# Representing functions as power series

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1. Nov 15, 2014

### Abner

1. The problem statement, all variables and given/known data
I have this function $f(x) = \frac{6}{1+49x^2}$, and i suppose to represent this function as a power series $\displaystyle f(x) = \sum_{n=0}^\infty c_n x^n$. Then i need to find the first few coefficients in the power series.

2. Relevant equations

3. The attempt at a solution
After attempting to find a solution to this problem, i get this series of $\displaystyle f(x) = \sum_{n=0}^\infty 6(-1)^n 49x^{2n}$
I don't know if that is correct, since the only coefficient that i get correct is $c_0 = 6$, and the rest of the coefficients are wrong. Also, i get $R = 1/7~\text {since a} = 6, \text{r} = 49x^2 \lt 1$

2. Nov 15, 2014

### throneoo

you get c0=6 by calculating f(0)

now consider f'(x)=∑cnxn-1 from n=1 to infinity
to get c1 , you can try calculating f'(0) .
similarly , cm where m>1 can be found via higher derivatives of f(x)

3. Nov 15, 2014

### Staff: Mentor

You don't say how you got your Maclaurin series, but it might be easier to just do long division of 1 + 49x2 into 1 (then multiply that answer by 6). I'm guessing they DON'T want you to do this the long way.

4. Nov 15, 2014

### Abner

$\frac{6}{1+49x^2} = \frac{6}{1-(-49x^2)} = \displaystyle \sum_{n=0}^\infty 6(-1)^n (49x^2)^n = \displaystyle \sum_{n=0}^\infty 6(-1)^n 49x^{2n}$

5. Nov 15, 2014

### throneoo

how did you arrive at this ?

6. Nov 15, 2014

### Abner

i was just following some examples from the book.

7. Nov 15, 2014

### Panphobia

if you separate 49 and x^2 you need to distribute the power so, 49^n * x^2n

8. Nov 15, 2014

### Abner

Honestly, i don't understand how to solve this problem at all. I spent at least 3 hours in this problem, that i don't know what i'm doing.

9. Nov 15, 2014

### Panphobia

Basically in power series, you are given a function that looks similar to the formula for the sum of an infinite geometric series
so
a/(1-r), you're trying to go backwards and get the actual series now, so to try to make it look as close as possible to that formula.
so
$\frac{6}{1+49x^2} = \frac{6}{1-(-49x^2)} = \displaystyle \sum_{n=0}^\infty 6(-1)^n (49x^2)^n = \displaystyle \sum_{n=0}^\infty 6(-1)^{n} 49^{n}x^{2n}$
your a value is 6, and r = -49x^2, other than that what don't you understand?

10. Nov 15, 2014

### Abner

When calculating to find the first 5 coefficients, i get everything wrong except for the first one which = 6.

11. Nov 15, 2014