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MarcLeclair
Group Title
How can you rewrite the function as a power series?
f(x) = x/ (1 + 2x^2)
I know I have to take out the x so x ( 1 / (1 + 2x^2)
however I'm stuck afterwards
 one year ago
 one year ago
MarcLeclair Group Title
How can you rewrite the function as a power series? f(x) = x/ (1 + 2x^2) I know I have to take out the x so x ( 1 / (1 + 2x^2) however I'm stuck afterwards
 one year ago
 one year ago

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Jhannybean Group TitleBest ResponseYou've already chosen the best response.0
@terenzreignz
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{1}{1x}=1+x+x^2+x^3+...\] \[\frac{1}{1+x}=1x+x^2x^3+...\] \[\frac{1}{1+2x^2}=12x^2+(2x^2)^2(2x^2)^3+...\]
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
So would I write the sum as \[\sum_{n=1}^{\infty} (1)^{n} (x)^{2n+1}\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
you need a 2 inside the parentheses
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
Oh yeah my bad. But how can I find the radius of convergence if I can't use the ratio test afterwards? I mean there's no division going on. Only thing I can think of is the root test but its negative, so the root test won't work perfectly.
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
I thought power series was just sum x^n
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
well the representation
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
Well I took the x and added it to the 2x^(2n+1) because the x was "outside" of the sum. So i just added it and thats what I did. I don't feel like writting the equation. I don't know if you follow what I'm saying.It was my guess really.
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
converges if \(2x<1\) i.e if \(x<\frac{1}{2}\)
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
Ah So i just take it right away. Makes sense. Thanks a lot^^
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
also you should start your summation at \(n=0\) i think
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
to be more precise i guess if converges if \(2x^2<1\) but same answer
 one year ago
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