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A start: Note that the function $$f(x)=\frac{2}{b-a}\left(x-\frac{a+b}{2}\right)$$ maps the interval $[a,b]$ to $[-1,1]$ and does analogous things to $(a,b)$, $[a,b)$, and $(a,b]$. It is a nice function of the form $px+q$, so we need only check that it sends $a$ to $-1$ and $b$ to $1$. Now find examples of series in $t$ whose intervals of convergence are, ...

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If you know the Jordan canonical form of a matrix $A$, and if the set of roots $R=\{\lambda_{1},\cdots,\lambda_{k}\}$ of the characteristic polynomial for $A$ does not include $0$, then you can choose a branch of the square root $\sqrt{z}$ which is holomorphic on a neighborhood of $R$, and you can compute a square root using the Jordan form using a finite ...

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Long story short: this is fine, no monkey business necessary, as long as $|z|$ is sufficiently small. Long story long: This will certainly work for $z$ with a sufficiently small modulus. It comes down to the following: Let $\|\cdot\|$ be any sub-multiplicative matrix-norm. If $\sum a_n z^n$ is a power series with radius of convergence $R>0$, then ...

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Yes you are allowed to do that. The power series $$(1 + A)^{\frac12} = \sum_{n=0}^\infty \frac{(-1)^n(2n)!}{(1-2n)(n!)^2(4^n)}A^n = 1 + \textstyle \frac{1}{2}A - \frac{1}{8}A^2 + \frac{1}{16} A^3 - \frac{5}{128} A^4 + \dots,$$ converges at least in the region $\|A\|<1$. The convergence is locally uniform and "absolute", but the term "absolute" should ...

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Yes you can, but you have to see the radius of convergence of the series. In other worde for what values of $z$ the series will converge. For example the series $$\sum_{k=0}^{\infty} A^k t^k$$ will converge under the condition $$||tA||<1 \implies |t|<\frac{1}{||A||},$$ where $||A||$ is the norm of the matrix $A$.

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So your power series, for part (a)., we must conduct a test (Ratio test or Root test) to find the interval of values of x such that the series converges. So, using the root test: $$\lim_{n\to\infty} \left|\frac{x^n}{e^{\sqrt{\ln n}}}\right|^{1/n}$$ $$= \lim_{n\to\infty} \left|\frac{x}{e^{\frac{\sqrt{\ln n}}{n}}}\right|$$ The numerator of the fraction ...

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I would look first at the size of the terms. Note that $e^{\sqrt{\ln n}}$ grows more slowly than $e^{\ln n}$, that is, than $n$. So $\frac{|x^n|}{e^{\sqrt{\ln n}}}$ blows up if $|x|\gt 1$. Also, the denominators are $\ge 1$, so by comparison with the geometric series $\sum x^n$, we have convergence if $|x|\lt 1$. So the radius of convergence is $1$. ...

