# Expansions

• Dec 27th 2009, 08:22 PM
arze
Expansions
$\displaystyle f(x)=\frac{2x}{(x+1)^2(x^2+1)}=\frac{1}{x^2+1}-\frac{1}{(x+1)^2}$

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

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

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

Thanks
• Dec 27th 2009, 10:24 PM
CaptainBlack
Quote:

Originally Posted by arze
$\displaystyle f(x)=\frac{2x}{(x+1)^2(x^2+1)}=\frac{1}{x^2+1}-\frac{1}{(x+1)^2}$

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

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

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

Thanks

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

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

CB
• Dec 27th 2009, 11:59 PM
arze
Quote:

Originally Posted by CaptainBlack
You should know that for $\displaystyle |x|<1$:

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

CB

The log here is the base 10 log, I know this is for the natural log, but what about the base 10 one?
• Dec 28th 2009, 06:00 AM
CaptainBlack
Quote:

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:

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

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

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

CB
• Dec 29th 2009, 06:47 PM
arze
Ok I done that and checked the answer and it was $\displaystyle -2x-\frac{2}{3}x^2+x^4$ my answer would have been $\displaystyle \frac{1}{\ln10}(-2x-\frac{2}{3}x^2+x^4)$.
The next part of the question is:
Assuming the validity of the process, differentiate $\displaystyle log[\frac{f(x)}{2x}]$ as given to show that as far as the term in $\displaystyle x^3$, $\displaystyle \frac{1}{f(x)}\frac{d}{dx}f(x)=\frac{1}{x}-2-2x^2+4x^3$.
• Dec 29th 2009, 11:00 PM
CaptainBlack
Quote:

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

Just shows that natural logarithms were intended.

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

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

CB
• Dec 30th 2009, 12:05 AM
arze
Oops! I forgot to show my workings and I put in the wrong numbers.
$\displaystyle -2x-\frac{2}{3}x^3+x^4$ is the expansion of $\displaystyle log[\frac{f(x)}{2x}]$
differentiate that
$\displaystyle -2-2x^2+4x^3$

• Jan 18th 2010, 10:03 PM
soni12
hello
great puzzle its so simple logs are simple
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