New answers tagged

0

Let $$S_m=\sum_{k=m}^\infty kr^k=\sum_{k=m-1}^\infty (k+1)r^{k+1}=mr^m+r\sum_{k=m}^\infty (k+1)r^k.$$ Then $$S_m-mr^m-rS_m=r\sum_{k=m}^\infty r^k=\frac{r^{m+1}}{1-r}$$ and $$S_m=\frac{r^m(m-(m-1)r)}{(1-r)^2}.$$ So $$\sum_{k=1}^\infty kr^k=S_1=\frac r{(1-r)^2}$$ and $$\sum_{k=1}^n kr^k=S_1-S_{n+1}=\frac{r-r^{n+1}(n+1-nr)}{(1-r)^2}.$$


0

Here's an approach that requires the value of the geometric series $\sum_{n=0}^\infty x^n = \frac{1}{1-x}$ for $|x|<1$, and termwise differentiation of power series. To compute the value of \begin{align} S&= \sum_{n=1}^\infty n\left(\frac{1}{3}\right)^n, \end{align} define, for $|x|<1$ $$f(x) = \sum_{n=0}^\infty x^n = \frac{1}{1-x}$$ Then note ...


2

You need to solve the second sum first. It runs this way: you learnt in high school the factorisation formula $$1-x^n=(1-x)(1+x+\dots+x^{n-1}),$$ which can be rewritten as $$\frac1{1-x}=1+x+\dots+x^{n-1}+\frac{x^n}{1-x}.$$ From this you deduce that the sum $\;\displaystyle\sum_{i=0}^n x^i$ has a limit (‘the series converges’) if and only if $\lvert x\...


3

Here is an approach that relies on the relationships (i) $k=\sum_{\ell=1}^k(1)$ and (ii) $\sum_{k=\ell}^N r^k=\frac{r^{\ell}-r^{N+1}}{1-r}$ for $|r|<1$. Then, with $r=3^{-1}$ we have $$\begin{align} \sum_{k=1}^N \frac{k}{3^k}&=\sum_{k=1}^N 3^{-k}\sum_{\ell =1}^k(1)\\\\ &=\sum_{\ell =1}^N \sum_{k=\ell}^N 3^{-k}\\\\ &=\sum_{\ell =1}^N \frac{3^...


2

You need to know off by heart $$\frac{1}{1-x}=1+x+x^2+x^3+\dots$$ a formula that is constantly coming up in all areas of maths. You can easily prove it by comparing the sum $s$ with the sum $x\cdot s$. Ideally you would also remember $$\frac{1}{(1-x)^2}=1+2x+3x^2+4x^3\dots$$ which is also extremely useful. Since the tags suggest you have some calculus, the ...


0

Where does +c10^m come from? $$x^l={\overline{a\dots aba\dots a}}_{(10)}={\overline{a\dots a}}_{(10)}+(b-a)\cdot 10^m=a\frac{10^{n}-1}{9}+c\cdot 10^m$$ Why $b=1$ if $l \neq 3$, and $b\in \{1,8\}$ if $l = 3$? When $a=0$, we have $x^l=b\cdot 10^m$. Note here that $1^3=1,2^3=8,i^3\gt 10$ for $i\ge 3$, and that $1^4=1, j^4\gt 10$ for $j\ge 2$.


0

Which function of $x$, other than $x +c$, and Integral of $(\cos^2 x+\sin^2 x)$ have derivative $=1$ ?. They are an infinity of equivalent forms : Any integral of function which is an identity to $1$. For example : $\int (\cosh^2 x-\sinh^2 x)dx$ Any function which is an identity to $x$. For example : $\frac{x^2+x-1}{x+1}+\frac{1}{1+x}$ A more ...


0

The first half of the Fundamental Theorem of Calculus is that if $F(x)=\int_0^x f(y)\;dy$ and $f$ is continuous then $F'=f.$ The second half is that if $f$ is continuous and $F'=f$ then $G(x)=F(x)-\int_0^xf(y)\;dy$ is constant. The first half implies that $G'=F'-f=0,$ so to prove the 2nd half we must then show that if $G'=0$ then $G$ is constant: (1). ...