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First, we see that if $R(z_0) = \infty$ for some $z_0$, then $f$ is entire and hence $R(z) = \infty$ everywhere. So, assume that $R(z_1) < \infty$ (and hence $R(z_2) < \infty$ by the first remark). Suppose $R(z_1) > R(z_2)+|z_1-z_2|$. In particular, choose $\rho$ such that $R(z_2) < \rho < R(z_1)-|z_1-z_2|$. This means that $B(z_2, R(z_2)) ... 0 The Taylor Series states the following: One can define a function f(x) as an infinite series to approximate the function with a margin of error defined and bounded in Taylor's Theorem: This is the definition of the Taylor Series approximation centered at x0. Notice: The approximation is centered at x0, meaning that the margin of error is minimal near to x0 ... 1 MrWho, if you're looking for a proof, If order for the series to converge absolutely, we must have $$\sum_{k=1}^{\infty} \frac {(-1)^{k-1}}{k}$$ converge and $$\sum_{k=1}^{\infty} \frac {|(-1)^{k-1}|}{k} = \frac{1}{k}$$ converge as well. We can use the alternating series test to prove that the first series is convergent. Let$b_n$= \frac{1}{k} Then ... 1 This is a direct argument that this series sometimes converges and sometimes does not, depending on how you order the sum. The more standard thing to do is to define "conditionally convergent" as a series that converges, but not absolutely. In that case it is clear that this series converges by the alternating series test, and it is clear that it does not ... 0 The series$\sum_{k=1}^{\infty} \frac {(-1)^{k-1}}{k}$converges conditionally because it fulfills the Leibnitz Theorem ($\frac{1}{k}$is a decreasing zero-sequence) but does not converge absolutely because$\sum_{k=1}^{\infty} \frac {1}{k}$diverges. To understand why your series converges, you can maybe write the first terms out: $$\sum_{k=1}^{\infty} ... 3 The given series is convergent by the Leibniz's theorem but it isn't absolutely convergent since the harmonic series is divergent hence this series is conditionally convergent. 1 We can use a standard real example. If we want a non-real example, multiply everything by i. Consider the series$$1+\frac{1}{2}z+\frac{1}{3}z^2+\frac{1}{4}{z^3}+\frac{1}{5}z^4+\cdots.$$This converges for z=-1, but does not converge absolutely there, since the harmonic series 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots diverges. We can replace ... 1 Here is a non-elementary example arising from zeta functions. Consider the polylogarithm \text{Li}_s(x)=\sum_{n=1}^\infty \dfrac{x^n}{n^s}. Since n^{-s}=\exp(-s \ln n), differentiation with respect to s yields \partial_s\text{Li}_s(x)=-\sum_{n=1}^\infty \ln n\,\dfrac{x^n}{n^s}. Note in particular that x\to 1 gives the first derivative of the ... 0 References and personal experience suggest that the original author of this question, despite this temporary stall, has the extraordinary brainpower to overcome the difficulties of this problem! 1$$\sum_{n=0}^{\infty}a_{n}\times\sum_{n=0}^{\infty}b_{n}\times\sum_{n=0}^{\infty}c_{n}=\sum_{n=0}^{\infty}s_{n}$$where$$s_{n}=\sum_{r+s+t=n}a_{r}b_{s}c_{t}$$1 To express it as a single infinite series you can apply the Cauchy Product a second time to get$$ \sum_{m=0}^\infty e_m $$where$$ e_m = \sum_{l=0}^m c_l \sum_{k=0}^{m-l} a_kb_{m-l-k} $$2 As told in answers and comments, a series exists but it is quite tedious to derive. Writing$$\frac{\displaystyle\sum_{i=0}^{\infty}a_{i}x^{i}}{\displaystyle\sum_{j=0}^{\infty}(j-2)a_{j}x^{j}}=\sum_{i=0}^{\infty}b_{i}x^{i}$$the first coefficients are given by$$b_0=-\frac{1}{2}b_1=-\frac{a_1}{4 a_0}b_2=\frac{a_1^2-4 a_0 a_2}{8 a_0^2}$$... 2 Hint: Note that \displaystyle\sum_{j=0}^{\infty}(j-2)a_{j}x^{j}=(x\frac d{dx}-2)\sum_{i=0}^{\infty}a_{i}x^{i} 1 It's not true that F(z) = \sum_{n=0}^{\infty} a_n z^n, |z| < 1 must have a simple pole at z=1 after analytic continuation: it's not even true in general that F has a meromorphic continuation to a neighborhood of z=1 at all. Define a_n by a_n = \sum_{k=1}^n \frac{1}{k^2} and a_0 = 0; in other words, F(z) = \frac{1}{1-z} \mathrm{Li}_2(z), ... 0 \frac{1}{1-\cos(x)}=\frac{1+\cos(x)}{1-\cos^2(x)}=\frac{1+\cos(x)}{\sin^2(x)}=\csc^2(x)+\csc(x)\cot(x) That last one can be easily noticed to be the derivative of -(\cot(x)+\csc(x)). Now the question is a matter of finding the power series expansions of \cot(x) and \csc(x), which are more easily found, and differentiating. 0 In fact,$$ \sum_{n=0}^{+\infty}(n+1)x^n = \sum_{n=0}^{+\infty}\frac{d}{dx}(x^{n+1})= \frac{d}{dx}\sum_{n=0}^{+\infty}x^{n+1} = \frac{d}{dx}\biggl(\frac{x}{1 - x}\biggr) = \frac{1}{(1 - x)^2} $$For x = \frac{1}{3}, we have$$ \frac{9}{4} =\sum_{n=0}^{+\infty}(n+1)\frac{1}{3^n} = \sum_{m=1}^{+\infty}m\frac{1}{3^{m-1}} \quad \Rightarrow \quad ... 2 This is an expansion and follow up of my comment. As mentioned in comment.$\hspace0.5in$Without further restriction, there are no relation. An example is the function$f(x) = x^3$which is invertible over the real axis and yet its inverse function doesn't have a power series expansion at$x = 0$. In general, if your function is invertible only over ... 1 It's rational if it satisfies a linear equation with polynomial coefficients (that is, if it is a quotient of polynomials; that is, if it is a rational function). It's algebraic if it satisfies a polynomial equation with polynomial coefficients. For example: the generating function $$f(x)=2+3x+5x^2+9x^3+17x^4+33x^5+\cdots$$ is a rational series because it ... 1 Hint: You can use partial fractions then use the geometric series. 1 HINT: Assuming$|-z|<1,$$$\frac1{1-z+z^2}=\frac{1+z}{1+z^3}=(1+z)(1+z^3)^{-1}$$ 1 (a) Since the terms in the power series are non-zero for$x\neq 0$, the ratio test is adapted. (b) Yes: since$|x|\sin^2\theta\lt 1$we can write$1/(1-x\sin^2\theta)$as a power series and use the assumed result. 2 Note: The question changed dramatically after this answer was written. Since$\sum_n A_n$exists, we see that$A_n \to 0$and hence the radius of convergence of$\|A_n\|$is at least one, so the function$f(x) = \sum_n A_i x^n$is real analytic on$(-1,1)$. Furthermore, Abel's theorem (applied to each component) shows that$\lim_{x \uparrow 1} f(x) = ...

