Question

\[f(x)=ln(1+x)\]
\[ln(1+x)=\frac{d}{dx}(\frac{1}{1-x})\]
\[=\frac{d}{dx} \sum_{n=0}^{\infty} x^n\]
\[=\sum_{n=0}^{\infty}nx^{n-1}\]
\[|x|<1\]
-1 13 years ago

Answer 1

Please check, I think this is correct

Answer 2

Sorry , there is smthing wrong with your answer !

Answer 3

The problem reads "Please find a power series representation", and find the radius of convergence.

Answer 4

R=1

Answer 5

yeah ! that's right !
but in the first line...look at what you wrote !

Answer 6

so I guess I did it wrong because the function is
\[f(x)= ln(1+x)\]

Answer 7

sorry !!the second !!

Answer 8

so that would make it
\[=\frac{d}{dx} \sum_{n=0}^{\infty} (-1)^nx^n\]

Answer 9

and
\[=\sum_{n=0}^{\infty}(-1)^nnx^{n-1}\]

Answer 10

from 1 to infinity....But what you wrote still correct !

Answer 11

why is it from 1 to infinity instead of zero?

Answer 12

*instead of zero to infinity?

Answer 13

\[
\log (x+1)=\sum _{n=1}^{\infty }
\frac{(-1)^{n+1} x^n}{n}
\]

Answer 14

ok now I'm really confused

Answer 15

\[
\log (x+1)=\int_0^x \frac{1}{t+1}
\, dt
\]

Answer 16

I'm going to sleep ! I didn't pay attention to that ! this is what I was telling you from the start ! \[d(\ln(1+x))/dx)=1/(1+x)\] not what you wrote !

Answer 17

You have to integrate the series of \(\frac 1 {1+t}\) not differentiating it.

Answer 18

oh I see...

Answer 19

Did you get it?

Answer 20

what was your thought process from \[\int_0^x\frac 1 {1+t}dt\]
to
\[\sum_{n=0}^\infty (-1)^{n+1} \frac {x^{n+1} } { n+1 }\\\]

Answer 21

Ohhh....I forgot to read the last section on Paul's online notes
http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx

Answer 22

@eliassaab
\[1/(1+t)=\sum_{0}^{\infty} (-1)^nt^n\] isn't like this ?

Answer 23

that's what I initially thought....

Answer 24

so now we have to integrate
\[=\sum_{0}^{\infty} (-1)^nt^n\]

Answer 25

yes !

Answer 26

\[\frac{(-1)^{n+1}}{n+1}\frac{t^{n+1}}{n+1}\]

Answer 27

no ! \[\sum_{0}^{}(-1)^nt^ {n+1}/n+1 \]

Answer 28

It was for Mathsophia

Answer 29

http://www.wolframalpha.com/input/?i=Series+log%281%2Bx%29

Answer 30

so why is is \[(-1)^{n}\] instead of \[(-1)^{n+1}\]

Answer 31

in your final answer

Answer 32

\[
\frac 1 {1+t}= \sum_{n=0}^\infty (-1)^{n} t^n\\

\int_0^x\frac 1 {1+t}dt=\sum_{n=0}^\infty (-1)^{n} \frac {x^{n+1} } { n+1 }=\\
x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x
^6}{6}+\frac{x^7}{7}+O\left(x^8
\right)
\]

Answer 33

Did you get it?

Answer 34

My last post is a complete solution. I have to go to bed.

Answer 35

Thank you Eliassaab!!!

Answer 36

Yw.

Answer 37

\[\frac{d}{dx}\sum_{n=0}^\infty(-1)^nx^n\]

Answer 38

I thought it was the integral and not the derivative? I don't know anymore

Answer 39

it's the derivative you hve to integrate it

Answer 40

it is the integral -.-

Answer 41

gah this is confusing me too since i haven't done this in a while

Answer 42

http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx

Answer 43

its the integral of

\[\int \frac{dx}{x+1}\]

Answer 44

following elissab, the reason why it isn't

\[(-1)^{n+1}\frac{x^{n+1}}{n+1}\] is because you are integrating in respects to x and not n

Answer 45

that is why it is

\[(-1)^n\]

Answer 46

Victory!!!!

Answer 47

You're awesome!

Answer 48

I insist that my solution is the correct one. You need to integrate not differentiate.
See
http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeriesandFunctions.aspx
Example 5, where they expand log(5 -x) which is similar to your problem.

Answer 49

@Outkast3r09 , you are wrong about differentiating. See my post above.

Also
\[
\frac{d}{dx}\sum_{n=0}^\infty(-1)^nx^n= -\frac{1}{(x+1)^2}
\]
@experimentX @lalaly @dpaInc @mukushla

Answer 50

That is the final solution
\[
\frac 1 {1+t}= \sum_{n=0}^\infty (-1)^{n} t^n\\

\int_0^x\frac 1 {1+t}dt=\sum_{n=0}^\infty \int_0^x(-1)^{n} t^ndt=\\
\sum_{n=0}^\infty (-1)^{n}\left[\frac{ t^n } {n+1}\right]_0^x=

\\

\sum_{n=0}^\infty (-1)^{n} \frac {x^{n+1} } { n+1 }=\\
x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x
^6}{6}+\frac{x^7}{7}+O\left(x^8
\right)

\]

Answer 51

i go with @eliassaab

Answer 52

agree with @mukushla ...:)