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Math Help - Expansions

  1. #1
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    Expansions

    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
    Last edited by CaptainBlack; December 27th 2009 at 10:20 PM.
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  2. #2
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    Quote Originally Posted by arze View Post
    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
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    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?
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  4. #4
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    Quote Originally Posted by arze View Post
    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
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  5. #5
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    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.
    Last edited by arze; December 30th 2009 at 12:06 AM.
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  6. #6
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    Quote Originally Posted by arze View Post
    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
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  7. #7
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    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

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  8. #8
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    hello

    great puzzle its so simple logs are simple
    yoga retreat
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