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Multiplicity of root

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Tolaso J Kos
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Multiplicity of root

#1

Post by Tolaso J Kos » Mon Jan 11, 2016 1:54 pm

Given the function $f(x)=e^x-x-1$ prove that $0$ is a zero of $f$ of multiplicity $2$.
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Re: Multiplicity of root

#2

Post by Riemann » Wed Aug 01, 2018 9:40 am

Tolaso J Kos wrote:Given the function $f(x)=e^x-x-1$ prove that $0$ is a zero of $f$ of multiplicity $2$.
It suffices to prove that the limit $\displaystyle \lim \limits_{x \rightarrow 0} \frac{f(x)}{x^2}$ is finite. However,

\begin{align*}
\lim_{x\rightarrow 0} \frac{f(x)}{x^2} &= \lim_{x\rightarrow 0} \frac{e^x-x-1}{x^2} \\
&=\lim_{x\rightarrow 0} \frac{e^x-1}{2x} \\
&= \frac{1}{2}\lim_{x\rightarrow 0} e^x \\
&= \frac{1}{2}
\end{align*}

Done!
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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