examples on how to find Taylor series from other known series

examples on how to find Taylor series from other known series

In this section we present numerous examples that provide a number of useful procedures to find new

Taylor series

from Taylor series that we already know. However, we are only worried about “computing” and we don’t worry (for now) about the

convergence

of the series we find.

We know “by heart” the following series:

e

x

=

1

+

x

+

x

2

2

+

x

3

3

!

+

…

=

∑

n

=

0

∞

x

n

n

!

cos

⁡

x

=

1

-

x

2

2

!

+

x

4

4

!

-

x

6

6

!

+

…

=

∑

n

=

0

∞

(

-

1

)

n

⁢

x

2

⁢

n

(

2

⁢

n

)

!

sin

⁡

x

=

x

-

x

3

3

!

+

x

5

5

!

-

x

7

7

!

+

…

=

∑

n

=

0

∞

(

-

1

)

n

⁢

x

2

⁢

n

+

1

(

2

⁢

n

+

1

)

!

(

1

+

x

)

p

=

1

+

p

⁢

x

+

p

⁢

(

p

-

1

)

2

!

⁢

x

2

+

p

⁢

(

p

-

1

)

⁢

(

p

-

2

)

3

!

⁢

x

3

+

…

+

p

⁢

(

p

-

1

)

⋅

…

⋅

(

p

-

(

n

-

1

)

)

n

!

⁢

x

n

+

…

ln

⁡

(

1

+

x

)

=

x

-

x

2

2

+

x

3

3

-

x

4

4

+

…

=

∑

n

=

1

∞

(

-

1

)

n

+

1

⁢

x

n

n

Remark 1

.

Recall that the first three have

radius of convergence

R

=

∞

but for the last two

R

=

1

.

Example 1

.

Find the Taylor series about

x

=

0

for

sin

⁡

(

x

2

)

. If we try to take

derivatives

then we soon realize that consecutive derivatives get extremely hard to compute. However, one can do a simple trick. Since we know the Taylor series for

sin

⁡

(

x

)

we can evaluate it at

x

2

:

sin

⁡

x

=

x

-

x

3

3

!

+

x

5

5

!

-

x

7

7

!

+

…

=

∑

n

=

0

∞

(

-

1

)

n

⁢

x

2

⁢

n

+

1

(

2

⁢

n

+

1

)

!

sin

⁡

(

x

2

)

=

(

x

2

)

-

(

x

2

)

3

3

!

+

(

x

2

)

5

5

!

-

…

=

x

2

-

x

6

3

!

+

x

10

5

!

-

…

One can also use the

Σ

notation:

sin

⁡

(

x

2

)

=

∑

n

=

0

∞

(

-

1

)

n

⁢

(

x

2

)

2

⁢

n

+

1

(

2

⁢

n

+

1

)

!

=

∑

n

=

0

∞

(

-

1

)

n

⁢

x

2

⁢

(

2

⁢

n

+

1

)

(

2

⁢

n

+

1

)

!

.

Example 2

.

Find the Taylor series about

x

=

0

for

e

-

x

2

(this is a very important

function

, for example in probability theory). Again, we use the simple Taylor series of

e

x

:

e

x

=

1

+

x

+

x

2

2

+

x

3

3

!

+

…

=

∑

n

=

0

∞

x

n

n

!

e

-

x

2

=

1

+

(

-

x

2

)

+

(

-

x

2

)

2

2

+

(

-

x

2

)

3

3

!

+

…

=

1

-

x

2

+

x

4

2

-

x

6

3

!

+

…

Using the sigma notation we obtain:

e

-

x

2

=

∑

n

=

0

∞

(

-

x

2

)

n

n

!

=

∑

n

=

0

∞

(

-

1

)

n

⁢

x

2

⁢

n

n

!

.

Example 3

.

Series can also be multiplied by

x

. For example, we find the Taylor series for

x

⁢

e

x

:

x

e

x

=

x

(

1

+

x

+

x

2

2

+

x

3

3

!

+

…

)

=

x

+

x

2

+

x

3

2

+

x

4

3

!

+

…

)

or

x

⁢

e

x

=

x

⁢

(

∑

n

=

0

∞

x

n

n

!

)

=

∑

n

=

0

∞

x

n

+

1

n

!

Example 4

.

Series can also be divided by

x

provided

that the result has

only

non-negative

exponents

. For example, we find the Taylor series for

ln

⁡

(

1

+

x

)

x

:

ln

⁡

(

1

+

x

)

x

=

x

-

x

2

2

+

x

3

3

-

x

4

4

+

…

x

=

1

-

x

2

+

x

2

3

-

x

3

4

+

…

or

ln

⁡

(

1

+

x

)

x

=

∑

n

=

1

∞

(

-

1

)

n

+

1

⁢

x

n

n

x

=

∑

n

=

1

∞

(

-

1

)

n

+

1

⁢

x

n

-

1

n

Example 5

.

As well, we can multiply two Taylor series (term by term). Suppose we want to find the

Taylor polynomial

of degree

3

about

x

=

0

of

e

x

⁢

cos

⁡

x

. Then we can multiply the respective Taylor polynomials of degree

3

of

e

x

and

cos

⁡

x

and disregard any term higher than

3

:

(

1

+

x

+

x

2

2

+

x

3

6

)

⁢

(

1

-

x

2

2

)

=

(

1

-

x

2

2

)

+

(

x

-

x

3

2

)

+

(

x

2

2

)

+

(

x

3

6

)

+

…

Since every other term in the product is of degree higher than

3

we disregard them. Thus:

T

3

⁢

(

x

)

=

(

1

-

x

2

2

)

+

(

x

-

x

3

2

)

+

(

x

2

2

)

+

(

x

3

6

)

=

1

+

x

-

x

3

2

+

x

3

6

=

1

+

x

-

x

3

3

.

