Question

If you are trying to find a power series representation of a function f(x) and f'(x) is a known series, x^n then the power series rep of f(x) is the derivative of the series x^n, correct?

Answer 1

f'(x) - x^n
f(x)- 1/n+1 x^(n+1)

Answer 2

Yes, for only x^n other cases are different.
P.S. Remember to differentiate right. f(x) = x^n, f'(x) = nx^(n-1)

Answer 3

see thats what I got pk, but the answer my prof gave is what imran showed.. which would be the integral?

Answer 4

to go from f'(x) to f(x) you have to integrate , reverse of
f(x) to f'(x) where you differentiate

Answer 5

He is right, I misread your question.

Answer 6

oh so if I have to differentiate f(x) to get a known power series then i take the integral of that power series

Answer 7

and if i have to take the integral of f(x) to get a known power series then i differentiate the series?

Answer 8

well if they give you f'(x) and they want f(x) you would integrate

Answer 9

well the problem i was given was \[\ln (1+x^4)\] and I took the derivative of this.. and then found then differentiated the power series.. But now that I am looking at it I could have just integrated the known power series for 1/1-x and that would have gave me the power series for ln(1-x) and then i just plug in x^4; but I would assume both ways would have to give you equivalent series?

Answer 10

well id have to plug in -x^4

Answer 11

-ln(1-x)= 1/n+1 x^(n+1)

Answer 12

ln(1-x)=-1/(n+1) x^(n+1)

Answer 13

replace x with -x^4

Answer 14

\[\frac{-1}{n+1}\left(-x^4\right)^{n+1}\]

Answer 15

\[\sum _{n=0}^{\infty } \frac{-1}{n+1}\left(-x^4\right)^{n+1}\]

Answer 16

thanks, I see it now... sorta lol