Multivariable Calculus, ta

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Taylors Theorem

1. Introduction. Suppose f is a one-variable function that has n

+ 1 derivatives on an interval

about the point x

= a. Then recall from Ms. Turner’s class the single variable version of Taylor’s

Theorem tells us that there is exactly one polynomial p of degree

² n such that pÝaÞ = fÝaÞ,

p

v

ÝaÞ = f

v

ÝaÞ, p

vv

ÝaÞ = f

vv

ÝaÞ, u, p

ÝnÞ

ÝaÞ = f

ÝnÞ

ÝaÞ. This polynomial is given by

p

ÝxÞ = fÝaÞ + f

v

ÝaÞÝx ? aÞ +

f

vv

ÝaÞ

2!

Ýx ? aÞ

2

+ u + f

ÝnÞ

n!

Ýx ? aÞ

n

We also know the difference between f

ÝxÞ and pÝxÞ:

f

ÝxÞ ? pÝxÞ =

f

Ýn+1Þ

ÝYÞ

Ýn + 1Þ!

Ýx ? aÞ

n

+1

,

where

Y is somewhere between a and x.

The polynomial p is called the Taylor Polynomial of degree

² n for f at a.

Before we worry about what the Taylor polynomial might be in higher dimensions, we need to be
sure we understand what is a polynomial in more than one dimension. In two dimensions, a
polynomial p

Ýx, yÞ of degree ² n is a function of the form

p

Ýx, yÞ =

>

i,j

=0

i

+j=n

a

ij

x

i

y

j

.

Thus a polynomial of degree

² 2 (perhaps more commonly known as a quadratic) looks like

p

Ýx, yÞ = a

00

+ a

10

x

+ a

01

y

+ +a

11

xy

+ a

20

x

2

+ a

02

y

2

.

I hope it easy to guess what one means by a polynomial in three variables,

Ýx, y, zÞ, or indeed, in

any number of variables.

Now, how might we extend the idea of the Taylor polynomial of degree

² n for a function f at a

point a ? Simple enough. It’s a polynomial p

ÝxÞ of degree ² n so that

/

i

1

+u+i

q

f

ÝaÞ

/x

1

i

1

/x

2

i

2

u

/x

q

i

q

= /

i

1

+u+i

q

p

ÝaÞ

/x

1

i

1

/x

2

i

2

u

/x

q

i

q

,

for all i

1

, i

2

,

u, i

q

such that i

1

+ i

2

+ u + i

q

² n.

This looks pretty ferocious in general, so let’s see what it says for just two variables. In this case,
we have a

a, bÞ and the Taylor polynomial pÝx, yÞ at a becomes the polynomial such that

1

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/

i

+j

f

ÝaÞ

/

i

x

/

j

y

= /

i

+j

p

ÝaÞ

/

i

x

/

j

y

,

for all i

+ j

² n.

Example

Let f

Ýx, yÞ = cosÝx + yÞ, and let pÝx, yÞ = 1 ?

x

2

2

? xy ?

y

2

2

. Let’s verify that p is the Taylor

polynomial of degree

² 2 for f at Ý0, 0Þ. He we go.

f

Ý0, 0Þ = 1, and pÝ0, 0Þ = 1;

/f

/x

=

?sinÝx + yÞ, and /

p

/x

=

?x ? y;

/f

/y

=

?sinÝx + yÞ, and /

p

/y

=

?x ? y;

/

2

f

/x

2

=

?cosÝx + yÞ, and /

2

p

/x

2

=

?1,

/

2

f

/y

2

=

?cosÝx + yÞ, and /

2

p

/y

2

=

?1,

/

2

f

/x/y

=

?cosÝx + yÞ, and /

2

p

/x/y

=

?1.

Now it’s easy to see that

f

Ý0, 0Þ = 0 = pÝ0, 0Þ;

/f

/x

Ý0, 0Þ = 0 = /

p

/x

Ý0, 0Þ;

/f

/y

Ý0, 0Þ = 0 = /

p

/y

Ý0, 0Þ;

/

2

f

/x

2

Ý0, 0Þ = ?1 = /

2

p

/x

2

Ý0, 0Þ;

/

2

f

/y

2

Ý0, 0Þ = ?1 = /

2

p

/y

2

Ý0, 0Þ; and

/

2

f

/x/y

Ý0, 0Þ = ?1 = /

2

p

/x/y

Ý0, 0Þ.

