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The rearrangement appears to be okay. Assuming that $b$ is positive, if we start by switching variables to $z=-x$, then we're looking at $$f(z) = e^{-a(b+z)^2}$$ which is entire, so the series expansion $$f(z) = \sum_{n=0}^\infty \frac{(-a)^n}{n!}(b+z)^{2n}$$ converges absolutely for all $z$. In particular for positive $z$, binomial expansion on $(b+z)^{... 2 The same objection as before holds. If we consider $$f(z)=1-a_2 z+a_4 z^2-a_6 z^3 +\ldots$$ the fact that$\{a_{2n}\}_{n\geq 1}$is a positive decreasing sequence do not give that Newton's inequalities are fulfilled. If Newton's inequalities are not fulfilled,$f(z)$cannot have only real roots and the same applies to your original function. For instance, ... 0 Finally, I found the answers to my two questions. Let us start from the first. In this case, the answer is generally negative, as the following construction shows. Let us recursively build a sequence$(f_n)$of continuous functions$f_n: \mathbb{R} \rightarrow \mathbb{R}$,$f_n \geq 0$, with the following properties: (i) for every$n$, the support of$f_n$... 4 The radius of convergence is the smaller radius -- 3. The evens only converge if$|z| < 3$and the odds only converge if$|z| < 5$. Thus, both the evens and odds converge only if$|z| < 3$, and you need both of them to converge for the combined series to converge. Think about it this way, if you let$z = 4$(or any number between 3 and 5), then ... 2 Your answer is correct. One may recall that $$\frac1{1+u}=\sum_{n=0}^\infty (-1)^n u^n, \quad |u|<1,$$ giving, for$2x^2<1$, $$\frac1{1+2x^2}=\sum_{n=0}^\infty (-2)^n x^{2n}$$ that is $$\frac{x}{1+2x^2}=\sum_{n=0}^\infty (-2)^n x^{2n+1}, \quad x \in \left(-\frac{\sqrt{2}}2,\frac{\sqrt{2}}2\right).$$ 1 A more general method: Start with an alternating series$\sum_{k} (-1)^k a_k x^k$(for simplicity of exposition, suppose$a_1 > 0$) for which all the zeroes lie on the real line. Now, add a multiple of$x$, deforming the locations of any surviving zeroes away from the real line (because a zero at$z$now has a multiple of$z$as its value). Or, start ... 5 An analytic function may have coefficients with alternating signs and still violate Newton's inequalities, enforcing a complex root. For instance, $$f(x) = x^2+\sum_{k\geq 0}\frac{(-1)^k}{k!}x^k = e^{-x}+x^2$$ has complex roots at$2\cdot W\left(\pm\frac{i}{2}\right)\approx 0.325199\, \pm 0.785257\, i.$1 Consider $$f(z) = \frac{1-z+z^2}{4+z}= \frac14 - \frac{5}{16}z + \sum_{n>1} (-1)^n\frac{21}{2^{2n+2}}z^n$$ The terms in its expansion about$z=0$are of alternating signs. Yet it has two complex zeros, at $$\frac{1\pm\sqrt{3}i}{2}$$ A simpler example is $$g(z) = \frac{1-z+z^2}{1+z}= 1 - 2z + \sum_{n>1} (-1)^n3z^n$$ 3 1. No. 2. Yes. Consider the function$f$defined on$\mathbb{C}$by the power series $$f(z) = \sum_{n=0}^\infty (-1)^n a_n z^n$$ where$a_0=a_2=1$,$a_1=0$, and$a_n=0$for all$n\geq 2$(so that, indeed,$a_n \geq 0$for all$n\in\mathbb{N}$). We do have that this series is alternating... even though it is a bit trivial. Now, what are the zeroes of$f(z) =...

3

Consider $\sum_{n=1}^{\infty}(2^n\sin y)x^n.$ If $y\in \mathbb R \setminus \pi\mathbb Z,$ then $$\limsup_{n\to \infty}|2^n\sin y|^{1/n} = 2.$$ Hence for those values of $y,$ the radius of convergence is $1/2.$ On the other hand, if $y \in \pi\mathbb Z,$ then the series vanishes identically and the radius of convergence is $\infty.$

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Suppose $f(x)$ is a square wave. $f_n(x)$ is the first n terms of the Fourier series representing f(x). $f_n(x) = \frac 1b \sum_\limits{k=0}^{n} \frac {\sin(2\pi(2k+1) x)}{2k+1}$ with $b>a$ $f_n(x)$ is continuous for all $n.$ $f_n(x)$ converges. In a neigborhood of $0$ $f(x) = \begin{cases}-\frac{\pi}{4b} &x<0\\0&x=0\\\frac{\pi}{4b} &... 1 The sum of a geometric series is$\sum_{n=1}^{\infty}{x}^n$=$\frac{1}{1-x}$where$\mid {x} \mid\le $1 By diffrenciating and multiplying by$x^2$we get$\sum_{n=1}^{\infty}{nx}^{n+1}$=$\frac{x^2}{1-{x}^2}$By letting x =$\frac{1}{3}$, we get :$\sum_{n=1}^{\infty}\frac{n}{3^{n+1}}$=$\frac{1}{4}$And thus$\sum_{n=1}^{\infty}\frac{2n}...

