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0

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) = ...


-1

The general approach would consist in computing derivatives over and over, which is not fun with quotients. The better approach here is to start with the known power series of $1-\sin x$ and $\frac{e^x-1}{x}$ and compute their reciprocals term by term.


0

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

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.


3

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


0

$$-F(x)=\sum_{r=1}^\infty\dfrac{(ix)^{2r}}{2r}$$ Now $\ln(1+y)+\ln(1-y)=-2\sum_{r=1}^\infty\dfrac{y^{2r}}{2r}$


4

Differentiating, $F'(x) = x - x^3 + x^5 - x^7 + \cdots$. This is a geometric series, hence for $|x| < 1$ $$F'(x) = \frac{x}{1+x^2}$$ Now integrate to find $F$.


0

It can be proven that the power series and its derivative have the same radius of convergence. So consider the derivative of the above series $\displaystyle \sum_{n=1}^\infty na_nx^{n-1}, b_{n-1} = na_n$, and using root test: $L = \displaystyle \lim_{n\to \infty} \sqrt[n-1]{\left|b_{n-1}x^{n-1}\right|}= |x|\cdot \displaystyle \lim_{n\to \infty} ...


0

The root test will also show that $R=1$ since $n^{1/n} \to 1$ and $1101^{1/n} \to 1$.


0

$R=\lim {\dfrac{a_n}{a_{n+1}}}=\dfrac{na_n}{(n+1)a_{n+1}}\times \dfrac{n+1}{n}=1$ as $n\rightarrow \infty$


1

we are looking for when $\lim\limits_{n\rightarrow\infty}|x|\left|\frac{a_{n+1}}{a_n}\right|<1$ Now from the first equation, it follows that $\lim\limits_{n\rightarrow\infty}|a_n|=\lim\limits_{n\rightarrow\infty}\frac{1101}{n}$, and so, similarly, $\lim\limits_{n\rightarrow\infty}|a_{n+1}|=\lim\limits_{n\rightarrow\infty}\frac{1101}{n+1}$. Thus, ...


0

Hint. As $n \to \infty$, you have $$ |a_n| \sim \frac{1101}n $$ then, by the comparison test, your initial series has the same radius of convergence as $$ \sum_{n\geq 1} \frac{x^n}n =-\log (1-x) $$ that is $R=1$.


1

Now that I've had a chance to read Steven Stadnicki's answer, I see that my answer below is essentially the same, with some added detail. Since $\frac1{\cos(x)}$ is even, all the odd order terms are $0$. So we will look only at the even order terms. Since $$ ...


5

To expand my comment into a solution: consider the (formal) product $1=\cos(z)\cdot\frac1{\cos(z)}$. Expanding this out (using the already-established fact that $\frac1{\cos(z)}$ is even) and comparing powers of $x^2$, we get $a_0=1$ and $\displaystyle\sum_{i=0}^n\dfrac{a_{2i}}{(2i)!}\dfrac{(-1)^i}{(2n-2i)!}=0$; multiplying the latter by $(2n)!$ gives ...


1

This is not a full solution, but I hope that it might help you in some way. There is a nice characterization of the Maclaurin series of $ \sec $ by the Frenchman Désiré André, which he discovered while studying Bertrand’s Ballot Problem. Define a sequence $ (A_{n})_{n \in \mathbb{N}} $ of natural numbers by $$ \forall n \in \mathbb{N}: \quad A_{n} ...


1

Note: We know, that $\cos$ is an even function, i.e. $\cos(z)=\cos(-z)$ and the powers of $z$ with non-zero coefficients of the power series representation of $\cos$ are even. \begin{align*} \cos(z)=\sum_{n=0}^{\infty}(-1)^n\frac{z^{2n}}{(2n)!} \end{align*} Let's consider $f(z)=\frac{1}{\cos(z)}$. Since $\cos$ is even we obtain \begin{align*} ...


2

Hint: use $\sin 2u = 2 \sin u \cos u$; the power series of $\sin$ and term by term integration.


2

Hint: $$\int_0^x\sin t^2\cos t^2\;dt=\frac12\int_0^x\sin 2t^2\;dt$$ and then you may want to check Fresnel Integrals


2

The radius $r$ of convergence is such that $$\frac{1}{r} = \lim_{n\to \infty} \sqrt[n]{n^{n^{1/3}}}= \lim_{n\to \infty} n^{n^{-2/3}} = \lim_{n\to \infty} \exp(n^{-2/3}\log n) = \exp(0) = 1,$$ thus $r = 1$.


2

It's false. For instance, $\sum_{n=1}^{+\infty}\frac{z^n}{n^2}$ has ray of convergence $1$, but it does converge absolutely $\forall|z|\leq1$.


13

Assume the series $\sum_{k=0}^\infty a_kz^k$ converges for some $z\in{\mathbb C}$. Then it converges absolutely for any $z'$ that is nearer to the origin than $z$; see the proof below. From this fact it follows by mere logic, that when the series diverges at some $z\in{\mathbb C}$ it cannot converge at any point $z''$ farther away from the origin. Proof. ...


