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## New answers tagged power-series

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Hint : You should express the coefficients of $B$ using the Cauchy product of power series. Remark also that $$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^n$$

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Given $R=\lim\limits_{n\to\infty} \frac{a_n}{a_{n+1}}=\lim\limits_{n\to\infty} \frac{a_{n-1}}{a_{n}}$ Also, let $$\sum \limits_{n=0}^{\infty} \frac{a_n x^{n+1}}{n+1}=\sum \limits_{n=1}^{\infty} \frac{a_{n-1} x^{n}}{n}=\sum \limits_{n=1}^{\infty} b_{n} x^{n}$$ where $b_n=\frac{a_{n-1}}{n}$ $R_1=\lim\limits_{n\to\infty} ... 1$f(x)=\frac{x^7}{a^8-x^8}=\frac{x^7}{a^8}\frac{1}{1-(\frac{x}{a})^8}=\frac{x^7}{a^8}\sum_{i=0}^\infty(\frac{x}{a})^{8i}$with convergence interval as$|\frac{x}{a}|<1$1 Outline: Our function can be rewritten as $$f(x)=\frac{x^3}{2^3}\cdot \frac{1}{(1-x/2)^3}.$$ To find the expansion of$\frac{1}{(1-t)^3}$, note that for suitable$t$we have $$\frac{1}{1-t}=1+t+t^2+t^3+\cdots.$$ Differentiate twice with respect to$t$. 2 Hint. You may write $$\frac{nx^n}{4n^2-1}=\frac{x^n}{4 (2 n-1)}+\frac{x^n}{4 (2 n+1)}$$ and one may recall $$\sum_{n=0}^\infty\frac{u^{2n+1}}{2 n+1}=\frac12 \log\left(\frac{1+u}{1-u}\right), \qquad |u|<1.$$ 0 Consider the case where$b_1 = 1$and$b_j = 0$for all other$j$. Then your function is$f(z) = 1 - z^2$. Since$f(1) = f(-1) = 0$, this function doesn't have an inverse. Since the problem of finding the inverse cannot be solved for this specific case, it also can't be solved for the general case. You're going to need to tell us more about the ... 0 let$(a_n)$so $$y(x)=\sum_0^\infty a_nx^n$$ then $$y'(x)=\sum_1^\infty a_nn(x-a)^{n-1}=\sum_0^\infty a_{n+1}(n+1)x^{n}$$ $$(1+x)y'(x)=y(x) \iff (1+x)\sum_0^\infty a_{n+1}(n+1)x^{n}=\sum_0^\infty a_nx^n$$ $$\iff \sum_0^\infty a_{n+1}(n+1)x^{n}+\sum_0^\infty a_{n+1}(n+1)x^{n+1}=\sum_0^\infty a_nx^n$$ $$\iff \sum_0^\infty a_{n+1}(n+1)x^{n}+\sum_1^\infty ... 0 If you search online, you will find many sources and examples on the web. such as: Link 1 Link 2 2$$ \frac{x^{n-1}}{2n-1} = \frac{y^{2n-2}}{2n-1} \quad \text{where }y=\sqrt x $$and$$ \frac d {dy}\, \frac{y^{2n-1}}{2n-1} = y^{2(n-1)} = x^{n-1}.  \sum_{n=1}^\infty \frac{x^{n-1}}{2n-1} = \sum_{n=1}^\infty \frac{y^{2n-1}}{2n-1}. $$The derivative of this with respect to y is$$ \sum_{n=1}^\infty y^{2n-1} = \frac y {1-y^2} = \frac A {1-y} + \frac B ... 2 Hint:$\text{arctanh }x=\displaystyle\sum_{n=1}^\infty\frac{x^{2n-1}}{2n-1}$1 You can consider the geometric sum$\sum_{n=0}^\infty \left(-\frac {x^2}3\right)^n$This series is also$\sum_{k=2}^\infty \left(-\frac{x^2}3\right)^{k-1}=\sum_{k=1}^\infty \left(-\frac13\right)^{k-1} x^{2k-2}$and an integration from 0 to x give$\sum_{k=1}^\infty \left(-\frac13\right)^{k-1} \frac{x^{2k-1}}{2k-1}$. So you can calculate the sum of the ... 2 Consider $$f(x)=\sum_{n=1}^{\infty}\frac{(-x^2)^{n-1}}{2n-1}$$ Then $$f(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^{2n-2}}{2n-1}$$ $$x\, f(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^{2n-1}}{2n-1}$$ $$(x\, f(x))'=\sum_{n=1}^{\infty}{(-1)^{n-1}x^{2n-2}}=\sum_{n=1}^{\infty}{(-x^2)^{n-1}}=\sum_{n=0}^{\infty}{(-x^2)^n}=\frac{1}{1+x^2}$$ $$x\, f(x)=\int ... 8 Note that for suitable t,$$\frac{1}{1+t^2}=1-t^2+t^4-t^6+\cdots.$$Integrate term by term from 0 to x. We get$$\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots.$$Divide by x. We get$$\frac{\arctan x}{x}=1-\frac{x^2}{3}+\frac{x^4}{5}-\frac{x^6}{7}+\cdots.