0

Also $\sum_{n=0}^\infty\frac{x^n}{n!}$ has infinite radius of convergence, but of course $$\lim_{x\to\infty}\sum_{n=0}^\infty\frac{x^n}{n!}=\infty$$ It's true that your series has alternating signs, so the example is not of the same kind, but it's too week an argument. Note also that the integral $$\int_0^\infty\frac{\sin x}{x}\,dx$$ is only “...

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The geometric series $f_{z_0}$ has a series expansion with center $z_0=0$ and radius of convergence $R=|a|$ with $a$ a being a simple pole. \begin{align*} f_{z_0}(z)=\sum_{n=0}^\infty\left(\frac{z}{a}\right)^n=\frac{a}{a-z} \end{align*} When we consider a series expansion around another point $z_1$ we know that the radius of convergence is the distance from ...

2

For $|z/a|<1$ the sum is $\frac{1}{1-(z/a)} = \frac{a}{a-z}$. That is the continuation.

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Mean Fix $q \in (0,1)$ and consider any initial condition $X_0 \in \mathbb{R}$. From my above comment, the expectation is $E[X_i] = 2 + (E[X_0]-2)(1/2)^i$ for $i \in \{0, 1, 2, ...\}$. Hence, $$\boxed{E[X_0]=2 \implies E[X_i] = 2 \quad \forall i \in \{0, 1, 2, ...\}}$$ Second moment Fix $q \in (0,1)$ and assume $E[X_0]=2$, so $E[X_i]=2$ for all $i$. ...

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The problem is that Wolfram Alpha interprets "sin(A)" for a matrix A (or array of however many dimensions, or list of list of lists, or what have you) as meaning simply the result of applying sin component-wise. This is not what you intended, and you did your intended calculation perfectly fine.

3

Since \begin{align} \tan(\arctan(1+x)-\arctan(1)) &=\frac{(1+x)-1}{1+(1+x)\cdot1}\\ &=\frac x{2+x}\tag{1} \end{align} we have \begin{align} \arctan(1+x) &=\frac\pi4+\arctan\left(\frac{x}{2+x}\right)\\ &=\frac\pi4+\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\left(\frac x{2+x}\right)^{2k+1}\tag{2} \end{align} Expanding $(2)$ using the ...

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Hint: Some transformations to make the expression easier manageable. We obtain \begin{align*} \sum_{i=0}^{a-2}&(-1)^{a+b-i-2}(a+b-i-1)x^{a+b-i-2}\\ &=\sum_{i=0}^{a-2}(b+1+i)(-x)^{b+i}\tag{1}\\ &=\sum_{i=b+1}^{a+b-1}i(-x)^{i-1}\tag{2}\\ &=-D_x\sum_{i=b+1}^{a+b-1}(-x)^i\tag{3}\\ &=D_x\left(\frac{(-x)^{a+b}-(-x)^{b+1}}{1+x}\right)\tag{...

1

Applying the formula for the binomial series representation we obtain \begin{align*} \frac{x}{(1+6x)^2}&=x\sum_{n=0}^\infty\binom{-2}{n}(6x)^n\tag{1}\\ &=x\sum_{n=0}^\infty\binom{n+1}{1}(-6x)^n\tag{2}\\ &=\sum_{n=0}^\infty(n+1)(-6)^nx^{n+1}\\ &=\sum_{n=1}^\infty(-6)^{n-1}nx^n\qquad\qquad\qquad |x|<\frac{1}{6}\tag{3} \end{align*} ...

1

OK, let us start from the well-known formula $$\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^n, \quad |x|<1 \tag{1}$$ Next we obtain the series expansion \begin{align} \frac{1}{(1+x)^2} &= -\frac{d}{dx}\frac{1}{1+x} = - \frac{d}{dx} \sum_{n=0}^{\infty}(-1)^nx^n \\ &= -\sum_{n=0}^{\infty}n(-1)^n x^{n-1} = \sum_{n=0}^{\infty}n(-1)^{n+1}x^{n-1} \\ &... 0 Observe that : \dfrac{x}{(1+6x)^2}= x\dfrac{d}{dx}\left(-\dfrac{1}{6(1+6x)}\right),and you have \dfrac{1}{1+6x} = 1+(-6x) + (-6x)^2 +\cdots , |x| < \frac{1}{6} 1 So the distribution is a singular multi-fractal measure. This one looks close in appearance to the singular binomial measure. Here are some introductions to multi-fractals and the measure. Rice: Intorduction to Multifractals Rudolf H. Reidi-www.stat.rice.edu/~riedi/Publ/PDF/intro.pdf Rice: Binomial Multifractals-www.stat.rice.edu/~riedi/Publ/TALKS/... 1 You need to evaluate\lim_{k\to\infty}\sup\sqrt[k]{\left|\frac{c_k}{k+1}\right|}=\lim_{k\to\infty}\sup\sqrt[k]{\left|\frac{c_k}{k+1}\right|}=\lim_{k\to\infty}\sup\sqrt[k]{|c_k|}\cdot\lim_{k\to\infty}\sqrt[k]{\frac1{k+1}}$$The last equality is justified since both lim sup exist, though the right one is just limit as it exists and equals one. This is the ... 6 Impossible is nothing. By termwise integration of the Taylor series of \frac{1}{1+x^2} centered at x=1 we easily get the Taylor series of \arctan(x) centered at x=1. Its radius of convergence is \sqrt{2} since the closest singularities to x=1 lie at \pm i.$$\arctan(x)=\frac{\pi}{4}+\sum_{m\geq 0}\left(\frac{(-1)^m(x-1)^{4m+1}}{2^{2m+1}(4m+1)}-...