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The matrix $$A(x):=\sum_{n=0}^\infty A_n x^n$$ is invertible at $x=1$. The set of invertible matrices is an open set in the space of all matrices. Thus, an idea to solve the problem is to show that $x\mapsto A(x)$ is continuous. In order to prove that you need additional assumptions. One would be to assume that the power series $$\sum_{n=0}^\infty ... 0 I would just write$$ |S(x)| = |e^{x_n-n}-1| =|e^{(x_n-(n+1))+1}-1| \leq e^{|(x_n-(n+1))+1|}-1\leq e^{|x_n-(n+1)|+1} -1\leq e^{|x|+1}-1 $$Observe that |x|=x in your question. 1 Let me suggest to go back to elementary properties of the radius of convergence R, characterized by the following pair of implications. Consider some series \sum\limits_n\alpha_nx^n with radius of convergence R. If |x|\lt R then |\alpha_nx^n|\to0. If |x|\gt R then the sequence |\alpha_nx^n| is unbounded. In your first example, ... 1 Hint I suppose that you are very close since$$S (x) = \sum_{k=1}^{\infty} \frac{(x_n-n)^k}{k!}=\sum_{k=0}^{\infty} \frac{(x_n-n)^k}{k!}-1=e^{(x_n-n)}-1$$2 It suffices to show that the coefficients a_n cannot converge to 0. Now suppose that |a_k| \leq \varepsilon for k \geq m. Then for z\in\mathbb{D}$$|f(z)|\leq \frac{\varepsilon}{1-|z|}+\sum_{k=0}^{m-1}|a_k|.$$This implies that if a_k converges to 0 then \lim_{|z|\to 1}(1-|z|)|f(z)|=0. In particular f cannot have a pole on \mathbb{T} ... 0 I suppose the the pole is at z_0=1. Put \displaystyle f(z)=g(z)+\sum_{k=1}^s \frac{u_k}{(z-1)^k} where \displaystyle h(z)=\sum_{k=1}^s \frac{u_k}{(z-1)^k} is the singular part of f at 1 (u_s\not =0). Hence \displaystyle g(z)=\sum b_n z^n is analytic in a neighborhood of \overline{\mathbb{D}}, thus its radius of convergence is >1. Thus ... 0 If what you say you've proven is correct, then you are done. If the series converged at some z on the boundary, then \lim_{n\rightarrow \infty} a_n=0=g(1), contradicting that the order of the pole is k. 3$$\lim_{n \to \infty}\sqrt[n]{\frac{2^n+1}{n}} = \lim_{n \to \infty}\sqrt[n]{\frac{2^n}{n}} =\lim_{n \to \infty}\frac{2}{ \sqrt[n]{n}}=2$$Why can we omit the +1 term? Good question. ;) Write$$2^n+1 = 2^n(1+\frac{1}{2^n})$$As n \to \infty, the limit of both sides becomes equal. Therefore$$\lim_{n \to \infty} 2^n+1 = \lim_{n \to \infty} ...

