Question

I dont understand the question in this exercise, can someone help :)

Answer 1

Let f denote the power of the series sum function. What is x * 2 * f'' (x) + f (x)
for x in the open interval of convergence?

Answer 2

sorry x^2 * f'' (x) + f (x)

Answer 3

I found the interval of convergence in this series to \[\sum_{n=0}^{∞}(-1)^n \frac{ x^n }{ n^2-n+1 }\] [-1,1]

Answer 4

the problem wants to evaluate the \(x^2 f''(x)+f(x)\) so what is confusing for u? :)

Answer 5

\[f(x)=\sum_{n=0}^{∞}(-1)^n \frac{ x^n }{ n^2-n+1 }\]

Answer 6

this is your start point\[f'(x)=\sum_{n=0}^{∞}(-1)^n \frac{ n x^{n-1} }{ n^2-n+1 }\]

Answer 7

So
\[f''(x)\sum_{n=0}^{∞}\frac{ (n-1)nx^{n-2} }{ n^2-n+1 }\]

Answer 8

and for course (-1)^n

Answer 9

*of

Answer 10

yes and what is final answer?

Answer 11

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

Answer 12

\[\left| x \right|<1\]
?

Answer 13

right :)

Answer 14

What means with "the open interval"?

Answer 15

well any open interval like \((a,b)\) in the open interval of convergence, because as u can see at the end nodes like x=1 the last expression is indeterminable

Answer 16

\[\sum_{n=0}^{∞}(-1)^n 1^n=1-1+1-1+1-... \ \ \ \ !!!\]

Answer 17

Ok so it tells me that I don't need to check the endpoints.

Answer 18

sorry i was out...yes

Answer 19

And its a alternating series. Thank you @mukushla

Answer 20

yw :)