Series 215 problem 06.mws

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 f(x) = sum((-1)^n*x^(2*n),n = 0 .. infinity) , when abs(x) < 1 . This fact is immediate because the series is a geometric series with ratio r = -x^2 and a = 1 . Maple can also evaluate this sum:

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

1/(x^2+1)

Note that Maple did not ask whether abs(x) < 1 , 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 arctan(x) ? We know that Int(1/(1+x^2),x) = arctan(x)+C .

Integrating the series for 1/(1+x^2) term by term, we can write arctan(x) = sum(Int((-1)^n*x^(2*n),x),n = 0 .. infinity) . We can ask Maple to find this:

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

Sum((-1)^n*x^(2*n+1)/(2*n+1),n = 0 .. infinity)

Thus arctan(x) = Sum((-1)^n*x^(2*n+1)/(2*n+1),n = 0 .. infinity) + C, and knowing that arctan(0) = 0 , we must have arctan(x) = Sum((-1)^n*x^(2*n+1)/(2*n+1),n = 0 .. i... when abs(x) < 1 .

Submission:

Consider the function given by f(x) = 1/(1+x) .

a) Show that if abs(x) < 1 , then f(x) = sum((-1)^n*x^n,n = 0 .. infinity) .

b) We know that the indefinite integral F(x) = ln(1+x) of f satisfies F(0) = 0 . By integrating the power series for f(x) term by term, find a power series representation for F(x) .

Submission worksheet:

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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 f(x) = sum((-1)^n*x^(2*n),n = 0 .. infinity) , when abs(x) < 1 . This fact is immediate because the series is a geometric series with ratio r = -x^2 and a = 1 . Maple can also evaluate this sum:

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

1/(x^2+1)

Note that Maple did not ask whether abs(x) < 1 , 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 arctan(x) ? We know that Int(1/(1+x^2),x) = arctan(x)+C . Using our theorem and our observation above we can write arctan(x) = sum(Int((-1)^n*x^(2*n),x),n = 0 .. infinity) . We can ask Maple to find this:

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

Sum((-1)^n*x^(2*n+1)/(2*n+1),n = 0 .. infinity)

So what we have is arctan(x) = Sum((-1)^n*x^(2*n+1)/(2*n+1),n = 0 .. infinity) + C, and knowing that arctan(0) = 0 , we must have arctan(x) = Sum((-1)^n*x^(2*n+1)/(2*n+1),n = 0 .. i... when abs(x) < 1 . We can use this fact to estimate arctan(1/2) . The series we just computed is alternating and decreasing when x = 1/2 and limit((-1)^n*(1/2)^(2*n+1)/(2*n+1),n = infinity) = ... , so the series passes the Alternating series test. Thus if we wish to estimate arctan(1/2) with an error less than .001, we need only to determine an n such that (1/2)^(2*(n+1)+1)/(2*(n+1)+1) < .1e-2

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

2.065782740

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));

.4634672619

> arctan(.5);

.4636476090

Submission:

Consider the function given by f(x) = 1/(1+x) . Show that if abs(x) < 1 , then f(x) = sum((-1)^n*x^n,n = 0 .. infinity) .

We know that the indefinite integral F(x) = ln(1+x) of f satisfies F(0) = 0 . By integrating the power series for f(x) term by term, find a power series representation for F(x) . Show that the power series for F that you obtained is an alternating series passing the alternating series test (when abs(x) < 1 ). Find an n such that the sum s[n] of the first n terms of the series is within .001 of the sum for x = 1/2 . Use this result to estimate ln(3/2) to within .001. Then compute ln(3/2) using Maple and compare the results.

Submission worksheet:

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

We can use the power series in problem 1 to estimate arctan(1/2) . The series we computed is alternating and decreasing when x = 1/2 and limit((-1)^n*(1/2)^(2*n+1)/(2*n+1),n = infinity) = ... , so the series passes the Alternating series test. Thus if we wish to estimate arctan(1/2) with an error less than .001, we need only to determine an n such that (1/2)^(2*(n+1)+1)/(2*(n+1)+1) < .1e-2

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

2.065782740

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));

.4634672619

> arctan(.5);

.4636476090

Submission worksheet:

a) Show that the power series for F(x) = ln(1+1) that you obtained in activity 1 is an alternating series passing the alternating series test (when abs(x) < 1 ).

b) Find an n such that the sum s[n] of the first n terms of the series is within .001 of the sum for x = 1/2 .

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:

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