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0

Your calculation is fine, up to the very last step. That is, once, you have $\left | x-1 \right |\leq 2e$, you're done because the series is centered at $x_{0}=1$, so $R=2e$


-1

Actually, you got the answer right. The disk is just not centered at the origin, since the series is centered at 1.


0

The power series around zero is $$y(x) = \sum_{k=0}^\infty a_k \,x^k.$$ Therefore $$ y' = \sum_{k=0}^\infty k \,a_{k} \,x^{k-1} =\sum_{k=1}^\infty k \,a_{k} \,x^{k-1}=\sum_{k=0}^\infty \,(k+1) \,a_{k+1} \,x^{k}, $$ and $$ y'' = \sum_{k=0}^\infty \, k\,(k+1) \, a_{k+1} \,x^{k-1}= \sum_{k=1}^\infty \, k\,(k+1) \, a_{k+1} \,x^{k-1}= \sum_{k=0}^\infty \, ...


3

We should note that this differential equation can be solved quickly since it is a separable differential equation: $$y'/y = 2x \implies \ln y = x^2 + C \implies y = e^C e^{x^2}$$ and the initial condition $y(0)=1$ yields $y = e^{x^2}$. The power series method idea is to assume the solution can be expressed as a power series (there are theorems to justify ...


3

$$y’=2xy\Longleftrightarrow$$ $$\frac{dy}{dx}=2xy\Longleftrightarrow$$ $$dy=2xydx\Longleftrightarrow$$ $$\frac{1}{y}dy=2xdx\Longleftrightarrow$$ $$\int \left(\frac{1}{y}\right)dy=\int (2x) dx\Longleftrightarrow$$ $$\ln|y|=x^2+C\Longleftrightarrow$$ $$y=e^{x^2+C}$$ $------$ $$y=e^{x^2+C}\Longrightarrow$$ $$1=e^{0^2+C}\Longleftrightarrow$$ ...


4

The idea is that if two power series agree on some region of convergence, then they are the same. That is if $\sum_n f_n x^n = \sum_n g_n x^n$ for $x$ is some region, then $f_n = g_n$. So suppose $y(x) = \sum_n y_n x^n$. Then we can think of $y$ as the coefficients of its power series, that is $(y_0, y_1,...)$. Then $y'(x) = \sum_n n y_n x^{n-1}$. This ...


9

By differentiating term by term and changing the index we find from the equation $y'=2xy$ $$\sum_{n=0}^\infty (n+1)a_{n+1}x^n=2\sum_{n=1}^\infty a_{n-1}x^n$$ so we get $a_1=0$ and then $a_{2n+1}=0$ for all $n$ and $$a_n=\frac2n a_{n-2},\quad n\ge2 $$ so since $y(0)=a_0=1$ and by induction we get $$a_{2n}=\frac{2^n}{2n(2n-2)\cdots 2}a_0=\frac1{n!}$$ ...


3

$$\text{Consider }S = \sum_{i=0}^N r^i$$ $$(1 - r)S = \big(1 + r + r^2 + \ ... \ + r^N\big) - \big(r + r^2 + ... + r^{N+1}\big) = 1 - r^{N+1}$$ So $$S = \frac{1 - r^{N+1}}{1 - r} = \frac{r^{N+1} - 1}{r - 1}$$ In the case of $3$, $S_3 = \frac{1}{2}(3^{N+1} - 1)$


2

If $$f(x)=\sum_{n=1}^\infty \frac{x^{n}}{ n(2n-1)}$$ $$f'(x)=\sum_{n=1}^\infty \frac{x^{n-1}}{ (2n-1)}$$ $$f''(x)=\sum_{n=1}^\infty \frac{(n-1)x^{n-2}}{ (2n-1)}$$ So $$g(x)=2x f''(x)+f'(x)=2\sum_{n=1}^\infty \frac{(n-1)x^{n-1}}{ (2n-1)}+\sum_{n=1}^\infty \frac{x^{n-1}}{ (2n-1)}=\sum_{n=1}^\infty x^{n-1}=\frac{1}{1-x}$$ Edit The long (and stupid) way would ...


