- El_Arrow

Help! I need to find a power series representation for the function

- jamiebookeater

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- El_Arrow

f(x)=x^2/(1-7x)^2

- El_Arrow

- El_Arrow

this is what i have so far |dw:1437181408215:dw|

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## More answers

- El_Arrow

dont know if its right or not

- anonymous

That's not correct. You cannot use the geometric series on that since you still have the denominator squared. You would have to eliminate that squared denominator before you go to that.

- El_Arrow

how do i do that?

- anonymous

You can ignore the \(x^{2}\) and integrate \(\frac{1}{(1-7x)^{2}}\)

- El_Arrow

so have |dw:1437182143406:dw|

- El_Arrow

what do i do after that?

- anonymous

Thats what you got for the power series representation of the antiderivative?

- El_Arrow

oh no

- anonymous

Yeah, basically what I'm having you do is:
\[\int\limits_{}^{}\frac{ 1 }{ (1-7x)^{2} }dx = \sum_{n=0}^{\infty}??\]

- El_Arrow

i got 1/7 ln(1-7x)

- anonymous

Shouldn't have ln in your answer.

- El_Arrow

|dw:1437182482351:dw|

- anonymous

Right, there ya go. Now of course you need + C, but that won't be an issue with what we are doing. So what is the power series representation of that function?

- El_Arrow

would it be |dw:1437182579940:dw|

- anonymous

Yep. I like to separate everything with power series, though, makes it easier to see simplification a lot of the time. So I would have it written like:
\[\sum_{n=0}^{\infty}7^{n-1}x^{n}\]
Okay, so this series is equal to the antiderivative of what we started with. SO we want to go backwards and get back to where we started, which means I need to differentiate my result. You've seen how to differentiate and integrate series?

- El_Arrow

no

- El_Arrow

so this is the power series representation |dw:1437182937373:dw|

- El_Arrow

so we need to find the radius now

- anonymous

Well, you essentially take the derivative as normal, you worry about the variable terms while constants just stay as is. So kind of informally, the derivative would be something like this:
\[\sum_{n=0}^{\infty}7^{n-1}\frac{ d }{ dx }(x^{n})\] Where n is a constant. So what's rhe derivative of \(x^{n}\)?

- El_Arrow

nx^n

- El_Arrow

right?

- anonymous

\(nx^{n-1}\)

- El_Arrow

why minus 1?

- El_Arrow

oh nvm

- anonymous

But one thing we must also do is raise the index from 0 to 1. This is because, as is, letting the first n be 0 would make the series 0, so we fix that by raising the index by one. Thus taking the derivative would give us:
\[\sum_{n=1}^{\infty}7^{n-1}nx^{n-1}\]
Now just multiply in the \(x^{2}\)

- El_Arrow

like this |dw:1437183623856:dw|

- El_Arrow

cause the x^-1 and x^2 add each other right?

- anonymous

Yep. Just have to make the index 1 on your series.

- El_Arrow

so using the power series representation how do i determine its radius of convergence?

- El_Arrow

@dumbcow is the radius 1/7?

- dumbcow

yes the radius of convergence is 1/7

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