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0

The series you present is not a Taylor series for G(z) about the origin. A Taylor series would be of the form $G(x)= a_0+a_1 x+a_2 x^2+...$. It would be easier to compute the radius of convergence of $1-2 G(x)=(1-4abz^2)^{-1/2}$ as it's the same radius as for $G(x)$. A special Taylor series discovered by Newton, is the generalized binomial theorem : If ...


1

Consider the series for $\cosh (2z)$ and divide it by $z$....


3

HINT: $$z\cdot S=z\sum_{k=0}^\infty \frac{2^{2k}z^{2k-1}}{(2k)!}=z\sum_{k=0}^\infty\dfrac{(2z)^{2k}}{(2k)!}$$ Now $e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$ $2\cosh(y)=e^y+e^{-y}=?$


1

Let me just add one thought: We can view this as looking at the real analytic $f(x,y) = \sum a_{mn}x^my^n$ along the line $(t,1-t), t \in \mathbb R.$ Can such an $f$ vanish on a line without being identically $0,$ i.e., without having all $a_{mn} = 0?$ Sure, happens all the time. For example the function $xy$ vanishes on the axes. It's thus clear that for ...


4

Oops. The answer is trivially no - the functions are not linearly independent! (I've been thinking about why one of them is not in the closed span of the others; in fact one of them is in the span of the others.) If you're allowing $m=0$ and $n=0$ then $(-1) + (x) + (1-x)$ is a counterexample (all but three of the coefficients vanish). If you're assuming ...


0

Here is yet another answer. Your sum is: $$\sum_{j=0}^n2^j=\frac{1}{2}(2)\sum_{j=0}^n2^j$$ $$=\frac{1}{2}\sum_{j=0}^n2^j(2)$$ $$=\frac{1}{2}\sum_{j=0}^n2^{j+1}$$ $$=\frac{1}{2}\sum_{k=1}^{n+1} 2^k$$ $$=\frac{1}{2}\left(-2^0+2^0+\sum_{k=1}^n2^k+2^{n+1}\right)$$ $$2\sum_{j=0}^n2^j=\sum_{k=0}^n2^k+2^{n+1}-2^0$$ $$\sum_{j=0}^n2^j=2^{n+1}-1$$


0

Let $S$ this sum, then by difference of $2S$ and $S$ we have: \begin{array}{cccccc} 2S&=& & &2&+&4&+&8&+&\ldots&+&2^n&+&2^{n+1}\\ -S&=&-1&-&2&-&4&-&8&-&\ldots&-&2^n&\\ \hline ...


1

One can write $$S=1+2+\cdots+2^n,$$ and note that $$S=2S-S.$$ Hence, since $$2S=2(1+2+\cdots+2^n)=2+4+\cdots+2^{n+1}=S-1+2^{n+1},$$ we have $$S=2S-S=S-1+2^{n+1}-S=2^{n+1}-1.$$


0

Well this is a GP with ratio $2$ Having $n+1$ terms So the sum will be $$\frac{1\{2^{(n+1)}-1\}}{(2-1)}= 2^{(n+1)}-1$$


5

In binary you get the number $\underbrace{11\dots 11}_n$ which is the number that goes before $\underbrace{100\dots 00}_{n+1}=2^{n+1}$


2

\begin{align} S&=1+2+4+\dotsb+2^n\\ 1+S&=1+1+2+4+\dotsb+2^n\\ &=2+2+4+\dotsb+2^n\\ &=4+4+\dotsb+2^n\\ &=8+\dotsb+2^n\\ &\dotsb\\ &=2^n+2^n\\ 1+S&=2^{n+1}\\ S&=2^{n+1}-1 \end{align} Write it out for a specific example to understand this better.


1

There are two ways to compute the power series. In each way you need to compute the series for $g(z) = 1/z$ using the geometric approach. $$g(z) = \frac{1}{z} = -\frac{1}{i - (z-i)} = i \frac{1}{1 - (\frac{z-i}{i})} = \sum_{k=0}^{\infty} (-i)^{k-1} (z-i)^k$$ for $|z - i| < 1$. Now you have 2 options (I personally prefer option number 2): Use the ...


1

You can simplify your expression. For example, $$\frac{(n!)^2}{(n+1)!^2} = \left(\frac{n!}{(n+1)!}\right)^2 = \left(\frac{n!}{n!\cdot (n+1)}\right)^2$$ which you can easily simplify...


