Representing functions as power series

[Abner]

Homework Statement

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.

Homework Equations

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

 

[throneoo]

you get c₀=6 by calculating f(0)

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

 

[Mark44]

Homework Statement

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.

Homework Equations

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

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.

 

[Abner]

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.

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

 

[throneoo]

<br /> \displaystyle \sum_{n=0}^\infty 6(-1)^n (49x^2)^n =<br /> \displaystyle \sum_{n=0}^\infty 6(-1)^n 49x^{2n}<br />

how did you arrive at this ?

 

[Abner]

how did you arrive at this ?

i was just following some examples from the book.

 

[Panphobia]

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

 

[Abner]

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

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.

 

[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)} =<br /> \displaystyle \sum_{n=0}^\infty 6(-1)^n (49x^2)^n =<br /> \displaystyle \sum_{n=0}^\infty 6(-1)^{n} 49^{n}x^{2n}<br />
your a value is 6, and r = -49x^2, other than that what don't you understand?

 

[Abner]

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)} =<br /> \displaystyle \sum_{n=0}^\infty 6(-1)^n (49x^2)^n =<br /> \displaystyle \sum_{n=0}^\infty 6(-1)^{n} 49^{n}x^{2n}<br />
your a value is 6, and r = -49x^2, other than that what don't you understand?

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

 

[Panphobia]

No it is right, http://www.wolframalpha.com/widgets/view.jsp?id=f021a5566d8509939615e02a20f267e3
put in the values on wolfram, and you'll see it is right. Also, you shouldn't have got the first coefficient right in your summation since you didn't distribute the power n, you would be left with 49*6, the one without the error I posted is the correct one.