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you say this appears in your textbook. It is only wise to learn all convergence criteria in the corresponding chapter or section, for the exercises in the end are based thereon. Raabe's convergence criterion applies for this series. Here's a nice exposé. http://en.wikipedia.org/wiki/Ratio_test One need calculate, for a series $\sum a_n$ (here $a_n \geq ... 2 You have$\displaystyle G(x)=(1+x+x^2+x^3)^5=\big(\frac{1-x^4}{1-x}\big)^5=(1-x^4)^5(1-x)^{-5}\displaystyle=\big(1-5x^4+\binom{5}{2}x^8-\binom{5}{3}x^{12}+\cdots\big)\big(\sum_{k=0}^{\infty}\binom{k+4}{4}x^4\big)$, so now you just have to find the coefficient of$x^{12}$in this expression. 4 If$x\leqslant1$, then$a_n\geqslant1$for every$n$hence the series$\sum\limits_na_n$diverges. If$x\gt1$, then the expansion$x^{1/n}=\exp((\log x)/n)=1+(\log x)/n+o(1/n)$yields $$\log(a_{n+1}/a_n)=\log(2-x^{1/n})\sim-(\log x)/n,$$ hence$\log a_n\sim-(\log x)\cdot\log n$. Thus: If$x\gt\mathrm e$, one can pick$1\lt y\lt\log x$then ... 3 A rigorous way to define the sine function is to consider it as the solution to the IVP: $$\begin{cases} y^{\prime \prime} + y = 0\\ y(0) = 0 \\ y^{\prime}(0) = 1 \end{cases}$$ 2 Some other important formulas regarding$\sin(x)$that you didn't mention are the infinite product $$\sin(x) = x \prod_{k=1}^{\infty}\Big( 1 - \frac{x^2}{\pi^2 k^2} \Big)$$ and the partial fractions decomposition $$\frac{1}{\sin(x)^2} = \sum_{k=-\infty}^{\infty} \frac{1}{(x-\pi k)^2}, \; \; x \notin \pi \cdot \mathbb{Z},$$ although I guess the latter only ... 0 When$c=0$, the recurrence doesn't determine$a_0$or$a_1$. There is a two-parameter family of solutions, and you can use$a_0$and$a_1$as the parameters. So you can choose$a_0$and$a_1$to be whatever you like. It's convenient to take the two basic solutions to be the ones with$(a_0,a_1) = (1,0)$and$(0,1)$; every solution is then a linear ... 1 Hint: Write out the solution for$c=1$(which necessarily has no constant term and leading term$a_0 x$) and compare with what you get for your solution at$c=0$if you don't assume$a_1=0$. 4 Let$w = \frac{z+1}{z-1}$. Then you have a power series in$w$, centered at$0$. Find its radius of convergence, call that$R$. Then find which$z$correspond to$\lvert w\rvert < R$. The map$z \mapsto \frac{z+1}{z-1}$can be explicitly inverted. 3 By the root test we have $$\left|2^n\frac{(4z-8)^n}n\right|^{1/n}\xrightarrow{n\to\infty}2|4z-8|<1\iff|z-2|<\frac18\iff z\in B\left(2,\frac18\right)$$ so the radius is$\frac18$. 2 Your ratio is incorrect. You should have: $$\frac{c_n}{c_{n+1}}=\frac{2^{3n}(n+1)}{2^{3(n+1)}n}=\frac{2^{3n}(n+1)}{2^{3n+3}n}=\frac{n+1}{2^3n}$$ 2 Because$|c_n 4^n|\rightarrow 0$when$n\rightarrow \infty$, and$|c_n (-2)^n|=|c_n 2^n|=|c_n 4^n|\cdot \frac{1}{2^n}\leq \frac{1}{2^n} $when$n$is verylarge, so by M-test,$\sum c_n(-2)^n$is convergent. 2 Outline: Since the first series converges, the terms$|c_n|4^n$approach$0$. It follows by Comparison (geometric series) that the series$\sum (-1)^n c_n 2^n$converges absolutely,and hence converges. 2 Notice that$(|c_n|4^n)_{n\geqslant 1}$is a bounded sequence, hence$|c_n|\leqslant M 4^{-n}$for a universal constant$M$, and we deduce$|c_n|2^n\leqslant M2^{-n}\$.

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