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0

$x=4$ is not a singular point. $\color{blue}{x=0}$ is the only singular point of this equation (and it is a regular singular point). It's important to get that right since the next step is to appeal to the following theorem regarding the radius of convergence for power series solutions: Theorem. If $x_0$ is a regular singular point of ...


1

An idea: Using the fact that $1=T_n(x)f(x)$, you can get a recurrence relation on the coefficients $(b_k)_{k}$ of $f\colon x \mapsto \sum_{k=0}^\infty b_k x^k$ by applying a Cauchy product: $$ 1 = \sum_{k=0}^\infty a_k x^k \sum_{k=0}^\infty b_k x^k = \sum_{k=0}^\infty \left(\sum_{\ell=0}^k a_{k-\ell}b_\ell\right)x^k $$ where $a_k = \begin{cases} \frac{1}{k!} ...


1

I can offer another way to see what Winther is getting at, and maybe provide a different direction you can take this problem. Using the fact that sine is an odd function, you know that for all $n \in \Bbb{N}$ that $$a_{-n}\frac{\sin(-nL)}{-n} = a_{-n}\frac{-\sin(nL)}{-n} = a_{-n}\frac{\sin(nL)}{n}$$ and hence ...


1

Generally, given a power series $f(x) = \sum_{k-0}^\infty f_k x^k$, there is a radius of convergence $R$ associated with the terms $f_k$. Then if $|x|<R$, the series converges absolutely, if $|x|>R$ then the series diverges and, without knowing anything more, we can say nothing about convergence for $|x|=R$. The radius of convergence is given by a ...


2

$e^{t}=1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+ \cdots$, then $$e^t-1-t=\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+\cdots=\sum_{2}^{\infty}\frac{t^k}{k!}$$


1

We have $$\sum_{k=1}^n\frac{t^{k+1}}{(k+1)!}=\sum_{k=2}^{n+1}\frac{t^{k}}{k!}=\sum_{k=0}^{n+1}\frac{t^{k}}{k!}-1-t\xrightarrow{n\to\infty}e^t-1-t$$


1

Write out the first 5 or so terms of the series for the exponential and the series you have here. What you'll find is that the series you have here is exactly the series for the exponential but starting at the $t^2$ term. Thus if you subtract the first two terms from the exponential you get the series that you have here. To transform one series into the ...


4

It's easier to shift indices, so that the summand looks the same between the sums you are comparing. Your sum is $$\sum_{k=2}^{n+1} \frac{t^k}{k!}.$$ You can now "add and subtract" the first two terms of the sum for the exponential: $$\sum_{k=2}^{n+1} \frac{t^k}{k!} = \sum_{k=0}^{n+1} \frac{t^k}{k!} - \sum_{k=0}^1 \frac{t^k}{k!} \\ = \sum_{k=0}^{n+1} ...


0

Checking your expansion I got: $$ 2+\frac{x}{2}+\frac{7 x^2}{16}+...+$$ for the Maclaurin expansion. This agrees well (when plugging in x = .1 )with the actual answer and is less than your error estimate which is just as it should be.


1

If we have a function $h: [-L,L]\to \mathbb{R}$ then (under some conditions) we can expland it in a Fourier series $$h(x) = \sum a_n \sin(2\pi x/L) + b_n \sin(2\pi x/L)$$ where the coeficients $a_n,b_n$ are given by some well known relations. In your case you don't have a function to expand in a Fourier series (which it seems like you are trying to do). ...


0

I suppose that you can make the expresion shorter using $(1+a)(1-a)=1-a^2$ so $$A=\frac{1}{(1+x)(1−x)(1+x^2)(1−x^2)(1+x^3)(1−x^3)\cdots}=\frac{1}{(1−x^2)(1−x^4)(1-x^6)\cdots}$$ and now use the fact that $$\frac{1}{1-y}=\sum_{i=0}^{\infty}y^i$$ and replace successively $y$ by $x^2$, $x^4\cdots$,$x^{2n}$ before computing the overall product. Doing so, for a ...


1

$$\frac{1}{1-x^k}=1+x^k+x^{2k}+x^{3k}\cdots$$ $$\frac{1}{1+x^k}=1-x^k+x^{2k}-x^{3k}\cdots$$


1

If you set $$f(x)=\prod_{n=0}^\infty (1-x^n)^{-1}=\sum_{n=0}^\infty p(n)x^n$$ we see that yours is just $$f(x^2)=\sum_{n=0}^\infty p(n)x^{2n}.$$


0

This appears in the oeis, wherein it is given that the generating function is $$\prod_{k>0}1-x^{2k-1}=\prod_{k>0}\frac{1}{1+x^k}$$ There are also many references there, I highly recommend that link.


