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## Hot answers tagged roots

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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 $$|f(z)-g(z)| = \left| \... • 112k 3 votes Accepted ### Roots and extrema of the polynomial P_n(x)=\sum_{k=0}^n\binom{n+k}{2k}(-x)^k. Let S_n(x) = \sum_{k=0}^n P_k(x), and U_n(y) = S_n(2 - 2y), it is easy to prove that U_0(y) = 1, U_1(y) = 1 + 1 - (2-2y) = 2y and$$U_{n+1}(y) = 2y U_n(y) - U_{n-1}(y)U_n is then the ... • 12.8k 2 votes Accepted ### 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} ... • 165k 2 votes ### 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 ... • 447k 1 vote ### 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 ... • 1,874 1 vote ### 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}... • 58.5k 1 vote ### 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. ...
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