0

Any function in the form : $$f(x) = x+C$$ Why? because it is the solution of the following differential equation: $$\dfrac{df}{dx}=1 \Rightarrow df = dx$$ by integration we get: $$f(x) = x + C$$ This is the only possible solution (see the uniqueness of solution for a differential equation) NOTE: this holds only for continuous functions. EDIT: due to the ...


1

An elegant method is to use the Expansion Theorem in Umbral Calculus. Below is a typical statement, from p. 18 of Roman's book The Umbral Calculus.


1

$$e^{-1}+4e^{-9}+9e^{-25}+16e^{-49}\approx 0.36837308051278053458657911933771842$$ should be good enough. The truncation error is $$\sum_{i=4}^\infty(i+1)^2e^{-(2i+1)^2}=\sum_{i=4}^\infty(i+1)^2e^{-4i^2-4i-1}<\sum_{i=4}^\infty(i+1)^2e^{-64-4i-1}<2\cdot10^{-34}$$ as can be computed analytically. You can obtain this estimate from $$\sum_{i=n}^\infty ...


1

The Taylor expansion of $\exp\sin x$ around zero is $1+x+x^{2}/2+O(x^{4})$. Therefore, the error is \begin{align*} \left|\exp\sin x-(1+x+x^{2}+x^{3})\right| & =\left|-x^{2}/2+O(x^{3})\right|\\ & \leq|x^{2}|/2+|O(x^{3})|\\ & \approx|x^{2}|/2 & \text{for }|x|\text{ small}. \end{align*}


0

As you noted the Taylor series for $f(x) = e^x$ is $f(x) \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$. Now plug in $\sin x$ and use the fact that for values close to $x$ we have $\sin x \approx x$. So we have: $$e^{\sin x} \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$$ Now the error of the first approximation is around $\frac{x^2}{2} + \frac{5x^3}{6}$. ...


2

$$\arctan(x)=\sum_{n\geq 0}\frac{(-1)^n}{2n+1} x^{2n+1}\tag{1} $$ $$\arctan(x)-x=\sum_{n\geq 1}\frac{(-1)^n}{2n+1} x^{2n+1}\tag{2} $$ $$\frac{\arctan(x)-x}{x^3}=\sum_{n\geq 1}\frac{(-1)^n}{2n+1} x^{2n-2}=\color{red}{\sum_{n\geq 0}\frac{(-1)^{n+1}}{2n+3}x^{2n}}\tag{3} $$ The radius of convergence (i.e. $1$) is left unchanged by our manipulations.


1

HINT: Note that we have $$\begin{align} \frac{d}{dx}\arctan(x)&=\frac{1}{1+x^2}\\\\ &=\sum_{n=0}^\infty (-1)^n x^{2n} \tag 1 \end{align}$$ for $|x|<1$. Then, integrate term by term to arrive at a series for $\arctan(x)$. SPOILER ALEERT: Scroll over the highlighted area to reveal the solution


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You can use this equation to get all kinds of power series expansions for Ln(x+y+1). The equation adapts itself because you may chose to hold x or y constant and integrate it. Study the equation and play with it. It yields all kinds of neat results, Including some very important results. Hint let x=X^2. It provides insight into series. $$\sum_{n=0}^\...


3

Where did you find that equation? It's quite different from what I got, which I shall explain now. First, a common power series is $$ \frac{1}{1-x} = \sum_{i\geq0} x^{i}. $$ Using the substitution $x=-t^{n}$, $$ \frac{1}{1+t^{n}} = \sum_{i\geq0} (-t^{n})^{i} = \sum_{i\geq0} (-1)^{i}t^{in}. $$ Then, $$ \int_{0}^{x} \frac{1}{1+t^{n}} dt = \int_{0}^{x} \...


0

To compute this series, let us look at $$ \sum_{k=1}^{\infty} \frac{k^{k-1}}{k!} (- z e^z)^k = \sum_{k=1}^{\infty} \frac{(-1)^k k^{k-1}}{k!} z^k e^{kz} $$ as a formal power series. This series can be rewritten in the form $$ \sum_{m=1}^{\infty} \alpha_m z^m. $$ where \begin{align*} \alpha_m &= \sum_{k=1}^m \frac{(-1)^k \, k^{k-1}}{k!} (\text{coef. by }...


