# Tag Info

8

Using the binomial theorem and the Cauchy convolution we have: $$\begin{eqnarray*}[z^h]\left(\frac{1+z}{1-z}\right)^{1/3}&=&(-1)^h\sum_{l\leq h}\binom{-1/3}{l}\binom{1/3}{h-l}(-1)^l\\ &=& (-1)^h\binom{-1/3}{h}\cdot\phantom{}_2 F_1\left(-\frac{1}{3},-h;\frac{2}{3}-h,-1\right).\end{eqnarray*}\tag{1}$$ Since: $$\lim_{h\to+ \infty}\phantom{}_2 ... 7 Note that$$\frac{1}{i!}\left(\frac{e^2-1}{2}\right)^i \prod_{j=0}^i (x-2j) = x \prod_{j=1}^i \frac{\frac{x}{2}-j}{j} (e^2-1)^i = x\binom{\frac{x}{2}-1}{i}(e^2-1)^i,$$so$$\sum_{i=0}^\infty \frac{1}{i!}\left(\frac{e^2-1}{2}\right)^i \prod_{j=0}^i (x-2j) = x\sum_{i=0}^\infty \binom{\frac{x}{2}-1}{i}(e^2-1)^i = x\left(1+(e^2-1)\right)^{\frac{x}{2}-1} = x ...

5

Hint: For which values of $y$ does $$\sum_{n=1}^{\infty} y^n$$ converge? Now let $$y = \frac{1}{1+|z|^2}.$$

5

Hint Start with $$\frac{1}{1-x}=\sum_{i=0}^{\infty}x^i$$ and integrate with respect to $x$. You then have $$\log(1-x)=-\sum_{i=1}^{\infty}\frac{x^i}{i}$$ So, $$x^n \log(1-x)=-\sum_{i=1}^{\infty}\frac{x^{n+i}}{i}$$ Added later to this answer If you want to start with the derivative and then integrate, there is no problem. If $$f(x)=x\log(1-x)$$ ...

4

Let $w = \frac{z+1}{z-1}$. Then you have a power series in $w$, centered at $0$. Find its radius of convergence, call that $R$. Then find which $z$ correspond to $\lvert w\rvert < R$. The map $z \mapsto \frac{z+1}{z-1}$ can be explicitly inverted.

4

Write $$\left(\dfrac{1+x}{1-x}\right)^{1/3} = \frac{(1+x)^{1/3} - 2^{1/3}}{(1-x)^{1/3}} + \frac{2^{1/3}}{(1-x)^{1/3}} =: f(x)+g(x).$$ The function $f$ is continuous on the disk $|z| \leq 1$, so by Darboux's method (Thm. VI.14 in Analytic Combinatorics) we have $$[x^n] f(x) = o\left(\frac{1}{n}\right).$$ The coefficients of the second series are given ...

4

If $x\leqslant1$, then $a_n\geqslant1$ for every $n$ hence the series $\sum\limits_na_n$ diverges. If $x\gt1$, then the expansion $x^{1/n}=\exp((\log x)/n)=1+(\log x)/n+o(1/n)$ yields $$\log(a_{n+1}/a_n)=\log(2-x^{1/n})\sim-(\log x)/n,$$ hence $\log a_n\sim-(\log x)\cdot\log n$. Thus: If $x\gt\mathrm e$, one can pick $1\lt y\lt\log x$ then ...