0

By ratio test you can find that the series converges if $\left|x\right|<1 $. In this case this series admits a closed form in terms of special functions. In fact we have $$\sum_{n\geq1}\frac{x^{n}}{\sqrt{n+1}}=x\sum_{n\geq0}\frac{x^{n}}{\sqrt{n+1}}=x\Phi\left(x,\frac{1}{2},x\right) $$ where $\Phi\left(x,s,a\right) $ is the Lerch Trascendent. For $x=1 ...


3

Here is a useful finite evaluation: $$ 1+u+u^2+...+u^n=\frac{1-u^{n+1}}{1-u}, \quad |u|<1. \tag1 $$ Then by differentiating $(1)$ you get $$ 1+2u+3u^2+...+nu^{n-1}=\frac{1-u^{n+1}}{(1-u)^2}+\frac{-(n+1)u^{n}}{1-u}, \quad |u|<1, \tag2 $$ and by making $n \to +\infty$ in $(1)$ and $(2)$, using $|u|<1$, gives $$ 1+u+u^2+...+u^n+...=\frac{1}{1-u} \tag3 ...


0

An often easier way is that the radius of covergence is the least upper bound of the $r\ge 0$ such that $a_nr^n\to 0$ as $n\to\infty$. Here, $\,\dfrac{a_n}{r^n}=\Bigl(\dfrac r2\Bigr)^n$ and it is well known it tends to $0$ if and only if $\,\dfrac r2<1\iff r<2$, hence $R=2$.


2

To complement the answer by CPM: You can also use the formula of Chauchy-Hadamard. It states, that the series $\sum a_n (x-c)^n$ has the radius of convergence $$R=\frac{1}{\lim \sup_{n\to\infty} \sqrt[n]{|a_n|}}$$ Thus $$R=\frac{1}{\lim \sup_{n\to\infty} \sqrt[n]{\frac{1}{2^n}}}=\frac{1}{\lim \sup_{n\to\infty} \frac{1}{2}} = 2$$


2

The center of your interval of convergence is $x=1$, but not the radius. To find the radius of convergence you need to use the ratio test. $$\lim_{x\to \infty}\left| \frac{(x-1)^{n+1}}{2^{n+1}}\cdot \frac{2^{n}}{(x-1)^{n}}\right|=\lim_{x \to \infty} \left|\frac{x-1}{2}\right| $$ To converge, this needs to be less than $1$. Thus $|x-1|<2$. So your ...


2

From Calculus: 8th Edition by Larson: [A]n infinite series of the form $$ \sum_{n = 0}^\infty a_n(x-c)^n$$ is called a power series centered at c, where c is a constant. So here $c = -4$.


0

As you say, by the ratio test the radius of convergence as a power series about $0$ is indeed $R=1$, so the series must converge in $D=\{z \in \mathbb{C} : |z|<1 \}$. Lets see what happens if $z \in \partial D=\{ z: |z|=1\}$. Consider the set $$ E=\{ e^{2\pi i \cdot r} : r \in \mathbb{Q} \} \subset \partial D $$ For every $z \in E$, we have $$ ...


1

Taylor expanding $\cos z$ about $z=\frac{\pi}{2}$ one finds that: \begin{equation} \cos z=-\left(z-\frac{\pi}{2}\right)+\frac{1}{6}\left(z-\frac{\pi}{2}\right)^3+\ldots \end{equation} Similarly, Taylor expanding $\frac{1}{z+\frac{\pi}{2}}$ about $z=\frac{\pi}{2}$ one finds that: \begin{equation} ...


0

Let $U$ be the larger domain. Certainly $f$ is analytic in $U\setminus \{1\}.$ But $f$ is also analytic in $D(1,1).$ Why? Because it equals a power series centered at $1$ there. Thus for any point in $U,$ $f'(z)$ exists. Hence $f$ is analytic in $U.$


7

Lemma: in any ring $R$, any power series in $R[[x]]$ with constant term $1$ is invertible. Explanation. Every power series with constant term $1$ can be written as $1-xf(x)$ for some other power series $f(x)$. Then $(1-xf(x))(1+xf(x)+x^2f(x)^2+\cdots)=1$. The geometric series can be simplified to a power series in $x$ by expanding all of the powers $f(x)^n$ ...


