Question

I think I get it now :)
Find a power series for f(x)=xln(1+x)
\[f(x)=xln(1+x)\]
\[\frac{d}{dx}xln(1+x)\]
\[=\frac{x}{x+1}+ln(x+1)\]
\[=\sum_{n=0}^{\infty}(-1)^nx^n+\frac{d}{dx}ln(x+1)\]
\[\sum_{n=0}^{\infty}(-1)^nx^n+\sum_{n=0}^{\infty}(-1)^n\frac{x^{n+1}}{n+1}\]

Answer 1

How did I do?

Answer 2

ok
since

Answer 3

multiply the above series with x

Answer 4

I keep making that error.

Answer 5

Why is the
\[ln(1+x)=\sum (-1)^{n-1}\frac{x^n}{n}\] and not
\[\sum_{n=0}^{\infty}(-1)^n\frac{x^{n+1}}{n+1}\]

Answer 6

I've worked on this problem before and I just keep getting it wrong. What's your thought process?

Answer 7

ok let me do this completely so that you will understand it .

Answer 8

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

Answer 9

The thing you get and the standard one are equivalent, yours is just shifted 1 back.

Answer 10

i think they are the same right?

Answer 11

Yeah, they are.

Answer 12

how did you do this shift

Answer 13

You can leave it with the shift, but it's just simpler to use the standard form and start the series from n = 1.

Answer 14

Here is another one of my problems. when do I have n=0 and when do I have n=1?

Answer 15

when n=0 doesn't yield a power series

Answer 16

elaborate @Outkast3r09
Does that mean the last n=0 statement above is incorrect?

Answer 17

no say if you get

\[x^{-1}\]

Answer 18

you'd move it up, or in this.. it's undefined at n=0

Answer 19

ok so when were multiplying x to ln(x+1) how would that effect
\[\sum_{n = 0}^{\infty} (-1)^n \frac {x^{n+1}}{n + 1}\]
?

Answer 20

*we're

Answer 21

it'd effect \[x^{n+1}\]

Answer 22

it is simply just you are adding to the power of x

Answer 23

\[x^nx^1x^1\]=\[x^{n+2}\]

Answer 24

oh so...
\[\sum_{n = 0}^{\infty} (-1)^n \frac {x^{n+2}}{n + 1}\]

Answer 25

yes

Answer 26

what if it's ln(x^2+1) that would make \[\int\frac{2x}{x^2+1}\]

Answer 27

just to make clarification the series i found start at n=1
you started it at n=0 it is shifted at n=0

Answer 28

\[\sum\int\frac{2x}{x^2+1}\]
\[2\sum\int\frac{x}{x^2+1}\]

Answer 29

it is different now.

Answer 30

So I wouldn't start off by taking the derivative of ln(x^2+1)
hmmmmm lets see

Answer 31

you need to expand this in power series formula
so you should have the form \[\Large \frac{1}{1-x}\]

Answer 32

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

Answer 33

yes

Answer 34

\[2x\sum(-x^2)^n\]
\[2x\sum(-1)^n(x^{2n})\]

Answer 35

\[2\sum(-1)^nx^{3n}\]

Answer 36

\[\sum(-2)^nx^{3n}\]

Answer 37

brb

Answer 38

you here ?

Answer 39

I'm back

Answer 40

so how did I do?

Answer 41

could I write the sum from n=0 in this case? because that would make the first number a 1

Answer 42

I did it wrong again.
:(

Answer 43

i hope you got it now :) and you can do any log series now :)

Answer 44

my sum makes perfect sense to me.

Answer 45

Is the idea that I replace x by x^2 in \[(-1)^n\frac{x^{n+1}}{n+1}\]
which gives me
\[(-1)^n\frac{(x^2)^{n+1}}{n+1}\]

Answer 46

yes

Answer 47

what happens to the 2 that was upfront?

Answer 48

never mind

Answer 49

THank you @sami-21 and everyone that helped!

Answer 50

you are welcome:)