HELP WITH CALCULUS POWEr SERIES!?

Question

use a power series to find a power series representation of the given function. then determine the radius of convergence of the resulting series.

f(x)= x / (9-x^2)

PLEASE HELP!

Answer 1

a/(1-h) = a + ah + ah^2 + ah^3 + ah^4 + ...

x/(9-x^2) = {x/9} / {1 - (x^2/9)}
thus this is a power series and also a geometric series whose first term is x/9 and whose common ratio is x^2/9

f(x) = x/9 + x^3/9^2 + x^5/9^3 + x^7/9^4 + ... + x^(2n-1)/9^n + ...

the convergence is from the center which is 0 to a ratio not exceeding 1 unit...

|x^2/9| < 1
x^2 < 9
-3 < x < 3 ... interval of convergence.. ... . radius is 3.

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Answer 2

f(x) = x/9 * 1/(1-x^2/9)
The second factor is the sum of a geometric series with r =x^2/9
convergent if Ix^2/9I<1 so -3 The power series is
x/9( 1+x^2/9 +x^4/81 +(x^2/9)^3++++

Answer 3

The nth coefficient of a power series is given by the value of the nth derivative at the center of the series (x=0 in this case)of the function divided by n factorial your function has nth derivatives of the form 2(n+1)!(5^n)/(1-5x)^(n+2) assuming n starts at 0 and the series is centered at 0 so the bottom of the fraction goes to 1((1-5(0))^(n+2)=1) so the value of the nth coefficient is 2(n+1)!(5^n)/(n!)= 2(n+1)5^n plugging in for n=0,1,2,3,4, gives the correct answer that you provided by the way recognizing the general pattern of the values of the derivative just comes with practice there is no real certain method for recognizing the pattern