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Math Help - Taylor expansion of composite function

  1. #1
    Member Jones's Avatar
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    Taylor expansion of composite function

    Hi,

    If i want to use Taylor expansion on a function like:

    \frac{ln(sin(x))}{cosx}

    Do i have to substitute the expansion of sin x, in to the expansion of ln x?
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  2. #2
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    Quote Originally Posted by Jones View Post
    Hi,

    If i want to use Taylor expansion on a function like:

    \frac{ln(sin(x))}{cosx}

    Do i have to substitute the expansion of sin x, in to the expansion of ln x?
    Yes, that would be a good idea. Then divide everything by the expansion for \cos{x}.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Jones View Post
    Hi,

    If i want to use Taylor expansion on a function like:

    \frac{ln(sin(x))}{cosx}

    Do i have to substitute the expansion of sin x, in to the expansion of ln x?
    For certain interval restrictions \frac{\ln(\sin(x))}{\cos(x)}=\frac{\ln\left(\sqrt{  1-\cos^2(x)}\right)}{\cos(x)}=\frac{\ln\left(1-\cos^2(x)\right)}{2\cos(x)} and since \left|\cos^2(x)\right|<1 (with the proper interval restrictions) we see that \frac{\ln\left(1-\cos^2(x)\right)}{\cos(x)}=\frac{1}{\cos(x)}\sum_{  n=1}^{\infty}\frac{\cos^{2n}(x)}{n}=\sum_{n=1}^{\i  nfty}\frac{\cos^{2n-1}(x)}{n}
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