. munotes.in
Page 51
51Chapter 4: Applications of Derivatives If all the principal minors are positive, then the associated T uadratic form is positive
definite.
If the principal minors are alternately positive and negative, starting Zith a negative
minor for k 1, then the associated Tuadratic form is negative definite.
If a principal minor of order k is negative, Zhen k is an even number, then Q(x)
taNes both positive a nd negative values.
We noZ use these facts about Tuadratic forms to derive the seco nd derivative test.
A definition first.
Definition 4. : If f is a C2 function from Rn to R, then the symmetric matri[ A =
H(x) = ൬డమሺ࢞ሻ
డ௫డ௫ೕ൰ is called the +essian matri[ of f at x. Thus,
A = H( x) = ۉۈۇడమ
డ௫భమడమ
డ௫భడ௫మڮడమ
డ௫భడ௫ڭڰڭ
డమ
డ௫భడ௫డమ
డ௫మడ௫ڮడమ
డ௫మیۋۊ .
If a א
ࡾnǡ݂ࢇ ݂ሺࢇ
ࢎሻȁȁࢎȁȁ݂ሺࢇࢎሻൌ ݂ሺࢇሻ݂ࢺሺࢇሻxࢎ +R 1(c).
If a is a critical point, then ݂ࢺሺࢇሻ 0, and therefore Ze get
f(a + h) − f(a) = R 1(c).
1oZ, ܴଵሺܿሻ ൌଵ
ଶǨσσܦ݂ሺࢇࢎߠሻ݄ ݄, Zhere Ͳ൏ߠ൏ͳ
ൌଵ
ଶǨܪࢎሺࢇࢎߠሻࢎ௧Ǥ We Zrite
צࢎצଶȁܧሺࢇǡࣂሻȁ ൌ
ʹǨหσσ൛ܦ ݂ሺࢇࢎߠሻെܦ ݂ሺࢇሻൟ݄ ݄ห
ͳ
ʹǨσσหܦ݂ሺࢇࢎߠሻെܦ ݂ሺࢇሻห צࢎצ ଶ
Therefore, ȁܧሺࢇǡߠሻȁ ଵ
ଶǨσσหܦ݂ሺࢇࢎߠሻെܦ ݂ሺࢇሻหǡ when ࢎ് ǤǥǥǥǥǤ ሺͶǤሻ
ࢎ ՜ ǡ
ܥא݂ ʹǡ
Ǥ
ǡܧሺࢇǡߠሻ՜ ǡࢎ ՜ Ǥଵ
ଶǨܪࢎሺࢇࢎߠሻݐࢎ ൌ
ଵ
ଶǨܪࢎሺࢇሻݐࢎȁȁࢎȁȁʹܧሺࢇǡߠሻǡܧሺࢇǡߠሻ՜ ǡࢎݏܽ ՜ 0. munotes.in
Page 52
52A1ALଶǨܪࢎሺࢇሻݐࢎȁȁࢎȁȁʹܧሺࢇǡߠ .) ................. 4.7
Theorem 7. : If f is a function from Rn to R, and has continuous second order
partial derivatives in a ball % a; r around a stationary point a of f, then
i) f has a relative minimum at a, if H(a) is positive definite
ii) f has a relative ma[imum at a, if h(a) is negative definite
iii f has a saddle point at a, if H(a has both positive and negative eigen
values.
Proof : Using the notations that Ze have used in the discussion Mus t before this
theorem,
Ze can Zrite f(a + h) – f(a) = ଵ
ଶǨܪࢎሺࢇሻݐࢎหȁࢎȁหʹܧሺࢇǡߠሻǤ
ܧሺࢇǡߠሻ՜ ǡ ࢎ՜ ǡ
݂ሺࢇࢎሻȂ݂ሺࢇሻଵ
ଶǨ hH(a)ht .
i This value Zill be positive for all h, if H(a is positive definite. Hence, f(a + h)
– f (a) ! 0 for all h, such that 0 __ h|| < r . This tells us that f (a + h)
݂ሺࢇሻאࢎ % a; r , that is , a i s a r e l a t i v e m i n i m u m p o i n t
of f.
The argument for proving ii and iii are e[actly similar, and Ze are sure you can
Zrite those.
Remark 4.2 : i If an even principal mi nor, that is a principal minor of even order
is negative, then the point is a saddle point.
ii If detH( a) 0, the test is inconclusive, and a i s c a l l e d a degenerate
stationary point of f.
*o through the folloZing e[amples carefully, they illustrate ou r discussion
here.
E[ample 4.2 : Locate and classify the stationary points of the functions gi ven by
i x2 + xy + 2x + 2y + 1, ii x3 + y3 – 3xy, iii (x − 1)exy.
Solution : i Let f(x, y) = x2 + xy + 2x + 2y + 1. Then fx = 2x + y + 2, f y = x + 2. fx
= f y = 0 ฺ ݔ ʹ ൌ Ͳ ǡ ʹ ݔ ݕ ʹ ൌ Ͳ ฺ x = − 2 and y = 2 .
Therefore, f has only one stationary point, ( − 2, 2). Now, fxx = 2, f yy = 1, and fxy = 0.
Thus , H(( −2, 2)) = ቀʹͳ
ͳͲቁ, and det (H (( − 2, 2))) = −1.
Therefore, f has a saddle point at ( − 2, 2).
ii Let f(x, y) = x3 + y3 – 3xy. Then, fx = 3x2 – 3y, f y = 3y2 – 3x.
fx = f y = 0 ฺ ݕ ൌ ݔ ʹ ǡ ݔ ൌ ݕ ʹ ฺ ݔ ൌ ݕൌ Ͳ ǡݔݎ ൌ ݕ ൌ ͳ Ǥ
ǡ ሺͲǡͲሻሺͳǡͳሻǤ munotes.in
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53Chapter 4: Applications of Derivatives ǡݔݔ݂ ൌ ݔǡݕݕ݂ ൌ ݕǡݕݔ݂ ൌെ͵Ǥ
ǡܪሺሺͲǡͲሻሻൌቀͲെ ͵
െ͵ Ͳቁ.
det(H 0, 0 - 9 0, and 0, 0 is a saddle point.
H 1, 1 ቀെ ͵
െ͵ ቁ . The principal minors are 6, a nd 27. %oth are positive, and
hence f has a local minimum at 1, 1 .
iii Let f(x, y) = (x - 1)exy. Then fx = exy(xy – y + 1), f y = x(x - 1)exy
fx = 0 ฺ ݕݔ Ȃ ݕ ͳ ൌ Ͳ ǡ ݕ݂ ൌ Ͳ ฺ ݔ ሺ ݔ െ ͳ ሻൌ Ͳ ฺ x = 0, or x = 1 .
x 0 ฺ y 1, and x 1 contradicts fx = 0. So, 0, 1 is the only stationary point.
fxx = exy(y + xy2 – y2 + y), f xy = exy(x – 1 + x2y – xy + x), f yy = x2(x - 1)exy.
Therefore, H 0, 1 ቀͳെ ͳ
െͳ Ͳቁ . det(H 0, 1 - 1 0.
Hence, 0, 1 is a saddle point.
E[ample 4.3 : Locate and classify the stationary points of f(x, y, z) =
i) xyz ݁ି௫మି௬మି௭మ,
ii) x2y + y2z + z2 - 8ξʹx, iii) x2 – xy + yz3 – 6z.
Solution : i fx = yz ݁ି௫మି௬మି௭మെʹݔʹ݁ݖݕି௫మି௬మି௭మൌ݁ି௫మି௬మି௭మyz(1 – 2x2)
fy = ݁ି௫మି௬మି௭మݖݔሺͳȂʹݕʹሻǡݖ݂ ൌ ݁ି௫మି௬మି௭మݕݔሺͳȂʹݖʹሻǤ
ǡ ǡ
ሺܽǡͲǡͲሻǡሺͲǡܾǡͲሻǡሺͲǡͲǡܿሻǡሺേଵ
ξଶǡേଵ
ξଶǡേଵ
ξଶ , Zhere a, b, c are real
numbers, as the stationary points.
fxx = − 4xyz ݁ି௫మି௬మି௭మെʹݖݕݔሺͳȂʹݔʹሻ݁ି௫మି௬మି௭మ
fxy = z(1 – 2x2)݁ି௫మି௬మି௭మെʹݕʹݖሺͳȂʹݔʹሻ݁ି௫మି௬మି௭మ,
fyz = ݁ି௫మି௬మି௭మݔሺͳȂʹݕʹሻȂʹݖݔʹ݁ି௫మି௬మି௭మ(1 – 2y2).
We have indicated the procedure. We are sure noZ you Zill be ab le to get fxz, fyy,
and fzz. Evaluating these second order partial derivatives at the statio nary points,
Ze find,
H((a, 0, 0 ൭ͲͲ Ͳ
ͲͲ݁ܽ ିమ
Ͳ݁ܽ ିమͲ൱ detH((a, 0, 0 0. Therefore, a, 0, 0 is a
degenerate point of f. Similarly, 0, b, 0 and 0, 0, c are also degenerate points.
H ଵ
ξଶǡଵ
ξଶǡଵ
ξଶሻሻൌ൮െξʹ݁ିయ
మͲͲ
Ͳെξʹ݁ିయ
మͲ
ͲͲ െ ξʹ݁ିయ
మ൲Ǥ munotes.in
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54A1ALଶǡʹ
െ͵ǡെʹξʹ݁ିଷ
ଶǤǡ ൬ͳ
ξʹǡͳ
ξʹǡͳ
ξʹ൰
Ǥ
Ǥ
ሺଵ
ξଶǡିଵ
ξଶǡିଵ
ξଶሻǡሺିଵ
ξଶǡଵ
ξଶǡିଵ
ξଶሻǡሺିଵ
ξଶǡିଵ
ξଶǡଵ
ξଶሻǡ
ሺଵ
ξଶǡିଵ
ξଶǡଵ
ξଶሻǡሺଵ
ξଶǡଵ
ξଶǡିଵ
ξଶሻǡሺିଵ
ξଶǡଵ
ξଶǡଵ
ξଶሻǡሺିଵ
ξଶǡିଵ
ξଶǡିଵ
ξଶ .
ii fx = 2xy - 8 ξʹǡݕ݂ ൌ ݔʹʹݖݕǡݖ݂ ൌ ݕʹ
ʹݖǤǡݕݔ ൌ Ͷ ξʹǡݔ ʹ ൌ െ ʹ ݖݕ ǡ ݕ ʹ ൌ
െʹݖǤ ݔǡݕǡݖെǡݔ ൌ ʹ ξʹǡݕൌʹǡݖൌ
െʹǤǡሺͲǡͲǡͲሻሺʹ ξʹ, 2, − 2).
saddle point.
iii fx = 2x – y, f y = - x + z3, fz = 3yz2 – 6. ETuating these to ]ero, Ze get 1, 2, 1
as the stationary point. ChecN that H 1, 2, 1 ൭ʹെ ͳͲ
െͳ Ͳ ͵
Ͳ͵ ͳ ʹ൱, and the principal
minors are 2, - 1, - 6. Hence, 1, 2, 1 is a saddle point.