2

If $(a_n)_{n\in\Bbb N}$ is a sequence of complex numbers such that the series $\sum_{i=0}^na_n$ is convergent in the usual sense then certainly the sequence $(a_n)_{n\in\Bbb N}$ is bounded (it must even converge to$~0$); it follows that for any $z\in\Bbb C$ with $|z|<1$ the series $\sum_{i=0}^na_nz^n$ is absolutely convergent, and therefore convergent. ...


2

This is because of Abel's lemma, which is the basis for analytic functions: Abel's lemma. – If $r_0$ is positive real number such that the sequence $(\lvert a_n\rvert r_0^n)$ is bounded from above, then the series $\sum a_nz^n$ is absolutely convergent for $\lvert z\rvert<r_0$. Hence if $R=\sup\bigl\{\lvert z\rvert\:;\: \sum ...


1

I think that in this case, an example is worth more than a detailed answer. Consider the power series $$ \sum_{i=1}^\infty\frac{(-1)^{i+1}}{i2^i}x^i $$ (this is $\log(1+x/2)$ for $|x|<2$). Now, apply the ratio test (but keep the $x$) to get $$ ...


1

For power series, the area of convergence really is the interior of a circle, perhaps with some points on the boundary. Here is a quote from Rudin's Principles of Mathematical Analysis, second edition (also called Baby Rudin), page 60, referring to power series of the complex number $z$. More specifically, with every power series there is associated a ...


2

If we have that $\sum_{k\geq0}\left(-1\right)^{k}a_{k} $ such that $\lim_{k\rightarrow\infty}a_{k}=0 $ and $a_{k+1}<a_{k} $ holds $$\left|R_{N}\right|=\left|\sum_{k=N+1}^{\infty}\left(-1\right)^{k}a_{k}\right|<a_{N+1}. $$ So you have to note that $$\frac{0.5^{5}}{5!5}\simeq0.0000520 $$ so you can approximate your value with ...


0

Making use of the fact that this is an alternating series to bound the error. $$f(x)={\frac{\sin(x)}{x}}$$ $$\int f(x) \, dx \approx x-\frac{x^3}{18}+\frac{x^5}{600}-\frac{x^7}{35280}+...-$$ Truncating the series at the $x^5$ term. $$x-\frac{x^3}{18}+\frac{x^5}{600} \text{ with } x = \frac{1}{2} \approx .49310763888$$ Using the last term as an error ...


1

Yes, you have done this correctly, and you are correct that $\dfrac{1}{1+z}=1-z+z^2-z^3+\cdots$ for $\vert z\vert<1$. If you are willing to take the formula for the sum of a geometric series for granted, then that can lead to a slightly quicker answer (although you have essentially derived the formula in your argument). The series $1-z+z^2-z^3+\cdots$ is, ...


3

You are already 90% of the way there; using term-by-term integration or any of the other methods described at Taylor series of $\arctan$, you can derive $$\arctan x=\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{2n+1},\qquad |x|<1,$$ but you cannot claim directly that this is true for $x=1$ because this is out of the range of the equation. (You can't do ...


-2

It is easy to answer your second question with Maple: dsolve(((D@@2)(w))(z)+sin(z)*(D(w))(z)+(z^2+1)*w(z) = 0, w(z), series, order = 10); $$ w \left( z \right) =w \left( 0 \right) +\mbox {D} \left( w \right) \left( 0 \right) z-{\frac {w \left( 0 \right) }{2}}{z}^{2}-{\frac { \mbox {D} \left( w \right) \left( 0 \right) }{3}}{z}^{3}+{\frac {w \left( 0 ...


3

You get that the coefficient for $x^5$ is $\dfrac {1}{5!} - \dfrac {1}{4!} + \dfrac {1}{2!2!} - \dfrac {1}{3!2!}$, which is indeed $\dfrac2{15}$. So, no need to change method.


3

You may exploit the fact that $\tan x$ is an odd function, hence in a neighbourhood of the origin: $$ \tan x=\sum_{n\geq 0} a_{2n+1}\, x^{2n+1}\tag{1} $$ as well as: $$ \frac{d}{dx}\tan x=\frac{1}{\cos^2 x}=1+\color{purple}{\tan^2 x} = \sum_{n\geq 0}(2n+1)\,a_{2n+1}\,x^{2n} \tag{2}$$ from which it follows that $a_1=1$ and: $$ a_{2n+1} = ...


3

The following way may be easier. Since $\tan x$ is an odd function, it can be represented as $\tan x=ax+bx^3+cx^5+\cdots.$ Since $\sin x=\cos x\tan x$, one has $$x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots=\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots\right)\left(ax+bx^3+cx^5+\cdots\right).$$ Comparing the coefficients gives you $$1=a$$ ...


0

We have to find out the power series of $\frac{x^3}{(3x+4)^2}$. Recall that if $|x|<1$ then $$\frac{1}{(1+x)^2}=(1+x)^{-2}=\sum_{n\geq 0}(-1)^n(n+1)x^{n}.$$ here we have $\frac{1}{(3x+4)^2}=(3x+4)^{-2}=(3x)^{-2}(1+\frac{4}{3x})^{-2}$ if $|\frac{4}{3x}|<1$ i.e. $|x|>4/3$ and ...


