# Tag Info

4

I do not know how this is teached but I think that you make the problem difficult changing the starting value of the index in the summation. Let me try $$y=\sum_{n=0}^{\infty}a_n x^n$$ $$y'=\sum_{n=0}^{\infty}n a_n x^{n-1}$$ $$y''=\sum_{n=0}^{\infty}n(n-1) a_n x^{n-2}$$ Rewrite the equation as $$(x^2 + 1)y'' - 6xy' + 10y =x^2y''+y''-6xy'+10y=0$$ and include ...

4

It's actually not so difficult. Using: $$\frac{1}{1-x-y}=\sum_{n,m\ge 0}{n + m\choose n}x^ny^m$$ You get $$\frac{x}{1-x-x^3}=x\cdot\frac{1}{1-x-x^3}=x\cdot \sum_{n,m\ge 0}{n + m\choose n}x^n(x^3)^m=\sum_{n,m\ge 0}{n + m\choose n}x^{n+3m+1}$$ And after a last manipulation you can get a formal power series of the classical form $\sum a_nx^n$

3

$$\frac{1}{1-x}=1+x+x^2+x^3+x^4+...$$ $$x(\frac{1}{1-(x+x^3)})=x(1+(x+x^3)+(x+x^3)^2+(x+x^3)^3+.....)$$

3

$$e^{x^2-1}=\frac1e\sum_{n=0}^\infty\frac{(x^2)^n}{n!}=\cdots$$

3

Using Stirling Numbers of the Second Kind, we have $\newcommand{\stirtwo}[2]{\left\{{#1}\atop{#2}\right\}}$ $$k^m=\sum_{j=0}^m\stirtwo{m}{j}\binom{k}{j}j!\tag{1}$$ Therefore, \begin{align} \sum_{k=0}^\infty k^mz^k &=\sum_{k=0}^\infty\sum_{j=0}^m\stirtwo{m}{j}\binom{k}{j}j!z^k\\ &=\sum_{j=0}^m\stirtwo{m}{j}j!\sum_{k=j}^\infty\binom{k}{j}z^k\\ ... 3 For the sake of simplicity, assume that  f  is entire. Then \begin{align} \forall (x,y) \in \mathbb{R}^{2}: \quad f(x + i y) & = \sum_{m = 0}^{\infty} a_{m} (x + i y)^{m} \\ & = \sum_{m = 0}^{\infty} \left[ a_{m} \sum_{n = 0}^{m} \binom{m}{n} x^{n} (i y)^{m - n} \right] \\ & = \sum_{m = 0}^{\infty} \left[ \sum_{n = 0}^{m} a_{m} i^{m ... 2 EDIT: Sorry, the previous answer was not correct. You have \Bigg|\frac{x^8-1}{3}\Bigg| <1 \Leftrightarrow |{x^8}-1|<3 \Leftrightarrow |x|<4^{1/8} =\sqrt[4]{2}. $$In these examples you typically want to work without getting rid of the absolute value, since the x in your series may also be considered as complex number; and those have no ... 2 Suppose we seek to evaluate$$\sum_{k\ge 1} k^m z^k.$$Put$$k^m = \frac{m!}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{m+1}} \exp(kw) \; dw \\ = \frac{m!}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{m+1}} \sum_{q=0}^k {k\choose q} (\exp(w)-1)^q \; dw \\ = \frac{m!}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{m+1}} \sum_{q=0}^k \frac{k!}{(k-q)!} ...

2

Start from: $$\sum_{k\geq 1}z^k = \frac{z}{1-z}$$ then apply the operator $\varphi: f(z)\to z\cdot f'(z)$ to both terms $m$ times. That gives that for any $|z|<1$ we have: $$\sum_{k\geq 1}k\,z^k = \frac{z}{(1-z)^2},\quad \sum_{k\geq 1}k^2 z^k=\frac{z+z^2}{(1-z)^3}$$ and so on.