See if you can solve these e[ercises noZ.
E[ercises:
1 Find the stationary points of f(x, y) = i) 42 2y xx ii) (x + y)exy. 42 2y xx
2 Find the e[treme values of f(x, y) = x2 + y3 + 3xy2 – 2x.
3 Is 0, 0 an e[treme point of 2cos x + y) + exy?
4 Locate and classify th e stationary points of
i f(x, y) = (2 - x)(4 - y)(x + y - 3), ii f(x, y, z) = 4xyz – x4 – y4 – z4,
iii f(x, y, z) = 64x2y2 – z2 + 16x + 32y + z , iv f(x, y, z) = xyz(x + y+ z – 1).
4.4 Lagrange’s Multipliers
LooN at these situations:
i A rectangular cardboard sheet is given. We have to maNe a cl osed bo[ out of it.
What is the ma[imum volume that is possible" munotes.in
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55Chapter 4: Applications of Derivatives ii Temperature varies on a metal surface according to some for mula. Where do the
ma[imum and minimum temperature occur on the surface"
In both these problems Ze have to ma[imi]e or minimi]e a certai n function:
volume in the first case, and temperature in the second. So the se are ma[-min.
Problems. %ut there is a difference betZeen these and the probl ems considered in
the last section. Here, an additional constraint or condition i s imposed. The given
cardboard sheet has a fi[ed area. The ma[imumminimum temperatu re points are
to be on the given surface.
In this section Ze shall see hoZ such problems are solved. A ve ry useful method
Zas developed by -oseph Louis Lagrange. This method gives a nec essary condition
for the e[treme points of a function. We noZ state the theorem and then illustrate
its use through some e[amples.
Theorem 4.7 : Let f : Rn ՜ ࡾ ǡ ܥא݂ ͳ Ǥ ݃ ͳ ǡ݃ ʹ ǡǤǤǤǡ݉݃ ሺ ݉ ൏
݊ሻ
ܥͳǡ
ࡾnǤ אࢇ
݂ǡ݃ߘ ଵሺࢇሻǡ݃ߘ ଶሺࢇሻǡǤǤǤǡ݃ߘ ሺࢇሻ
ߣଵ݃݅ܦͳሺࢇሻߣ ଶ݃݅ܦʹሺࢇሻǤǤǤߣ Digm(a) = 0, i 1, 2, . . . , n.
We can also Zrite the vector eTuation ݂ሺࢇሻ σߣ݃ߘ
ଵ(a) = 0.
When Ze Zant to find the e[treme values of a function f : Rn ՜ ࡾ ǡא݂
ܥͳǡ
ǡ ݃ͳሺݔͳǡݔʹǡǤǤǤǡ݊ݔሻൌ
Ͳǡ݃ʹሺݔͳǡݔʹǡǤǤǤǡ݊ݔሻൌ ͲǡǤǤǤǡ݉݃ሺݔͳǡݔʹǡǤǤǤǡ݊ݔሻൌ Ͳǡ ൏
ǡ݂݅ܦሺࢇሻߣ ଵ݃݅ܦͳሺࢇሻ
ߣଶ݃݅ܦʹሺࢇሻǤǤߣ Digm(a) = 0, i 1, 2, . . . , n.
These n eTuations, along Zith the m eTuations, g1(x1, x2, . . . ,x n) = 0, g 2(x1, x2, . .
. ,xn) = 0, . . . , g m(x1, x2, . . . ,x n) = 0, are then solved to get the values of the n m
unNnoZns, x1, x2, . . . ,xn, ߣଵǡߣଶǡǤǤǤǡߣ. The solutions x = (x 1, x2, . . . ,x n) are the
stationary points, and contai n the e[treme points of f .
ߣଵǡߣଶǡǤǤǤǡߣ are called Lagrange’s Multipliers . We use one multiplier for each
constraint.
To analytically classify these stationary points into local ma[ imum, minimum, or
saddle, is a very complicated process. It is usually easier to looN at the physical or
geometrical aspect of the problem to arrive at any conclusion. We noZ solve a feZ
problems, so that the entir e process is clear to you.
E[ample 4.4 : Find the dimensions of the bo[ Zith ma[imum volume that can be
made Zith a cardboard sheet of si]e 12 cm2.
Solution : If the dimensions of the bo[ are x, y, z cms, then its volume V = xyz c.
cms. And surface area is 2 xy + yz + xz) sT. cms. Here Ze have to ma[imi]e V, munotes.in
Page 56
56A1ALxyz, and
g(x, y, z) = xy + yz + xz – 6. Hence,
݂ࢺሺݔǡݕǡݖሻ݃ࢺߣሺݔǡݕǡݖሻൌ ฺ
fx + ݔ݃ߣ ൌ Ͳ ฺ ݖݕߣሺݕݖሻൌ Ͳǡ ݕ݂ݕ݃ߣ ൌ Ͳ ฺ ݖݔߣሺݔ
ݖሻൌ Ͳǡݖ݂ݖ݃ߣ ൌ Ͳ ฺ ݕݔߣ (x + y) = 0.
xyz = െߣሺݕݔݖݔሻൌെߣሺݕݔݖݕሻൌെߣሺݖݔݖݕሻǤ ߣ ൌ Ͳǡܸ ൌ
Ͳǡ
Ǥ ߣ ് 0, then xy + xz = xy + yz = xz + yz . That
is, x = y = z unless, of course, x = y = z = 0 .
Therefore, xy + yz + xz = 6 ฺ ͵ ݔ ʹ ൌ ฺ ݔൌ ξʹ
Ǥǡܸ ൌ ʹ ξʹ c.
cms. is the ma[imum volume.
E[ample 4. : Find the e[treme values of the function given by f(x, y, z) = 2x + y
+ 3z, subMect to x2 + y2 = 2, x +z = 5.
Solution : Let g1(x, y, z) = x2 + y2 – 2 0, and g2(x, y, z) = x + z – 5 0. Then
݂ࢺ ߣ ଵ݃ࢺଵߣଶ݃ࢺଶൌฺ
fx + ߣଵ݃ͳݔߣ ଶ݃ʹݔ ൌ Ͳ ฺ ʹʹߣ ଵݔߣ ଶ = 0
fy + ߣଵ݃ͳݕߣ ଶ݃ʹݕ ൌ Ͳ ฺ ͳʹߣ ଵݕ = 0
fz + ߣଵ݃ͳݖߣ ଶ݃ʹݖ ൌ Ͳ ฺ ͵ߣ ଶൌͲǤǡߣ ଶൌെ͵ǡʹߣ ଵݔൌ
ͳǡʹߣ ଵݕ =− 1.
ߣଵൌͲฺߣ ଶൌെʹǤɉ ଶൌ
െ͵Ǥɉ ଵ
Ǥ
ǡݔ ൌଵ
ଶఒభǡݕ ൌ
ିଵ
ଶఒభǤݔʹݕʹൌʹǡɉ ଵൌ
േଵ
ଶǤǡݔൌേͳǡݕൌט 1. Hence, the stationary points are 1, - 1, 4 and
- 1, 1, 6 , and the e[treme values are 13 and 17.
E[ample 4. : Find the minimum distance of a point on the intersection of the
planes,
x + y – z 0, and x + 3y + z 2 from the origin.
Solution: The distance of P(x, y, z) from the origin is
ඥݔଶݕଶݖଶǤǡ݂ሺݔǡݕǡݖሻൌݔଶݕଶݖଶ, subMect to
g1(x, y, z) = x + y – z = 0, and munotes.in
Page 57
57Chapter 4: Applications of Derivatives g2(x, y, z) = x + 3y + z – 2 = 0.
݂ࢺ ߣ ଵ݃ࢺଵߣଶ݃ࢺଶൌฺ
fx + ߣଵ݃ͳݔߣ ଶ݃ʹݔ ൌ Ͳ ฺ ʹݔߣ ଵߣଶ = 0
fy + ߣଵ݃ͳݕߣ ଶ݃ʹݕ ൌ Ͳ ฺ ʹݕߣ ଵ͵ߣଶ = 0
fz + ߣଵ݃ͳݖߣ ଶ݃ʹݖ ൌ Ͳ ฺ ʹݔȂߣ ଵߣଶൌͲǤ
ǡݔ ൌିሺఒభାఒమሻ
ଶǡݕൌିሺఒభାଷఒమሻ
ଶ ,
z = ሺఒభିఒమሻ
ଶǤݔݕȂݖൌͲǡɉ ଵɉଶ 0.
Therefore, x 0 and y = z . Using this in x + 3y + z – 2 0, Ze get y = z ò.
Thus, the stationary point is
0, 12, 12 . The distance of this point from the origin is ଵ
ξଶ .
*eometrically, the constraints are eTuations of tZo planes. The re is no ma[imum
to the distance of a point on their line of intersection from t he origin. So, the
stationary point is a minimum point.
Here are some problems you can try.
1 Find the e[treme values of the function f(x, y) = xy o n t h e s u r f a c e
2 82 2y x 1.
2 Find the e[treme values of z 32yx on the unit circle in the xy-plane.
3) Find the distance of the poi nt (10, 1, − 6) from the interse ction of the planes,
x + y + 2z 5 and 2x – 3y + z 12.
4. Summary
In this chapter we have introduced Taylor’s theorem for functio ns of several
variables. We have also seen hoZ to get Taylor polynomials of a given order for a
given function. Of course, to be able to do this, the function must have continuous
partial derivatives of higher orders.
We have then discussed the location of ma[ima and minima of a r eal-valued
function of several variables. This has tremendous applications in diverse fields of
study. In particular, Ze have proved that the e[treme points of a function are located munotes.in
Page 58
58A1ALpoints at Zhich all the first order partial derivatives are ]er o. The classification of
these points into ma[ima, minima, or saddle points depends on t he signs of the
principal minors of the Hessian matri[.