3

Almost certainly not: if there were a "general formula" for $\sum_{i=1}^r p_i^n$ for even a single $n$, you'd also have a "general formula" for the prime numbers by $$p_r=\sqrt[n]{\left(\text{“general formula” for }\sum_{i=1}^r p_i^n\right) - \left(\text{“general formula” for }\sum_{i=1}^{r-1}p_i^n\right)}$$


2

We have: $$\begin{align}&f(x)=\sum_{n=0}^\infty\binom\alpha n x^n\\{}\\ &f'(x)=\sum_{n=1}^\infty\binom\alpha n nx^{n-1}\end{align}$$ Perhaps the trick you're looking for is $$n\binom\alpha n=\alpha\binom{\alpha-1}{n-1}$$ Added on request: $$\alpha f(x)=\sum_{n=0}^\infty \alpha\binom\alpha ...


1

One may recall that, as $x \to 0$, $$ \begin{align} e^x&=\sum_{n=0}^{\infty}\frac{x^n}{n!}, \quad x \in \mathbb{C}, \tag1\\\\ \frac{1}{1-x^2}&=\sum_{n=0}^{\infty}x^{2n}=\frac1{2}\sum_{n=0}^{\infty}(1+(-1)^n)x^{n}, \quad |x|<1, \tag2 \end{align} $$ then using the Cauchy product we get $$ ...


1

One method of evaluating $\sum_{n=0}^\infty(1+n)x^n$ can be like this, we take the generating function f = $\sum_{n=0}^\infty x^n $, then $$\sum_{n=0}^\infty (n+1)x^n = (xD + 1) f $$ $$ \frac{x}{(1-x)^2} + \frac{1}{1-x} = \frac{1}{(1-x)^2}$$here D means differentiation w.r.t x


0

Hint Factor $x^3$ out to get $1-x^3+(-x^3)^2 + (-x^3)^3 \pm \ldots$ which is a geometric series. You can decompose the original one directly as well, using $-x^3$ as the common multiple.


3

The number of solutions of the equation equals the coefficient of $z^n$ in the expression $$ \frac{1}{(1-z)(1-z^2)(1-z^3)\ldots }=\sum_{n}p(n)z^n, $$ where $p(n)$ is the Euler partition function, so that the number of combinations is $p(n)$. (Christoph): There is a bijection between the partition function and the desired function, since from a partition ...


1

A very simple proof of this actually involves complex analysis. The set of complex numbers extends to the reals. $e^{z}=\sum_{0}^ \infty \frac{z^{k}}{k!} $ This is a series for $e^{z}$ expanded at $\alpha=0$. In complex analysis in order for a number to be inside the disc of convergence it has to follow the following conditions: Let $\beta$ be the ...


1

We will use the fact that $$ \sum_{n=0}^\infty\frac{x^n}{n!} $$ converges absolutely for all $x\in\mathbb{R}$. This can easily be shown using the ratio test and means that for any $x$, there is an $N$ so that $$ \sum_{n=N+1}^\infty\frac{|x|^n}{n!}\le\epsilon $$ Find an $N$ so that we have $$ ...


0

This is known as Series Reversion. The analytical expression for the inverse coefficients can be written out, but it gets quite complex for higher $n$.


5

Pick any $B>0$. Then, if $|x|\leq B$, then one has that $|f_k(x)|=|x|^k/k!\leq B^k/k!$ for each non-negative integer $k$. Since $$\sum_{k=0}^{\infty}\frac{B^k}{k!}=\exp(B)$$ is convergent, Weierstrass's $M$-test reveals that the series $\sum_{k=0}^{\infty} x^k/k!$ converges uniformly for $x\in[-B,B]$. Together with the continuity of the partial sums ...