$$Finally, let x=\frac{1}{\sqrt{3}}. 3 In complex analysis, this is often called "analytic continuation". The function \frac{1}{2-x} you are describing only has a single pole x = 2 on the complex plane, and otherwise well defined. But to write an arbitrary function in terms of power series, then such formalism has limitations, namely one would in general decrease the actual radius of ... 1 Such an F does not exists. There is no anti-derivative that is an elementary function. Since e^{-x^2} is an analytic function, you can still integrate the power series term by term to get an analytic anti-derivative, but this will not be an elementary function. (Summarized from the comments.) 2 No, certainly not. For example,$$ \sum_{k=0}^\infty (f(z))^k $$converges if and only if |f(z)| < 1 and that set is (in general) not a disc. For a concrete example with an interesting domain of convergence, take$$ \sum_{k=0}^\infty \sin^k z. $$0 While not a rigorous proof, I can offer some insight into why this converges when p>1. Note that for any constant n>0, the value of x^\frac{1}{n} always surpasses \ln(x) for all x that are sufficiently large. This is because \begin{gather} x^\frac{1}{n}\geq\ln(x)\\ e^\frac{x}{n}\geq x\\ \frac{e^x}{x}\geq \frac{1}{e^\frac{1}{n}} \end{gather} ... 0 Outline: The fact that we have divergence if p\le 1 ahould be obvious. To show that we have convergence when p\gt 1, let p=1+\delta. Do a Limit Comparison with \sum_2^\infty \frac{1}{k^{1+\delta/2}}. The key fact is that a positive power of k in the long run grows faster than any power of \ln k. 1 First rearrange x^2y''+y=0 to$$y''+\frac{1}{x^2}y=0$$Since the equation has a regular singular point a likely plan of attack assumes y = \sum\limits_{n= 0}^\infty c_n x^{n+p} where p is yet to be determined. Taking derivatives and substituting above gives$$\sum\limits_{n= 0}^\infty (n+p)(n+p-1)c_n x^{n+p-2}+\frac{1}{x^2}\sum\limits_{n= 0}^\infty ... 3 This is not an answer (I don't know the formal proof) but a comment because the power of the Euler-summation for series like this is much impressive but often not really known. Here is a table of the progression to the final value without and with Euler-summation. Euler-summation can have "orders", which intuitively means, iterates (but can be ... 0 lim sup$ | {\frac {1}{n}}|^{\frac {1}{n} } = 1.$hence the series is convergent for all$ x \in (-1,1) $.Also it is clear that the sum converges for$x=1$but not for$ x = -1$.Hence, the series is convergent for all$x \in [-1,1)$. 1 If we assume that the series continues as$y=\sum\limits_{k=1}^\infty x^k, we have $$y=\frac{x}{1-x}\tag{1}$$ then we can compute $$x=\frac{y}{1+y}=\sum_{k=1}^\infty(-1)^{k-1}y^k\tag{2}$$ However, if we don't make this assumption, we can invert the series using Lagrange Inversion: \begin{align} \left[\,y^k\,\right]x ... 1 All sums are over n\in\mathbb{Z}. IfR(r)=\sum a_nr^n\text{,}$$then you have$$\frac{dR}{dr}=\sum na_nr^{n-1}\qquad\frac{d^2R}{dr^2}=\sum n(n-1)a_nr^{n-2}\text{.}$$So$$\begin{align} \frac{d}{dr}\left[r^2 \frac{dR}{dr}\right]-\ell(\ell+1)R &=0\\ 2r\frac{dR}{dr}+r^2\frac{d^2R}{dr^2}-\ell(\ell+1)R &=0\\ 2r\sum na_nr^{n-1}+r^2\sum ... 0 First differentiate so we can write the ODE in a more workable form: $$r^2R'' + 2rR' - l(l+1)R=0.$$ AssumingR(r) = \sum_{n\in\mathbb Z}a_nr^n, we compute \begin{align} R'(r) &= \sum_{n\in\mathbb Z}(n+1)a_{n+1}r^n\\ R''(r) &= \sum_{n\in\mathbb Z}(n+1)(n+2)a_{n+2}r^n. \end{align} Hence we have $$r^2\sum_{n\in\mathbb Z}(n+1)(n+2)a_{n+2}r^n + ... 