We pointed out that there are some situations, Zhere Ze need to find the e[treme
values subMect to certain constraints. Such problems, and the m ethod of tacNling
them is also discussed, and illustrated through some e[amples.
munotes.in
Page 59
UNIT 3
IN9ERSE AND IMPLICIT )UNCTION
T+EOREMS
Unit Structure
5.0 ObMectives
5.1 Introduction
5.2 Inverse Function Theorem
5.3 Implicit Function Theorem
5.4 Summary
.0 Objectives
After reading this chapter, you should be able to
x state and prove Inverse Functi on Theorem for fun ctions of sever al variables
x checN if some simple functions are locally invertible
x state and prove Implicit Function Theorem for functions of seve ral variables
.1 Introduction
In this chapter Ze introduce tZo very important theorems. across these theorems even for functions of a single variable. In each case, Ze shall
first discuss the single variable case, and then e[tend the con cept to functions of
several variables. A Zord of caution : these theorems are not e asy. To help you
understand them better, Ze are going to prove some smaller resu lts, and then use
them in the proof of the theorems. Do study this chapter carefu lly and Ze are sure
you Zould have no difficulty in digesting the concepts.
.2 Inverse )unction Theorem
The inverse function theorem is a very important theorem in Cal culus. familiar Zith its one dimensional version. %efore Ze introduce the theorem for
functions from Rn to Rn, Ze shall recall some results about functions of one
variable: 59munotes.in
Page 60
60A1ALa, b@ R is continuous, and f (c) ! 0 for some c (a, b), then
such that
c ) (a, b), and f(x) ! 0 (c ). In other
Zords, Ze can alZays find a neighbourhood of the point c, in Zhich f(x) has the
same sign as f(c).
2 If f : >a, b@ R is a continuously differentiable function, and for
some
c (a, b) , then using 1 Ze can prove that such that f is an inMective
function on
c ) (a, b . Further, f-1: f(c )
(c ) is differentiable at f(c) ,
The statement in 2 is the inverse function theorem. 1ote that Ze do not NnoZ
Zhether the inverse of f e[ists on > a, b@. %ut Zhat this theorem tells us, is that if
, then f is “locally invertible” at c. For e[ample, Ze NnoZ that the
function f : >0, 2 @ R, f(x) = sinx does not have an inverse. %ut
is a continuous function, and . So, the theorem says
that f is locally invertible at . That is, Ze can find a neighbourhoo d N of , such
that f restricted to N has an inverse. ChecN that f is inMective Zhen restricted to N
, a n d h e n c e h a s a n i n v e r s e o n N.
We shall noZ see if this theorem e[tends to functions of severa l variables. Let us
start Zith a definition.
Definition .1 : Let f : E Rn, Zhere E Rn. If f C1, f is said to be locally
invertible at a E , i f t h e r e e [ i s t s a n e i g h b o u r h o o d N1 of a, N 1 E , a n d a
neighbourhood N2 of f(a), such that f(N1) = N 2, f is inMective on N1, and f-1 : N 2
N1 is a C1 function.
We shall soon state and prove the inverse function theorem. In the proof, Ze are
going to use some minor results. chapters of this course. 1e[t Ze state and prove one other resu lt, Zhich Zill be
useful to us.
Theorem .1 : Let f = (f 1 f2, . . . , f n) : E Rn, Zhere E is an open set in Rn.
Suppose f C1. If the -acobian of f, J(a) 0 f o r s o m e a E , t h e n f is inMective
on a neighbourhood of a in E.
)
)
)
)
)
munotes.in
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61Chapter 5: Inverse and Implicit Function Theorems Proof : If X1, X2, . . . , Xn E, Ze consider a point X = (X1, X2, . . . , Xn)
, Zhose first n coordinates are the coordinates of X1, the ne[t n are the coordinates
of X2, and so on. We define a function, M, such that
j(X) = det[D jfi(Xi)] = det .
1oZ, the function j, being an nîn determinant, is a polynomial of its n2 entries, and
each entry, is a continuous function, since f C1. Thus, j is a continuous
function on its domain. We Zrite A = (a, a, . . . , a . Then j(A) = det[D jfi(a)] = J( a)
0. 1oZ, since f C1, all the entries of j(A) are continuous, and hence, j(A) is
also continuous. The continuity of j(A) ensures that there e[ists a neighbourhood N
of A, such that j(X) 0 , if X N.
In other Zords, there e[ists a conve[ neighbourhood Na of a, such that j(X) 0 , if
X = (X1, X2, . . . , Xn) is a point, for Zhich Xi N a for every i 1, 2, . . . , n.
...... .... 5.1
This Na is the reTuired neighbourhood. We have to shoZ that f is inMective on Na.
For this, suppose x, y Na , such that f(x) = f(y . Then fi(x) = f i(y) for every i 1,
2, . . . , n.
Then, using the Mean Value Theorem for scalar fields See Remar N 6.2 ii . , Ze
get
fi(x) − fi(y) = f i(ci) ( x − y) f i(ci) ( x − y) 0 for some ci on the line
segment Moining x and y. So, if x – y 0, then fi(ci) 0 for some ci on the line
segment Moining x and y, that is, in the neighbourhood Na, since Na is conve[. This
means, Djfi(ci 0 for every M, 1 . Thus, if C = (c1, c2, . .
. , cn), then j(C) = det[D jfi(ci)] 0. %ut this contradicts 5.1 . So, Ze conclude that
x – y 0, Zhich proves that f is inMective on Na.
Remark .1 : i A function may not be inMective on its entire domain. %ut if its
-acobian is non-]ero at a point, then it is inMective on a neig hbourhood of that point.
In other Zords, it is locally inMective.
ii If the -acobian is non-]ero, then the linear transformation Df, Zhich represents
the derivative of f, is non-singular, and hence, is a linear isomorphism.
E[ample .1 : a Consider the function f(x, y) = (excosy, exsiny). This function is
not inMective, since f(x, 0) = f(x, 2 . %ut,
X
f
f
f
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Page 62
62A1ALHere Ze have a function, Zhich is locally inMective at every po int of its domain, but
is not inMective on the domain.
b Consider the function f(x, y) = (x3, y3), defined on R2. The -acobian of this
function is ]ero at 0, 0 . %ut the function is locally inverti ble at 0, 0 . In fact, it is
an invertible function.
Theorem .2 The Inverse Function Theorem : Let f = ( f 1, f2, . . . , f n) C1,
f: E Rn , Zhere E is an open set in Rn. Let T f E . Suppose J(a 0 for some
a E . T h e n t h e r e e [ i s t s a u n i T u e f u n c t i o n f-1 from < to ;, Zhere ; is open in E,
< is open in T, such that
i a ;, f(a) <, ii < f ; , iii f is inMective on ;, iv f-1: < ;, f-1 <
;, v f-1 C1 on <.
Proof : Using Theorem 5.1, Ze can conclude that f is inMective on a neighbourhood
N of a in E. So, f : N f(N) is biMective, and hence has an inverse, f-1 : f(N) N.
Let r ! 0 be such that N. Since is compact in Rn , Ze use Theorem
3.4.1 to conclude that f is also compact in Rn . 1oZ f is continuous and
inMective on the compact set . Hence, using Theorem 3.4.2, Ze can say that
f-1 is continuous on f .
1oZ, % a, r is an open set in , and therefore,
% a, r is open in f . T h a t i s , f % a, r is open in f .
Also, f(a) f % a, r . Therefore, there e[ists a ! 0, such that % f(a),
f % a, r .
TaNe ; f-1 % f(a), , and < % f(a), . T h e n ; a n d < s a t i sfy i , ii , iii and
iv in the statement of the theorem.
To prove the last assertion v in the statement, Ze have to sho Z that all the partial
derivatives of all the component functions of f-1 are continuous on <. For this Ze
first define the function j(X) = det[D jfi(xi)] , as in Theorem 5.1. Here X = (X1, X2,
. . . , X n). Then, as before, there is a neighbourhood Na of a, such that j(X) 0,
Zhenever each Xi N a. We can assume that the neighbourhood N N a. This
ensures that j(X) 0, Zhenever each Xi .
NNN
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63Chapter 5: Inverse and Implicit Function Theorems1oZ Ze first prove that D if-1 e [ i s t s o n < . L e t y <, and consider tyf teyfi 1 1 ,
Zhere ei is the ith c o o r d i n a t e v e c t o r , a n d t i s a s c a l a r . L e t x = f-1(y), and
= f-1(y + tei). Then
f() – f(x) = tei. Thus, fi() – fi(x) = t, and fj() – fj(x) = 0, Zhen i M.
%y applying Mean Value Theorem RemarN 6.2 ii , Ze can Zrite
txf xfm m
fm xm) txx
, m 1, 2, . . . , n. Here xm is a point on the
line segment Moining x and .
So, Ze get a system of n eTuations for the n values of m . Th e left hand side of an
eTuation in this system is 1, if m i, otherZise it is 0. The right hand side is of the
form
D1fm(xm) tx x1
1 D2fm(xm) tx x2
2 . . . Dnfm(xm) tx xn n
, m 1, 2, . . . , n.
The determinant of this system of linear eTuations is j(X), Zhich Ze NnoZ is non-
zero. Hence we can solve it by Cramer’s rule and get the variab les tx xj j
as the
Tuotient of tZo determinants. Then, as t tends to ]ero, approaches x, and hence,
each xm also approaches x. The determinant in the denominator, j(X) det[D jfi(xi)]
then approaches J(x), the -acobian of f at x, Zhich is again non-]ero. Thus, as t
tends to ]ero, the limit of tx xj j
e [ i s t s . T h a t i s , tyf teyfi
t lim1 1
0
o
e[ists. Thus, Dif-1(y) e[ists for all i, and for all y in <.
We have obtained the partial derivatives of the components of f-1 as Tuotients of
tZo determinants. The entries in these determinants are partial derivatives of the
components of f, Zhich are all continuous. Since a determinant is a polynomial of
its entries, Ze conclude that the partial derivatives of f-1 are continuous on <.
E[ample .2 : ShoZ that the function f: R2 R2, f(x, y) = (2xy, x2 – y2) is not
invertible on R2, but is locally invertible at every point of E ^ x, y _ x ! 0`. Also
find the inverse function at one such point.
Solution : Here f 1, 1 f( − 1, − 1) = (2, 0). Therefore f is not inMective, and hence
is not invertible on R2. On the other hand, if x, y E, then
((((
)
(((((((
)
((((
)
f
.
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Page 64
64A1AL f is locally invertible.
Suppose f(x, y) = (u, v). If x, y E, then y xu
2, and v = x2 . Therefore, 4x4 -
4x2v – u2 0. Thus, x2 22 2uv v , and x 22 2uv v 12,
y u 2v 2 −12
.3 Implicit )unction Theorem
If x2 + y2 0, find . graduate classes. Here, Ze taNe f(x, y) = x2 + y2, and find fx = 2x , and fy = 2y . Then dxdy 2x/2y = x/y. Of course, y cannot be ]ero.
