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| visits | member for | 8 months |
| seen | 1 hour ago | |
| stats | profile views | 172 |
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Determining $\sin(15)$, $\sin(32)$, $\cos(49)$, etc. added 162 characters in body |
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answered | Determining $\sin(15)$, $\sin(32)$, $\cos(49)$, etc. |
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Alternating functional Series Convergence SOS… It converges for every fixed value of $x.$ |
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Alternating functional Series Convergence SOS… For $x<0$ denoting $t=-x$ we have that partial sums $f_N(-t) = \sum_{k=0}^N \frac{t^k \sqrt{k}}{k!}>\sum_{k=0}^N \frac{t^k }{k!}\to e^t$ as $N\to\infty,$ therefore, $\sum_{k=0}^\infty \frac{t^k \sqrt{k}}{k!}$ is unbounded as $t\to\infty.$ |
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Alternating functional Series Convergence SOS… As noted by Robert Israel, the series doesn't converge uniformly on whole $\mathbb{R}$ and is not uniformly bounded: $(\nexists M>0):\;\;(\forall{x}\in\mathbb{R}) \ \ \ \ |f(x)|< M.$ |
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Alternating functional Series Convergence SOS… It's a standard fact of the theory of power series: power series converges absolutely inside the interval (or disk) of convergence and converges uniformly on every compact subset of that interval. |
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Two problems on analytic function and Mapping of elementary functions Zeroes of analytic function cannot have an accumulation point. |
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Alternating functional Series Convergence SOS… We can state that power series $\sum_{k=0}^\infty (-1)^k \frac{x^k \sqrt{k}}{k!}$ with infinite radius of convergence converges uniformly on every compact $K\subset\mathbb{R}.$ |
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Alternating functional Series Convergence SOS… added 171 characters in body |
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answered | Alternating functional Series Convergence SOS… |
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convergence of series question For the series $\sum\limits_{n=1}^{\infty}{a_n}$ its general term is $a_n.$ |
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Understanding the (Partial) Converse to Cauchy-Riemann Yes, continuity of all partial derivatives $u_x,\ u_y,\ v_x,\ v_y$ is given. |
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Understanding the (Partial) Converse to Cauchy-Riemann Yes, it is definition of differentiability for multivariate functions. |
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answered | Understanding the (Partial) Converse to Cauchy-Riemann |
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convergence of series question In general, no. $\dfrac{1}{n}\to{0}$ however the series $\sum\limits_{n=1}^{\infty}{\dfrac{1}{n}}$ is divergent. |
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convergence of series question If series $\sum\limits_{n=1}^{\infty}{a_n}$ converges then $\lim\limits_{n\to\infty}{a_n}=0.$ |
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answered | convergence of series question |
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Find for all value of constant $a>0$; the interval of convergence of the power series $\sum_{n=0}^{\infty}\frac{1}{1+a^{n}}x^{n}$ 1. Case $a>1.$ The series $\sum\limits_{n=0}^{\infty}\frac{1}{1+a^{n}}x^{n}$ converges for $x\in (-a,\ a).$ 2. Case $a\leqslant{1}.$ The series $\sum\limits_{n=0}^{\infty}\frac{1}{1+a^{n}}x^{n}$ converges for $x\in (-1,\ 1).$ |
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Find for all value of constant $a>0$; the interval of convergence of the power series $\sum_{n=0}^{\infty}\frac{1}{1+a^{n}}x^{n}$ No. The series diverges at the endpoints in both cases since general term of the series does not tend to zero as $n\to\infty.$ |
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Find for all value of constant $a>0$; the interval of convergence of the power series $\sum_{n=0}^{\infty}\frac{1}{1+a^{n}}x^{n}$ I mean a necessary condition for convergence, not the ratio test. |
