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Tricky Application of Rouche's Theorem

For “small” $z$ is $e^z \approx 1 + z $ or $e^z - z \approx 1 \ne 0$. That suggests to apply Rouché's theorem to the functions $f(z) = e^z-z$ and $g(z) =1$: For $|z| = 1$ is $$ |f(z)-g(z)| = \left| \...
Martin R's user avatar
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Roots and extrema of the polynomial $P_n(x)=\sum_{k=0}^n\binom{n+k}{2k}(-x)^k$.

Let $S_n(x) = \sum_{k=0}^n P_k(x)$, and $U_n(y) = S_n(2 - 2y)$, it is easy to prove that $U_0(y) = 1$, $U_1(y) = 1 + 1 - (2-2y) = 2y$ and $$U_{n+1}(y) = 2y U_n(y) - U_{n-1}(y)$$ $U_n$ is then the ...
Kroki's user avatar
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Show that $a_n \underset{(+\infty)}{\sim} n$ with $a_n$ solution of $e^{-x}\sum_{k=0}^{n}\frac{x^k}{k!}=\frac{1}{2}$

Let $N_x$ denote a random variable having the Poisson distribution with rate $x$. Then \begin{align*} f_n(x) = \mathbf{P}(N_x \leq n). \end{align*} Moreover, by realizing the family $(N_x)_{x\geq 0}$ ...
Sangchul Lee's user avatar
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Tricky Application of Rouche's Theorem

You can try the symmetric version of Rouché': If $f$ and $g$ are analytic in a neighbourhood of $K$ and $|f(z) - g(z)| < |f(z)| + |g(z)|$ for $z \in \partial K$, then $f$ and $g$ have the same ...
Robert Israel's user avatar
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Roots and extrema of the polynomial $P_n(x)=\sum_{k=0}^n\binom{n+k}{2k}(-x)^k$.

We have the identity: $$\sin\left(\frac{x}{2}\right)P_n(2\cos(x) +2) = (-1)^{n+1}\sin\left(\frac{2n+1}{2}x\right)$$ This holds by induction on the recursion you provided. Now this should answer your ...
Lee Fisher's user avatar
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Rouché's Theorem: How many roots does $\lambda-z=\frac{1}{3}e^{z^2}$ have in the strip Re$(z)\in [-1,1]$?

On the circle $|\lambda-z|=\frac{1}{3}e\;$we have \begin{align*} & |\lambda-z| < 1 \\[4pt] \implies\;\;& |\text{Re}(\lambda-z)| < 1 \\[4pt] \implies\;\;& |\text{Re}(\lambda)-\text{Re}...
quasi's user avatar
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Paradox: Roots of a polynomial require less information to express than coefficients?

I do not think any of the current answers actually address the question. So, let me add my two cents. The keyword is Commutativity. It kills the information of the order of multiplication: $$ab=ba.$$ ...
Bumblebee's user avatar
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