While doing this e[ercise, actually Ze have used a theorem, the implicit function
theorem. To recall, in this setting, a function Zhich can be Zr itten as y = g(x , is
called an e[plicit function, and one Zhich can be e[pressed onl y as f(x, y) 0, is
called an implicit function. The implicit function tells us tha t under certain
conditions, Ze can e[press an implicit function as an e[plicit one, and then Ze can
use this e[pression to find dxdy .
In this section Ze are going to discuss this implicit function theorem for functions
of several variables. %efore Z e state and prove the general cas e, Ze first prove the
case for functions involving only tZo variables, x and y.
Theorem .3 : Let f be a real-valued C1 function, defined on the product ,
Zhere and are tZo intervals in R. Let a, b , and f(a, b) 0, but
fy(a, b) 0 . T h e n t h e r e e [ i s t s a n i n t e r v a l I i n R, containing a, and a C1 function
g : I R, such that g(a) = b, and
f(x, g(x)) 0 for all x I.
Proof : We consider a function, h: R2, given by h(x, y) = (x, f(x, y)). If
Ze Zrite
h , the -acobian matri[ of h is
R
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65Chapter 5: Inverse and Implicit Function TheoremsJh(x, y) . The determinant of this matri[, is not ]ero at
a, b . Thus, h is a C1 function, Zith a non-]ero -acobian at a, b . Therefore, by the
inverse function theorem, Theorem 5.2 , Ze can conclude that h is locally invertible
at a, b . Let u be the local inverse of h. x for all x and y in R. That is,
u(x, y) = (x, (x, y)) for all x and y in R. We noZ define g as, g(x) = (x, 0),
and shoZ that it has all the reTuired properties.
1oZ, since h(a, b) = (a, 0), u(a, 0) = (a, b). This means, a, 0 b. Thus, g(a)
= b.
Also, (x, 0) = h(u(x, 0)) = h(x, (x, 0)) = h(x, g(x)) = (x, f(x, g(x))). This implies
that
f(x, g(x)) 0.
Since u is a C1 function, g is also C1. Differentiating f(x, g(x)) 0 Zith respect to
x using chain rule, Ze get D1f(x, g(x)) + D 2f(x, g(x)) (x) 0, and thus,
(x) , ,
21
xgxfDxgxfD, since D2f(x, g(x)) 0.
%asically, this theorem tells us that under certain conditions, the relation
f(x, y) = 0, betZeen x and y can be e[plicitly Zritten as y = g(x).
Remark .2 : If instead of fy(a, b) 0 , Z e t a N e t h e c o n d i t i o n fx(a, b) 0, then
Ze can e[press x as an e[plicit function of y.
E[ample .3 : Can f(x, y) = x3 + y3 – 2xy be e[pressed by an e[plicit function
y = g(x) in a neighbourhood of the point 1, 1 "
Solution : 1ote that f 1,1 0, and fy = 3 y2 – 2x 1 at 1, 1 . Further, f i s a
C1 function on R2. Therefore, Ze can apply Theorem 5.3, and conclude that there
e[ists a uniTue function g, defined on a neighbourhood of 1, such that g 1 1.
Also, (x) x yy x
2 32 3
22
in this neighbourhood.
E[ample .4 : ChecN Zhether Theorem 5.3 can be applied at all points, Zhere
f(x, y) = x2 – y2 0.
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Page 66
66A1AL= −2y, and fx = 2x. At the point 0, 0 , fx and fy are both ]ero, and hence Ze cannot
apply the theorem. At all the remaining points, the function sa tisfies all the
conditions of Theorem 5.3, and hence it can be applied. of these points, Ze Zill get either
g(x) = x , or g(x) = − x.
We noZ go a step further, and co nsider a real-valued function o f several variables.
Theorem .4 : Let f be a real-valued C1 function, defined on an open set, U, in Rn.
Let
a = (a 1, a2, ... , a n-1) Rn-1, such that a, b U, f(a, b) 0, and Dnf(a, b) 0 .
Then there e[ists a uniTue C1function g, defined on a neighbourhood N of a, such
that g(a) = b , and
f(x, g(x)) 0 for all x N.
Proof : We consider a function h : U Rn−1 [ R, defined by h(x, y) = ( x, f(x, y)).
If Ze Zrite h = (h 1, h2, ... , h n), then hi(x, y) = x i, for 1 i n – 1, and hn(x, y) =
f(x, y). Therefore, the -acobian matri[ of h is given by
The determinant of this matri[ is Dnf, Zhich is non-]ero. Therefore, Ze can apply
the inverse function theorem Theorem 5.2 , and conclude that h i s l o c a l l y
invertible at a, b . If u is the local inverse of h, and Ze Zrite u = (u1, u2), then you
Zill see that u1(x, y) = x for all x, y . Thus, u(x, y) = ( x, u2(x, y)) for all x, y . We
noZ define g(x) = u 2(x, 0 , and shoZ that this has the reTuired properties.
1oZ, u(a, 0 a, b . This gives g(a) = u 2(a, 0) = b.
Also, (x, 0) = h(u(x, 0)) = h(x, ( x, 0)) = h(x, g(x)) = ( x, f(x, g(x))). This implies
that
f(x, g(x)) = 0.
E[ample . : E[amine Zhether the function f(x, y, z) = x2 + y2 – 4 can be
e[pressed as a function y = g(x, z) in a neighbourhood of the point 0, -2, 0 .
munotes.in
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67Chapter 5: Inverse and Implicit Function TheoremsSolution : We note that f(0, −2, 0) = 0, and D2f = 2y = − 4 at (0, −2, 0). So, applying
the implicit function theorem, there exists the required neighb ourhood of (0, −2, 0).
In fact, you can checN that in the neighbourhood, N = B((0, − 2, 0), 1), we can
e[press the function as y = − (4 – x2)12 .
Here are some e[ercises that you should try :
1 Determine Zhether the folloZing functions are locally invert ible at the given
points :
i) f(x, y) = (x3y + 3, y2) at 1, 3
ii f(x, y, z) = (excosy, exsinz, z) at 1, 1, 1 .
2 For each of the folloZing functions, shoZ that the eTuation f(x, y, z) 0 defines
a continuously differentiable function z = g(x, y), in a neighbourhood of the given
point:
i f(x, y, z) = x3 + y3+ z3 – xyz – 2 , 1, 1, 1
ii f(x, y, z) = x2 + y3 – xysinz , 1, - 1, 0 .
That brings us to the end of this chapter. We hope you have stu died the concepts
carefully, and have understood them.
.4 Let Us Sum Up
In this chapter Ze have discussed tZo very important theorems: the inverse function
theorem, and the implicit function theorem. The proofs of these theorems are a little
complicated. So Ze have tried to go step by step from functions of one variable to
functions of many variables.
The Inverse Function Theorem: gives the conditions under Zhich a function, even
though not invertible on its domain, is seen to be locally inve rtible. The -acobian
of the function being non-]ero at a point ensures the local inv ertibility of the
function in a neighbourhood of that point.
The Implicit Function Theorem: gives the conditions, under Zhic h an implicit
relationship betZeen variables can be e[pressed in an e[plicit manner. Here, again,
the -acobian plays an important role.
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UNIT 4
RIEMANN INTE*RAL - I
Unit Structure
6.1 Introduction
6.2 Partition
6.3 Riemann Criterion
6.4 Properties of Riemann Integral
6.5 RevieZ
6.6 Unit End E[ercise
.1 Introduction
The Riemann integral dealt Zith in calculus courses, is Zell su ited for computations
but less suited for dealing Zith limit processes.
%ernhard Riemann in 1868 introduced Riemann integral. He need t o prove some
neZ result about Fourier and trigonometric series. Riemann inte gral is based on
idea of dividing. The domain of function into small units over each such unit or
sub-interval Ze erect an appro[imation rectangle. The sum of th e area of these
rectangles appro[imates the area under the curve.
As the partition of the interval becomes thinner, the number of s u b - i n t e r v a l
becomes greater. The appro[imating rectangles become narroZer a nd more
precise. Hence area under the curve is more accurate. As limits of sub-interval tends
to ]ero, the values of the sum of the areas of the rectangles t ends to the value of an
integral. Hence the area under cu rve to be eTual to the value o f the integral.
%efore going for e[act definition of Riemann e[plained the foll oZing definitions.
.2 Partition
A closed rectangle in n
is a subset A of n
of the forms. 68munotes.in
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69Chapter 6: Riemann Integral - I>@ >@ >@11 2 2,, . . . . ,nn Aa b a b a b uu u Zhere iiab
. 1ote that 12,, . . . . ,n xx x A iff iiiaxb idd .
The points 12, ,....,n xx x are called the partition points.
The closed interval >@>@ > @10 1 21 2 1 ,, ,, . . . . . . , ,nn n Ix x Ix x Ix x are called the
component internal of >@,ab.
Norm : The norm of a portion P is the length of the largest sub-intern al of P and is
denoted by P.
)or e[ample : Suppose that 10 1 ,, . . . .k Pt t t is a partition of >@11,ab a n d 20 ,....,r PS S is a partition of >@22,ab. Then the partition 12. PP P of >@ >@11 2 2,,ab ab u divides the closed rectangle >@ >@11 2 2,,ab ab uinto .r-gub
rectangles.
In general if iP divides >@,iiab i n t o ik s u b - i n t e r v a l t h e n 1,....n PPP
>@ >@11,. . . .,nn ab ab uu i n t o 12.....n Kk k k s u b - r e c t a n g l e . T h e s e s u b - r e c t a n g l e s a r e
called sub-rectangles of the partition p.
Refinement :
Definition : Let A be a rectangle in n
and :fAo
be a bounded function and
P be partition of A for each sub-rectangles of the partition. ^`
>@1inf :
.. . ,ssms f f x x S
glbo f f o n x x
^`
>@1sup :
... ,ssMs f f x x S
lub o f f o n x x
Zhere 1, 2, ....,Sn
The loZer and upper sums of ffor ‘p’ are defined by ,s
sLfp m f s Q ¦ and ,s
sUfp M f s Q ¦
Since ssmM Ze have ,,Lfp Ufp d munotes.in
Page 70
70M.SC. MATHEMATICS Refinement of a partition : Let 12,, . . . ,n PP P P and
1,...,n PPP be partition
of a rectangle A in n
. We say that a partition
P is a refinement of P if
PP.
If 1P and 2P are tZo partition of A then 12 PPP
is also a partition of A is called
the common refinement of 1P and 2P.
A function :fAo
is called integrable on the rectangle A in n
if
fis
bounded ..glb? of the set of all upper sum of
f and ..lub of the set of all loZer
sum of
f e[ist.
