Power Series

Example For what values of does the series converge?

Solution We consider two cases:

Case 1: . Then so by the divergence test diverges.

Case 2: . Then let and we have .

Ratio Test Suppose

Then converges.

By the ratio test, converges.

In fact, by the geometric series formula we have

∑𝑛=0

∞

𝑥𝑛¿11−𝑥

Recall the ratio test:

Summary

∑𝑛=0

∞

𝑥𝑛converges to 11−𝑥 for

However, diverges for

Theorem (Radius of convergence) Given a power series there are three possibilities:

(1) The series converges for all (2) The series converges only when (3) The series converges on an interval and diverges for Remark: in case (3), the series may or may not converge at the endpoints and .

Example Find the radius of convergence of

Solution Apply the ratio test: let and we have

𝑎𝑛+1

𝑎𝑛=

(𝑛+1 ) !𝑥𝑛+1

𝑛 ! 𝑥𝑛¿𝑛𝑥→ { 0 𝑥=0

𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑠 𝑥 ≠0

By the ratio test the series converges only when .

Which case of the radius of convergence theorem is this?

Example Find the radius of convergence of∑𝑛=0

∞ (−1 )𝑛2𝑛𝑥𝑛

3𝑛√𝑛

Solution

And investigate the convergence at the endpoints if applicable.

We apply the ratio test with Then

𝑎𝑛+1

𝑎𝑛=

(−1 )𝑛+12𝑛+1𝑥𝑛+1

3𝑛+1√𝑛+1(−1 )𝑛2𝑛 𝑥𝑛

3𝑛√𝑛

¿− 2 𝑥3

√𝑛√𝑛+1

→− 2𝑥3

And when so the radius of convergence is .

Now when the series becomes ∑𝑛=0

∞ (−1 )𝑛

√𝑛which converges by the alternating series test.

On the other hand, when the series beomces ∑𝑛=0

∞ 1√𝑛

which diverges since it is a series

So the interval of convergence is

Using Series to Represent Functions

Earlier, we saw our first example of using a power series to represent a function.

Formula (Geometric Power Series)

Actually, most familiar functions have a power series representation, and this can often be useful.

Example Express as a power series and find the interval of convergence.

Solution Let and this becomes . So

11+𝑥2

=1

1− 𝑦¿∑𝑛=0

∞

𝑦 𝑛¿∑𝑛=0

∞

(−𝑥2 )𝑛¿∑𝑛=0

∞

(−1 )𝑛 𝑥2𝑛

For the interval of convergence, as usual, we let

𝑎𝑛+1

𝑎𝑛=

(−1 )𝑛+1𝑥2𝑛+2

(−1 )𝑛𝑥2𝑛 ¿−𝑥2

So we have convergence when , i.e., the interval .

Exercise Check that the series does not converge when or

Example (Representing a function as a power series) Find a power-series representation of .

Solution Our series will have the form .

𝑔 (𝑥 )=𝑐0+𝑐1𝑥+𝑐2𝑥2+… 𝑔 (0 )=𝑐0 𝑓 (𝑥 )=sin 𝑥 𝑓 (0 )=0 Therefore

𝑔 ′ (𝑥 )=𝑐1+2𝑐2 𝑥+3𝑐3 𝑥2+… 𝑔 ′ (0 )=𝑐1 𝑓 ′ (𝑥 )=cos 𝑥 𝑓 ′ (0 )=1 Therefore 𝑔 ′ ′ (𝑥 )=2𝑐2+6𝑐3𝑥+12𝑐4 𝑥2+…𝑔 ′ ′ (0 )=2𝑐2𝑓 ′ ′ (𝑥 )=− sin𝑥 𝑓 ′ ′ (0 )=0 Therefore 𝑔 ′ ′ ′ (𝑥 )=6𝑐3+24𝑐4 𝑥+… 𝑔 ′ ′ ′ (0 )=6𝑐3𝑓 ′ ′ ′ (𝑥 )=−cos 𝑥𝑓 ′ ′ ′ (0 )=−1 Therefore

… … … … …

𝑔 (2𝑛 ) (𝑥 )=(2𝑛)! 𝑐2𝑛+…𝑔 (2𝑛 ) (0 )=(2𝑛)!𝑐2𝑛

𝑓 (2𝑛) (𝑥 )=(−1 )𝑛sin 𝑥𝑓 (2𝑛) (0 )=0

Therefore

𝑔 (2𝑛+1 ) (𝑥 )=(2𝑛+1 ) !𝑐2𝑛+1+…

𝑔 (2𝑛+1 ) (0 )=(2𝑛+1 ) !𝑐2𝑛𝑓 (2𝑛+1 ) (𝑥 )=(−1 )𝑛cos 𝑥

𝑓 (2𝑛+1 ) (0 )= (−1 )𝑛Therefore

sin 𝑥=𝑥− 13 ! 𝑥3+15 ! 𝑥

5− 17 ! 𝑥7+19 ! 𝑥

9−…Conclusion: if can be represented as a power series, then

Experiment Use graphing software to check if

So is a razor-sharp approximation of the sine function for sufficiently small

Exercise Derive the power series for

Example (Power Series Integration)

Write the power series for the integral∫ 𝑑𝑥1−𝑥7

Solution This is an extremely difficult integral, but we don’t have to do it. Instead, write the integrand as a power series.

11−𝑥7

=1+𝑥7+(𝑥7 )2+ (𝑥7 )3+…

Now integrate term-by-term!

∫ 11−𝑥7

𝑑𝑥=𝑥+18 𝑥

8+115 𝑥

15+122 𝑥

22+…

Lesson: sometimes a function may be difficult to integrate, but its power series, if you can find it, is easy to integrate.

Example Find a power series representation and the radius of convergence of the power series, for .

Solution First differentiate . 𝑓 ′ (𝑥 )= 15−𝑥

Now rewrite it in the geometric series form:

¿15

11− 𝑥 /5

𝑓 ′(𝑥 )=15 (1+ 𝑥5 +( 𝑥5 )

2

+(𝑥5 )3

+…)Now integrate term-by-term.

𝑓 (𝑥 )=∫ 15 (1+𝑥5 +( 𝑥5 )

2

+( 𝑥5 )3

+…)𝑑𝑥¿ 15 (𝑥+ 𝑥2

2 ⋅5 + 𝑥3

3 ⋅52+…)

¿ 15∑𝑛=0

∞ 𝑥𝑛+1

(𝑛+1 )5𝑛

We identify the pattern and write in summation form:

Exercise Use the ratio test to show that the radius of convergence is .