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The Taylor series allows you to find a series which approximates your function, given that you know the derivatives at a particular point. In your case, you have found a Taylor series because you know $f(0), f'(0), f''(0), ...$ etc You can shift the Taylor series by a constant, say $\Delta x$, so that you can use $f(\Delta x), f'(\Delta x), f''(\Delta ... 0 We have:$f(x) = f(2) + f'(2)(x-2)+\dfrac{f''(2)}{2!}(x-2)^2 + \cdots$. Thus we need to find a formula for$f^{(n)}(x)$. We have:$f'(x) = \dfrac{1}{x}$,$f''(x) = -1x^{-2}$,$f'''(x) = (-1)(-2)x^{-3}$. Thus$f^{(n)}(x) = (-1)(-2)\cdots (-(n-1))x^{-n}= (-1)^{n-1}\cdot (n-1)!x^{-n}$. You can take it from here. 2 I don't think this is what you are after, but I would consider $$S_1 = e^x - \sum_0 ^b \frac{x^k}{k!}$$ a closed form for your sum. 2 The Maple code sum(x^n/(4*n+1), n = 1 .. infinity); outputs $$\frac 1 4\,x{\it LerchPhi} \left( x,1,5/4 \right) ,$$ where the$ {\it LerchPhi} $function is described here. 15 Hint. You may consider the following derivative $$\left(\sum_{n=0}^{\infty}\frac{x^{4n+1}}{4n+1}\right)'=\sum_{n=0}^{\infty}x^{4n}=\frac1{1-x^4},\quad |x|<1.$$ Then, integrating the power series termwise for$|x|<1(which is allowed), you get $$\sum_{n=0}^{\infty}\frac{x^{4n+1}}{4n+1}=\int_0^x\frac1{1-t^4}dt=\int_0^x\left(\frac{1}{4 ... 2 In one solution let A=1,B=144,C=289, and in the other let A=81,B=100,C=289. In both cases let R=246 and D=9. For both cases then your equation holds, yet for the first one A+B+C=434 and for the second one A+B+C=470. So the sum cannot be determined. 0 To apply the ratio test we have to consider$$r_n=\frac{x^{n+1}(1-x^{n+1})}{n+1}\frac{n}{x^n(1-x^n)}=x\frac{1-x^{n+1}}{1-x^n}\ .$$If |x|<1 then r_n\to x and the series converges. If |x|>1 then$$r_n=x^2\frac{1-x^{-(n+2)}}{1-x^{-n}}\to x^2$$and the series diverges. If x=1 then the series vanishes (and converges). If x=-1 the series is ... 2 The ratio test does work here:$$\begin{align} \left|\frac{a_{n+1}}{a_n}\right| &= \left|\frac{x^{n+1}(1-x^{n+1})}{x^n(1-x^n)}\right|\cdot\frac{n}{n+1} \\ &= \left|x\cdot\frac{1-x^{n+1}}{1-x^n}\right|\cdot\frac{n}{n+1} \end{align}$$If |x|<1 then as n\to\infty, x^n\to 0 and that ratio tends to x. If |x|=1 then the series obviously ... 0 Use the ratio test: \displaystyle \lim_{n \to \infty} \frac{\frac{x^{n+1}(1-x^{n+1})}{n+1}}{\frac{x^n(1-x^n)}{n}} = \lim_{n \to \infty} \frac{n (x-x^{n+2})}{(n+1) (1-x^{n})} = \lim_{n \to \infty} \frac{ x-x^{n+2}}{ 1-x^{n}} = \lim_{n \to \infty} \frac{ x ( 1-x^{n+1})}{ 1-x^{n}} We now see that the numerator is similar to \ x times the denominator. ... 1 If |x|>1 so |x^n(1-x^n)|/n does not converge to 0, and if |x|<1 we can take the sum separeted. 0 Split it into two series$$ \sum_n\frac{x^n}{n}-\sum_n\frac{x^{2n}}{n} $$Both of these series are individually amenable to the ratio test. 0 CASE 1: |x|<1 We recall that the series for -\log (1-x) is$$-\log (1-x)=\sum_{n=1}^{\infty}\frac{x^n}{n}$$for |x|<1. For |x|<1, the series of interest is$$\begin{align} \sum_{n=1}^{\infty}\frac{x^n(1-x^n)}{n}&=\sum_{n=1}^{\infty}\frac{x^n}{n}-\sum_{n=1}^{\infty}\frac{(x^2)^n}{n}\\\\ &=-\log(1-x)+\log(1-x^2)\\\\ ... 