Let ^`inf , Uf Ufp
^`sup , Lf Lfp
If Uf Lf is called
f is R-integrable over A. ?if can be Zritten as
AUf Lf f ³.
Theorem :
Let Pand Pc be partitions of a rectangle A in n
. If Pc refines P then shoZ that ,,Lfp LfP cdand ,, Uf P Ufp cd.
Proof :
Let a function :fAo
is bounded on A P
P are tZo partition of A and Pc
is retinement to P.
Any subrectangle S of Pc is union of some subrectangles 12,, . . . . ,k ss s of Pc and 12 .....k VS Vs Vs Vs .
1oZ ^` ^` inf inf simf f xxs f xxs d 1, ....,
i ssmf m f i k?d
,s
spLfp m fVs
¦ 1....ss kmf V s mfV s V s?
1 1.....
k ss kmf V s mf V sd munotes.in
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71Chapter 6: Riemann Integral - IThe sum of LHS for all subrectangle is of Pc Zill get ,LfP c. 1,,Lfp Lfp?d
1oZ, ^`sup sMf f x x S
^`sup i fxx S t 1, ...,s siMf M f i K t ,s
spUfp m f V s
¦
1oZ, 12 ....ikMs f V S Ms f V S V S V S
12..... ....ss k Ms f V s M f V s M f V sd
TaNing the of L.H.S. for all subrectangle iS of Pc Zill get ,, , Uf P Uf P Uf P cc?t.
Theorem :
Let 1P 2P be partitions of rectangle A :fAo
be bounded function. ShoZ
that 21,,LfP UfP d 12,,LfP fP d
.
Proof :
Let a function :fAo
be a bounded find 1P 2P a r e a n y t Z o p a r t i t i o n
of A.
Let 12 PP P
P? is a refinement of both 1P 2P
1 ,, Uf P Uf P d……….. (I)
2 ,, Uf P Uf P d……….. (II)
1 ,,LfP LfP t……….. (III)
2 ,,LfP LfP t……….. (IV) ? We get 12,, , , Uf P Uf P Lf P Lf P ttt.
Hence 12,, Uf P Lf P t
Similarly, 22 1,, , , UfP Uf P Lf P Lf P ttt.
Hence, 21,, Uf P Lf P t munotes.in
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72M.SC. MATHEMATICS Theorem :
Let a function :fAo
be bounded on A then for any 0,! a partition P on A
such that , Uf P Uf and ,LfP Lf !
Proof :
Let a function :fAo
be bounded on A ^`inf , Uf Uf P a n d ^`sup , Lf LfP f o r a n y 0,! partitions 1P 2P o f A s u c h t h a t 1, Uf P Uf 2,LfP Lf ! .
Let 12 PPP
the common refinement of 1P and 2P.
1
2,,
,,Uf P Uf P Uf
LfP LfP Lfdd
t! ? , Uf P Uf ,LfP Lf !
.3 Riemann Criterion
Let A be a rectangle in n
A bounded function :fAo
is integrable iff for
every 0!, there is a partition P of A such that ,, Uf P Lf P .
Proof :
Let a function :fAo
is bounded. ^`inf , Uf Uf P ^`sup , Lf LfP
Let f be integrable of A Uf Lf?
for any 0,! a partition P on A s u c h t h a t ,2 Ufp Uf and ,2Lfp Lf ! . ,2 Ufp Uf? ,2Lfp Lf . munotes.in
Page 73
73Chapter 6: Riemann Integral - I ,, 2 2 Ufp Lf P Uf Lf? . , Ufp Lf?
Conversely,
Let for any 0,! a partition P on A such that ,, Ufp Lf P . ,, UPf U f U f Lf Lf LfP ªº ª º ª º¬¼ ¬ ¼ ¬ ¼
Since ,, Uf P Uf o t
Uf Lf o t
and , Lf LfP o t ?Ze have, oUf L fd
Since is arbitrary, Uf Lf ?f is integrable over A.
E[ample 1
Let A be a rectangle in n
and :fAo
be a constant function. ShoZ that f is
integrable and .
AfC V A ³ for some C
.
Solution : fx C x A f? is bounded on A
Let P be a partition of A ^`
^`inf
sup s
smf f xxs C
Mf f x x s C
,s
SSLfP m fVS C VS C VA? ¦¦ ,s
SSUf P M f V S C V S C VA ¦¦ munotes.in
Page 74
74M.SC. MATHEMATICS Uf Lf C VA? f? is integrable over A. ? by Reimann criterion, 0 s.t. .
AfC V A ³ for some C
.
E[ample 2 :
Let >@>@: 0,1 0,1FX o
,1oi f x i s r a t i o n a lfx yif x is irrational ® ¯
Show that ‘ f’ is not integrable.
Solution :
Let P be a partition of >@ >@0,1 0,1u into S subport of P.
TaNe any point 11,xy S such that x is rational. ,fx y o? and 11,xy S such that 1x, is irrational 11,1fxy? ^`
^`inf 0
sup 1s
smf f xxS
Mf f x x S?
,0
,1
1, 0s
S
s
SLfP m fVS
Uf P M f V S
Uf Lf
Uf Lf ? ?
? z¦
¦ f? is not integrable >@ >@0,1 0,1u
.4 Properties of Riemann Integral
1 Let :fAo
be integrable and gf e[cept at finitely many points shoZ
that g is integrable and AAfg ³³.
Proof :
Since f is integrable over A. munotes.in
Page 75
75Chapter 6: Riemann Integral - I? by Riemann Criterion, a partition P of A.
Such that ,, Uf P Lf P ……… (I)
Let Pc be a refinement of P, such that
1 xA Zith fx g x z, it belongs to 2n subrectangles of Pc
2 12nVSdu
Where d numbers of points in A at Zhich fgz
^` ^`
^` ^`sup inf
inf supxAxA
xAxAug x f x
gx f x
?Pc is refines P, Ze have
,, , ,
,, , ,LfP LfP UfP UfP
Uf P Lf P Uf P Lf Pccddd
cc?d
1oZ
,, Ug P U fP cc
1d
ij ij ij
iMs g Ms f V s
¦¦
On other rectangle, fg and so ij ijMs g Ms f .
^`supij
xAMs g g x
d ^` ^` inf infij ijxA xAMs f f x Ms f f x
t d
ij ijMs g Ms f u d 2
11,,nd
ij
ijUg P U fP uVS
§·cc?d ¨¸
©¹¦¦
Let ^` 2
11
11sup , , 2 .nd
n
ij
ijVV S U g P U f P u V d u v
d d d ¦¦ ……. (II)
1oZ similarly Ze get 11,, 2nLgP L fP d V t
……... (III) by (II) & (I II Ze
get. munotes.in
Page 76
76M.SC. MATHEMATICS
11 1 1
1,, , 2 , 2
22
2
22 2 2nn
n
n
nUg P Lg P U fP d u LfP d
du V
du
du--
d
d
d
11,, Ug P Lg P?
% y R e i m a n n C r i t e r i o n * i s i n tegrable by eTuation II
11
11,, 2
,, 2n
nUg P U fP d u v
Ug P U fP d u
1ote that 11,, 2n
AgU g P U f P du - dd³
1,22nLfP d u -d
1
12,2 2n
nduLfPdu
1
1,22
,
ALfP
LfP
f
³
This is true for any 0!
AAgfd³³ ………………….. (IV)
1oZ ,,2
AgL g P L f P cctt³
,
2
AAUf P
ffct
t!³³ ^`inf ,
2A
AAfU f P
gf?
? ! ³
³³ munotes.in
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77Chapter 6: Riemann Integral - I?This is true for any 0!
AAgf?t³³……… (V) ?from IV V Ze get
AAgf ³³
2 Let :fAo
be integrable, for any partition P of A and sub-rectangle S,
shoZ that
i sssmf m g mf g d and
ii sssMf M g Mfg t
Deduce that
,, ,LfP L g P Lf g P d and
,, , Uf g P Uf P U g P d
Solution :
Let P be a partition of A and S be a Subrectangle ^`
inf s
smf f xxS
mf f x xS?
d
Similarly smg g x xS d ssmf m g f x g x xS?d ssmf m g is loZer bound of ^` ^` fx g xx S f g xx S ssmf m g is loZer bound of ^` ^` fx g xx S f g xx S ^`
inf ss
smf m g f gxxS
mf gd
sssmf m g mf g?d munotes.in
Page 78
78M.SC. MATHEMATICS ii ^` Ms f sub f x x s
Ms f f x x st
Similarly Ms g g x x S t Ms f Ms g f x g x x S?t Ms f Ms g is upper bound of ^` ^` fx g xx S f g xx S sup Ms f Ms gt ^` fg x x S M s fg Ms f Ms g Ms f g?t
Hence,
,,
,sp
spLfP L g P M sf M sg VS
Ms f g V S
Lf g P
d
¦
¦
,, ,
,,
,s
sLfP L g P Lf g P
Uf P U g P M sf M s gV S
Ms f g V S
Uf g P?d
t
t ¦
¦ ,, , Uf P U g P Uf g P t Proved.
3 Let :fAo
be integrable, :gA o
integrable than shoZ that fg
is integrable and
AA Afg f g ³³ ³.
munotes.in
Page 79
79Chapter 6: Riemann Integral - IProof :
Let P be any partition of A then ,, , , , , Uf g P Lf g P Uf P U g P Lf P L g P d ªº¬¼ ,,, , Uf P U g P Lf P L g Pd…………………….. (I) f? is integrable.
%y Rieman interion for given 0,! a partition P, of A such that 11,,2Uf P Lf P ……………………….………….… (II)
Similarly g
is integrable for 0,! a partition 2P of A such that 22,,2Ug P LfP ……………………………………… (III)
Then
12 PP P
is a refinement of both 12 PP.
1,, LfP LfP?d
1,, Uf P Uf P t
2,, LgP L fP d
2,, Ug P Ug P t………………………………………….. (IV)
11 2, , , ,Uf P Lf P Uf P Lf P?! t
22 2, , , ,Ug P Lg P Ug P Lg P ! t ……………….. (V)
The eTuation I is true for any partition P of A.
In general, it is true for partition
P of A
,,
,,,,
22Uf g P Lf g P
Uf P Lf P U g P L g P?
d
,, Uf g P Lf g P?
%y Riemann Criterian fg is integrable.
Let 0 since ^`sup ,
Aff P ³ so a partition P such that 1,2
Aff P ³.
munotes.in
Page 80
80M.SC. MATHEMATICS Similarly a partition 23,, . . . .n PP P of A S 2
3
4,2
,2
,2A
A
AgL g P
Uf P f
Ug P g
³
³
³
Let 1234 PPP P P
.
