# Maclaurin Series of xln(1+2x)?

• March 25th 2010, 11:43 AM
Fox555
Maclaurin Series of xln(1+2x)?
Hi,

Can someone suggest a short way of solving this? I'm doing it the regular way, that is, I took it's derivatives up to seventh degree, then put 0 into all of them, wrote a few terms from the Taylor series formula, then came up with a rather ugly general term (with 4 individiual terms before the sigma).

Is there a better-easier way of solving this?

Thanks.
• March 25th 2010, 11:50 AM
chisigma
Is...

$\ln (1+ \xi) = \xi - \frac{\xi^{2}}{2} + \frac{\xi^{3}}{3} - \dots$ (1)

... so that all what you have to do is setting in (1) $2x$ instead of $\xi$ and then multiply by $x$...

Kind regards

$\chi$ $\sigma$
• March 25th 2010, 11:52 AM
General
Quote:

Originally Posted by Fox555
Hi,

Can someone suggest a short way of solving this? I'm doing it the regular way, that is, I took it's derivatives up to seventh degree, then put 0 into all of them, wrote a few terms from the Taylor series formula, then came up with a rather ugly general term (with 4 individiual terms before the sigma).

Is there a better-easier way of solving this?

Thanks.

All what you need is knowing the Maclaurin Series for $ln(1+t)$ ..
• March 25th 2010, 12:14 PM
Fox555
Ok, thanks to both of you, but after putting 2x in place of t in the expansion of ln(1+t), how am I going to multiply this by x? Shall I just write the whole sigma formula multiplied by x? Would it be OK like that?

Sorry if this is obvious, but my mind seems to have stopped working after many hours of studying. :) Hope you understand.

Thanks.
• March 25th 2010, 12:19 PM
Lafexlos
Quote:

Originally Posted by Fox555
how am I going to multiply this by x? Shall I just write the whole sigma formula multiplied by x? Would it be OK like that?

yes, it'll be OK if you multiply whole sigma formula.
• March 25th 2010, 12:34 PM
Fox555
Quote:

Originally Posted by Lafexlos
yes, it'll be OK if you multiply whole sigma formula.

Ok, got it. Here is the very result then (in sigma notation) .

xln(1+2x) = http://i39.tinypic.com/2ylqk36.jpg

Right? It seems somewhat off to me though...
• March 25th 2010, 01:06 PM
Lafexlos
if you send $x$ to the inside of sigma, you will get something like this :

$\sum_{n=1}^\infty {(-1)}^{n-1} \frac {{2^n}{x^{n+1}}}{n}$
• March 25th 2010, 01:11 PM
Fox555
Thank you again Lafexlos.