1

The width of a fret whose number is $n$ is given by \begin{align} d(n)-d(n-1)&=\left(s-\frac{s}{2^{\frac{n}{12}}}\right)-\left(s-\frac{s}{2^{\frac{n-1}{12}}}\right)\\ &= \frac{s}{2^{\frac{n}{12}}}\left(2^{\frac{1}{12}}-1\right). \end{align} Thus, the percentage width, relative to the length of the string, is given by \begin{align} \frac{...

0

Denote $a_n=\frac{(-5)^n}{n\sqrt n}x^n$. Then $\frac{a_{n+1}}{a_n}=5|x|\frac{n\sqrt n}{(n+1)\sqrt {n+1}} \rightarrow 5|x|$. Thus the radius of convergence is $(-\frac{1}{5},\frac{1}{5})$. Note, however, that when $|x|=\frac{1}{5}$, the series is absolutely convergent (recall that $\sum_{n=0}^{\infty}\frac{1}{n \sqrt n}$ converges). So the interval includes ...

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Using the cauchy hadamard theorem for a power series $\sum_{n=0}^{\infty}a_nx^n$: $$1/R=\lim_{n\rightarrow \infty}\sup |a_n|^{1/n}$$ You have for your problem: $$1/R=\lim_{n\rightarrow \infty}\sup |\frac{(-5)^n}{n\sqrt{n}}|^{1/n}=5\lim_{n\rightarrow \infty}n^{-1/n^2}$$ $$=5*1=5\Rightarrow R=1/5$$ The last limit I evaluated using the log.

0

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$\mathrm{sech}(x)$ is defined everywhere on the real axis, but in the complex plane it has singularities at $z=\pm \frac{i\pi}{2}$. So the radius of convergence cannot be larger than $\frac{\pi}{2}$.

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Easiest reason is in complex analysis. $\mathrm{sech}(x)$ has a pole at $i\pi/2$, so the radius of convergenge cannot extend beyond that point. And moreover, there are no singularities strictly closer to $0$, so the radius of convergence is exactly $\pi/2$.

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Hint: use the fact that if the series converges absolutely, then it converges conditionally and the ratio test.

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Note that $\left(\frac{1+ z}{1- z}\right)^2$ is defined everywhere except at z= 1. Since the "center" is z= 0, the radius of convergence is obviously 1.

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Using algebraic manipulation and the geometric series you get $$\left(\frac{1+z}{1-z}\right)^2 = \frac{(1+z)^2}{(1-z)^2}=\frac{1+2z+z^2}{1-z)^2}=\frac{1-2z+z^2 + 4z}{(1-z)^2} = 1+ \frac{4z}{(1-z)^2}$$ $$=1 + 4z\frac{d}{dz}\left(\frac{1}{1-z}\right) =1 + 4z\frac{d}{dz}\left(\sum_{k=0}^{\infty}z^{k}\right) =1 + 4z\sum_{k=1}^{\infty}kz^{k-1} = 1 + \sum_{k=1}^{... 2 As suggested by @lab bhattacharjee, one may use the ratio test, as n \to \infty, giving$$ \frac{u_{n+1}}{u_n}=\frac{(n+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{n^n}=\left(1+\frac1n\right)^n \to e.  Can you take it from here?

3

The ring $\mathbb{F}_p[t]$ consists of polynomials with coefficients in the finite field of order $p$: these are finite expressions of the form $\sum_{i=0}^n a_it^i$, with $a_i \in \mathbb{F}_p$. In contrast, the ring of formal power series allows for infinite expressions: a typical element looks like $\sum_{i=0}^\infty a_i t^i$. So this ring strictly ...

0

As rightly answered first by Nick and as reflected by others, Taylor's Polynomial (T) is a polynomial which looks just like the given function f(x) i.e. resembles/ approximates the real function as close as possible. T about a point A is the value given by it when x=a. Since the T is just an approximation polynomial to the real function f(x), there will be ...

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