2

Because the $B(w_i,r_{w_i})$'s cover $frB(0,R)$, the $B(w_i,r_{w_i})$'s and $B(0,R)$ cover the clausure of $B(0,R)$. This is a open covering so it also covers $B(0,R+\varepsilon)$ for some $\varepsilon>0$. Now you have by definition that the radius of convergence of $F$ is $S>R$. This leads to a contradiction with the Cauchy-Hadamard formula because ...


0

You reach this equation: $$ \sum_{n=0}^\infty(b-n)a_nx^n=0 $$ So $(b-n)a_n=0$ for every n. Hence, if b is not a natural number, then all $a_n$ are 0. And if $b$ is an integer then $a_b$ can be chosen arbitrarly so the solution would be: $$ y=\left\{\begin{array}{cc}a_bx^b & b\in\mathbb{N}\\0&\text{otherwise} \end{array}\right. $$


2

$$\arctan(x) = \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)}x^{2n+1} $$ for any $x\in(-1,1)$, so: $$ \frac{1}{10}-\frac{1}{3000}<\arctan\left(\frac{1}{10}\right) < \frac{1}{10}-\frac{1}{3000}+\frac{1}{500000}$$ and we may take $k$ as: $$ k = \left\lfloor \frac{100}{3}\cdot 299\right\rfloor = \color{red}{996}.$$


1

You're on the right track, but a few things need to be mentioned. First, if the series involves complex numbers then you're actually looking for the disk of convergence, not the interval. Second, your notation is very inappropriate. When you apply the root test, you're applying it to $a_n$ only. Including the summation symbol is incorrect. Third, your ...


1

This is meant to elucidate the A given By Arctic Char because of the long chat with the OP that followed it. $f_n\to f$ uniformly on a domain $D$ iff $$\lim_{n\to \infty}\|f-f_n\|=0,$$ where $$\|f-f_n\|=\sup_{x\in D}|f(x)-f_n(x)|.$$ Suppose $f_n\to f$ uniformly on $D$ then also $\lim_{n\to \infty}\|f_{n+1}-f\|=0.$ Then we have $$\|f_{n+1}-f_n\|\leq \|f-...


2

Being a finite sum, this one converges for every $x \in \Bbb C$, so its radius of convergence is $\infty$.


4

This is basically just a guess, but I would expect the limit to be $\infty$. Since $n \to \infty$, $\frac 1 n \to 0$, so we can ignore that term, giving us: $$\lim_{n \to \infty} \frac{n^n}{n^n} t^n=\lim_{n \to \infty} t^n=\infty$$


3

Hint Write $$\sum_{n=0}^\infty\frac {(-1)^n x^n}{(y+1)^{n+1}}=\sum_{n=0}^\infty\frac {(-x)^n }{(y+1)^{n+1}}=\frac 1{y+1}\sum_{n=0}^\infty\left(\frac {-x}{y+1}\right)^n=\frac 1{y+1}\sum_{n=0}^\infty a^n$$ where $a=-\frac {x}{y+1}$. I am sure that you can take it from here.


5

Let's make it look nice. ∑[(x^(2n+1))/((x^2+1)^(n+1))]*[(2n!!)/(2n+1)!!] You say $\arctan(x) =\sum\dfrac{x^{2n+1}}{(x^2+1)^{n+1}}\dfrac{(2n)!!}{(2n+1)!!} $ Since $(2n+1)!! =\prod_{k=1}^n (2k+1) =\dfrac{\prod_{k=1}^n (2k)(2k+1)}{\prod_{k=1}^n (2k)} =\dfrac{(2n+1)!}{2^nn!} $ and $(2n)!!=2^nn! $, this becomes $\arctan(x) =\sum\dfrac{x^{2n+1}}{(x^2+1)^{n+1}}...


3

This is the unfolding by the generalized binomial theorem of $$ \Bigl(1-\frac{x}{1+x}\Bigr)^{-p-1} = (1+x)^{p+1}$$


2

The generalized binomial theorem states that $(1+z)^a =\sum_{n=0}^{\infty} \binom{a}{n} z^n $. Therefore $\sum_{n=0}^{\infty} \binom{-p-1}{n} (\frac{-x}{1+x})^n =(1-\frac{x}{1+x})^{-p-1} =(\frac{1}{1+x})^{-p-1} =(1+x)^{p+1} $. I am not worrying about convergence.


