Power Series Representations of Functions

[mammothrob]

My question is this

FInd a Power Series representaion of a function for


f(x)= (x^2)ln(1+x^2)


E <--------------------sigma
(x^2) <--------- means x squared

These are my last two steps and and solution.

x^2 E(-1)^n (x^2)^(n+1) /(n+1) =
x^2 E(-1)^N (X^(2n+2))/n+1

The books answer is

E(-1)^N (X^(2n+4))/n+1


I just done see how they got x raised to (2n+4),
distributing the X^2 through n+1 should get me x raised to (2n+2)

I hope this makes sense... i would have posted more of my solution but i want to make sure my question is understood first
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[pka]

You seem to have done this correctly.
\(\displaystyle \L
\begin{eqnarray}
\frac{1}{{1 + x}} & = & \sum\limits_{k = 0}^\infty {\left( { - 1} \right)^k x^k } \\
\ln (1 + x) & = & \sum\limits_{k = 0}^\infty {\frac{{\left( { - 1} \right)^k x^{k + 1} }}{{k + 1}}} \\
\ln (1 + x^2 ) & = & \sum\limits_{k = 0}^\infty {\frac{{\left( { - 1} \right)^k \left( {x^2 } \right)^{k + 1} }}{{k + 1}}} \\
x^2 \ln (1 + x^2 ) & = & \sum\limits_{k = 0}^\infty {\frac{{\left( { - 1} \right)^k x^{2k + 4} }}{{k + 1}}} \\
\end{array}\)

 

[mammothrob]

thanks you for the repost.

I cant believe i missed that. I wasnt multipling these right

( X^2)^n+1 * x^2

its funny how long I was staring at that one. ahhh

thank you

 

[soroban]

Hello, mammothrob!

You didn't specify any limits for those sums.
\(\displaystyle \;\;\)I believe your limits are incorrect . . .

Find a Power Series representaion of a function for: \(\displaystyle \,f(x)\:=\:x^2\cdot\ln(1\,+\,x^2)\)

The book's answer is: \(\displaystyle \L\,\sum \frac{(-1)^nx^{2n+4}}{n+1}\)

We know: \(\displaystyle \L\:\ln(1\,+\,u)\;=\; u\,-\,\frac{u^2}{2}\,-\,\frac{u^3}{3}\,+\,\frac{u^4}{4}\,-\,\cdots\;=\;\sum^{\infty}_{n=1}(-1)^{^{n+1}}
\left(\frac{u^n}{n}\right)\)

Then: \(\displaystyle \L\,\ln(1\,+\,x^2)\;=\;\sum^{\infty}_{n=1}(-1)^{n+1}\left(\frac{x^{2n}}{n}\right)\)

Hence: \(\displaystyle \L\:x^2\cdot\ln(1\,+\,x^2)\;=\;x^2\cdot\sum^{\infty}_{n=1}(-1)^{n+1}\left(\frac{x^{2n}}{n}\right) \;=\;\sum^{\infty}_{n=1}(-1)^{n+1}\left(\frac{x^{2n+2}}{n}\right)\)

That should have been your answer . . .



Evidently, the book began with \(\displaystyle n\,=\,0\)

\(\displaystyle \;\;\)so their answer is: \(\displaystyle \L\:\sum^{\infty}_{n=0}(-1)^n\left(\frac{x^{2n+4}}{n+1}\right)\)

 

[pka]

That should have been your answer . . .
Evidently, the book began with \(\displaystyle n\,=\,0\)
\(\displaystyle \;\;\)so their answer is: \(\displaystyle \L\:\sum^{\infty}_{n=0}(-1)^n\left(\frac{x^{2n+4}}{n+1}\right)\)

Not at all, I expect his book expected it done the way I did it above.