1 Yes, your method is correct. Basically, to find a power series solution, we would assume the existence of\{a_n\}$, such that$y(x) = \sum a_n x^n$. The next thing to do would be finding a recurrent form of$a_n$, which leads to finding a closed form of it (in case it exists). What you have done is that throughout your method, you found the expression of ... 2 You can find a formula on Wikipedia:$\sum \limits _{n=1} ^\infty H_n ^{(m)} x^n = \frac {\mathrm{Li}_m (z)} {1-z}$, where$\mathrm{Li}_m$is the polylogarithm. I doubt that you will be able to find something involving more elementary functions. -1 Since $$x^2-2x+4=(x-1)^2+3=(x-2u)(x-2\bar{u}),$$ with $$u=e^{i\frac\pi3},$$ we have \begin{eqnarray} f(x)&=&\frac{2x-2}{(x-2u)(x-2\bar{u})}=\frac{1}{x-2u}+\frac{1}{x-2\bar{u}}=-\frac{\bar{u}}{2}\cdot\frac{1}{1-\frac{\bar{u}}{2}x}-\frac{u}{2}\cdot\frac{1}{1-\frac{u}{2}x}\\ ... 2 Given $$f(x) = \frac{ 2 x - 2 }{ x^2 - 2 x + 4 }. \tag 1$$ Assuming you want to expand$f(x). Let $$\phi_\pm = 1 \pm \mathtt{i} \sqrt{3}. \tag 2$$ We can write (1) as $$f(x) = \frac{ 1 }{ x - \phi_+ } + \frac{ 1 }{ x - \phi_- }. \tag 3$$ Whence $$f(x) = - \sum_{k=0}^\infty \left( \frac{1}{{\phi_+}^{k+1}} + ... 0 Given any function f, if we restrict to where f is nonzero (so that either \log f or \log -f exists), we can write \frac{f'}{f}=\frac{d}{dt}\log |f|. If f=c \prod (x-\alpha_i)^{k_i} is a rational function, \log f = \log c + \sum k_i \log (x-\alpha_i) and so$$\frac{f'}{f}=\frac{d}{dt}\log |f| = \sum k_i \frac{1}{x-\alpha}.Now, use the ... 2 The given f(x) is given by \begin{align} f(x) = \frac{2(x-1)}{x^2 - 2x + 4} \end{align} and can be seen as, where a = 1 + \sqrt{3} i and b = 1 - \sqrt{3} i, such that ab=4, \begin{align} f(x) = \frac{-2 \, (1 - x)}{ab \, \left(1 - \frac{x}{a}\right) \left( 1 - \frac{x}{b}\right)} \end{align} for which \begin{align} \ln f(x) &= \ln\left(- ... 0 You could try noting\frac{1}{(x-2)^{2}}=\frac{1}{2}\frac{d}{dx}\left(\frac{1}{1-(x/2)}\right)$$Then$$f(x)=2(x-1)\frac{1}{2}\frac{d}{dx}\left(\frac{1}{1-(x/2)}\right)$$lends itself to a power-series expansion immediately... 0 O.K. That's right you did exactly . 3 Multiply numerator and denominator by \frac{2a}{\log x}:$$ \frac{2a}{1-2a\frac{\log\log x}{\log x}}. $$a<1/2 and \log(x) < x; so 2a\log\log x/\log x < 1, and$$ f(x)=2a\sum_{n=0}^\infty \left(2a\frac{\log\log x}{\log x}\right)^n. $$So, f(x)\sim 2a. 2 Let m=\lceil -\mathrm ep\rceil. Then the ratio of magnitudes of successive terms beyond m is less than 1/\mathrm e. Now choose n such that$$ \frac{|p|^m}{m!}\frac1{1-1/\mathrm e}\mathrm e^{-(n-m)}\le\frac12\mathrm e^p\;. $$Since the m-th term has magnitude |p|^m/m! and the magnitudes of the remaining terms are bounded by a geometric sequence ... 1 They are Genocchi numbers, when you multiply by 2. In particular even Genocchi numbers are related to Bernoulli numbers:$$G_{2k}=2(1-2^{2k})B_{2k}.$$0 The answer is no. You have$$x \longmapsto f(x)=\frac1x, \quad x \in D=(0,\infty),$$which is differentiable on the interval D, but it does not admit a series expansion of the form \sum^{\infty}_{n=0}a_{n}x^{n} on D. 3 Hint: Note that$$\displaystyle f(z) = \frac{\pi}{\sin \pi z} - \frac{1}{z} + \frac{1}{z-1}+\frac{1}{z+1}$$is holomorphic in the disc D'(0,2) and can be expanded in Taylor Series, with the formula for the n-th coefficient f_n = \dfrac{1}{n!}\left.\dfrac{d^nf(z)}{dz^n}\right\vert_{z = 0}. You know the Laurent series for the \displaystyle -\frac{1}{z} ... 