1

A convergent series gets closer to the sum as more terms are taken. A asymptotic series initially gets closer to the sum but, after a while, gets increasingly further away and usually ultimately diverges. In many asymptotic series, the error is less than the last term. Since the terms eventually get large, the usual practice is to stop the series when the ...


1

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


1

$$\frac{1}{e^{x}-1}=\frac{e^{-x}}{1-e^{-x}}=\sum_{n=0}^{\infty} e^{-(n+1)x}$$ Can you finish the rest?


2

A necessary condition for convergence is that the terms go to $0$.


0

Answer. The radius of convergence is $R=1$. Explanation. We use the Root Test. We have the series $\sum a_nx^n$, where $$ a_n=\left\{\begin{array}{lll} (k!)^2 & \text{if} & n=k^2, \\ 0 & \text{if} & \sqrt{n}\not\in\mathbb N.\end{array}\right. $$ Hence, $$ \lvert a_n\rvert^{1/n}=\left\{\begin{array}{lll} \big((k!)^2\big)^{1/k^2} & ...


3

The radius of convergence is $R=1/\limsup_{m\longrightarrow\infty}|c_m|^{\frac{1}{m}}$ (see Wikipedia). In your case $c_m=0$ if $m$ is not of the type $n^2$ $c_{n^2}=n!^2$ If you compute $\lim_n(n!^2)^{\frac{1}{n^2}}$, it is 1, because $n^2$ is very strong. Then $\limsup_{m\longrightarrow\infty}|c_m|^{\frac{1}{m}}=\max(0,1)=1$ and $R=1$.


0

Taylor's theorem is the generalization of linear approximations. Let $f(x)$ be a differentiable function and $p$ some fixed point. Define the linear function $L(h) = f(p) + f'(p)h$. This is function approximates $f(p+h)$, when $h$ is a small number. To be more precise by what "approximates", we are saying that, $f(p+h) = L(h) + \varepsilon(h)$, where ...


1

Substitute $y=x^{3}$ $$S = 1-y+y^{2}-y^{3}+\cdots$$ $$S = \frac{1}{1+y} = \frac{1}{1+x^{3}}$$


2

Hint: if $|f(x)| < 1$ then $$\frac{1}{1+f(x)} = 1 - f(x) + f(x)^2 - f(x)^3 + \dotsb$$


1

Hint: remember that for $\lvert x\rvert <1$, $$\frac{1}{1-x} = \sum_{n=0}^\infty x^n = 1+x+x^2+x^3+\dots$$ This gives you the power series expansion at $0$. For the one at $-1$, write $x=-1+\varepsilon$ and $$\frac{1}{1-x} = \frac{1}{2-\varepsilon} = \frac{1}{2}\frac{1}{1-\frac{\varepsilon}{2}}$$ and use the same formula.


1

$$y=y(a)+y'(a)(x-a)+\frac{y''(a)}{2!}(x-a)^2$$ $$y'=6x-6$$ $$y''=6$$ at x=a $$y'(a)=6a-6$$ $$y''(a)=6$$ subsitute in the general series to get $$y=3a^2-6a+5+(6a-6)(x-a)+3(x-a)^2$$ $$y=3a^2-6a+5+6ax-6a^2-6x+6a+3x^2-6ax+3a^2=3x^2-6x+5$$ that means the function not change for any value of $a$


0

There is an a way to do this in which it will be clear that the power series is equal to $f(x)$. Let $x=y+a$, then $$\begin{align} f(x) & =3(y+a)^2-6(y+a)+5=3y^2+6ay+3a^2-6y-6a+5\\ & =3y^2+(6a-6)y+(3a^2-6a+5) \end{align}$$ and since $y=x-a$, $$f(x)=3(x-a)^2+(6a-6)(x-a)+(3a^2-6a+5).$$


0

You made some computational mistakes. $$f(x) = f(a) + f^{\prime}(a)(x-a) + f^{\prime\prime}(a) (x-a)^{2}/2! + \dots$$ In your case, $$f(x) = 3a^{2} -6a + 5 + (6a-6)(x-a) + 3(x-a)^{2}$$ with all higher derivatives equal to zero. Simplify the RHS from here.


0

A power series, following Wikipedia, is an infinite sum $$ \sum_{n=0}^{\infty} a_nx^n, $$ where $a_n$ are the coefficients of the $n$-th term. If for some $n\in \mathbb{N}$ it holds that $a_m=0$ when $m\geq n$, and $a_n\neq 0$, then you obtain a polynomial of degree $n$.


2

this equation can't be solved in the known elementary functions the solution of the equation above contains the Error-function $$w(x)=c_1 e^{\frac{x^2}{2}}+\sqrt{\frac{\pi }{2}} e^{\frac{x^2}{2}+\frac{1}{2}} \text{erf}\left(\frac{x+1}{\sqrt{2}}\right)$$



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