- 1. Functions determined by power series.
- 2. More with functions that are determined by a power series.
- 3. Using power series to estimate the value of a function.

**1. **
**Functions determined by power series.**

Given a power series representing a function on its interval of convergence, we can find power series representations of the integral and derivative of this function. For example, consider the function below.

`> `
**f:=x->1/(1+x^2):**

We know that , when . This fact is immediate because the series is a geometric series with ratio and . Maple can also evaluate this sum:

`> `
**sum((-1)^n*x^(2*n),n = 0 .. infinity);**

Note that Maple did not ask whether , so it is telling us something dangerous if we accept it at face value.

How can we use the power series for
*f*
above to find the power series associated to
? We know that
.

Integrating the series for term by term, we can write arctan(x) = . We can ask Maple to find this:

`> `
**Sum(int((-1)^n*x^(2*n),x),n=0..infinity);**

Thus arctan(x) = + C, and knowing that , we must have when .

**Submission:**

Consider the function given by .

a) Show that if , then .

b) We know that the indefinite integral
of
* f*
satisfies
. By integrating the power series for
term by term, find a power series representation for
.

**Submission worksheet:**

`> `

**2.**
** More with functions that are determined by a power series.**

We know that if a function is defined by a power series over an interval of convergence, then the function is both differentiable and integrable over that interval of convergence by differentiating or integrating term by term.

With the aid of this fact (it is in fact a theorem) we can use functions and their known power series to discover power series associated to other functions. As an example, let us consider the function

`> `
**f:=x->1/(1+x^2):**

We know that , when . This fact is immediate because the series is a geometric series with ratio and . Maple can also evaluate this sum:

`> `
**sum((-1)^n*x^(2*n),n = 0 .. infinity);**

Note that Maple did not ask whether , so it is telling us something dangerous if we accept it at face value. How can we use the above to find the power series associated to ? We know that . Using our theorem and our observation above we can write arctan(x) = . We can ask Maple to find this:

`> `
**Sum(int((-1)^n*x^(2*n),x),n=0..infinity);**

So what we have is arctan(x) = + C, and knowing that , we must have when . We can use this fact to estimate . The series we just computed is alternating and decreasing when and , so the series passes the Alternating series test. Thus if we wish to estimate with an error less than .001, we need only to determine an n such that

`> `
**solve((1/2)^(2*(n+1)+1)/(2*(n+1)+1) = .1e-2);**

Astoundingly, this says that we need only use the first 3 terms to get the desired estimate. Let's check this.

`> `
**evalf(sum((-1)^n*(1/2)^(2*n+1)/(2*n+1),n = 0 .. 3));**

`> `
**arctan(.5);**

**Submission:**

Consider the function given by . Show that if , then .

We know that the indefinite integral
of
* f*
satisfies
. By integrating the power series for
term by term, find a power series representation for
. Show that the power series for F that you obtained is an alternating series passing the alternating series test (when
). Find an n such that the sum s[n] of the first n terms of the series is within .001 of the sum for
. Use this result to estimate ln(3/2) to within .001. Then compute ln(3/2) using Maple and compare the results.

**Submission worksheet:**

`> `

**3. Using power series to estimate the value of a function.**
** **

We can use the power series in problem 1 to estimate
. The series we computed is alternating and decreasing when
and
, so the series passes the Alternating series test. Thus if we wish to estimate
with an error less than .001, we need only to determine an
*n*
such that

`> `
**solve((1/2)^(2*(n+1)+1)/(2*(n+1)+1) = .1e-2);**

Astoundingly, this says that we need only use the first 3 terms to get the desired estimate. Let's check this.

`> `
**evalf(sum((-1)^n*(1/2)^(2*n+1)/(2*n+1),n = 0 .. 3));**

`> `
**arctan(.5);**

**Submission worksheet:**

a) Show that the power series for that you obtained in activity 1 is an alternating series passing the alternating series test (when ).

b) Find an
*n*
such that the sum
of the first
terms of the series is within .001 of the sum for
.

c) Use the
*n*
you found in b) to estimate ln(3/2) to within .001.

d) Estimate ln(3/2) using Maple and compare this answer with your result in c).

**Submission worksheet:**

`> `

`> `

`> `