The Theory Behind Taylor Series

The Theory Behind Taylor Series

The Theory Behind Taylor Series
Swapnil Sunil Jain
April 23, 2006

The Theory Behind Taylor Series

We first make a safe

assumption

that any

function

f(x) can be approximated with the help of an nth degree

polynomial

p(x). Thus,

f

⁢

(

x

)

≈

p

⁢

(

x

)

=

a

n

x

n

+

a

n

-

1

x

n

-

1

+

.

.

.

+

a

4

x

4

+

a

3

x

3

+

a

2

x

2

+

a

1

x

+

a

0

(1)

Now all we need to do in order to

know

this function is to figure out the values of the coefficients

a

n

,

a

n

-

1

,

.

.

.

,

a

4

,

a

3

,

a

2

,

a

1

,

a

0

of the polynomial p(x).

Getting

a

0

is easy, we just set x = 0, and we get

a

0

=

p

⁢

(

0

)

.
However, getting the rest of the coefficients requires some trick. But before doing anything else, let’s just take the first few

derivatives

of the polynomial p(x):

p

′

⁢

(

x

)

=

n

a

n

x

n

-

1

+

.

.

.

+

4

⋅

a

4

x

3

+

3

⋅

a

3

x

2

+

2

⋅

a

2

x

+

a

1

p

′′

⁢

(

x

)

=

n

(

n

-

1

)

a

n

x

n

-

2

+

.

.

.

+

4

⋅

3

⋅

a

4

x

2

+

3

⋅

2

⋅

a

3

x

+

2

⋅

a

2

p

3

⁢

(

x

)

=

n

(

n

-

1

)

(

n

-

2

)

a

n

x

n

-

3

+

.

.

.

+

4

⋅

3

⋅

2

⋅

a

4

x

+

3

⋅

2

⋅

a

3

p

4

⁢

(

x

)

=

n

(

n

-

1

)

(

n

-

2

)

(

n

-

3

)

a

n

x

n

-

4

+

.

.

.

+

4

⋅

3

⋅

2

⋅

a

4

.

.

.

Now getting the next coefficient

a

1

is easy! We just set x = 0 and we get

p

′

⁢

(

0

)

=

a

1

. Similarly, setting x = 0 for all the rest of the derivatives we get:

p

′′

⁢

(

0

)

=

2

⋅

a

2

p

3

⁢

(

0

)

=

3

⋅

2

⋅

a

3

p

4

⁢

(

0

)

=

4

⋅

3

⋅

2

⋅

a

4

.

.

.

Hmmm… Do you see a pattern? The coefficient in front of the nth term seems to be n!. By this logic, the nth derivative of p(x) (evaluated at 0) should look like the following:

p

n

⁢

(

0

)

=

n

!

⋅

a

n

Solving for

a

n

, we get:

a

n

=

p

n

⁢

(

0

)

n

!

By using this

formula

we can figure out all the coefficients (i.e.

a

n

,

a

n

-

1

,

.

.

.

,

a

2

,

a

1

,

a

0

) of the polynomial p(x) (and be able to approximate f(x)!). Thus, our original function (1), which was approximated by the polynomial p(x), could be written (in the form of a polynomial) as:

f

⁢

(

x

)

≈

p

⁢

(

x

)

=

p

n

⁢

(

0

)

n

!

⁢

x

n

+

p

n

-

1

⁢

(

0

)

(

n

-

1

)

!

⁢

x

n

-

1

+

…

+

p

3

⁢

(

0

)

3

!

⁢

x

3

+

p

2

⁢

(

0

)

2

!

⁢

x

2

+

p

1

⁢

(

0

)

1

!

⁢

x

+

p

0

⁢

(

0

)

0

!

or in summation form as:

f

⁢

(

x

)

≈

∑

k

=

0

n

f

k

⁢

(

0

)

k

!

⁢

x

k

(2)

This is the nth-order

Taylor series

expansion of f(x) about the point x = 0. However, this is only an approximation. To get the

exact

form, we have to have an infinite series summing over all the possible integer values of k
from 0 to

infinity

(i.e. an

infinite

degree polynomial)

f

⁢

(

x

)

=

∑

k

=

0

∞

f

k

⁢

(

0

)

k

!

⁢

x

k

(3)

Now, one more thing that we can do is to change the point of evaluation from x = 0 to x = a for some real value a. Thus, the Taylor series expansion of f(x) about a point x = a becomes the following (Note that setting a = 0 gives us back our above expression):

f

⁢

(

x

)

=

∑

k

=

0

∞

f

k

⁢

(

a

)

k

!

⁢

(

x

-

a

)

k

(4)

Title

The Theory Behind Taylor Series

Canonical name

TheTheoryBehindTaylorSeries1

Date of creation

2013-03-11 19:24:28

Last modified on

2013-03-11 19:24:28

Owner

swapnizzle (13346)

Last modified by

(0)

Numerical id

1

Author

swapnizzle (0)

Entry type

Definition

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