Then 1,,22
Aff P L f P d ³
Similarly ,2
AgL g P ³ ,2
AUf P f ³ and ,2
AUg P g ³ ,, ,
AA Afg L f P L g P L f g P f g d d³³ ³
,
,,
22
AA
AAUf g P
Uf P U g P
fg
fgd
d
³³
³³
AA A AAfg f gfg? ³³ ³ ³³
This is true for any 0! AAA AA A AAfgf gfg f gfg?d d ³³³ ³³ ³ ³³
4 Let :fAo
be integrable for any constant C, shoZ that
AACf C f ³³
Proof :
Let C
munotes.in
Page 81
81Chapter 6: Riemann Integral - ICase 1
Let 0! and suppose 0C!.
Let P be a partition of A and S be a subrectangle of P.
^`
^`
^`
sup
sup
sup sM Cf Cf x x S
Cf x x S
Cf x x S
CMs f
Similarly, s ms Cf Cm f
,
,SSUC fP M sC fvS C M sfvS
CU f P?
¦¦
Similarly ,, LC f P C L f P f? i s i n t e g r a b l e f o r a b o v e 0, a partition P of A such that ,, Uf P Lf P C
,, , ,
,,U Cf P L Cf P C U f P C L f P
CU fP L fP
CCC?
ªº¬¼
u
%y Riemann Criteria. Cfis integrable
for 0,a ! partition P of A such that
,,
,
,AA
A
A
AA A AC f C f C L f P L Cf PC
Cf U Cf P
CU f P C fC
f Cf C f C fCC§· ¨¸
©¹
d d
§· ¨¸
©¹
§· §·? ¨¸ ¨¸
©¹ ©¹³³
³
³
³³ ³ ³ munotes.in
Page 82
82M.SC. MATHEMATICS This is true for any 0
AA A
AAC f Cf C f
Cf C fdd? ³³ ³
³³
Case II
1oZ suppose 0C
Let P be a partition of A and S be any subrectangle in P. Ms Cf C Ms f? and smC f C M sf ,, LC f P C U f P? and ,, UC fP C LfP f? is integrable for above 0,! a partition P of A such that ,, Uf P Lf PC
,, , ,
,,U Cf P L Cf P C L f P C U f P
CU fP L fP
CC?
ªº¬¼
%y Riemann Criteria Cfis integrable.
for 0,! a partition P of A such that AAAC f Cf C f ³³³.
This is true for every 0! AA A
AAC f Cf C f
Cf C f d
? ³³ ³
³³
munotes.in
Page 83
83Chapter 6: Riemann Integral - IE[ample 3:
Let ,:fgA R o be integrable suppose fgd shoZ that AAfgd³³.
Solution :
%y definition ^`inf ,
AfU f P ³ and ^`inf ,
AgU g P ³.
Let P be any partition of A S be any subrectangle in P
as fgd
^` ^`,,
inf , inf ,ssmf m g
Uf P U g P
Uf P U g Pd?d
d
This is true for any partition AAfg?d³³
E[ample 4:
If :fAo
is integrable shoZ that if is integrable and AAffd³³.
Solution : Suppose f is integrable first Ze have to shoZ that f is integrable.
L e t P b e a p a r t i t i o n o f A S b e s u b r e c t a n g l e o f P t h e n
^`
^`
^`
sup
sup
sup Ms f f x x S
fxxS
fx x S
Ms f
Similarly Ms f Ms f
,
,
,,ss
SS
s
S
ss ss
PPUfP M f V S M fV S
LfP m fV S
Mf mfV S Mf mfV S
Uf P Lf P
? d
d ¦¦
¦
¦¦ f? is integrable, for 0,! a partition P such that ,, Uf P Lf P . ,, , , UfP LfP U f P L f P?d
munotes.in
Page 84
84M.SC. MATHEMATICS ?%y Riemann criteria f is integrable over
.
1oZ ^`inf ,
P
AFU f P ³
inf
inf
infsPSP
sP
sPPMf V S
Mf V S
Mf V S
¦
¦
¦
^`inf
inf ,s
PMf V S
UfPd
¦ AAff? ³³
E[ample :
Let :fAo
and P be a partition of A shoZ that f is integrable iff for each sub-
rectangle S the function f
sZhich consist of f restricted to S is integrable and that
in this case S ASffs ¦³³. Suppose :fAo
is integrable.
Let P be a partition of A S be a sub-rectangle in P.
1oZ to shoZ that fSso
is integrable.
Let 0,! a partition Pc o f A s u c h t h a t ,, Uf P Lf P c f? i s
integrable
Let PP Pcc
then 1P is refinement of both P Pc. 1 ,, Uf P Uf P c?t 1 ,,LfP LfP cd 11,, ,, Uf P Lf P Uf P Lf P cc?d ………………… (I) 1P
is refinement of P S
is union of some subrectangle of 1P say 1iSU s i
. munotes.in
Page 85
85Chapter 6: Riemann Integral - I
111,,ss
SPUf P Lf P M f m f V S
?! ¦ for all rectangle.
1
,,ik
is
iMs f m f V S
ffUP LPSS t
¦ ?%y Riemann Criterion
f
S? is integrable.
Conversely, Suppose f
Sis integrable for each SP.
To shoZ that f is integrable.
Let 0,! partition SP of S such that ,,SSffUP LPkss ………………………………. (II) f
S? is integrable for each SP Zhere K is number of rectangle in P.
Let 1P be the partition of A obtained by taNing all the subrectangle defined in the
partition SP.
There is a refinement 1
SP of SP containing subrectangles in 1P. 11,,SS Ufs P Lfs P k? …………………………… (III) 11
1111 1,,SS
SPUf P Lf P M f m f V S
? ¦
1
1111
11,,
,Ss
SP SP
SS
SP
SPMs f m f V S
Ufs P Lfs P
k
kk
§·
¨¸¨¸©¹
¦¦
¦
¦ ?%y Riemann Criterian f is integrable. munotes.in
Page 86
86M.SC. MATHEMATICS Let 0!
111,
SS
SP SP S
s
SP SPfS k LfSP
mf V S
§· ¨¸
©¹
§·c ¨¸¨¸©¹¦¦³
¦¦
Let 1P be a partition of A, obtained by taNing allthe subrectangle de fined in SP.
1
11
1
11
1
111
11
1
1,,s
SP SP S
A
s
SP
s
SP SPfS k m f VS
LfP f UfP
Mf V S
Mf V S
§·? ¨¸
©¹
§· ¨¸©¹¦¦³
³
¦
¦¦ ,S
SP SP S
SP SP AC UfS P fSk
fS C f fS
§·? ¨¸
©¹? ¦¦ ³
¦¦³³ ³
This is true for all 0! SP SP S
SP ASfS f fS
ff S
?dd? ¦¦³³³
¦³³
E[ample :
Let :fAo
be a continues function shoZ that f is integrable on A.
Solution :
Let :fAo
be a continuous function to shoZ that f is integrable.
Let 0!, since A is closed rectangle it is closed and bounded in n
. A? is compact. f
is continuous function on compact set fis uniformly continuously on
. ?for the above 0, 0G! ! such that ,,xg A xy f x f y V A G . munotes.in
Page 87
87Chapter 6: Riemann Integral - ILet P be a partition of A such that side length of each subrect angle is less than nG.
I f ,xy Sfor some subrectangles S then 22
11 ....nn xy x y x y
2
SnnG§· ¨¸©¹ fx fy VA
S
is compact f? is continuous f? attains its bound in S.
Let 12,, . . . . . ,k SS S be the subrectangle in A. Then for 1, ,ii i ik x y S such that ii s iiMs f f x m f f y .
1,,
ik
is i
iUf P Lf P M s f m f V S
? ¦
1
1k
ii i
i
kk
ii
iV Afx fy VS
VS VSVA VA
VAVA
¦
¦¦ ?%y Riemann Criterion f is integrable.
munotes.in
Page 88
88M.SC. MATHEMATICS . Review
After reading this chapter you Zould be NnoZing.
Defining R-integral ove r a rectangle in n
Properties of R-integrals
R-integrabal functions
Continuity of functions using
-intervals.
. Unit End E[ercise
I Let >@ >@ 0,1 0,1f uo
be defined by
1 ,0 03
1313fx y i f y
if y d d
d d
s h o Z t h a t f is integrable.
II Let Q be rectangle in n
fQo
be any bounded function.
a ShoZ that for any partition P of Q ,,LfP UfP
b ShoZ that upper integral of function f e[it.
III Let f be a continuous non-negative function on >@0,1 and suppose there e[ist >@0 , xa bsuch that 00 fx ! shoZ that
0fx d x a!³.
IV Let f be integrable on >@,ab and >@:,Fa b o
and 1Fx f x t h e n
prove that
afx d x F b F a ³
V Which of the folloZing functions are Riemann integrable over >@0,1. -ustify
your ansZer.
a T h e c h a r a c t e r i s t i c f u n c t i o n o f t h e s e t o f r a t i o n a l n u m b e r i n >@0,1.
b sin fx x xy for 01x
03f munotes.in
Page 89
89Chapter 6: Riemann Integral - IVI Prove that if f is
-integrable then f is also R-integrable is the converse
true" -ustify your ansZer.
VII ShoZ that a monotone function defined on an interval >@,ab is R-inegrable.
VIII A function >@ 0,1f o
is defined as 1111 1
33 3nn nfx x d Z h e r e n
00f
s h o Z t h a t f is R-integrable on >@0,1 calculate 1
0fx d x³.
I; fx x x «»¬¼ >@1, 3xZhere x«»¬¼ denotes the greatest integer not greater
than x shoZ that f is R-integrable on >@1, 3.
; A function >@ ,fa b o
is continuous on >@,ab 0 fx t >@, xa b and 0b
afx d x ³ shoZ that 0 fx >@, xa b.
munotes.in
Page 90
90M.SC. MATHEMATICS UNIT 4
7
MEASURE =ERO SET
Unit Structure
7.1 Introduction
7.2 Measure ]ero set
7.3 Definition
7.4 Lebesgue Theorem only statement
7.5 Characteristic function
7.6 FUBIN’s Theorem
7.7 RevieZs
7.8 Unit End E[ercises
7.1 Introduction
As Ze have seen, Ze cannot tell if a function is Riemann integr able or not merely
by counting its discontinuities one possible alternative is to looN at hoZ much space
the discontinuities taNe up. Our Tuestion then becomes : i Ho Z can one tell
rigorously, hoZ much space a set taNes up. Is there a useful de finition that Zill
concide Zith our intuitive understanding of volume or area"
At the same time Ze Zill develop a general measure theory Zhich serves as the
basis of contemporary analysis.
In this introductory chapter Ze set for the some basic concepts of measure theory.