0

Hint 1: What is the generating function for the Fibonacci numbers? Hint 2: How would you get every other term of a power series? (Big hint: What does $f(x)\pm f(-x)$ do?)


3

Hint: Show that if $A(x) = a_0 + a_1 x + a_2 x^2 + \cdots $, then: $$\frac{A(x) + A(-x)}{2} = a_0 + a_2 x^2 + \cdots $$ Can you apply this to your series?


3

Let $f(x) = 1+2x+3x^2+5x^3+8x^4+13x^5+21x^6+\cdots$. You probably already know a closed form for $f(x)$. Then, $f(-x) = 1-2x+3x^2-5x^3+8x^4-13x^5+21x^6-\cdots$. Do you see how to get the series you want from $f(x)$ and $f(-x)$? To get a closed form for $f(x)$ try combining the following equations in a way that leaves a finite number of terms on the ...


5

Continued fractions: \begin{align} \frac{11235}{100000}&=\frac{2247}{20000}\\ &=\cfrac{1}{8+\cfrac{2024}{2247}}\\ &=\cfrac{1}{8+\cfrac{1}{1+\cfrac{223}{2024}}}\\ &=\cfrac{1}{8+\cfrac{1}{1+\cfrac{1}{9+\cfrac{1}{13+\cfrac{2}{17}}}}}\\ &=\cfrac{1}{8+\cfrac{1}{1+\cfrac{1}{9+\cfrac{1}{13+\cfrac{1}{8+\cfrac{1}{2}}}}}}\\ &=[0;8,1,9,13,2] ...


12

Like you hint, we can observe that the value $$0.11235955056\ldots = \sum_{k = 1}^{\infty} 10^{-k} F_k,$$ where $F_k$ is the $k$th Fibonacci number (with the convention that $F_0 = 0$, $F_1 = 1$), is simply the value of the series $$F(x) := \sum_{k = 1}^{\infty} F_k x^k = x + x^2 + 2x^3 + 3x^4 + 5x^5 + 8x^6 + 13x^7 + \cdots$$ at $x = \frac{1}{10}$. Now, ...


2

An exhaustive search (by computer) of all fractions with numerator and denominator $< 100$ shows there is only one whose decimal representation starts 0.11235: $$ \frac{10}{89} $$ The decimal expansion continues $\ldots 955056179775\ldots$ (I haven't bothered finding aprogram that can calculate enough that we can see the period), that's completely ...


0

These are fractions $f$ with $11235/100000\le f<11236/100000.$ A direct, if slow, method for counting these would be $$ \sum_{d=1}^{100}\left\lfloor\frac{100000d}{11235}\right\rfloor-\left\lfloor\frac{100000d}{11236}\right\rfloor $$ or sum(d=1,100,100000*d\11235-100000*d\11236)


2

Your reindexed summation is correct. Since the sums (outer and inner) for the second series start at $n = 2$ resp. $k = 2$ while in the first they start at $0$, combining the two power series is not totally direct. One way is to split the sums, treat $n = 0$ and $n = 1$ separately, and in the remaining series, for the inner sum $k = 0$ and $k = 1$. A more ...


0

One way is to look at $$ 1+2x^2+3x^4+4x^6+5x^8+\dots $$ as $$ 1+2t+3t^2+4t^3+5t^4+\dots $$ where $t=x^2$. The last series is the derivative of $$ 1+t+t^2+t^3+t^4+t^5+\dots=\frac1{1-t} $$ Therefore, $$ 1+2t+3t^2+4t^3+5t^4+\dots=\frac1{(1-t)^2} $$ and $$ 1+2x^2+3x^4+4x^6+5x^8+\dots=\frac1{(1-x^2)^2} $$ Now, just multiply by $1+x$: $$ \begin{align} ...


1

I would go about this by first splitting the series up: $$1+x+2x^2+2x^3+3x^4+3x^5+...=(1+x)(1+2x^2+3x^4+...)$$ Letting $s=1+2x^2+3x^4$ we can do a few tricks: $$s-x^2s=\begin{array}{c} 1&+2x^2&+3x^4+... \\ &-x^2&-2x^4-...\end{array}$$ $$=1+x^2+x^4+...$$ Which converges to $\frac{1}{1-x^2}$ for $-1 < x < 1$ (proving this is not hard, ...