4

You essentially solved it; just make the substitution $$f(x) = \frac{1}{2+3x} = \frac{1}{2}\left(\frac{1}{1-\left(\frac{-3x}{2}\right)} \right) = \frac{1}{2}\sum_{n=0}^\infty \left(\frac{-3x}{2}\right)^n.$$ Note that $1/(1-x)$ converges when $|x| < 1$, though, so in this problem the radius of convergence is $$\left|\frac{-3x}{2}\right| < 1 \implies ... 3 I would really like to know the closed form given in the book. The following answer was given for the original post. The only thing I have been able to do to arrive to something is to replace n! by Stirling approximation, that is to say$$n! \simeq \sqrt{2 \pi } e^{-n} n^{n+\frac{1}{2}}$$and using this, the result of the summation is given by ... 3 A rigorous way to define the sine function is to consider it as the solution to the IVP:$$ \begin{cases} y^{\prime \prime} + y = 0\\ y(0) = 0 \\ y^{\prime}(0) = 1 \end{cases} $$3$$\lim_{n \to \infty}\sqrt[n]{\frac{2^n+1}{n}} = \lim_{n \to \infty}\sqrt[n]{\frac{2^n}{n}} =\lim_{n \to \infty}\frac{2}{ \sqrt[n]{n}}=2$$Why can we omit the +1 term? Good question. ;) Write$$2^n+1 = 2^n(1+\frac{1}{2^n})$$As n \to \infty, the limit of both sides becomes equal. Therefore$$\lim_{n \to \infty} 2^n+1 = \lim_{n \to \infty} ...

3

By the root test we have $$\left|2^n\frac{(4z-8)^n}n\right|^{1/n}\xrightarrow{n\to\infty}2|4z-8|<1\iff|z-2|<\frac18\iff z\in B\left(2,\frac18\right)$$ so the radius is $\frac18$.

3

You can't apply the limit test because you don't know the limits $|\frac{a_n}{a_{n+1}}|$ or $|\frac{b_n}{b_{n+1}}|$. If these limits exist, you can use them to find a RoC, but knowing the radius doesn't imply that the limits exist.* To show the answer is $3$, recall that a power series converges absolutely (and uniformly) inside it's radius of convergence. ...

3

Note that $S'(x)=\displaystyle\sum\limits_{k=0}^\infty \frac{(-1)^k x^{2k}}{(2k)!}$. The first term here is $1$, so when you differentiate, it dies. Then $$S''(x)=\sum_{k\geqslant 1}\frac{(-1)^k x^{2k-1}}{(2k-1)!}$$ i.e. we start the sum at $k=1$. Can you continue? All you need to do now is shift an index. Note that we repeatedly used $(x^n)'=nx^{n-1}$ and ...

2

For one direction, if $f$ has a removable singularity or a pole of order $k$ in $0$, what can you say about the function $z\mapsto z^m f(z)$ for large enough $m$? For the other direction, note that if $f$ has an essential singularity in $0$, then so has $z \mapsto z^m f(z)$ for all $m \in \mathbb{Z}$. Appeal to the Casorati-Weierstraß theorem if necessary.

2

Hint: Note that $\displaystyle\sum_{j=0}^{\infty}(j-2)a_{j}x^{j}=(x\frac d{dx}-2)\sum_{i=0}^{\infty}a_{i}x^{i}$

2

As told in answers and comments, a series exists but it is quite tedious to derive. Writing $$\frac{\displaystyle\sum_{i=0}^{\infty}a_{i}x^{i}}{\displaystyle\sum_{j=0}^{\infty}(j-2)a_{j}x^{j}}=\sum_{i=0}^{\infty}b_{i}x^{i}$$ the first coefficients are given by $$b_0=-\frac{1}{2}$$ $$b_1=-\frac{a_1}{4 a_0}$$ $$b_2=\frac{a_1^2-4 a_0 a_2}{8 a_0^2}$$ ...

2

You have $\displaystyle G(x)=(1+x+x^2+x^3)^5=\big(\frac{1-x^4}{1-x}\big)^5=(1-x^4)^5(1-x)^{-5}$ $\displaystyle=\big(1-5x^4+\binom{5}{2}x^8-\binom{5}{3}x^{12}+\cdots\big)\big(\sum_{k=0}^{\infty}\binom{k+4}{4}x^4\big)$, so now you just have to find the coefficient of $x^{12}$ in this expression.

2

Some hints: When $\sum_{k=1}^\infty c_k$ converges then $\sum_{k=1}^\infty c_k\>\rho^k$ converges absolutely for every positive $\rho<1$. Distinguish the cases $|z|<1$ (easier case) and $|z|>1$. Solution: When $\sum_{k=1}^\infty c_k$ converges then there is an $M>0$ with $|c_k|\leq M$ for all $k\geq1$. Assume $|z|=:\rho<1$. As ...