7.2 Measure =ero Set
Definition :
A subset ‘A’ of n
said to have measure ‘O’ if for every 0! there is a cover ^`12,. . . .UU of A by closed rectangles suc h that the total volume
1ivU if
¦. 90munotes.in
Page 91
91Chapter 7: Measure Zero Set
Theorem :
A function ‘f’ is Riemann integrable iff ‘f’ is discontinuous o n a set of Measure
]ero.
A function is said to have a property of Continuous almost ever yZhere if the set on
Zhich the property does not hold has measure ]ero. Thus, the st atement of the
theor em is that ‘f’ is Riemann integrable if and only if it is conti nuous atmost
everyZhere.
Recall positive measure : A measure function >@:0 ,uM of s u c h t h a t
1 1ii
i iVu V uf f
§· ¨¸©¹¦
.
E[ample 1:
1 “Counting Measure” : Let X be any set and MP X the set of all subsets :
If EX is finite, then EEPK if EX is infinite, then EP f
2 “Unit mass to 0x - Dirac delta function” : Let X be any set and MP X
choose 0xXset.
0
01
0Ei f x E
if x EP
E[ample 2:
ShoZ that A has measure ]ero if and only if there is countable collection of open
rectangle 12,, . . . .VV such that i AV
and iVv ¦.
Solution :
Suppose A has measure ]ero.
For 0,! countable collection of closed rectangle 12,, . . . .VV such that 1i
iAVf
and
1 2i
iVVf
¦.
For each i, choose a rectangle iusuch that iiuv and 2iiVu Vv d. munotes.in
Page 92
92M.SC. MATHEMATICS Then 11ii
iiAv uff
and
1112ii i
iiiVu Vu Vvfff
dd¦¦¦
1222i
ivuf
d ¦
1ote that : iu are open rectangles in n
conversely,
Suppose for 0,! countable collection of open rectangles 12,, . . . .uu such that 1i
iAuf
and
1i
iVuf
¦.
For each ,i consider iiVu then iV is a closed rectangle and iiVv Vu .
Then 11ii
iiAu vff
and
11ii
iiVv Vuff
¦¦.
A has measure ]ero.
Note : Therefore Ze can replace closed rectangle Zith open rectangles in definition
of measure ]ero sets.
E[ample 3:
ShoZ that a set Zith finitely many points has measure ]ero.
Solution :
Let ^`1,....,m Aa a be finite subset of n
.
Let 12 0, , ,.....,ii i i naa a a ! and 11
11 1111,. . .22 22nn
ii iiVi a aªº§· §·«» u¨¸ ¨¸©¹ ©¹«»¬¼ 11
1111... ,22 22nn
in in iiaaªº§· §·«»u ¨¸ ¨¸©¹ ©¹«»¬¼
Then 1
11
122nn
ii
iVV i
§· ¨¸©¹
Clearly iaV i for 1imdd 1m
iAV i
?
and 11
11 111
22 2mm
ii
ii iVV if
¦¦¦ ?%y definition of measure of ]ero ? A has measure of ]ero. munotes.in
Page 93
93Chapter 7: Measure Zero SetE[ample 4:
If 123 .... AA A A
a n d e a c h Ai has measure ]ero, then shoZ that A has
measure ]ero.
Solution :
Let 0!and 12 .... AA A
Zith each Ai has measure ]ero.
Each Ai has measure ]ero for 1, 2, ....i a cover ^`12,, . . . . ,ii i nuU U of Ai
%y closed rectangle such that
1,1 , 2 , . . . .2ii i
iVu if
¦
Then the collection of iiU is cover A
11 2i i
iiVVff
? ¦¦
Thus 12 ....n AA A A
has measure ]ero.
E[ample :
Let nA
be a Rectangle shoZ that A does not have measure ]ero. %ut Aw has
measure ]ero.
Proof :
Suppose A has measure ]ero.
A is a rectangle in n
0 VA?!
Choose 0! such that VA …………………….. (I)
A has measure ]ero countable collection of open rectangle ^`iu such that 1i
iAuf
and iVu ¦
A is compact
This open cover has a finite subcover after renaming. We may as sume that ^`12,, . . . .k uu u is subcover of the cover ^`iu. munotes.in
Page 94
94M.SC. MATHEMATICS
1i
iAuf
?
.
Let P be partition of A that contains all the vertices all
1iusi t o N . L e t 12,, . . . . ,n SS S denote the subrectangle of partitions.
11 1nk
jii
ji iVA VS Vu Vuf
? d ¦¦¦
Zhich is a contradiction to I ? A does not have measure ]ero.
1 o t e t h a t Aw i s a f i n i t e u n i o n o f s e t o f t h e f o r m >@ >@ >@11,, . . . . . , ,ii nn Ba b a b a b uu u . % can be covered by are closed rectangle. >@ > @ >@11,. . . . ., . . . . .,ii nn Ba b a a a bGG u u u u.
Then VBG depend on G and 0 VBGoas 0 Go. BG? has measure ]ero ?%oundary of A Aw is finite union of measure ]ero. A?w has measur5e ]ero.
E[ample :
Let nA
Zith Aqz. ShoZ that A does not measure ]ero.
Solution :
LetnA
, Zith Aqz
Let xAq ? 0r!, such that ,,Bx r A d %ut ^`
1,
n
ii
iBx r y A y x r
yA yx r
½ ®¾¯¿¦ munotes.in
Page 95
95Chapter 7: Measure Zero Set A? does not have measure ]ero.
E[ample 7:
ShoZ that the closed interval >@,ab does not have measure ]ero.
Solution :
Suppose ^`1 iiu be a cover of >@,ab by open intervals. >@,ab
is compact this open cover has a finite subcover.
After renaming, Ze may assume ^`12,, . . . . ,n uu uis the subcover of ^`iu of >@,ab
We may assume each iuintersect >@,ab otherZise replace iuZith >@,iua b
Let 1n
i
iuu
If u is not connected then >@,ab is contained in one of connected component of u. >@,i ab u for some i >@,j ab u?
for ijz
Which is not possible u? is connected u is an open interval say , uc d Then as > ,,ab u cd iVu d c b a ! ¦
In particular Ze cannot find an open cover of >@,abZith total length of the cover 2ba . >@,ab? does not have measure ]ero.
munotes.in
Page 96
96M.SC. MATHEMATICS E[ample :
If >@0,1 A is the union of all open intervals ,iiab such that each rational number
in 0,1 is contained in some ,iiab. If
11
iTb i a if
¦ t h e n s h o Z t h a t t h e
boundary of A does not have measure ]ero.
Solution :
We first shoZ that >@0,1 ?AAw
1ote that ? AA Aw q A
is open AAq
Also >@0,1 QA
>@0,1 QA?
>@0,1 A?
%ut >@ >@ 0,1 0,1 AA >@
>@0,1
0,1 ?A
AA? ?w
Let 10T !
If Aw has measure ]ero then since 0,! a cover of Aw Zith open intervals such
that sum of length of intervals 1T Aw
is closed and bounded Aw is compact finite subcover ^`1n
iiu for Aw 1iuT? ¦
1ote that ^`1 1 ,ii i iui n a bf
dd c o v e r >@0,1 a n d s u m o f l e n g t h s o f t h e s e o p e n
intervals is less than 11TT Z h i c h i s n o t p o s s i b l e a s >@ ^`10,1 1 ,ii i iui n a b Af
d d ?w
does not have measure ]ero. munotes.in
Page 97
97Chapter 7: Measure Zero Set7.3 Definition
A subset ‘A’ of n
has content ‘O’ if for every 0!, there is a finite cover ^`12,, . . . . . ,n uu u of A by closed rectangles such that
1n
i
iVu
¦
Remark :
1 If A has content O, then A clearly has measure O.
2 Open rectangles can be used instead of closed rectangles in the definition.
E[ample :
If A is compact and has measure ] ero then shoZ that A has conte nt ]ero.
Solution :
Let A be a compact set in n
Suppose that A has measure ]ero ? a cover ^`12,, . . . .uuof A such that
1i
iVuf
¦ for every 0!. A
is compact, a finite number 12,, . . . . . ,n uu u of iu a l s o c o v e r s A a n d
11n
ii
iiVu Vuf
¦¦ A? has content ]ero.
E[ample 10 :
*ive one e[ample that a set A has measure ]ero but A does not h ave content ]ero.
Solution :
Let >@0,1AQ
Then A is countable A has measure ]ero
1oZ to shoZ that A does not have content ]ero.
Let > ^`, 1iiab i n dd be cover of A >@ >@,. . . .,ii nn Aa b a b?
munotes.in
Page 98
98M.SC. MATHEMATICS >@ >@11,. . . .,nn Aa b a b?
%ut >@0,1A >
1,1n
ii
iab
?!¦
In particular, Ze cannot find a finite cover for A such that
11 ,2n
ii
iab
¦
A? does not have content ]ero.
E[ample 11:
ShoZ that an unbounded set cannot have content ]ero.
Solution :
Let nA
be an unbounded set.
To shoZ that A does not have content ]ero
Suppose A has content ]ero for 0,! finite cover of closed rectangles ^`1k
iiu of
A such that 1k
i
iAu
and
1k
i
iVu
¦.
Let >@>@11,. . . .,ii i i n i nua b a b u u
Let ^`12 min , ,.....ii i k iaa a a
^`12 ma[ , ,.....ii i k ibb b b
then >@ >@11,. . . .,in nua b a bu u
>@ >@11,. . . .,nn Aa b a b?u u A? is bounded
Which is contradiction A? does not have content ]ero.
munotes.in
Page 99
99Chapter 7: Measure Zero SetE[ample 12: :fAo
i s n o n - n e g a t i v e a n d 0
Af ³ Z h e r e A i s r e c t a n g l e , t h e n s h o Z t h a t ^` 0 xA f xz has measure ]ero.
Solution :
For ^`1 , n nA x A f xn
1ote that ^` ^` ,0 0 xA f x xA F xz ! ^f
is non-negative` ^`
111 n
nnxA f x Anff
!
We have to shoZ that nA has measure ]ero 0
Af ³
and ^`inf , 0
P
AfU f P ³for 0,! a partition P such that , Uf P n
Let S be a subrectangle in P
if 1
ns SA Mfnz d
clearly ^` n SP SAz
covers nA and
11
,ss
SP SP
nVS M fVS M fnn
fP n
VS
SA
sp§· ! ¨¸©¹
?
z
¦¦
¦
%y definition nAhas content ]ero nA has measure ]ero ^`,0 xA f x?zis countable union of measure ]ero set. ^` 0 xA f x?z has measure ]ero. munotes.in
Page 100
100M.SC. MATHEMATICS
Oscillation ,ofa of ‘f’ at a ? for 0G!, Let ^` ,, s u p Maf fx x A x a GG ^` ,, i n f maf f x x A x a GG
The oscillation ,ofa of f at a defined by ,l i m , , , ,
oofa Maf m af
GGG
o
This limit alZays e[ist since ,, ,, Maf m af GG decreases as G decreases.