Exercises

1. Verify that the polynomial in the Example is also the Taylor polynomial for f at (0,0) of degree
² 3.

2. Let f

Ýx, yÞ = sinÝx + yÞ.Which Which of the following is the Taylor polynomial of degree ² 2 for

f at (0,0)? Explain.
a) p

Ýx, yÞ = 1 + x

2

+ y

2

b) p

Ýx, yÞ = xy

2

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c) p

Ýx, yÞ = x

2

+ xy + 2y

d) p

Ýx, yÞ = x + y

2. Derivatives. Prior to finding a general recipe for the Taylor polynomial, we need look at finding
higher order derivatives of certain composite functions. Let f be a real-valued function defined on a
subset of R

q

. Suppose that in a neighborhood of the point x, the function f has a lot of continuous

partial derivatives. Define the function g by

g

ÝtÞ = fÝa + thÞ,

where a

= Ýa

1

, a

2

,

u, a

q

Þ and h = Ýh

1

, h

2

,

u, h

q

Þ. We know from the chain rule that g

v

ÝtÞ is

given by

g

v

ÝtÞ = 4fÝa + thÞ 6 h

=

/f

/x

1

, /f

/x

2

,

u, /f

/x

q

6 Ýh

1

, h

2

,

u, h

q

Þ

= h

1

/

/x

1

+ h

2

/

/x

2

+ u + h

q

/

/x

q

f

Ýa+thÞ

In keeping with our general practice of restricting ourselves to dimensions one, two, or three, let’s
look first at the case q

= 2. As usual, we’ll write x x, yÞ and h = Ýh, kÞ. The expression for g

v

ÝtÞ

now looks like:

g

v

ÝtÞ = h /

/x

+ k /

/y

f

Ýx+thÞ

We are now in business, for we have a nice recipe for higher order derivatives of g :

g

ÝmÞ

ÝtÞ = h /

/x

+ k /

/y

m

f

Ýx+thÞ

For example,

g

vv

ÝtÞ = h /

/x

+ k /

/y

2

f

= h

2

/

2

/x

2

+ 2hk /

2

/x/y

+ k

2

/

2

/y

2

f

= h

2

/

2

f

/x

2

+ 2hk /

2

f

/x/y

+ k

2

/

2

f

/y

2

Example
Suppose f

Ýx, yÞ = x

2

y

3

+ y

2

. Let’s find the second derivative of the function

g

ÝtÞ = fÝ1 + 3t, ?2 + tÞ

3

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First,

g

vv

ÝtÞ = 3 /

/x

+ /

/y

2

f

= 9 /

2

f

/x

2

+ 6 /

2

f

/x/y

+ /

2

f

/y

2

Now,

/f

/x

= 2xy

3

, and

/f

/y

= 3x

2

y

2

+ 2y, and so

/

2

f

/x

2

= 2y

3

,

/

2

f

/y/x

= 6y

2

, and

/

2

f

/y

2

= 6x

2

y

+ 2.

Thus,

g

vv

ÝtÞ = 18Ý?2 + tÞ

3

+ 36Ý

?2 + tÞ

2

+ 6Ý1 + 3tÞ

2

Ý?2 + tÞ + 2

Exercises
3
. Let f

Ýx, yÞ = xe

y

. Find the derivative of g

ÝtÞ = fÝ1 + t, 3 ? 4tÞ.

4. Find the second derivative of the function g defined in Problem 3.

5. Let F

Ýu, vÞ = u

3

v

+ v

2

. Find the second derivative of R

ÝzÞ = FÝz, 3zÞ.

6. Find g

vvv

ÝtÞ, where g is the function defined in the Example.

3. The Taylor polynomial. To find the Taylor polynomial for a function f of several variables at a
point a, we shall simply apply the one-dimensional results to the function

g

ÝtÞ = fÝa + thÞ.

Thus,

g

ÝtÞ =

>

m

=0

n

g

ÝmÞ

Ý0Þ

m!

t

m

+

g

Ýn+1Þ

ÝYÞ

Ýn + 1Þ!

t

n

+1

,

where

Y is a number between 0 and t. Next, substitute t = 1 into the above:

g

Ý1Þ = fÝaÞ =

>

m

=0

n

g

ÝmÞ

Ý0Þ

m!