1

Note that $$(x+2)^{1/2}=(4+(x-2))^{1/2}=2\left(1+\frac{x-2}{4}\right)^{1/2}.$$ Now use what you know about the expansion of $(1+y)^{1/2}$.


0

$$ f_n(x):=n^2\sin{x\over n^4}\;\forall\; n $$ Claim: $\;0<M<\infty\implies\sum_{n=1}^{\infty}f_n$ converges uniformly in $[-M,M]$. Proof: Suppose $x\in[-M,M]$. $$ |\sum_{n=p}^q f_n(x)|\le\sum_{n=p}^q|f_n(x)|\le\sum_{n=p}^q n^2{|x|\over n^4}\le\sum_{n=p}^q{M\over n^2}=M\sum_{n=p}^q {1\over n^2}$$ So given $\varepsilon>0$ we can find $k(\...


0

Using the Taylor series directly is not a very motivated or clean way of showing the additivity of $\log$, but here's one approach. It's clearly sufficient to prove that $\exp(x + y) = (\exp x)(\exp y)$. The function $y = \exp(x)$ is the unique solution of the differential equation $y' = y$ with $y(0) = 1$. (The uniqueness result here is standard and easy ...


5

$$\sum_{n\geq 1}\frac{(2n-1)!!}{(2n)!!}x^n = \sum_{n\geq 1}\frac{(2n-1)!}{4^n n!^2}x^n \color{blue}{=} \sum_{n\geq 1}\binom{-1/2}{n}(-1)^n x^n = \color{red}{\frac{1}{\sqrt{1-x}}-1}.$$ Details of $\color{blue}{=}$: $$\binom{-1/2}{n}(-1)^n = \frac{(-1/2)(-3/2)\cdots(-(2n-1)/2)}{n!}(-1)^n = \frac{(2n-1)!!}{2^n n!}=\frac{(2n-1)!!}{(2n)!!}.$$ As an alternative ...


4

If it converges uniformly, then $$\sup_{x\in \mathbb R} \left|n^2 \sin(x/n^4)\right|\to 0$$ as $n\to \infty$. This is clearly false.


0

Any counterexample can do, $$-\log(1-0)\ne\log(1+e^0),$$ $$0\ne\log(2).$$


2

The relation surely doesn't hold for every $x$: for instance, if $x=-1$ you have $$ -\log(1-(-1))=-\log2<0, \qquad \log(1+e^{-1})>0 $$ It may be interesting looking for what values of $x$ equality holds. First we have to assume $1-x>0$, that is, $x<1$, for the left-hand side to exist. Once ensured this, we can write the equality as $$ \log\frac{...


1

It is clearly false because $\log(1+\exp(x))$ exists for all real $x$ while $-\log(1-x)$ does not.


0

For another solution approach using beta function, see my detailed solution at Prove $\sum_{n=0}^{\infty}{2^n(n^2-n\pi+1)(n^2+n-1)\over (2n+1)(2n+3){2n\choose n}}=1$ . The same steps can be followed with little changes in the coefficients. Using Beta function, we can express $${2^n n[n(\pi^3+1)+\pi^2](n^2+n-1)\over (2n+1)(2n+3){2n\choose n}} = {n[n(\pi^3+1)+...


1

$\textbf{Claim}$: Suppose $f$ is analytic in a domain $\Omega$. Then, $f$ has a primitive in $\Omega$ iff $\int\limits_{\gamma}{}f=0$ for every simple closed curve $\gamma \in \Omega$. In your case, every simple closed curve $\gamma$ in $\Omega=\{z:|z|>1\}$ is homotopic to either a contractible loop or a loop around $0$, which we can assume to be the ...


3

You are right. $\sin(z)$ is not only an analytic function, but an entire function: $$ \sin(z)=\sum_{n\geq 0}\frac{(-1)^n z^{2n+1}}{(2n+1)!} \tag{1}$$ holds as an identity for any value of $z\in\mathbb{C}$. It follows that: $$ \sin(x^3 y^2) = \sum_{n\geq 0}\frac{(-1)^n x^{6n+3}y^{4n+2}}{(2n+1)!}\tag{2}$$ holds as an identity for every value of $(x,y)\in\...