Example 6

.

Find the first three terms of the Taylor series for

1

+

2

⁢

sin

⁡

x

. Since

1

+

x

=

(

1

+

x

)

1

/

2

we know that:

1

+

x

=

1

+

x

2

-

x

2

8

+

…

Thus:

1

+

2

⁢

sin

⁡

x

=

1

+

2

⁢

sin

⁡

x

2

-

(

2

⁢

sin

⁡

x

)

2

8

+

…

Moreover,

sin

⁡

x

=

x

-

x

3

3

!

+

…

:

1

+

2

⁢

sin

⁡

x

=

1

+

2

⁢

(

x

-

x

3

3

!

+

…

)

2

-

(

2

⁢

(

x

-

x

3

3

!

+

…

)

)

2

8

+

…

By disregarding other than the first term in

sin

⁡

x

we obtain the first three terms of the series are:

1

+

2

⁢

sin

⁡

x

=

1

+

2

⁢

x

2

-

(

2

⁢

x

)

2

8

+

…

=

1

+

x

-

x

2

2

+

…

Example 7

(Differentiation).

Notice that we can deduce the series of

sin

⁡

x

from the series for

cos

⁡

x

by differentiating. Indeed

d

d

⁢

x

⁢

cos

⁡

x

=

-

sin

⁡

x

and we

differentiate

(term by term) the Taylor series of

cos

⁡

x

we obtain the Taylor series of

sin

⁡

x

(DO IT!).

Another example. Let us deduce the Taylor series of

1

(

1

-

x

)

2

. Notice that

d

d

⁢

x

⁢

1

(

1

-

x

)

=

1

(

1

-

x

)

2

. Since:

1

1

-

x

=

1

+

x

+

x

2

+

x

3

+

…

by deriving both sides we obtain:

1

(

1

-

x

)

2

=

0

+

1

+

2

⁢

x

+

3

⁢

x

2

+

…

=

∑

n

=

0

∞

(

n

+

1

)

⁢

x

n

and if we derive again we obtain:

2

(

1

-

x

)

3

=

2

+

6

⁢

x

+

12

⁢

x

2

⁢

…

Thus,

1

(

1

-

x

)

3

=

1

+

3

⁢

x

+

6

⁢

x

2

+

10

⁢

x

3

+

…

=

∑

n

=

0

∞

(

n

+

2

)

⁢

(

n

+

1

)

2

⁢

x

n

.

Example 8

(Integration).

Finally, we will deduce the Taylor series for

arctan

⁡

x

using integration (term by term). Notice that:

∫

1

1

+

x

2

⁢

𝑑

x

=

arctan

⁡

x

+

C

Moreover, since

1

/

(

1

-

x

)

=

1

+

x

+

x

2

+

x

3

+

…

by substituting

x

by

-

x

2

we obtain:

1

1

+

x

2

=

1

-

x

2

+

x

4

-

x

6

+

…

=

∑

n

=

0

∞

(

-

1

)

n

⁢

x

2

⁢

n

Thus, the Taylor series of

arctan

⁡

x

can be constructed integrating the previous one:

arctan

⁡

x

=

∫

1

1

+

x

2

⁢

𝑑

x

=

∫

(

1

-

x

2

+

x

4

-

x

6

+

…

)

⁢

𝑑

x

=

x

-

x

3

3

+

x

5

5

-

x

7

7

+

…

In

Σ

notation:

arctan

⁡

x

=

∫

1

1

+

x

2

⁢

𝑑

x

=

∫

∑

n

=

0

∞

(

-

1

)

n

⁢

x

2

⁢

n

⁢

d

⁢

x

=

∑

n

=

0

∞

(

-

1

)

n

⁢

∫

x

2

⁢

n

⁢

𝑑

x

=

∑

n

=

0

∞

(

-

1

)

n

⁢

x

2

⁢

n

+

1

2

⁢

n

+

1

Example 9

.

As an application of the previous example, we compute

π

. Indeed:

arctan

⁡

x

=

∑

n

=

0

∞

(

-

1

)

n

⁢

x

2

⁢

n

+

1

2

⁢

n

+

1

converges

between

-

1

≤

x

≤

1

and in particular

arctan

⁡

1

=

π

4

by the definition of

tan

⁡

x

and

arctan

⁡

x

. Thus:

π

4

=

∑

n

=

0

∞

(

-

1

)

n

⁢

1

2

⁢

n

+

1

2

⁢

n

+

1

=

∑

n

=

0

∞

(

-

1

)

n

2

⁢

n

+

1

and

π

=

4

⁢

(

∑

n

=

0

∞

(

-

1

)

n

2

⁢

n

+

1

)

=

4

⁢

(

1

-

1

3

+

1

5

-

1

7

+

1

9

-

…

)

For example, if one adds up to the

1

/

9

term, one obtains the

approximation

π

≈

3.33

. Unfortunately, the convergence is very slow. If you want to have about

m

correct digits then you have to add about

10

m

/

2

terms. For example, if you add

10

3

/

2

=

500

terms we get

3.143588

⁢

…

.

Title

examples on how to find Taylor series from other known series

Canonical name

ExamplesOnHowToFindTaylorSeriesFromOtherKnownSeries

Date of creation

2013-03-22 15:05:47

Last modified on

2013-03-22 15:05:47

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

8

Author

alozano (2414)

Entry type

Example

Classification

msc 41A58

Related topic

GettingTaylorSeriesFromDifferentialEquation

Related topic

TaylorSeriesOfArcusSine

Related topic

TaylorExpansionOfSqrt1x

Related topic

TaylorSeriesOfArcusTangent

Related topic

ApplicationOfLogarithmSeries

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