Expansions

**arze** · December 27th 2009, 08:22 PM

$f(x)=\frac{2x}{(x+1)^2(x^2+1)}=\frac{1}{x^2+1}-\frac{1}{(x+1)^2}$

Find the seires expansion of $\log \left[ \frac{f(x)}{2x}\right]$ up to $[x^4]$

$\log\left[\frac{f(x)}{2x}\right]=\log\left[\frac{1}{(x+1)^2(x^2+1)}\right]$

I suppose then we would have $-2\log(x+1)-\log(x^2+1)$ but i don't know how to proceed after this.

Thanks

**CaptainBlack** · December 27th 2009, 10:24 PM

Originally Posted by arze

$f(x)=\frac{2x}{(x+1)^2(x^2+1)}=\frac{1}{x^2+1}-\frac{1}{(x+1)^2}$

Find the seires expansion of $\log \left[ \frac{f(x)}{2x}\right]$ up to $[x^4]$

$\log\left[\frac{f(x)}{2x}\right]=\log\left[\frac{1}{(x+1)^2(x^2+1)}\right]$

I suppose then we would have $-2\log(x+1)-\log(x^2+1)$ but i don't know how to proceed after this.

Thanks

You should know that for $|x|<1$ :

$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...$

CB

**arze** · December 27th 2009, 11:59 PM

Originally Posted by CaptainBlack

You should know that for $|x|<1$ :

$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...$

CB

The log here is the base 10 log, I know this is for the natural log, but what about the base 10 one?

**CaptainBlack** · December 28th 2009, 06:00 AM

Originally Posted by arze

The log here is the base 10 log, I know this is for the natural log, but what about the base 10 one?

If you are doing this question you should already know the change of base relation for logarithms which here is:

$\log_{10}(u(x))=\frac{\log_e(u(x))}{\log_e(10)}$

If you do not know the change of base relation then stop doing this and go back and learn the propoerties of the $\log$ functions.

Also always specify the base of a $\log$ if it is not a natural logarithm. You teacher may think they are teaching a convention when they tell you $\log$ denotes $\log_{10}$ and $\ln$ denotes $\log_e$ but its not a universal convention.

CB

**arze** · December 29th 2009, 06:47 PM

Ok I done that and checked the answer and it was $-2x-\frac{2}{3}x^2+x^4$ my answer would have been $\frac{1}{\ln10}(-2x-\frac{2}{3}x^2+x^4)$ .
The next part of the question is:
Assuming the validity of the process, differentiate $log[\frac{f(x)}{2x}]$ as given to show that as far as the term in $x^3$ , $\frac{1}{f(x)}\frac{d}{dx}f(x)=\frac{1}{x}-2-2x^2+4x^3$ .

**CaptainBlack** · December 29th 2009, 11:00 PM

Originally Posted by arze

Ok I done that and checked the answer and it was $-2-2x^2+4x^3$ my answer would have been $\frac{1}{\ln10}(-2-2x^2+4x^3)$ .

Just shows that natural logarithms were intended.

The next part of the question is:
Assuming the validity of the process, differentiate $log[\frac{f(x)}{2x}]$ as given to show that as far as the term in $x^3$ , $\frac{1}{f(x)}\frac{d}{dx}f(x)=\frac{1}{x}-2-2x^2+4x^3$ .

Well first do the differentiation suggested and see what you get.

CB

**arze** · December 30th 2009, 12:05 AM

Oops! I forgot to show my workings and I put in the wrong numbers.
$-2x-\frac{2}{3}x^3+x^4$ is the expansion of $log[\frac{f(x)}{2x}]$
differentiate that
$-2-2x^2+4x^3$

**soni12** · January 18th 2010, 10:03 PM

great puzzle its so simple logs are simple
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