4 The taylor series you put in the comment is wrong. You should get$$ \ln\left(1+\frac{x^2}{4}\right)= \frac{x^2}{4}-\frac{x^4}{32}+\frac{x^6}{192}-\frac{x^8}{1024}+O(x^{10}) $$Integrating up to the 2nd power you get$$ \int_0^{1/2} \frac{x^2}{4}dx = \frac{1}{96} \simeq 0.010417 $$Integrating up to the 4th power you get$$ \int_0^{1/2} ... 1 The integral can be computed analytically:x \ln \left(\frac{1}{4} \left(x^2+4\right)\right)-2 x+4 \tan ^{-1}\left(\frac{x}{2}\right)$and with the limits becomes$\frac{1}{2} \left(-2+\ln \left(\frac{17}{16}\right)+8 \tan ^{-1}\left(\frac{1}{4}\right)\right)$or$0.010227$. 2 That is correct. In order to prove it's correct, one must deal with the following. From first-semester calculus you know that $$\frac d {dz}(A+B+C+\cdots +Z) = \frac {dA}{dz} + \cdots +\frac{dZ}{dz}$$ when there are only finitely many terms. But it doesn't always work with infinitely many terms. However, it does work with power series as long as you ... 2 The relevant theorem is the following. Theorem: Let$\sum a_nz^n$be a complex power series. Suppose that, for some$w\in\mathbb C$, the series$\sum a_nw^n$converges. Then for all$z\in\mathbb C$such that$|z|<|w|$, the series$\sum a_nz^n$converges absolutely. Proof. Suppose that$\sum a_nw^n$converges. We shall in fact only need to use a ... 0$a<<d\iff \dfrac ad<<1, \dfrac{a\pi}d<<1, $let$\dfrac{a\pi}d=2x$$$F=\dfrac{e^{2x}}{(e^{2x}-1)^2}=\dfrac1{(e^x-e^{-x})^2}$$ Use$e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$$$F=\dfrac1{4\left[x+\dfrac{x^3}{3!}+O(x^5)\right]}=\dfrac1{4x^2\left[1+\dfrac{x^2}6+O(x^4)\right]}$$ As$1+\dfrac{x^2}6+O(x^4)\approx1+\dfrac{x^2}6,$Use Binomial ... 0 For any series$S(z) =\sum_{n=1}^{\infty} z^{a_n} $where$a_n$is a increasing sequence of positive integers, the radius of convergence is one, since it converges for$|z| < 1$since$|z|^{a_n} \le |z|^n $since$a_n \ge n$and diverges for$|z| > 1$since$|z|^{a_n} \ge |z|^n$for the same reason. At$|z| = 1$, things can get complicated. However, ... 0 Instead of converting a sequence$A=(a_n)_{n=0}^{\infty}$into some kind of function, I would just specify what transformation you want to do on the series itself. For example, a translation by$m$could be written$T_m(A) =(0,0,...(m\ 0's), a_0, a_1, ...) $or, via the indices,$T_m(A)_i =0 \ if\ i < m; \ otherwise\ A_{m-i} $. For the "dilation", ... -3 (c-3, c+ 2] has midpoint (c-3+ c+2)/2= c- 1/2. It can be written as ((c-1/2)-5/2, (c- 1/2)+ 5/2] so has "radius" 5/2. 5 While it is true that in complex analysis, power series converges on discs (hence the name 'radius of convergence'), this is not necessary to see why real power series converge on a symmetric interval about their centre. A power series with real coefficients centred at the point$c$can be written as $$\sum_{n=0}^{\infty} a_n (x-c)^n,$$ and it will ... 0 In complex analysis, the radius of convergence is an actual radius. In real analysis we only have one axis, so the radii look like intervals. Draw this on paper. 2 As a start:$\begin{array}\\ \sum_{c=1}^{n-1} \sum_{k=c}^n \sum_j \frac{\rho(n,k)}{j!(k-c-j)!(c-j)!} &= \sum_{k=1}^n \sum_{c=1}^{k}\sum_j \frac{\rho(n,k)}{j!(k-c-j)!(c-j)!}\\ &= \sum_{k=1}^n \rho(n,k) \sum_{c=1}^{k}\sum_j \frac{1}{j!(k-c-j)!(c-j)!}\\ &= \sum_{k=1}^n \rho(n,k) \sum_{c=1}^{k}\frac1{c!}\sum_j \binom{c}{j}\frac{1}{(k-c-j)!}\\ ...

