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Math Help - taylor series

  1. #1
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    taylor series

     \log(1-x) = -x-\frac{x^2}{2}-\frac{x^2}{3}-\frac{x^4}{4}- \ldots -\frac{x^n}{n} - R_n for  -1 \leq x < 1 .

    Why isn't this the case for  -1 < x < 1 ? Instead, why do we replace  x with  -x to get  \log(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4} +\ldots + (-1)^{n-1} \frac{x^n}{n} - R_{n}' ?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by particlejohn View Post
     \log(1-x) = -x-\frac{x^2}{2}-\frac{x^2}{3}-\frac{x^4}{4}- \ldots -\frac{x^n}{n} - R_n for  -1 \leq x < 1 .

    Why isn't this the case for  -1 < x < 1 ? Instead, why do we replace  x with  -x to get  \log(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4} +\ldots + (-1)^{n-1} \frac{x^n}{n} - R_{n}' ?

    Ok

    I will go in depth with this,

    We know (hopefully) that

    \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n

    We also know

    \int\frac{-dx}{1-x}=\ln(1-x)

    So now making a substitution we can see that

    \ln(1-x)=-\int\sum_{n=0}^{\infty}x^n=-\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}

    So now we know that integrating/differentiating a power series only changes the endpoint behavior

    So we know that since

    \sum_{n=0}^{\infty}x^n converges \forall{x}\in(-1,1)

    So now we must check the endpoints.

    At x=-1 we have

    \sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}

    Which diverges by the Alternating series test.

    Since \frac{1}{n+2}\leq\frac{1}{n+1}\quad\forall{n}\in\m  athbb{N}

    Now we need to check at

    x=1

    We have

    -\sum_{n=0}^{\infty}\frac{1}{n+1}

    Which diverges by the integral test

    \int_0^{\infty}\frac{dx}{x+1}=\ln(x+1)\bigg|_{0}^{  \infty}=\infty


    So we have that

    \ln(1-x)=-\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}\quad\forall  {x}\in[-1,1)
    -------------------------------------------------------------------------

    Now for your second question we can see the by direct substitution into either the log or the \frac{1}{1-x} equation, or seeing that

    \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^{n}

    This implies that

    \ln(1+x)=\int\frac{dx}{1+x}

    is found by

    \int\sum_{n=0}^{\infty}(-1)^nx^n=\sum_{n=0}^{\infty}\frac{(-1)^nx^{n+1}}{n+1}


    Now once again we must check the endpoints,

    so at x=-1

    We have

    -\sum_{n=0}^{\infty}\frac{(-1)^n\cdot(-1)^n}{n+1}=-\sum_{n=0}^{\infty}\frac{1}{n+1}

    Which as I showed earlier is divergent by the integral test.

    But at

    x=1

    We have

    \sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}

    Which as was shown earlier is convergent by the alternating series test.


    Hope this helps.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Mathstud28 View Post

    So now we must check the endpoints.

    At x=-1 we have

    \sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}

    Which diverges by the Alternating series test.

    Since \frac{1}{n+2}\leq\frac{1}{n+1}\quad\forall{n}\in\m  athbb{N}
    It converges, and that is what the alternating series test tells us.

    RonL
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    It converges, and that is what the alternating series test tells us.

    RonL
    Yeah, I that is what I meant to say (as you can see by it's inclusion in the Radius of Convergence)
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  5. #5
    Math Engineering Student
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    -\ln (1-x)=\int_{0}^{x}{\frac{dy}{1-y}}=\sum\limits_{n\,=\,0}^{\infty }{\int_{0}^{x}{y^{n}\,dy}}=\sum\limits_{n\,=\,0}^{  \infty }{\frac{x^{n+1}}{n+1}}=\sum\limits_{n\,=\,1}^{\inf  ty }{\frac{x^{n}}{n}}.
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