0

Sine and cosine are immediately seen to be odd and even, respectively. Also, it's not too hard to show that sine and cosine are continuous. It's not hard to prove the sum rules for $\sin(A+B)$ and $\cos(A+B)$, with the series definition. (Complex numbers are recommended but not necessary.) Note that, using the sum rules, we have: ...


1

Here's an approach that avoids differential equations. Euler's identity $e^{ix}=\cos x+i\sin x$ is proved by simple inspection of the power series of the functions involved. The sum-product formula $e^{a+b}=e^ae^b$ can be proved by series manipulation: $$ e^{a+b}=\sum_{n=0}^\infty \frac{(a+b)^n}{n!} = \sum_{n=0}^\infty\sum_{k=0}^n ...


0

$2a_n-a_{n-1}=4$. From here: $2a_nx^n-a_{n-1}x^n=4x^n$. By summation: $$ \begin{align} \sum_{n=1}^\infty (2a_nx^n-a_{n-1}x^n)&=\sum_{n=1}^\infty 4x^n\\ &=\frac{4x}{1-x}\,\,\,\, for\,\, |x|\lt 1 \end{align} $$ For the left side of above equation, we can write: $$ \begin{align} \sum_{n=1}^\infty (2a_nx^n-a_{n-1}x^n)&=2(-a_0+\sum_{n=0}^\infty ...


2

$$\sum_{n=1}^\infty a_n=+\infty \tag{since $a_n\geq2/3$}$$ So the radius is $\leq1$. On the other hand, if $0<x<1$, then: $$\sum_{n=1}^\infty a_nx^n\leq \sum_{n=1}^\infty\frac{5}{3}x^n<\infty$$ So the series converges for $|x|<1$ Therefore, the radius is $1$.


1

The space $\ell_2$ is the space of infinity sequences square additive. You can map a vector in $\ell_2$, for example $(\xi_1, \xi_2 ,\cdots, \xi_k \, \cdots)$ to a power series $\xi_1 + \xi_2 z + \cdots, \xi_k z^k + ...$. This is known as the Z transform. think about $z=\exp(i \omega \Delta t)$ where $\omega = 2 \pi f$, and $f=1/\Delta t$ is the frequency. ...


1

You absolutely want to use Maclaurin series. Let me help you remember how that goes. If $f(x)$ is a sufficiently nice function then $$ f(x) = f(a) + f'(a)(x-a) + \frac12 f''(a)(x-a)^2 + \cdots + \frac{1}{n!}f^{(n)}(a)(x-a)^n + \cdots. $$ For your particular problem, let's say that $f(x)$ is solution to the ODE, and let's pretend that we know $f(0)$. ...


1

Make the change of variable y=1/x. The integral is from o to 1/x while the integrand is f(1/x).(1/x^2). So the required value is (1/x).lim(f(1/x)/x^2), the limit being taken near 0 and we need this to exist.


0

$(2/3)^{1/n}\le a_n^{1/n} \le (5/3)^{1/n}.$ Both the left side and right side of this inequality $\to 1.$ Thus $a_n^{1/n}\to 1.$ By the root test, the ROC is $1.$


0

Generally we use ratio test to find radius of convergence . Now first of all we find limit of sequence as a tends to infinity = 4/3 .Then we use ratio test to get x<1 for series to converge and hence radius of convergence is 1.


2

HINT: For the first one, suppose that $a_mx^m$ is the lowest-degree non-zero term of $A(x)$ and $b_nx^n$ is that of $B(x)$. What’s the coefficient of $x^{m+n}$ in $A(x)B(x)$? Be sure to consider all contributions to this term. For the second, if $(A(x))^2=(B(x))^2$, you have an infinite system of equations: $$\sum_{k=0}^na_ka_{n-k}=\sum_{k=0}^nb_kb_{n-k}$$ ...


1

If you know that a convergent power series converges to a holomorphic function in the interior of the region of convergence, then it's a matter of finding the region in which the series converges. The ratio test proceeds as follows: $$ \lim_{n\to\infty} \left|\frac{z^{n+1}/(2(n+1))}{z^n/(2n)}\right| = \lim_{n\to\infty} |z|\frac{n}{n+1} = |z|. $$ So the ...


