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MarcLeclair
 one year ago
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
MarcLeclair
 one year ago
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

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

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

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

MarcLeclair
 one year ago
Best ResponseYou've already chosen the best response.0Oh 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.

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

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

MarcLeclair
 one year ago
Best ResponseYou've already chosen the best response.0Well 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.

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

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

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

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