Theorem :
Let A be a closed rectangle and let :fAo
be a bounded function such that , Ofx f o r a l l xA shoZ that there is a partition P of A Zith ,, Uf P Lf P VA .
Proof :
L e t ,l i m , ,, ,
OxA U f x M x f m x f
GGG
o ? a closed rectangle xu containing x in its interior such that uuxxMM by
definition of oscillation. ^` xux A? is a cover of A A? is compact This cover has a finite subcover say ^`12,, . . . . ,xx x kuu u 1k
ixiAu
?
.
Let P be a partition for A such that there each subrectangle ‘S ’ of P is contained in
some xiu then ssMf mf for each subrectangle ‘S’ in f
,,ss
SP
SPUf P Lf P M f m f V S
VS
VA
?
¦
¦ munotes.in
Page 101
101Chapter 7: Measure Zero Set7.4 Lebesgue Theorem (Only Statement)
Let A be a closed rectangle and :fAo
is bounded function. Let ^Bx f is
not continuous at [`. Then f is integrable iff % is a set of me asure ]ero
7. Characteristic )unction
Let nC
. The characteristics function cFof C is defined by 1cxi f x CF
0if x C
If CAZhere A is a closed rectangle and :fAo
i s b o u n d e d t h e n Cf³ i s
defined as c
CfF³ provided cfF³is integrable >i.e. if f and cF are integrable@
Theorem :
Let A be a closed rectangle and CA. ShoZ that the function :cAFo
is
integrable if and only if Cw has measure ]ero.
Proof :
To shoZ that :CAFo
is integrable iff Cw has measure ]ero.
%y Lebesgue theorem, it is enough to shoZ that ^:c Cx A F w is discontinuous`
Let aCq an open rectangle ‘u’ containing a such that uC 1cnn UF? cF is continuous at a.
Let aE x t c E[terior of C
>%y definition union of all open sets disMoints from C@
E[t C is an open set an open rectangle u containing such that UE x t c 0cnn uF? munotes.in
Page 102
102M.SC. MATHEMATICS cF is continuous at a
If acw then cF is continous at a ……………………. (I)
Let acw for any open rectangle U Zith a in its interior contains a poi nt yCq
a point nzc
1 0ccyzFF? cF? is not continuous at a ^:c cx A F?w is discontinuous at x` ?%y Lebesgue Theorem. cF is interrable if and only if cw has measure ]ero.
Theorem :
Let A be a closed rectangle and CA
If C is bounded set of measure ]ero and c
AF³e[ist then shoZ that 0c
AF ³.
Proof : CAbe a bounded set Zith measure ]ero.
Suppose c
AF³e[ist cFis integral
To shoZ that 0c
AF ³
Let P be a partition of A and S be a subrectangle in P. S
does not have measure ]ero SC xS but xC
0
0c
scx
mF
F?
This is true for any subrectangle S in P munotes.in
Page 103
103Chapter 7: Measure Zero Set ,0cs cLP m V CFF? ¦
This is true for any partition P ^sup , cc
ALP P FF? ³ is partition of` c
AOF ³
7. )ubini ’s Theorem
Fubini’s Theorem reduces the computation of integrals over clos ed rectangles in ,1nn!
to the computation of integrals over closed intervals in
. Fubini’s
Theorem is critically important as it gives us a method to eval uate double integrals
over rectangles Zithout having to use the definition of a doubl e integral directly.
If :fA Ro is a bounded function on a closed rectangle then the least upp er bound
of all loZer sum and the greatest loZer bound of all upper sums e[ist. They are
called the loZer integral and upper integral of f and is denote d by ALF³ and AUF³
respectively.
Fubini’s Theorem
Statement : Let nA
and nB
be closed rectangles and let :fABuo
be integrable for xA, Let :xgB o
be defined by ,xgy F x y and let
,
,x
BB
x
BBx L g L f x y dy
u x U g U f x y dy
³³
³³
Then
and
are integable on A and
AB A A BfL L f x d y d x
u§· ¨¸
©¹³³ ³ ³
,
AB A A Bfu x d x U f x y d y d x
u§· ¨¸
©¹³³ ³ ³
Proof :
Let AP be a partition of A and BP be a partition of %. Then ,AB PP P is a partition
of ABu
Let AS be a subrectangle in AP and BS be a subrectangle in BP munotes.in
Page 104
104M.SC. MATHEMATICS Then by definition, AB SS S u is a subrectangle in P
1
AB
BBs
SP
ss A B
SPLf P m fVS
mf V S S
u
u¦
¦
AB
AA BBss B A
SP SPmf V S V Su
§· ¨¸
©¹¦¦…………………. (I)
For ,
AB B As s s xxSm f M gu ?For ,A xS AB B
BBss A B s x B
SPmV SV S m g V Su
?d ¦¦
,xB x
BLg P Lg Lx d ³
This is true for any xA
,
AB
AA BB
A
AAss B A
SP SP
sA
SPLfP m fVS VS
mL x V Su
§·? ¨¸
©¹
d¦¦
¦
,A Lx P
……………………………………… (II) ?From I II ,,A LfP L x P
………………………………………… (III)
1oZ ,S
SPUfP M fV s
AB
AA
BBSS A B
SP
SPMf V S S
AB
AA BBSS B A
SPSPMf V S V S
…………….. (IV) munotes.in
Page 105
105Chapter 7: Measure Zero SetFor ,
AB B AS S S x xS M f M g
For AxS
,
,AB B
BB BBSS B S x B
SP SP
xB x
BMf V S M g V S
ug P u g x
This is true for any xA
. AB
AA BBSS B A
SPSPMf V S V S
A
AASA
SPMu x V S
,A ux P
……………………………………….. (V)
from IV V
,,A Uf P U u xP
……………………………. (VI)
%y III VI ,, ,AA LfP L x P uL x P
,,A ux P U f P
………………………… (VII)
Also ,, , ,AA A LfP L x P L x P u x P
……… VIII f
is integrable sup , inf ,
sup , inf ,
BAPPAB
AAPPABLfP UfP f
Lx P ux P f
x
is integrable ,
AB A A Bf x L f x y dx
………………………. (IX)
munotes.in
Page 106
106M.SC. MATHEMATICS Also by VIII I; sup , inf ,
AAAAPPABLLx P Uux P f
ux
is integrable. ,
AB A A Bf u x dx U f x y dx
Hence Proved
Remark :
The Fubini’s theorem is a result which gives conditions under w hich it is possible
to compute a double integral using interated integrals, As a co nseTuence if alloZs
the under integration to be changed in iterated integrals. ,
,AB B B
BAfL f x y d x d y
Uf x y d x d y
These integrals are called iterated integrals.
E[ample 13:
Using Fubini’s theorem show that 12 21Df Df
i f 12Df and 21Df a r e
continuous.
Solution :
Let AR
and :fA
continuous
T.P.T 12 21Df Df
Suppose 12 21Df Df
00,xy
in domain of f such that 12 21 0 Dfa Dfa
Zithout loss of generality, 12 21 0 Dfa Dfa
or 12 21 0 Df Df a
………………………………….. (I) munotes.in
Page 107
107Chapter 7: Measure Zero Set12 21 ,0
ADf Df x g
Let ,, Aa bc d
By Fubini’s Theorem 21 21
22,,
,,
,,,,db
Ac a
d
cD f x y D f x y dx dy
D f b y D f g y dy
fb d fb c fa d fa c
Similarly, 12
21 12
21 12,, , ,,
,,
,0A
AA
ADfx y fb d fb c fa d fa c
Dfx y Dfx y
Df Df x y
Which is contradiction to I 12 21Df Df
proved
E[ample 14:
Use Fubini’s Theorem to compute the following integrals.
1 211
22
00.
1xdy dxIxy
Solution :
2
2
211
22
00
1 1
22
00
11
1
22
0 0.
1
1
1tan
11x
x
xdy dxIxy
dydxxy
ydx
xx
munotes.in
Page 108
108M.SC. MATHEMATICS 1
2
01..4 1dx
x
1
2
0
1
2
04 1
log 14
log 14dx
x
xx
x
ii) 11 2
0sin2yxI dy dx
Solution :
, 1 , 0 1 Cx y y x y
By Fubini’s Theorem 11 2
0
1 2
00
1 2
0
0
1 2
0sin2
sin2
sin2
sin2y
x
xxId x d y
xdxdy
xyd x
xxd x
Put 2
,2xt
x
1 t 0 2
munotes.in
Page 109
109Chapter 7: Measure Zero Set 2
2xdx dt
dtxdx
22
2
0
0011sin sin cos
1101dtIt t d tt
7.7 Reviews
After reading this chapter you Zould be NnoZing.
x Definition of Measure ]ero set and content ]ero set.
x Oscillation , Ofa
x Find set contain measure ]ero on content ]ero
x Statement of Lebesgue Theorem
x Definition of characteristic function its properties.
x Fubini’s Theorem & its examples.
7. Unit End E[ercises
1. If BA
and A has measure ]ero then sho Z that has measure ]ero.
2. ShoZ that countable set has measure ]ero.
3. If A is non-empty open set, then sh oZ that A is not of measure ]ero.
4. *ive an e[ample of a bounded set C if measure ]ero but C
does not have
measure ]ero.
5. ShoZ by an e[ample that a set A has measure ]ero but A does not have
content ]ero.
6. Prove that 11,. . . .,nn ab ab
does not have content ]ero if iiab
for each i.
7. If C is a set of content ]ero shoZ that the boundary of C has c ontent ]ero. munotes.in
Page 110
110M.SC. MATHEMATICS 8. *ive an e[ample of a set A and a bounded subset C of A measure ]ero such
that c
A
does not e[ist.
9. If f g are integrable, then shoZ that gf is integrable.
10. Let 0,1 U
be the union of all open intervals ,iiabsuch that each rational
number in 0,1 is contained in some ,iiab. ShoZ that if c f
e[cept on
a set of measure ]ero, then f is not integrable on 0,1.
11. If :, ,fa b a b
i s c o n t i n u o u s t h e n s h o Z t h a t ,,bb bb
ax axfx y d x d y fx y d y d x
12. Use Fubini’s theorem, to compute 22
00sinxdy dxxy
13. Let1,1 0, 2 A
and :fA
defined by ,s i nxfx y x y y e
compute Af
14. Let ,, s i nfx y z z x y
and 0, , 0,122A
computer Af
.
munotes.in