2

I think that the easiest way to handle $\frac{x}{e^x-1}$ is to write out the Taylor expansion of $e^x$, and use long division to compute the coefficients term by term. For $\frac{1}{1-\sin(x)}$, one needs to avoid points such that $\sin(x) =1$, but supposing that $|\sin(x)|<1$, we can use the geometric series and the Taylor expansion of $\sin(x) = ... 2 By ratio test $$\lim_{n\rightarrow\infty}\left|\frac{x^{n+1}n!\left(n+1\right)^{3}}{\left(n+1\right)!x^{n}n^{3}}\right|=\lim_{n\rightarrow\infty}\left|\frac{x}{n+1}\right|=0$$ so the radius of convergence is$\infty$and the interval of convergence is$\left(-\infty,\infty\right).$2 Just as in the expansion for$e^x$you can expand$e^{f(x)}$to $$e^{f(x)} \;\; =\;\; \sum_{n=0}^\infty \frac{(f(x))^n}{n!}.$$ For safety, I would just use functions$f(x)$which are differentiable. 2 I would say this is a very difficult thing to do (from which you can deduce that this is not going to be a very satisfactory answer...). Having just the power series about one point, the radius of convergence tells you the distance from the point of expansion of the first point at which the function fails to be analytic. This essentially divides into poles, ... 2 To answer both your old and your new question at the very same time, we can consider a surprising relationship between power series and recurrence relations through generating functions. As a simple example, consider representing$\frac{1}{1-x}$as a power series. In particular, we want to discover an$f_n$such that ... 2 The coefficient of$x^n$in the product $$(1+x)(1+x^2)(1+x^3)\cdot\ldots$$ gives the number of ways of writing$n$as a sum of distinct natural numbers, while the coefficient of$x^n$in the product: $$\frac{1}{1-x}\cdot\frac{1}{1-x^3}\cdot\ldots = (1+x+x^2+\ldots)(1+x^3+x^6+\ldots)\cdot\ldots$$ gives the number of ways to write$n$as a sum of different ... 1 Hint: Developing the two expressions with just enough terms to get the exact$x^7$coefficient, $$\frac1{1-x}\frac1{1-x^3}\frac1{1-x^5}\frac1{1-x^7}\\ =(1+x+x^2+x^3+x^4+x^5+x^6+x^7)(1+x^3+x^6)(1+x^5)(1+x^7)\\ =\color{green}1+\color{green}1x+\color{green}1x^2+\color{green}2x^3+\color{green}2x^4+\color{green}3x^5+\color{green}4x^6+\color{green}5x^7\cdots$$ ... 1$h(x) = \sum_{k=1}^n \sum_{v=1}^{\min\{k,j\}} \frac{(-1)^{n-k}k!}{(k-v)!} {n \brack k}f(x)^{k-v} B_{n,v}^f(x) $If$n \le j$, since$k \le n$, the sums become$\sum_{k=1}^n \sum_{v=1}^{k} = \sum_{v=1}^{n} \sum_{k= v}^n $. If$n > j, the sums could be split into two parts like this: \begin{align} \sum_{k=1}^n \sum_{v=1}^{\min\{k,j\}} ... 1 Your series is centered aboutx_0=1$. Since $$r=\lim_{n \to \infty} \left|\frac{a_n}{a_{n+1}}\right|$$ where$a_k$are the coefficients, plugging in your formula gives $$r=\lim_{n\to\infty}\left|\frac{\frac{(-1)^nn}{2^n}}{\frac{(-1)^{n+1}(n+1)}{2^{n+1}}}\right|=\lim_{n\to\infty}\frac n{2^n}\cdot\frac{2^{n+1}}{n+1}=2\lim_{n\to\infty}\frac n{n+1}=2$$ so the ... 1 Your recursive formula should be $$a_{n+1} = \frac {1-n}{n+1}a_n$$ You just made a small sign error. Then just plug in$a_0=1$and find the next few values by hand. $$a_0 = 1 \implies a_1=1 \implies a_2 =0$$ Then we don't need to go any further because we can see by our recursive formula that every$a_n$for$n\ge 2$must equal$0$as well. So your ... 1 I don't think you need a recurrence relation to find the generating function: it's simply $$\underbrace{(x+x^2+\dots+x^6)\dots(x+x^2+\dots+x^6)}_{4\text{ times}} = (x+x^2+\dots+x^6)^4.$$ Each$x^k$term in the resulting expansion corresponds to one way to obtain$k$from four dice. 1 So you are trying to prove that your function is equal to its power series representation, and as you said to do this you will use Taylors theorem and inequality.$\mathbf{Thereom:}$If$f(x)=T_n(x)+R_n(x) , $where$T_n$is the$n^{th}$degree taylor polynomial of$f$at$a$and$\lim_{n\to \infty} R_n(x)=0$for$|x-a| \lt R$, then$f$is equal to the ... 1 The Maple command dsolve((x^2+1)*(diff(y(x), x, x))-6*x*(diff(y(x), x))+10*y(x)) produces a polynomial solution $$y \left( x \right) ={\it \_C1}\, \left( -5\,{x}^{2}+1 \right) +{\it \_C2}\, \left( {x}^{5}+10\,{x}^{3}-15\,x \right) .$$ 1 I like this problem! I will have to assign this sometime. Apply the root test. The denominator satisfies$\lim_{n \to \infty} \sqrt[n]{2n+1} = 1$(use L'Hopitals). The limit in the numerator simplifies to$\lim_{n \to \infty} |x-2|^{n} ,$which is 0 when$|x-2|<1.$Therefore the preliminary interval of converge is$1<x<3.$The series is ... 1 $$\sum_{n=0}^d x^n= \frac{x^{d+1}-1}{x-1}$$ So we have: $$\sum_{m=1}^{\log_2 N} 2^m=\sum_{m=0}^{\log_2 N} 2^m-2^0=\sum_{m=0}^{ \log_2 N } 2^m-1 = \frac{2^{\lfloor \log_2N \rfloor +1}-1}{2-1}-1 \leq \frac{N \cdot 2-1}{1}-1 \\=N \cdot 2-2$$ 1 For the first part, you have established that$-5$and$-3$are not part of the radius of convergence, therefore the interval is$(-5,\,3)$, and not$[-5,\,3]$(which would include$-5$and$-3$). Same logic for the second part. You've shown that the series converges for$-5 < x < -3$but diverges for$-5$and$-3\$, therefore, it is the open interval ...

1

We can write $$x+x^2+\cdots+x^6=\frac{x-x^7}{1-x}$$ and then we can expand the numerator with a binomial expansion: $$(x+x^2+\cdots+x^6)^4 = \frac{(x-x^7)^4}{(1-x)^4} = (1-x)^{-4} \sum_{k=0}^4 {4 \choose k} (-1)^k x^{6k+4}$$ $$=\left( \frac16 \sum_{n=0}^\infty (n+3)(n+2)(n+1) x^{n} \right) \left( \sum_{n=0}^4 {4 \choose n} (-1)^n x^{6n+4} \right)$$ ...

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