# Homework Help: Find a power series for the function

1. Feb 23, 2012

### arl146

1. The problem statement, all variables and given/known data
Find a power series representation for the function and determine the radius of convergence.

f(x) = arctan(x/3)

2. Relevant equations
not really any equations ... just intervals

3. The attempt at a solution

here's what i did:

f'(x) = $\frac{1}{1+(x/3)^2}$ = $\frac{1}{1+(x^2/9)}$

arctan(x/3) = $\int\frac{1}{1+(x^2/9)}dx$ = $\int\frac{1}{1-(-x^2/9)}dx$ = ∫ (Ʃ (-1)n* $\frac{x^(2n)}{9}$)dx = ∫ (1/9 - x2/9 + x4/9 - x6/9 + ...)dx

= C + x/9 - x3/27 + x5/45 - x763 + ...

To find C:

make x = 0 so that C = arctan(0) = 0.
so, arctan(x/3) = x/9 - x3/27 + x5/45 - x763 + ...

= Ʃ (-1)n*(x^(2n+1) / (18n+9))
from n=0 to infinity

is this all right ??

2. Feb 23, 2012

### Dick

In your power series it isn't just (x^2)^n. It's (x^2/9)^n.

3. Feb 23, 2012

### arl146

I'm not sure if I know where/what you mean. Can you be a little more specific?

4. Feb 23, 2012

### Dick

1/(1+(x^2/9)) is equal to the sum of (-x^2/9)^n. Not the sum of (-x^2)^n/9.

5. Feb 23, 2012

### arl146

Oh ok. So you mean the 9 has an exponent of n too?

So you get arctan(x/3) = x - x^3/27 + x^5/405 - x^7/5103 + ...

so it would be summation (-1)^n * x^(2n+1) / ?
I don't know about the bottom part I can't figure it out

6. Feb 23, 2012

### Dick

Just keep track of how you are getting the numbers on the bottom. There is another error as well. arctan(x/3) isn't equal to $\int \frac{1}{1+(\frac{x}{3})^2} dx$. To do the integral you do a substitution of u=x/3. There's another factor coming from the du.

7. Feb 23, 2012

### arl146

oh =[

ok so when you have arctan(x/3) = ∫ $\frac{1}{1+(x/3)^2}$dx and then you do the u=x/3 and then 3*du=dx so then you have 3*∫ $\frac{1}{1+u^2}$dx
what do i do next?

8. Feb 23, 2012

### Dick

You don't have arctan(x/3)=$\int \frac{1}{1+(\frac{x}{3})^2} dx$. That's what I'm telling you. You need to fix that before you proceed. Use the u substitution to figure out what the integral actually is. Then use that to write arctan(x/3) correctly as an integral.

Last edited: Feb 23, 2012
9. Feb 23, 2012

### arl146

is the correct integral just the same thing just with just a 3 in front of the integral?

10. Feb 23, 2012

### Dick

If $\int \frac{1}{1+(\frac{x}{3})^2} dx$=3*arctan(x/3), what's arctan(x/3)?

11. Feb 23, 2012

### arl146

there'd be a 1/3 times the integral

12. Feb 23, 2012

### Dick

Yes, the 1/3 is the factor you are missing in your power series.

13. Feb 23, 2012

### arl146

so its this summation (-1)^n * x^(2n+1) / ?
just with a 3 on the bottom? but there has to more to the bottom, i cant figure that out, there doesnt seem to be a pattern that i can catch for the bottom

14. Feb 23, 2012

### Dick

There are going to be three things on the bottom, the 3, a power of 3^2 and something that comes from integrating x^(2n). This is really not that hard.

15. Feb 23, 2012

### arl146

So there should be the 3, 2n+1, and I don't get the other part

16. Feb 24, 2012

### Dick

I was thinking of the 9^n. What do you get? Compare with the series you gave in post 5 after you add the extra 1/3.

Last edited: Feb 24, 2012
17. Feb 24, 2012

### arl146

Wait, are the numbers on the bottom still right? Or does the 1/3 affect that?

18. Feb 24, 2012

### Dick

No, they are not right yet. You have to multiply by 1/3.

19. Feb 26, 2012

### arl146

ok so just

arctan(x/3) = x/3 - x^3/81 + x^5/1215 - x^7/15309 + ...

so like we said, there is a 3, 2n+1, and you said something about something that is related to the 9^n is on the bottom too?

20. Feb 26, 2012

### Dick

I really think you can work this out by yourself if you starting thinking about it.