2

Hint: Use $$2 - \sin x \leq 3$$ and then compare with a geometric series.

2

Here is a variant of this question asked on MathOverflow: http://mathoverflow.net/q/49395/. In particular, your question is answered by the cited result that $D$ can be any $G_\delta$ subset of the circle (such a set can be uncountable with dense complement).

2

The thought process should probably begin with: let's look at some function with a simple pole on the unit circle, and with a power series that I can find. This should remind of the example $$\frac{1}{1-z} = \sum_{n=0}^\infty z^n$$ Here the coefficients are all equal to $1$, so they are bounded. More generally, we can place a simple pole at any point $a$ ...

2

Hint. Check that $$\sum_{k=0}^{n} \binom{n}{k} (-1)^{k} \frac{n^{2}+n+1}{nk+n+1} = \prod_{k=1}^{n} \left( 1+\frac{1}{nk} \right)^{-1}.$$ The starting idea is to write $$\sum_{k=0}^{n} \binom{n}{k} (-1)^{k} \frac{n^{2}+n+1}{nk+n+1} = (n^{2}+n+1) \sum_{k=0}^{n} \binom{n}{k} (-1)^{k} \int_{0}^{1} x^{nk+n} \, dx.$$

2

Your ratio is incorrect. You should have: $$\frac{c_n}{c_{n+1}}=\frac{2^{3n}(n+1)}{2^{3(n+1)}n}=\frac{2^{3n}(n+1)}{2^{3n+3}n}=\frac{n+1}{2^3n}$$

2

Hint This is not really a question about complex numbers. Think geometric series. For the question as it stands now, you do not have to use the convergence radius expression. But to answer your question about $a_n$, if you will use the Ratio Test then you should use $a_n=\frac{1}{(1+|z|^2)^n}$. Note that our series is not a power series in the usual sense. ...

2

In general, to get the power series (or Taylor series at $x=0$) for $f(g(x))$ we need the Taylor series for $g(x)$ around $0$ and the Talyor series of $f(x)$ around $g(0)$. In the case of $\sqrt{\cos x}$, that means $f(x)=\sqrt{x}$ needs a Taylor series for $\sqrt{x}$ around $x=1$, which is: $$\sqrt{x} = \sum_{k=0}^\infty \binom{1/2}{k} (x-1)^k$$ Even if ...

2

Some other important formulas regarding $\sin(x)$ that you didn't mention are the infinite product $$\sin(x) = x \prod_{k=1}^{\infty}\Big( 1 - \frac{x^2}{\pi^2 k^2} \Big)$$ and the partial fractions decomposition $$\frac{1}{\sin(x)^2} = \sum_{k=-\infty}^{\infty} \frac{1}{(x-\pi k)^2}, \; \; x \notin \pi \cdot \mathbb{Z},$$ although I guess the latter only ...

2

HINT: Use binom expansion: $$(1+a)^{y}=1+a.y+\frac{a^2}{2!}.y(y-1)+\frac{a^3}{3!}.y(y-1)(y-2)+....$$ $y=\frac{x}{2}$ $$(1+a)^{\frac{x}{2}}=1+a.\frac{x}{2}+\frac{a^2}{2!}.\frac{x}{2}(\frac{x}{2}-1)+\frac{a^3}{3!}.\frac{x}{2}(\frac{x}{2}-1)(\frac{x}{2}-2)+....$$ ...

2

It suffices to show that the coefficients $a_n$ cannot converge to $0$. Now suppose that $|a_k| \leq \varepsilon$ for $k \geq m$. Then for $z\in\mathbb{D}$ $$|f(z)|\leq \frac{\varepsilon}{1-|z|}+\sum_{k=0}^{m-1}|a_k|.$$ This implies that if $a_k$ converges to $0$ then $\lim_{|z|\to 1}(1-|z|)|f(z)|=0$. In particular $f$ cannot have a pole on $\mathbb{T}$ ...

2

The given series is convergent by the Leibniz's theorem but it isn't absolutely convergent since the harmonic series is divergent hence this series is conditionally convergent.

Only top voted, non community-wiki answers of a minimum length are eligible