+

g

Ýn+1Þ

ÝYÞ

Ýn + 1Þ!

We know the value of g

ÝkÞ

from Section 2:

f

Ýa + hÞ =

>

m

=0

n

1

m!

h

1

/

/x

1

+ h

2

/

/x

2

+ u + h

q

/

/x

q

m

f

ÝaÞ

4

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+

1

Ýn + 1!

h

1

/

/x

1

+ h

2

/

/x

2

+ u + h

q

/

/x

q

n

+1

f

ÝcÞ

The point c lies somewhere on the line segment joining a and a

+ h.

The polynomial

p

ÝhÞ =pÝh

1

, h

2

,

u, h

q

Þ =

>

m

=0

n

1

m!

h

1

/

/x

1

+ h

2

/

/x

2

+ u + h

q

/

/x

q

m

f

ÝaÞ

is the Taylor polynomial of degree

² n for f at a; the last term is traditionally called the error term

or sometimes, the remainder term. Actually, if we let h

= x

? a, then qÝxÞ =pÝx ? aÞ is the thing

we called the Taylor polynomial in the first section.

This is pretty fierce looking. Let’s look at the two variable case:

f

Ýa

1

+ h, a

2

+ kÞ =

>

m

=0

n

1

m!

h /

/x

+ k /

/y

m

f

Ýa

1

, a

2

Þ

+

1

Ýn + 1!Þ

h /

/x

+ k /

/y

n

+1

f

Ýc

1

, c

2

Þ

where

Ýc

1

, c

2

Þ is on the line joining Ýa

1

, a

2

Þ and Ýa

1

+ h, a

2

+ kÞ.

Example

Let f

Ýx, yÞ = sinxsiny. For n = 2 and a = Ý0, 0Þ, Taylor’s polynomial becomes

p

Ýh, kÞ = fÝ0, 0Þ + h /f

/x

Ý0, 0Þ + k /f

/y

Ý0, 0Þ + h

2

2

/

2

f

/x

2

Ý0, 0Þ + hk /

2

f

/x/y

Ý0, 0Þ + k

2

2

/

2

f

/y

2

Ý0, 0Þ

We have

/f

/x

= cos x sin y;

/f

/y

= sin x cos y;

/

2

f

/x

2

=

?sinxsiny;

/

2

f

/x/y

= cos xcosy;

/

2

f

/y

2

=

?sinxsiny.

Thus,

p

Ýh, kÞ = hk.

Let’s get an estimate for how well this approximates sin x sin y near

Ý0, 0Þ. We know that

|sin x sin y

? xy| = 1

3!

x /

/x

+ y /

/y

3

f

ÝY, WÞ

where

ÝY, WÞ is one the segment joining Ýx, yÞ and the origin. Now,

5

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x /

/x

+ y /

/y

3

f

= x

3

/

3

f

/x

3

+ 3x

2

y /

3

f

/x

2

/y

+ 3xy

2

/

3

f

/x/y

2

+ y

3

/

3

f

/x

3

.

Next, let’s suppose that |x|

² c and |y| ² c for some constant c. Noting that all the partial

derivatives in the above expression are simply products of sine and cosines, we can estimate

x /

/x

+ y /

/y

3

f

² 8c

3

,

and so, at last,

|sin x sin y

? xy| ² 8c

3

6

= 4

3

c

3

Exercises

7. Find the Taylor polynomial of degree

² 1 for fÝx,yÞ = e

xy

at

Ý0, 0Þ.

8. Find the Taylor polynomial of degree

² 2 for fÝx,yÞ = e

xy

at

Ý0, 0Þ.

9. Find the Taylor polynomial of degree

² 3 for fÝx,yÞ = e

xy

at

Ý0, 0Þ.

10. Find the Taylor polynomial of degree

² 1 for fÝx,yÞ = e

x

cos y at

Ý0, 0Þ.

11. Use Taylor’s Theorem to find a quadratic approximation of e

x

cos y at the origin.

12. Estimate the error in the approximation found in Problem 11 if |x|

² 0.1 and |y| ² 0.1.

6


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