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Math Help - Taylor series of a logarithm

  1. #1
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    Taylor series of a logarithm


    why we do tailor around 1 and not 0
    ?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by transgalactic View Post

    why we do tailor around 1 and not 0
    ?
    Hint:
    Spoiler:
    Assuming that f is analytic at 0 then f(x)=\sum_{j=0}^{\infty}\frac{f^{(j)}(0)x^j}{j!} where (and this is meant as emphasis) f^{(0)}(0)=f(0)


    P.S. I don't see any fabric.
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    Quote Originally Posted by transgalactic View Post

    why we do tailor around 1 and not 0
    ?
    Because log z is singular (not defined) at zero.
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    the question states how to get over loss of significance problem in here
    f=ln x -ln y?

    f=ln x -ln y=ln z
    the tailor series for ln z is:
     \sum_{k=1}^{\infty}\frac{(-1)^{k+1}(Z-1)}{k}
    but how this expression equals z-1 ?
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by transgalactic View Post
    the question states how to get over loss of significance problem in here
    f=ln x -ln y?

    f=ln x -ln y=ln z
    the tailor series for ln z is:
     \sum_{k=1}^{\infty}\frac{(-1)^{k+1}(Z-1)}{k}
    but how this expression equals z-1 ?
    It doesn't. But, \ln(z){\color{red}\overset{z\to1}{\sim}}z-1
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  6. #6
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    why??

    how you got to this conclution that it close to z-1
    ?
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by transgalactic View Post
    why??

    how you got to this conclution that it close to z-1
    ?
    Your expression is equal to \ln(z). So, \lim_{z\to1}\frac{\ln(z)}{z-1}=\lim_{x\to0}\frac{\ln(1-x)}{x}=1. Another easier way is to write out the terms of your series and notice that \left|(z-1)-\sum_{n=1}\frac{(-1)^n(z-1)^n}{n}\right|<\varepsilon by taking |z-1|<\delta for a suitable \delta
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