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    If you have a function in the form y=ln[1+cos(x)], do you just expand it as if it were ln(1+x) but replace the x with cos(x)?

    Also, could you do this for any other function of x that isn't x, for example
    y=ln[1+(x^2)]?

    Thanks for any help!
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    (Original post by PhyM23)
    If you have a function in the form y=ln[1+cos(x)], do you just expand it as if it were ln(1+x) but replace the x with cos(x)?
    This will get you a series in \cos x. Instead, expand \cos x first - then you'll get \ln(1 + \text{polynomial})} and expand this normally.

    Also, could you do this for any other function of x that isn't x, for example
    y=ln[1+(x^2)]?

    Thanks for any help!
    In this case, you can just replace x with x^2 everywhere in the expansion and you're good to go.
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    (Original post by Zacken)
    This will get you a series in \cos x. Instead, expand \cos x first - then you'll get \ln(1 + \text{polynomial})} and expand this normally.



    In this case, you can just replace x with x^2 everywhere in the expansion and you're good to go.
    PRSOM

    Brilliant. Thank you for confirming this
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    (Original post by PhyM23)
    If you have a function in the form y=ln[1+cos(x)], do you just expand it as if it were ln(1+x) but replace the x with cos(x)?

    Also, could you do this for any other function of x that isn't x, for example
    y=ln[1+(x^2)]?

    Thanks for any help!
    Yes, you can, provided |f(x)|<1, though this is often not the most convenient way to find the Taylor series, and a little fiddling needs to be done if |f(x)|>1.
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    (Original post by joostan)
    Yes, you can, provided |f(x)|<1, though this is often not the most convenient way to find the Taylor series, and a little fiddling needs to be done if |f(x)|>1.
    Thank you for the reply!

    Please may you explain what you mean by this. Why isn't it the most convenient way, and what's the difference between what happens when |f(x)|<1 as opposed to when |f(x)|>1 ?
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    (Original post by PhyM23)
    Thank you for the reply!

    Please may you explain what you mean by this. Why isn't it the most convenient way, and what's the difference between what happens when |f(x)|<1 as opposed to when |f(x)|>1 ?
    The series for \ln (1 + x) is only valid for |x| < 1 (otherwise, your series would just diverge to infinity, this condition is given next to the series in your textbook or formula booklet). So if you replace x by f(x) your series is now only valid for |f(x)| < 1.
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    (Original post by PhyM23)
    Thank you for the reply!

    Please may you explain what you mean by this. Why isn't it the most convenient way, and what's the difference between what happens when |f(x)|<1 as opposed to when |f(x)|>1 ?
    Well normally one is interested in a power series of the form \displaystyle \sum_{n=0}^{\infty} a_nx^n, but \cos(x) is itself such an expansion, and you end up with:
    \displaystyle \sum_{n=0}^{\infty} a_n\cos^n(x)=\sum_{n=0}^{\infty}  \left(\sum_{m=0}^{\infty}\dfrac{  (-1)^mx^{2m}}{(2m)!}\right)^n.
    As I'm sure you can appreciate this is not very pleasant.

    For suitable f with |f(x)|>1 one would need to write \ln(1+f(x))=\ln(f(x))+\ln\left(1  +\dfrac{1}{f(x)}\right) in order to use the expansion you wanted.
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    (Original post by Zacken)
    The series for \ln (1 + x) is only valid for |x| < 1 (otherwise, your series would just diverge to infinity, this condition is given next to the series in your textbook or formula booklet). So if you replace x by f(x) your series is now only valid for |f(x)| < 1.
    Ah of course! I'm not sure why I got confused for a minute there

    I think I need to go to bed...
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    (Original post by PhyM23)
    Ah of course! I'm not sure why I got confused for a minute there

    I think I need to go to bed...
    See Joostan's (excellent) answer for how to deal with the case for |f(x)| > 1. :-)
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    (Original post by joostan)
    Well normally one is interested in a power series of the form \displaystyle \sum_{n=0}^{\infty} a_nx^n, but \cos(x) is itself such an expansion, and you end up with:
    \displaystyle \sum_{n=0}^{\infty} a_n\cos^n(x)=\sum_{n=0}^{\infty}  \left(\sum_{m=0}^{\infty}\dfrac{  (-1)^mx^{2m}}{(2m)!}\right)^n.
    As I'm sure you can appreciate this is not very pleasant.

    For suitable f with |f(x)|>1 one would need to write \ln(1+f(x))=\ln(f(x))+\ln\left(1  +\dfrac{1}{f(x)}\right) in order to use the expansion you wanted.
    That's really interesting! Thanks for this
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    (Original post by Zacken)
    See Joostan's (excellent) answer for how to deal with the case for |f(x)| > 1. :-)
    It is indeed an excellent answer. Yours and joostan's help is much appreciated
 
 
 
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