2

Hint: using $y=x^2$ and derivative in $y$: $$(1+x)(1+2x^2+3x^4+\ldots) $$ $$ =(1+x)(1+2y+3y^2+4y^3 +\ldots)$$ $$= (1+x)(y+y^2+y^3+y^4+\ldots)'$$ $$ = (1+x)\left( \frac{y}{1-y}\right)'$$ Edit: $$ = (1+x) \frac{1}{(1-y)^2} $$ $$ = \frac{1+x}{(1-x^2)^2} $$ $$ = \frac{1}{(1-x)(1+x^2)}.$$


0

Unless I'm mistaken, it is $$\sum_{n=1}^\infty nx^{2n-2} + \sum_{n=1}^\infty nx^{2n-1}$$ If you can compute one of the two terms, e.g. $\sum_{n=1}^\infty nx^{2n-2} = \sum_{n=0}^\infty (n+1)x^{2n} = \sum_{n=0}^\infty (n+1)(x^{2})^n$ (see e.g. this, then you'll also get the other term (by multiplying it by $x$), and thus the sum.


4

There are some way how to determine $R=1$, for example: Ratio test We have: $$\lim_{n \to \infty}\frac{|a_{n+1}|}{|a_n|}=\lim_{n \to \infty}\frac{(3n+1)x^3}{(3n+4)}=|x^3|$$ So series converges for $|x|<1$. Root test We have: $$\limsup_{n \to \infty}\sqrt[n]{|a_n|}=\lim_{n \to \infty}|x|^3\sqrt[n]{3n+1}=|x|^3$$


1

I hope this helps a little bit: $R - \varepsilon < S < R$ cannot be, since $\varepsilon = R - S$. However, I'm not quite sure why they dont just pick a $\rho$ such that $R - |z| < \rho < R$. I guess it's just to show the thought process: Since $|z| < R$ there exists a $S$ such that $|z| < S < R$. Afterwards, we can choose $\rho$ ...


6

For every $|x|\gt1$, the series converges since $|x^n-1|\geqslant\frac12|x|^n$ for every $n$ large enough, and $$F(x)=\sum_{n\geqslant1}\frac{(-1)^{n+1}}{x^n-1}=\sum_{n\geqslant1}(-1)^{n+1}x^{-n}\frac1{1-x^{-n}}=\sum_{n\geqslant1}(-1)^{n+1}x^{-n}\sum_{k\geqslant0}x^{-nk},$$ that is, $$F(x)=\sum_{n\geqslant1}\alpha_nx^{-n},\qquad\alpha_n=\sum_{d\mid ...


2

The answer between the two horizontal lines was to the original question, which specified $|x| <1$. This doesn't converge. If the sum is $$ \sum_{n=1}^{\infty}a_n $$ with $$a_n = \frac{(-1)^n}{1-x^n} $$ then, since $|x|<1$ we have $$\lim_{n\to\infty} x^n = 0 \qquad\Rightarrow\qquad \lim_{n\to\infty} 1-x^n = 1$$ This gives ...


2

As already mentioned, a possible approach consists in multiplying the sum by $x^{-m}$ which leads to a sum easy to express on exponential form. But what follows with this method is rather arduous, because further integration involves an incomplete Gamma function with negative integer parameter. In order to save time, we will use another method (a short-cut ...


1

Use common logarithms and a decent calculator. Using a fairly mediocre one on your example, I get $99\log_{10}99\approx197.5678842652$; subtracting $197$ and raising $10$ to the resulting power, I get $3.697296376497$, so the number must be about $3.697296376497\times10^{197}$. Added: More generally, for $a^b$ calculate $b\log_{10}a$, subtract the integer ...


2

For the first one : $$\sum\limits_{n=1}^{N} \ln \cos \frac{x}{2^n} = \ln \prod\limits_{n=1}^{N} \cos \frac{x}{2^n} = \ln \frac{\sin \frac{x}{2^N}\prod\limits_{n=1}^{N} \cos \frac{x}{2^n}}{\sin \frac{x}{2^N}} = \ln \frac{\sin x}{2^N\sin \frac{x}{2^N}}$$ where, we made use of the identity: $\sin y \cos y = \frac{1}{2}\sin 2y$ repeatedly. Taking limit as $N ...


3

For (1) Consider the product $P_N = \displaystyle\prod_{n = 1}^{N}\cos\dfrac{x}{2^n}$. Multiply by $\sin\dfrac{x}{2^N}$ and use the identity $\cos\theta\sin\theta = \dfrac{1}{2}\sin 2\theta$ repeatedly to get $P_N\sin\dfrac{x}{2^N} = \dfrac{1}{2^N}\sin x$. Therefore, $P_N = \dfrac{\sin x}{2^N\sin\frac{x}{2^N}} \to \dfrac{\sin x}{x}$ as $N \to \infty$. ...


1

Applying partial fractions, we find $$\frac{1}{(k+1)(k+3)}=\frac{\frac{1}{2}}{k+1}-\frac{\frac{1}{2}}{k+3}$$ Hence we have \begin{align}\sum\limits_{k=0}^\infty\frac{1}{(k+1)(k+3)}&=\frac{1}{2}\sum\limits_{k=0}^\infty\left(\frac{1}{k+1}-\frac{1}{k+3}\right) ...