4

For clarity put $w=z-i$. Then $$\frac{1}{1+z^2}=\frac{1}{w(w+2i)}=-\frac{i}{2w}\frac{1}{1-\frac{iw}{2}}$$ Now use the familiar expansion $$\frac{1}{1-\frac{iw}{2}}=1+i\frac{w}{2}-\frac{w^2}{4}-i\frac{w^3}{8}+\dots$$ to get $$-\frac{i}{2w}+\frac{1}{4}+i\frac{w}{8}-\frac{w^2}{16}-i\frac{w^3}{32}+\frac{w^4}{64}+i\frac{w^5}{128}-\dots$$


1

Another approach to this problem uses Poor man's Lagrange Inversion, which is the Cauchy Residue Theorem. We have $$(27x-4)T(x)^3+3T(x)+1 = 0$$ We thus obtain $$[z^n] T(z) = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} T(z) \; dz.$$ Solve for $z$ to get $$z = \frac{4T(z)^3-3T(z)-1}{27 T(z)^3}.$$ Therefore the substitution $w=T(z)$ yields $$...


2

By the Lagrange inversion theorem, the solution of $w^3-w=-x$ has the following Taylor series: $$ w(x)=\sum_{k\geq 0}\binom{3k}{k}\frac{x^{2k+1}}{2k+1} $$ whose radius of convergence is $\frac{2}{3\sqrt{3}}$. If we set $y(x)=\sqrt{\frac{3}{4-27 x}}\,w(x)$, by Lagrange inversion: $$ y(x)=\sum_{k\geq 0}\binom{3k}{k}\frac{3(-1)^k}{(2k+1)(4-27x)^{k+1}}\tag{1}...


2

I believe our definitions of the error function differ by a constant, but the following approach works anyway: $$\begin{eqnarray*}\sum_{k\geq 0}\frac{x^k}{k!!}&=&\sum_{n\geq 0}\frac{x^{2n}}{2^n n!}+\sum_{n\geq 0}\frac{x^{2n+1}}{(2n+1)!!}\\&=&e^{x^2/2}+\sum_{n\geq 0}\frac{2^n x^{2n+1}}{(2n+1)!}n!\\&=&e^{x^2/2}+\int_{0}^{+\infty}e^{-z}\...


1

If you don't want to leave the formal world, you still have a very simple alternative, namely to work in an extension of formal power series. A formal power series is a sequence of coefficients that is added and multiplied as if it were an (infinite-degree) polynomial in some indeterminate, say $X$. Let us denote this structure by $F^*[X]$, analogously to ...


1

You can do that without leaving the formal world but it is often harder. In general, the formula (or the definition) for the composition $f(g(x))$ when $g(x) = \sum_{n=0}^{\infty} b_n x^n$ and $f(x) = \sum_{n=1}^{\infty} a_n x^n$ is given by $$ f(g(x)) = \sum_{n=1}^{\infty} a_n \left( \sum_{m=1}^{\infty} b_m x^m \right)^n = \sum_{l=0}^{\infty} c_l x^l, \\ ...


1

We have $$f(g(x)) = \sum_{k = 0}^\infty \left( \dfrac x {1 - x} \right)^k .$$ Now, for $k \ge 1$, using $\dfrac 1 {(1 - x)^k} = \sum\limits_{m = 0}^\infty \binom{m + k- 1}{m} x^m$, we get \begin{align*} f(g(x)) & = 1 + \sum_{k=1}^\infty x^k \sum_m \binom{m + k - 1}{m} x^m\\ & = 1 + \sum_{k=1}^{\infty} \sum_m \binom{m + k - 1}{m} x^{m +k}. \end{...


1

One approach is to use the differential equation $$y'' + y = 0$$ satisfied by $y = \sin x$. The differential equation has unique solution $$y = y(0)\cos x + y'(0)\sin x$$ Now consider the power series $$f(x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \cdots$$ then $f(x)$ is defined for all $x$ (because the series is convergent for all $x$) and by ...



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