# Find power series representation

1. Jan 28, 2008

### rcmango

1. The problem statement, all variables and given/known data

Find a power series representation for the function and determine the radius of convergence.

heres the problem: http://img301.imageshack.us/img301/4514/30437250jj2.png [Broken]

2. Relevant equations

3. The attempt at a solution

i believe the derivative of arctant = 1/(1+t^2)

thats all i know for now, whats next?

Last edited by a moderator: May 3, 2017
2. Jan 28, 2008

### rcmango

so now it has been manipulated to look like a geometric series.

so |-t^2| < 1 converges when | t |< 1

taking square root of t, and both sides of the equation.

so interval of convergence = -1 < |t| < 1?

what i'm thinking: http://img166.imageshack.us/img166/4083/65255560ec7.png [Broken]

...also have we yet shown the power series representation above? Or must it look simliliar to something in the pic i posted?

thanks.

Last edited by a moderator: May 3, 2017
3. Jan 28, 2008

### Gib Z

The derivative of the function f(x) is easy if you know the fundamental theorem of Calculus. Apply that.

4. Jan 28, 2008

### rcmango

so now, just plug in 0?

wait, now getting slightly confused, i realised that the derivative of arctant is 1/(1+t^2)

but is that the derivative of the original problem that you've put at the end of your post?

is the problem almost done, or did we figure out a similiar example?

thanks.

5. Jan 29, 2008

### rcmango

what happened to all the work we have just done???

6. Jan 30, 2008

### Gib Z

Just start again shall we? Yes it's nice the derivative of arctan t is 1/(1+t^2).

The first derivative of that integral is given by the fundamental theorem of calculus. After that, derivatives are easy to computer with the product rule. At the end, replace all expressions of arctan by its series representation and presto.

7. Jan 30, 2008

### HallsofIvy

Staff Emeritus
You don't really need to calculate any derivatives.

By the formula for a geometric series,
$$\frac{1}{1+ t^2}= \frac{1}{1-(-t^2)}= \sum_{n=0}^{\infty}(-t^2)^n$$
Since arctan is the integral of that, and we can integrate power series term by term inside their radius of convergence,
$$arctan(t)= \sum_{n=0}^\infty(-1)^n\frac{1}{n+1}t^{n+1}$$
$$\frac{arctan(t)}{t}= \sum_{n=0}^\infty (-1)^n\frac{1}{n+1}t^n$$

Now integrate that term by term.