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Mathematica says that is equivalent to $$\sum_{c=1}^{n-1} \left(\sum_{k=c}^n \frac{U(-c,-2c+k+1,-1) \rho(n,k) (-1)^{-c}}{c!(k-c)!}\right)$$ where $U$ is Tricomi's (confluent hypergeometric) function. At least that gets rid of one explicit summation with the two cases $k-c \leq c$ and $k-c\geq c$.

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Note that if $x=1$ then you get $$1-\frac12+\frac13-\frac14+\cdots$$ which is convergent. This can be seen by rewriting it using $$\frac1n-\frac1{n+1}=\frac1{n(n+1)}$$ $$\left(\frac1{2}\right)+\frac16+\frac1{30}+\frac1{56}+\cdots<\frac1{2}+\frac14+\frac1{25}+\frac1{49}+\cdots<\frac1{2}+\sum^{\infty}_{n=2}\frac{1}{n^2}=\frac{\pi^2}{6}-\frac{1}{2}$$ ...

2

Both sources are correct, but your textbook is more comprehensive than Wolfram. Any power series has a radius of convergence, where the series converges for any number inside the radius and diverges for any number outside the radius. Wolfram correctly says that the radius of convergence is $1$. However, for real numbers, the two points at the radius of ...

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WA has an incomplete answer. Inasmuch as the McLaurin series for $\log (1+x)$ is $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}$$ we see that the series converges for $|x|<1$ from, say for example, the ratio test. Checking the end points, we see that for $x=1$ the series is the convergent alternating harmonic series while for $x=-1$ the series is the ...

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Wolfram is using the natural logarithm, $\log x = \ln x$ in general, unless a base is specified. And whilst Wolfram is correct, your textbook provides a more comprehensive reasoning, since at $x=1$, the series is the alternating harmonic series, which converges.

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Use integral test, since $1/n\ln n\to 0$ and is decreasing, the convergence of $\sum 1/n\ln n$ is the same as $\int_{2}^{+\infty} dx/x\ln x$...

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We have $$\int_0^1 \frac{1-e^{-tx}}{t} \, \mathrm{d}t= \int_0^1 x-\frac{tx^2}{2!}+\frac{t^2x^3}{3!} + \cdots + (-1)^{n-1}\frac{t^{n-1}x^n}{n!} + \cdots \, \mathrm{d}t$$ You can split the right hand side term by term to get $$\int_{0}^1 x\, \mathrm{d}t - \int_0^1 \frac{tx^2}{2!} \, \mathrm{d}t + \cdots + \int_0^1 (-1)^{n-1}\frac{t^{n-1}x^n}{n!}\, \mathrm{d}t ... 3 This answer is (a revision of) Simon Wadsley's comment posted here with the only goal to remove this question from the unanswered queue. Let k be a countable field, and take a map k[[x,y,z]] \to k[[u,v]] that sends x to u, y to uv and z to uf(v) for some f∈k[[v]]. It isn't hard to see that only for countably many choices of f can the ... 1 Note that \sum a_nx^{3n+1} converges if and only if \sum a_nx^{3n} converges. 0 We may notice that the original ODE is solved by x^{\alpha}=\exp(\alpha \log x), provided that:$$ \alpha(\alpha-1) -\alpha + p = 0,\tag{1} $$so two independent solutions are given by \exp\left[\left(1-\sqrt{1-p}\right) \log x\right] and \exp\left[\left(1+\sqrt{1-p}\right)\log x\right]. Assuming:$$ y(x) = \sum_{n\geq 0}a_n (x-1)^n \tag{2}$$we ... 1 Think of it this way: because \arctan{n} \to \pi/2 as n \to \infty, then the nth term of the sum approaches$$\left ( \frac{(\pi/2) x}{3 \cdot 2 \pi}\right )^n = \left (\frac{x}{12} \right )^n As a geometric sum over some $y$ (i.e., $\sum y^n$) has a radius of convergence of $1$, this sum has a radius of convergence of $12$.

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The radius is the limit of $\frac 1 {\sqrt[n] { {(6 \pi)} ^{-n} {(\arctan n)} ^n}}$, which is indeed $12$.

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$\Gamma(z)=\dfrac1z+\displaystyle\sum_{k=0}^\infty a_k\cdot z^k,~$ where the terms $a_k$ form this “beautiful” sequence here. :-$)$

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