Question

Find a power series representation for the rational function using differentiation or partial fractions.

f(x) = x^8 / (9+x^9 )^2

Answer 1

\[f(x)= \sum_{n=1}^{\infty} = ?\]

Answer 2

or maybe both?

Answer 3

they want the function represented as a power series with n starting at 1

Answer 4

\[\frac{x^9}{(9+x^9)^2}=\frac{1}{9+x^9}-\frac{9}{(9+x^9)^2}\]

Answer 5

that's what i started doing but apparently it might easier to begin with a simpler function like 1/ (9+x^9 )

Answer 6

and derive that

Answer 7

that one you can do like \(\frac{1}{1+x}\)

Answer 8

oh i see! i think that would make it easier

Answer 9

so for that one you'd get \[\sum_{n=0}^{\infty} (-1/9)^n x ^{9n}\]

Answer 10

oooh hold the phone
you have two different problems listed

Answer 11

oh sorry! it's the original problem with x^8

Answer 12

\[f(x) = \frac{ x ^{8} }{ (9+x^9)^2 }\] then integrate and get
\[-\frac{1}{9}\frac{1}{9+x^9}\]

Answer 13

write that one as a power series, then differentiate term by term

Answer 14

\[\sum_{n=0}^{\infty} (-1) (1/9)^{n+1}x ^{9n}\]

Answer 15

i think that's right, but i'm not sure how to make it so that n=1

Answer 16

@satellite73

Answer 17

i'm stuck on differentiation term by term

Answer 18

\[\frac{1}{9+x^9}=\frac{1}{9}\frac{1}{1+\frac{x^9}{9}}\]
\[=\frac{1}{9}\left (1-\frac{x}{9}+\frac{x^2}{9^2}-...\right)\]

Answer 19

the sigma notation is too hard for me