New answers tagged

1

If $a>1$. Let $\sqrt[n]{a}=1+h_n $, where $h_n>0$. Therefore $a=(1+h_n)^n=1+nh_n+\frac{n(n-1)}{1.2}h_{n^2}+ \dots +h_{n^n}$ . $>1+nh_n, \forall n$ $h<\frac{a-1}{n} , \forall n$ Let $\epsilon$ be any positive number then, $|h_n|=h_n<\frac{a-1}{n}<\epsilon$, then Let m be any positive integer,$>\frac{a-1}{\epsilon}$ $|\sqrt[n]{a}|=|...


0

A series is defined as $$\sum_{k=c}^{\infty}b_k=\lim_{n\to \infty}\sum_{k=c}^{n}b_k$$ And we can define a partial geometric sum as $$S=\sum_{k=0}^{n}ax^k=a\sum_{k=0}^n x^k$$ To see the relationship of your question we must notice that $$S=xS+a-ax^{n+1}\to S-xS=a(1-x^{n+1})\to S=a\frac{1-x^{n+1}}{1-x}$$ Then we have the relationship $$S=\sum_{k=0}^n ax^...


2

It is not true. It is a geometric series and for $x \gt 1$ we have $$\sum_{n=1}^\infty \frac{1}{x^{n}} = \frac1{x-1}$$ For $x$ large, the sum is clearly less than $1$, while your expression has the right side much larger than $1$ The question is changed, and is now equivalent to the above. The informal way to see it is $$S=\frac 1x + \frac 1{x^2}+\frac ...


0

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1

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 $\infty.$


-2

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 “...


1

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.


1

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 $i$. ...


1

Since $a_{n+2}=\frac{2n(3n+2)}{(n+3)(2n-5)}\,a_n$, with $a_1=0$, we find that all of the odd terms in the series are zero. Let $b_n=a_{2n}$ so that $y=\sum_{n=0}^\infty b_nx^{2n}$. Then, $b_{n+1}=\frac{4n(6n+2)}{(2n+3)(4n-5)}b_n$ and the ratio test gives $$\lim_{n\to \infty }\left(\frac{b_{n+1}x^{2n+2}}{b_nx^{2n}}\right)=3x^2<1$$ whenever $|x|<1\...


0

$$\frac{x}{15x^2+1}=\frac{x}{1+(\sqrt{15}x)^2}$$ now we can depend on the geometric series which is $$\frac{1}{1+x}=\sum_{n=0}^{\infty }(-1)^nx^n$$ let $x\rightarrow x^2$ $$\frac{1}{1+x^2}=\sum_{n=0}^{\infty }(-1)^nx^{2n}$$ to get the power series let $x\rightarrow \sqrt{15}x$,so $$\frac{x}{1+(\sqrt{15}x)^2}=x\sum_{n=0}^{\infty }(-1)^n(\sqrt{15}x)^{2n}=x\...


0

Rewrite the given function in the form of $$\frac{a}{1-r}$$ so you can express the representation as a geometric series. A geometric series takes the form of $$ \sum_{n=0}^{\infty } a(r)^n$$


0

It should be $$x\sum_{n=0}^\infty (-1)^n 15^n x^{2n}=\sum_{n=0}^\infty (-1)^n 15^n x^{2n+1}$$ and not $$\sum_{n=0}^\infty (-1)^n (15 x^{2})^{n+1}$$


1

By integrating the geometric series term by term, you can easily obtain: $$f(z) = -\log(1-z)$$ However, if you cannot sum a series into a closed form expression, you can sometimes still analytically continue the series to another series with a larger radius of convergence. E.g. if you substitute $z = \frac{w}{2-w}$ and re-expand in powers of $w$, what is ...


1

Another way to check:the Cauchy-Hadamard theorem simply gives a formula for the radius of convergence: https://en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard_theorem


0

Complex Analysis I think the subject becomes clear when viewed using complex analysis. In this Wikipedia article it is stated that The radius of convergence of a power series $f$ centered on a point $a$ is equal to the distance from $a$ to the nearest point where $f$ cannot be defined in a way that makes it holomorphic. This can be proven using Cauchy'...


1

Simplest bounds are: for $j > 0$ $$ \dfrac{n^{j+1}}{j+1} = \int_0^{n} x^j \; dx < \sum_{k=1}^n k^j \le \int_1^{n+1} x^j\; dx = \dfrac{(n+1)^{j+1}-1}{j+1}$$ Tighter bounds can be obtained by using a few terms of Faulhaber's formula. Thus if $j \ge 4$ $$ \sum_{k=1}^n k^j = \frac{n^{j+1}}{j+1} + \frac{1}{2} n^j + \frac{j}{12} n^{j-1} - \frac{j(j-1)(j-...


0

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2

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 ...


1

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 ...


0

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

Try using the Cauchy-Hadamard Theorem: https://en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard_theorem


0

$\mathrm{sech}(x)$ 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 $\frac{\pi}{2}$.


2

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 $\pi/2$.


0

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


1

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.


2

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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