0 Hint Consider$$\frac{dQ}{dx}=\displaystyle\sum_{k=1}^{\infty}{x^{k-1}}=\sum_{i=0}^{\infty}{x^{i}}$$You have now a well know series. Back to Q would give you the answer. But, by the way, Q is a very well known series. 1 The growth of x^n is called linear (n=1), quadratic (n=2), cubic (n=3), quartic (n=4), quintic (n=5)... You can find here a few names after that, but I don't expect anyone to say "octavic/octic" with a straight face. In general if you don't know n it's just called "polynomial growth". 0 For exponent 2, you say quadratic, for 3 and following, cubic, quartic, quintic... For general exponent, power function or power law. [In French you can use potentielle, but this can cause polysemic ambiguities.] The adverb quadratically can be freely used; I would abstain from "cubicly", "quarticly"... Also avoid the confusion with quadric, which ... 1 This topic is known as "inversion of power series". Do a search - you will get many interesting hits (it is not quite as bad as tvtropes). A very good reference is "generatingfunctionology" which can be downloaded for free here: https://www.math.upenn.edu/~wilf/DownldGF.html 1 This is not an answer but it is too long for a comment. Just as Will Jagy commented, you can inverse the series building one coefficient at the time. This is a quite tedious task but it is doable. Suppose that you have$$y=a_0+a_1+a_2x^2+a_3x^3+O\left(x^4\right)$$, replacing x by b_0+b_1+b_2y^2+b_3y^3, replacing, expanding and identifying one ... 7 Hint: If you take \sum_{n=1}^\infty \frac{1}{n^2} - \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2} what do you get? 0 Informally, it is sufficient to find the power series representation of the rational function$$f(x) = \frac{1}{1 - 0.5x - 0.25x^2} = \frac{4}{r_2 - r_1}\left(\frac{1}{r_2}\frac{1}{1 - \frac{x}{r_2}} - \frac{1}{r_1}\frac{1}{1 - \frac{x}{r_1}}\right).$$Now use the celebrated geometric sequence equality$$\frac{1}{1 - z} = \sum_{k = 0}^\infty z^k,$$for all ... 1 You have$$y=\sum_{n\ge 1}x^n=\frac{x}{1-x}\;,$$so (1-x)y=x, and hence$$x=\frac{y}{1+y}=y\sum_{n\ge 0}(-1)^ny^n\;.$$1 you have a geometric series. this has a simple recipe,$$ y = \frac{x}{1-x} $$for |x| < 1, that is -1 < x < 1. As it is a Mobius transformation, we can easily invert it,$$ x = \frac{y}{1 + y} $$0 It is true. Actually, this follows from a stronger result; from Knopp's Theory of functions Part 1, chapter 7, Theorem 2: If both power series$$\sum a_n(z-z_0)^n\quad \text{and}\quad \sum b_n(z-z_0)^n$$have a positive radius of convergence, and if their sums coincide for all points of a neighborhood of z_0, or only for an infinite number of such ... 1 This is true (at least if you mean Z(f) has no cluster point in (-R,R); zeroes might accumulate at \pm R) and is a standard theorem in complex analysis. Here is a sketch of a proof using only real methods. Suppose a\in (-R,R) is a cluster point of Z(f). It can be shown that f(x) also has a power series expansion centered at a which converges ... 0 The method used in this answer show that for every function f : \Bbb N \to \Bbb R there is a sequence (a_k) such that forall n, \sum_{k \ge 0} a_k k^n converges absolutely to f(n). 2 Using the generalized binomial coefficients (here), Taylor expansion is$$\sqrt{1+x}=\sum_{n=0}^\infty\binom{\frac{1}{2}}{n} x^n=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\frac{5 x^4}{128}+\frac{7 x^5}{256}+O\left(x^6\right)$$So, the numerator is$$\sum_{n=3}^\infty\binom{\frac{1}{2}}{n} x^n$$and the limit is just \binom{\frac{1}{2}}{3}=\frac 1 {16} ... 