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Thread: Leibniz rule and findamental theorem of algebra

  1. #1
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    Question Leibniz rule and fundamental theorem of algebra

    Hello,
    If 0<a<b, we have $\frac12\displaystyle\int_0^{2\pi} \frac{d\theta}{(b+a*cos\theta)^2}= \frac12\displaystyle\int_0^{2\pi} \rho(\theta)^2d\theta$.

    By differentiating under the integral sign we will deduce the fundamental theorem of algebra: a nonconstant polynomial $\mathbb{p}(\text{z})$ with coefficients in$\mathbb{C}$ has a root in $\mathbb{C}$

    Arguing by contradiction, assume $p(z)\not=0 $ for all $z\in\mathbb {C}$. For $ r\geq 0,$ consider the following integral around a circle of radius r centered at the origin:

    $I(r)=\displaystyle\int_0^{2\pi} \frac{d\theta}{\rho(r*e^{i\theta})}$.


    This integral makes sense since the denominator is never 0, so $\frac{1}{\text{p}(\text{z})}$ is continuous on $\mathbb{C}$. In case of area of ellipse, we can write $\rho=\frac{1}{b+a*cos \theta}$. But how to write $\rho$ in case of integral around a circle of radius r centered at the origin.

    I didn't understand what is the meaning of this integral? If any member of this MHF knows it, may reply with correct answer.
    Last edited by WMDhamnekar; Sep 27th 2019 at 07:15 AM. Reason: Spelling mistake corrected
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  2. #2
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    Re: Leibniz rule and findamental theorem of algebra

    Hello, Please read ''differentiating under integral sign".
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  3. #3
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    Re: Leibniz rule and findamental theorem of algebra

    There is a proof of the Fundamental Theorem of Algebra using Leibniz's rule for differentiation under the integral sign

    Consider the integral

    $$I(r)=\int_0^{2\pi } \frac{1}{p\left(r e^{i \theta }\right)} \, d\theta $$

    That's $p$ not $\rho$

    We take the derivative and show that $I(r)$ is constant and reach a contradiction.

    Now, I don't understand the part with $\rho $ and the ellipse
    Thanks from topsquark and WMDhamnekar
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