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Thread: Power Series Expansion

  1. #1
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    Power Series Expansion

    Find the power series expansion of $\displaystyle ln(1+x)$ and hence derive the formula discovered in 1705 by Wilhelm Leibniz which states that

    $\displaystyle ln2 = 1-1/2+1/3-1/4+1/5-1/6+...$


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  2. #2
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    $\displaystyle y = \ln(1+x)$

    $\displaystyle y' = \frac{1}{1+x}$

    $\displaystyle y' = \frac{1}{1+x} = \frac{1}{1 - (-x)} = 1 - x + x^2 - x^3 + x^4 - x^5 + ...
    $

    integrate term for term ...

    $\displaystyle y = C + x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...$

    at $\displaystyle x = 0$ , $\displaystyle \ln(1+x) = 0$ ... $\displaystyle C = 0$

    $\displaystyle \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...$

    $\displaystyle \ln(1+1) = \ln(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ...$
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Wildharmony View Post
    Find the power series expansion of $\displaystyle ln(1+x)$ and hence derive the formula discovered in 1705 by Wilhelm Leibniz which states that

    $\displaystyle ln2 = 1-1/2+1/3-1/4+1/5-1/6+...$


    ~Much appreciate the help
    Let $\displaystyle S_n=a+ar+ar^2+\cdots+ar^{n-1}$, so $\displaystyle rS_n=ar+ar^2+ar^3+\cdots+ar^n$

    So $\displaystyle S_n-rS_n=a-ar^n\implies{S_n=\frac{a-ar^n}{1-r}}$

    Now $\displaystyle S_n=a+ar+ar^2+\cdots+ar^{n-1}=\sum_{n=0}^{n-1}ar^n$

    So now consider $\displaystyle \lim_{n\to\infty}S_n=\sum_{n=0}^{\infty}ar^n=\lim_ {n\to\infty}\frac{1-ar^n}{1-r}$

    Now consdier that if $\displaystyle 1\leqslant|r|$ this limit diverges to infinity, but if $\displaystyle |r|<1$ then $\displaystyle \lim_{n\to\infty}S_n=\frac{a}{1-r}$

    $\displaystyle \therefore\forall{r}\backepsilon|r|<1~\sum_{n=0}^{ \infty}r^n=\frac{1}{1-r}$

    Now consider that $\displaystyle |r|<1\implies|-r|<1$

    So $\displaystyle \forall{r}\backepsilon|r|<1~\frac{1}{1-(-r)}=\frac{1}{1+r}=\sum_{n=0}^{\infty}(-r)^n=\sum_{n=0}^{\infty}(-1)^nr^n$

    So now consider

    $\displaystyle \begin{aligned}\int_0^1\frac{dx}{1+x}&=\lim_{\varp hi\to{1}^-}\int_0^{\varphi}\frac{dx}{1+x}\\
    &=\lim_{\varphi\to{1^-}}\int_0^{\varphi}\sum_{n=0}^{\infty}(-1)^nx^ndx\end{aligned}$

    Now we have shown that the interval of convergence of $\displaystyle S_{\infty}$ is $\displaystyle (-1,1)$, thus on thatinterval it is uniformly convergent, thus on $\displaystyle (0,\varphi)\subset(-1,1)$ the integral and summation be switched.

    So $\displaystyle \begin{aligned}\lim_{\varphi\to{1^-}}\int_0^{\varphi}\sum_{n=0}^{\infty}(-1)^nx^ndx&=\lim_{\varphi\to{1^-}}\sum_{n=0}^{\infty}\int_0^{\varphi}(-1)^nx^ndx\\
    &=\lim_{\varphi\to{1^-}}\sum_{n=0}^{\infty}\frac{(-1)^n\varphi^{n+1}}{n+1}\\
    &=\lim_{\varphi\to{1^-}}\sum_{n=1}^{\infty}\frac{(-1)^n\varphi^{n}}{n}\end{aligned}$

    Now notice that this series interval of convergence is $\displaystyle (-1,1]$, so by abel's theorem

    $\displaystyle \lim_{\varphi\to{1^-}}\sum_{n=0}^{\infty}\frac{(-1)^n\varphi^n}{n}=\sum_{n=0}^{\infty}\frac{(-1)^n}{n}$

    $\displaystyle \begin{aligned}\therefore\int_0^1\frac{dx}{1+x}&=\ ln(1+x)\bigg|_{x=0}^{x=1}\\
    &=\ln(2)\\
    &=\sum_{n=0}^{\infty}\frac{(-1)^n}{n}\quad\blacksquare\end{aligned}$
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