1

you can differentiate the equation repeatedly to get the following: $$y' = 2y, y'' = 2y' = 2^2y, y^{(3)}= 2^3y, \cdots, y^{n} = 2^ny, \cdots \tag 1$$ suppose $y(0) = a.$ then substituting $x = 0,$ in $(1),$ we get $$y'(0) = 2a, y''(0)=2^2 a,\cdots,y^{(n)} = 2^n a, \cdots \tag2 $$ now using the maclaurin series $$y =y(0)+xy'(0)+\frac{x^2}{2!}y''(0)+ ...


0

Assume $y = \sum_{i=0}^{\infty}a_nx^n$ is a solution. Then $y' = \frac{dy}{dx} = \sum_{i=1}^{\infty} na_nx^{n-1}$ Now plug this into the ODE and try and combine the two series by manipulating the index on the first term. Once you do this you should be able to factor out an $x^n$ to get a recurrence relation to help you solve the rest of the problem! Try and ...


1

The problem is not that simple. Well, it's not very complicated neither, but one needs to use Jacobi's imaginary transformation. Introducing a new variable $\tau$ by $x=e^{i\pi \tau}$, we obtain $$\vartheta_2(0|\tau)=\left(-i\tau\right)^{-\frac12}\vartheta_4\left(0|\tau'\right),\qquad \tau'=-\tau^{-1}.$$ Now as $x\to 1$, we have $\tau\to i0$, $\tau'\to ...


-4

Series solution of a first order differential equation? Also you have written $dy/dy = 2y$: Shurely shome mishtake mish Moneypenny? EDIT Why downvote? I know it's fine to use series solution for first order; wlog that is; but it's not really needed. Also the typo is important.


2

Assuming that the obvious pattern holds, the last number is $3458^{3458}$, the one before that is $3457^{3457}$, then $3456^{3456}$ and then $3455^{3455}$.


2

$$\sum_{i=1}^n p_1^i p_2^{n-i} = p_2^n \sum_{i=1}^n \left ( \frac{p_i}{p_2} \right )^i = p_2^n \frac{p_1}{p_2} \frac{\displaystyle 1-\left (\frac{p_1}{p_2}\right )^n}{\displaystyle 1-\frac{p_1}{p_2} }= p_1 \frac{p_2^n-p_1^n}{p_2-p_1}$$ Plugging in your values $p_1=1/n$ and $p_2=1-1/n$, the sum is then $$\frac1{n^{n}} \frac{(n-1)^n-1}{n-2}$$ When $n=2$, ...


5

Hint: Observe that $n^in^{n-i}=n^n$ is independent of $i$. This renders the summation a geometric summation.


4

If this problem is $$f(x)=\sum_{n=0}^{\infty}\dfrac{x^{2^n}}{1-x^{2^{n+1}}}$$ Hint:Let $x^{2^n}=t$,then $$\dfrac{x^{2^n}}{1-x^{2^{n+1}}}=\dfrac{t}{1-t^2}=\dfrac{1}{1-t}-\dfrac{1}{1-t^2}=\dfrac{1}{1-x^{2^n}}-\dfrac{1}{1-x^{2^{n+1}}}$$ so $$f(x)=\dfrac{1}{1-x}-\lim_{N\to\infty}\dfrac{1}{1-x^{2^{N+1}}}=\dfrac{1}{1-x}$$


1

You obtained $\sum_{n = 0}^\infty (n^2 + 7n)a_n x^{n+5} = 0$ which is correct. So, it is necessary that $a_n=0$ except in case of $(n^2+7n)=0$ which roots are $n=0$ and $n=-7$. The two remaining terms are $a_0 x^{0+5}$ and $a_{-7}x^{-7+5}$ Changing the symbols of coefficients, the general solution of the ODE is : $$y(x)=c_1x^5+c_2 x^{-2}$$


1

As closed a form as I could get I posted here. It looks pretty ugly... but I'm not sure how much prettier it can get.



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