154 Chapter 5 Let’s Learn About Taylor Expansions!

When a general function f(x) (provided it is differentiable infinitely many

times) can be expressed as

f x a a x a x a x a x

n

n

( )

= + + + + + +

0 1 2

2

3

3

... ...

the right side is called the Taylor expansion of f(x) (about x = 0).

Left side =

−

= −

1

1 2

1

Right side

= + + + + +

1 2 4 8 16 ...

It turns out that expression u is correct for all

x satisfying −1 < x < 1, which is the allowed interval

of a Taylor expansion. In technical terms, the inter-

val −1 < x < 1 is called the circle of convergence.

S? The two

sides are not

equal.

For example, substituting x = 2

in both sides of expreion u,

This means that f(x) perfectly coincides with an

infinite-degr polynomial in a definite interval

including

x = 0. It should be noted, however,

that the right side may become meaningle

because it may not have a single defined value

outside the interval.

How to Obtain a Taylor Expansion 155

How to Obtain a Taylor Expansion

When we have

v

f x a a x a x a x a x

n

n

( )

= + + + + + +

0 1 2

2

3

3

... ...

let’s find the coefficient a

n

.

Substituting x = 0 in the above equation and noting f(0) = a

0

, we find that

the 0th-degree coefficient a

0

is f(0).

We then differentiate v.

w

′

( )

= + + + + +

−

f x a a x a x na x

n

n

1 2 3

2 1

2 3 . .. ...

Substituting x = 0 in w and noting f ′(0) = a

1

, we find that the 1st-degree

coefficient a

1

is f ′(0).

We differentiate w to get

′′

( )

= + + + −

( )

+

−

f x a a x n n a x

n

n

2 6 1

2 3

2

... ...

Substituting x = 0 in , we find that the 2nd-degree coefficient a

2

is

1

2

0

′′

( )

f

.

Differentiating

, we get

′′′

( )

= + + −

( )

−

( )

+

−

f x a n n n a x

n

n

6 1 2

3

3

... ...

From this, we find that the 3rd-degree coefficient a

3

is

1

6

0

′′′

( )

f

.

Repeating this differentiation operation n times, we get

f x n n a

n

n

( )

( )

= −

( )

× × +1 2 1... ...

where f

(n)

(x) is the expression obtained after differentiating f(x) n times.

From this result, we find

nth-degree coefficient

a

n

f

n

n

=

( )

( )

1

0

!

n! is read “n factorial” and means

n n n× −

( )

× −

( )

× × ×1 2 2 1... .

156 Chapter 5 Let’s Learn About Taylor Expansions!

We, that

introduction

was a lile

t long...

So, why is our

company’s

predicament

the Taylor

expansion?

I mean that if

f(x) is a

function that describes

Burnham Chemical’s

advertising expenses,

their suort of

our paper could be

considered the third

term of a Taylor

expansion. f(x) = The Japan

Times + The Kyodo News +

TheAsagake Times

That’s right.

Actuay for

BurnhamChemical,

the amount of money

they spend for

us is only a very

sma amount—the

3rd-degr term,

obtained after

dierentiating

thrtimes.

Since it’s

insignificant for

them anyway,

they’ likely

suort us like

they did before

even if they

change their

executives.

Mr. Seki, where

did you go out

for drinks when

you worked at

the main oice?

What?

A the

way at

the end?

How to Obtain a Taylor Expansion 157

You know,

drinking with

your coeagues

after work,

talking about

sue stories...

Oh.

We’re done

with our work.

So, sha we

go out for a

drink?

Okay,

let’s go.

Yes!

Headliner’s Pub

(Open 24Hours)

158 Chapter 5 Let’s Learn About Taylor Expansions!

Nice

atmosphere,

isn’t it?

Ah, yes. Are a

these people

journalists?

Lk, that’s Ishizuka,

the photographer

who is the youngest

wier of the Japan

Photographic Prize.

And that’s Mr. Nakata,

a heavyweight in the

designers’ circle.

The guys over

there are from the

Sanda City Post.

Mr.

Calculus!?

Hey, Calculus,

long time no

s. Join us.

What a

nickname! But,

it certainly

fits.

Let’s s. I

hope I can

listen to their

profeional

discuion.

Bierman

got diabetes

recently.

Mr. Stack

now ss

a doctor

because of

high bld

preure...

Maybe I

should think

about geing

a medical

checkup

sn.

It’s just mile-

aged men’s talk!

Thisis usele.

Heo.

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