1 HINT:$$f(x)=(1+x)^{1/2}=1+\frac12 x-\frac18x^2+\frac1{16}x^3+O(x^4)$$0 The answer to the question is actually no. For example, by Weierstrass there are polynomials p_n(x) \to |x| uniformly on (-1,1) and |x| is not even differentiable at 0. On the point that the power series converges if |x-x_0|< d(x,\partial U), that's also false. For example f(x) = 1/(1+x^2) is real analytic on \mathbb {R} but its power ... 0 Here, the power series$$f(x) = \sum_{n = 1}^\infty \bigl(3^{\frac{1}{n^2}} - 1\bigr) x^n$$converges absolutely and uniformly on the compact interval [-1,1], and therefore defines a continuous function there. By the general theory of power series, f is differentiable on the open interval (-1,1), and the derivative is obtained by termwise ... 2 To get the series up to x^5: \begin{array}\\ \frac{1 - x + 2 x^2}{1 - 5 x^3 + 3 x^6} &=(1 - x + 2 x^2)\sum_{n=0}^{\infty} (5 x^3 - 3 x^6)^n\\ &=(1 - x + 2 x^2)(1+(5 x^3 - 3 x^6)+...)\\ &=(1 - x + 2 x^2)(1+5 x^3 )\\ &=(1 - x + 2 x^2)+(5x^3 - 5x^4 + 10 x^5)\\ &=1 - x + 2 x^2+5x^3 - 5x^4 + 10 x^5\\ \end{array} 1 HINT: Stirling's Formula states$$k! =\sqrt{2\pi k}\left(\frac ke\right)^k \left(1+O\left(\frac1k\right)\right)$$Then,$$\left(\frac{k^{2k+5}\,(\log k)^{10}\,\log (\log k)}{(k!)^2}\right)^{1/k}\sim e^2\,\left(\frac{k^{4}\,(\log k)^{10}\,\log (\log k)}{2\pi}\right)^{1/k}\to e^2$$1 The indicial must be at x=0 in this case. Dividing by the coefficient of the highest order derivative gives$$ y'' + \left(\frac{1}{x}+1\right)y'+\frac{1}{2x}y = 0. $$The indicial equation is found by keeping the \frac{1}{x}y' terms and \frac{1}{x^{2}}y terms:$$ y''+\frac{1}{x}y' = 0. $$Set y=x^{p}. The conditions to have ... 0 The expansion \log (1+x) =\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n} is valid for -1<x\le 1. Let x=1/t. Note that x\to 1/t maps 0<x\le 1 into t\le 1, and maps -1<x<0 into t>-1. Therefore, we see that the expansion \log (1+1/t) =\sum_{n=1}^{\infty}(-1)^{n-1}\frac{t^{-n}}{n} is valid for t\ge1 and t<-1. 1 This is an odd function and it is entire. The rad. is infinity. we just distribute mult. over addition of (1+x^2)(x-\frac{x^3}{3!}+\frac{x^5}{5!}-...) 1 Write$$\sum_{m=0}^\infty {\mathbb{E}(X^m)\,\theta^m\over m!}=M_X(\theta)={1\over (1-\theta/\lambda)^2}=\sum_{n=0}^\infty (n+1)\left({\theta\over\lambda}\right)^n. $$Matching the coefficient of \theta^4 gives$${\mathbb{E}(X^4)\over 4!}={5\over\lambda^4}. $$2 Note by integrating the series for \frac{1}{1+x^2} term by term, that for |x|\lt 1 we have$$\arctan x=\sum_1^\infty (-1)^{n-1}\frac{x^{2n-1}}{2n-1},$$and therefore$$\frac{\arctan x}{x}=\sum_1^\infty (-1)^{n-1}\frac{x^{2n-2}}{2n-1}$$Let x=\frac{1}{\sqrt{3}}. Our series has sum \frac{\pi}{2\sqrt{3}} 1 You have to show that the sequence of functions is actually dominated which is not hard (hint, just pick a constant function). 1 You need to use Abel partial summation theorem to prove that$$ \log(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}x^n $$is uniform convergent on [0,1] and thus continuous on on [0,1]. Hence$$ \lim_{x\to1}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}x^n=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}=\log{2} $$Let B_n=\sum_{k=m}^n \frac{(-1)^{k+1}}{k}. Since ... 4 You can prove that for all x\in(-1,1),$$\log(1+x)=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}x^k\$ and use Abel theorem.

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