Module 2

Limits, homogenous, Maxima Minima

2.1 LIMITS

Let f be a function of two variables that is defined in some circular region around (x_0,y_0). The limit of f as x approaches (x_0,y_0) equals L if and only if for every epsilon>0 there exists a delta>0 such that f satisfies

whenever the distance between (x,y) and (x_0,y_0) satisfies

Using limit

when the limit exists. The usual properties of limits hold for functions of two variables: If the following hypotheses hold:

and if c is any real number, then we have the results:

Linearity 1:

 

Linearity 2:

 

Products of functions:

 

Quotients of functions:

 

2.2 Continuity

A function f of two variables is continuous at a point (x_0,y_0) if

 exists

This definition is a direct generalization of the concept of continuity of functions of one variable.

Continuous functions of two variables satisfy all of the usual properties familiar from single variable calculus:

2.3 Differentiability of functions

Definition:

Suppose f is a (real valued) function defined on an open interval I and x0 ∈ I. Then f is said to be differentiable at x0 if limx→x0 f(x) − f(x0) x − x0 exists, and in that case the value of the limit is called the derivative of f at x0. The derivative of f at x0, if exists, is denoted by f 0 (x0) or df dx(x0).

Homogenous function:- A function is called a homogenous function of degree n if by putting the function becomes i.e.

For eg: is a homogenous function of degree 2 for putting

2.4 Homogenous Functions

2.4.1 Euler’s Theorem:-

If z is a homogenous function of two variables x and y of degree n then

.

i)       Variables x & y of degree n then

ii)    If u is a degree n then we can P.T.

3.     If z is homogenous function of degree n is x and y and then

4.     If z is homogenous function of degree n is x and y and z=f(u) then

Where

Q1.    If then verify Euler’s theorem for u.

Sol 1.               

By symmetry,

b)    Put x=xt, y=yt we get

Thus u is homogenous function of

Hence, by Euler’s Theorem

Q2.    If prove that

Sol 2.       Putting x=xt, y=yt we get

Q3.    If find value of

Sol 3.       Let

Let u=v+w

Putting x=xt, y=yt, z=zt

Thus v is homogenous function of degree 6.

Q4.    If find value at

Sol 4.       Let

Put

Similarly

By Euler’s theorem

By Euler’s theorem

Adding two result

L.H.S  

At n=1, y=2

L.H.S

If , prove that

Sol 5.       Putting x=xt, y=yt we get

Q5.    If P.T.

Sol 6.      

Put x=xt,           y=yt

Q6.    If Prove that

Sol 6.       Let

Putting x=xt, y=yt, we get

Z=f(u)=sin u is a homogenous function of x,y of degree 2/5

Hence, proved.

Q7.    If Prove that  

Sol 7.       Let

Putting x=xt, y=yt, we get

Thus is a homogenous function of x, y of degree 2.

Derivatives of

Hence, the required result.

2.5 Maxima & Minima

Condition for

to be Maximum & Minimum.

A function

is said to be maximum at

if

for small values of

and

, positive or negative.

A function

is said to be minimum at

if

for small values of

and

, positive or negative.

The necessary condition for

to be maximum or minimum at

are

and

For Maxima or Minima we should have at

If  or  thanis max at

If  or  thanis max at

First find

and

Then solve

simultaneously for

and

So that

for the values of

obtained in (ii).

When this condition is situation i

is maximum if

or

and

is maximum if

or

.

At obtained in (ii)

Let us above denote

, ,

For maxima or minima we must have

is maximum if

is maximum if

Step 1: Find and

Step 2: Solve the equation

Step 3: Find the value of at the roots obtained in Step II.

If

and

then

is max

If

and

then

is max

If

is neither max or min.

2.5.1 Type 1 Two independent variables: -

Q1.                            Discuss the maxima and minima of .

Sol 1.                     We have

 Step 1:    

 Step 2:     

Stationary Values

 Step 3:For

   And

is a maxima.

Minimum value

i)       For

is neither max nor min. It is a saddle point.

ii)    For

It is neither max nor min. It is a saddle point.

iii)  For

 Add

is  a maxima.

Max. value

Example: Find all stationary values of

Sol: We have

Step I:

Step II: We now solve

  i.e.

And       or

(i)        When

are stationary points.

(ii)      When

or .

and are stationary points.

Step III: (i) When

and

and

is minimum at.

The minimum value.

(ii)      When

and

is maximum at.

The maximum value

(iii)   When  

is neither maximum nor minimum.

(iv)    When  

is neither maximum nor minimum.

2.5.2 Type II : Three Independent Variable :

Example 1: If where , find the stationary value of u.

Sol. : Step I : We have to find the stationary value of

           ………………………(1)

with the condition that            ………………………(2)

Consider the Lagrange’s function

gives,          ………………………(3)

Step II : We have to eliminate x, y, z and from(1), (2) and (3).

From (3),                             ………………………(4)

Putting these values in (2),

Hence, from(4),

Hence, the stationary value of is

Example 2 : Find the minimum distance from the point to the cone

Sol : If   is any point on the cone, its distance from the point is

When the distance is minimum, its square is also minimum.

Step I : We have to find the stationary value of

                             ………………(1)

With the condition that                            ………………….(2)

Consider the Lagrange’s function

  gives

     ………………….(3)

Step II : We have to eliminate x, y, z from (1), (2) and (3).

From (3),

Hence,      

Hence, the required point is .

The distance between and is

Example 3 : If , find the values of x, y, z for which is maximum.

Sol. : We have to maximize with the condition that .

This means we have to maximize

i.e. we have to minimizei.e.

Step I : We have to minimize  ………………(1)

With the condition that     ……………….(2)

Consider the Lagrange’s function,

gives

But

and

2.6 Taylor’s Theorem in several variables

In Calculus II you learned Taylor’s Theorem for functions of 1 variable.

Here is one way to state it.

1 of 2: (Taylor’s Theorem, 1 variable) If g is defined on (a, b) and derivatives of order up to m and then

Where the remainder R satisfies

Here is the several variable generalizations of the theorem. I use the following bits of notation in the statement, its specialization to and the sketch of the proof:

Theorem 2 (Taylor’s Theorem) Suppose U is  a convex open set in and has continuous partial derivatives of all orders up to and including. Fix . Then

Taking m=4 gives

Where

Sketch of proof : Let

Use the Chain Rule repeatedly to get

Where the sum is over all ordered k tuples and for. Now use the one variable Taylor’s Theorem to write as a polynomial of degree m – 1 in plus a remainder, obtaining

Finally, do the combinatorics to rewrite the sum with no repetitions (so, for example, you group the terms

and

together.)

2.7 Maclaurin Series

Formulas

Q1.             Prove that

Sol 1.          

Diff wrt

Hence by Maclaurin’s Series

we get

Q2.             Show that

Sol 2.       Let

  [U.V Value of derivative]

 By Maclaurin Series

Q1.             Expand upto

Sol 1.       We have

Q2.             Show that

Sol 2.       We have

Q3.             Prove that

Sol 3.       We have

Q4.             Expand in Power of

Hence Prove that

Sol 4.