0

Hint: Have you seen this example? $$\begin{align} \sum_{k=1}^\infty{1\over k(k+1)}&=\sum_{k=1}^\infty\left({1\over k}-{1\over k+1} \right)\\ &=\left({1\over1}-{1\over2}\right)+\left({1\over2}-{1\over3}\right)+\left({1\over3}-{1\over4}\right)+\cdots\\ &=1-\left({1\over2}-{1\over2}\right)-\left({1\over3}-{1\over3}\right)-\cdots \end{align}$$ ...


0

Further hint: What is $\frac{1}{k+1} - \frac{1}{k+3}$ with a common denominator? How does that result relate to $\frac{1}{(k+1)(k+3)}$? What happens to each term because you're subtracting?


1

Hint: partial fractions, telescoping series...


1

It is rather strange, because Wolfram Alpha is perfectly happy to return a Laurent series for e.g. series of 1/(x+x^2) at x = 0 Somehow, $1/x$ is treated differently.


1

You may recall that $$ \sum_{k=0}^{\infty}r^n=\frac{1}{1-r},\quad |r|<1. \tag1 $$ Just substitute $r \rightarrow 1-x $ in $(1)$, you get, for the right hand side $$ \frac{1}{1-r}=\frac{1}{1-(1-x)}=\frac 1 x $$ as long as $|1-x|<1$ ($0$ being excluded).


1

Hint: When expanding the power series solution about an ordinary point $x_0$, the radius of convergence is at least as large as the distance from $x_0$ to the nearest singular point in the complex plane.


0

Hint:- $\dfrac{4}{x+2}=4(x+2)^{-1}$ Now expand by Binomial Theorem keeping in mind the restrictions on $x$.


0

Given the power series you have first written: $$\sum_{n=1}^{\infty} x^{n-1} = \frac{1}{1-x}$$ If we rewrite the fraction: $$\frac{4}{2+x} = \frac{2}{1+\frac{x}{2}}$$ Now we have in a form that is similar to the first power series above, $\frac{1}{1-x}$ Notice that: $$\frac{2}{1+\frac{x}{2}} = 2\left(\frac{1}{1-\left(-\frac{x}{2}\right)}\right) = ...


2

For $\;|x|<2\;$ we get $$\frac4{x+2}=\frac2{1+\frac x2}=2\left(1-\frac x2+\frac{x^2}{2^2}-\ldots\right)=2\sum_{n=0}^\infty(-1)^n\frac{x^n}{2^n}=\sum_{n=0}^\infty(-1)^n\frac{x^n}{2^{n-1}}$$


3

HINT $$ \frac{4}{2+x}=\frac{2}{1+x/2}=\sum_{n=0}^{\infty}(-1)^n x^n 2^{1-n}\qquad\text{for}\; |x|<2 $$


2

Use the fact that this is a power series with radius of convergence $r=1$. Thus, for $\lvert y\rvert < r$, $$ \sum_{k=0}^\infty (k+1)y^k = \frac{d}{dy} \sum_{k=0}^\infty y^{k+1} = \frac{d}{dy} \sum_{k=1}^\infty y^{k} $$ and $$ \sum_{k=1}^\infty y^{k} = \frac{y}{1-y}. $$ Now, you can compute the derivative of $\frac{y}{1-y}$, and evaluate it on $x^2$ ...


1

Try convert(expression, FPS, variable) or convert(expression, FPS, variable = a), e.g. convert(exp(x), FPS, x); $$ \sum_{k=0}^\infty \dfrac{x^k}{k!} $$ Of course it won't always work, but for many functions that have reasonably simple Taylor (or Laurent) series it will. EDIT: This might sometimes work for a bivariate series as well. Maybe not in your ...


0

For every $c$, $$n+c=\frac{\Gamma(n+c+1)}{\Gamma(n+c)},$$ hence the recursion you arrived at can be rewritten as $$a_{n+1}=\frac{-a_n}{4(n+r+1)(n+r+\frac12)}=a_n\frac{(-1)^n4^n\Gamma(n+r+1)\Gamma(n+r+\frac12)}{(-1)^{n+1}4^{n+1}\Gamma(n+r+2)\Gamma(n+r+\frac32)}a_n,$$ which immediately leads to $$ ...


1

Hint: $$f(z)=\frac4{4-z^2}+\frac{e^z-e^{-z}}2$$


1

I think for this one it is much easier to use logic. You have divided the sum into two parts which are manageable and known, which is good. The first is $$\sum_{n=0}^{\infty} \left(\frac{z^2}{4}\right)^n$$ Which converges when $|z|<2$ (the magnitude of the inside must be less than one). The other sum you got looks a lot like $e^z$. In fact, it is less ...



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