1

Remember that $(n+1)!=(n+1)n!$ and so on. Hence $$\eqalign{ {(2n+2)!\over(2n)!}{(n!)^2\over(n+1)!^2} &=\frac{(2n+2)(2n+1)(2n)!}{(2n)!}\frac{(n!)^2}{(n+1)^2(n!)^2}\cr &=\frac{(2n+2)(2n+1)}{(n+1)^2}\cr &=\frac{2(2n+1)}{n+1}\cr &=\frac{4+\frac2n}{1+\frac1n}\cr}$$ which tends to $4$ as $n\to\infty$. The case involving $(3n)!$ is very ...


2

Hint \begin{equation} \frac{(2n + 2)!}{(2n)!} \frac{(n!)^2}{(n + 1)!} = \frac{(2n + 2)(2n + 1)}{(n + 1)(n + 1)} = \frac{4n^2 + 6n + 2}{n^2 + 2n + 1} = \frac{4 + \frac{6}{n} + \frac{2}{n^2}}{1 + \frac{2}{n} + \frac{1}{n^2}} \end{equation}


2

Denote $\varphi=\arg{z}.$ Then $$\dfrac{\left(2e^{i\varphi}\right)^{3n}}{8^n (1-in)}=\dfrac{e^{3in\varphi}}{1-in}=\dfrac{e^{3in\varphi}(1+in)}{1+n^2}=e^{3in\varphi}\cdot\dfrac{1}{1+n^2}+ie^{3in\varphi}\cdot\dfrac{n}{1+n^2}.$$ The series $\sum\limits_{n=0}^{\infty}{e^{3in\varphi}\cdot\dfrac{1}{1+n^2}}$ is absolutely convergent, and ...


1

HINT: The series $\sum_{n=1}^\infty \frac{e^{inx}}{n}$ converges whenever $x\ne 2\ell \pi$ for integer-values of $\ell$. Here, the series of interest is asymptotically $i\sum_{n=1}^\infty \frac{e^{inx}}{n}$ with $x=3\arg(z)$.


-1

False as range(f) cannot have dimension one { Consequence of Open mapping theorem} unless f is a constant. True as f(n+1/n)=0 for all n implies f assumes value 0 in the nbd. of each n€N. By Identity theorem f must be identically zero. False as f(1/n)=0 for all n implies f assumes value 0 in the nbd. of zero. Use Identity theorem to get f is identically zero ...


0

Here's the input circle being slowly warped to the resulting output fractal. gif Code.


1

You're approach is fine (besides the missing integration constant according to @Winther's comment). Hint: Note, that you could also obtain a power series representation by applying the binomial series to your expression, since \begin{align*} \left(\frac{x}{2-x}\right)^3 =\left(\frac{x}{2}\right)^3\left(1-\frac{x}{2}\right)^{-3} =\ldots \end{align*} ...


1

Yet another way is to recognize that the single sum can be written as a double sum. To that end, we have $$\begin{align} \sum_{k=1}^{\infty}\frac{3k}{7^{k-1}}&=21\sum_{k=1}^{\infty}\frac{k}{7^k}\\\\ &=21\sum_{k=1}^{\infty}\frac{1}{7^k}\sum_{\ell=1}^{k}\,1\\\\ &=21\sum_{\ell=1}^{\infty}\sum_{k=\ell}^{\infty}\frac{1}{7^k}\\\\ ...


6

Let $S=\sum_{k\geq 1}\frac{k}{7^k}$. Then: $$ 6S = 7S-S = \sum_{k\geq 1}\frac{k}{7^{k-1}}-\sum_{k\geq 1}\frac{k}{7^k} = \sum_{k\geq 0}\frac{k+1}{7^k}-\sum_{k\geq 1}\frac{k}{7^k}=1+\sum_{k\geq 1}\frac{1}{7^k}=\frac{7}{6}$$ hence: $$ \sum_{k\geq 1}\frac{3k}{7^{k-1}} = 21\cdot S = 21\cdot\frac{7}{36} = \color{red}{\frac{49}{12}}. $$


7

Using the fact that $$\sum_{n=0}^\infty x^n=\frac 1{1-x},\quad |x|<1,$$ differentiate both sides to get $$\sum_{n=1}^\infty nx^{n-1}=\frac 1{(1-x)^2},\quad |x|<1.$$ Now, $$\sum_{n=1}^\infty\frac{3n}{7^{n-1}}=\frac 3{(1-1/7)^2}.$$


1

This solution is based on the identity $\sum_{r=0}^\infty\,\binom{n+r}{r}x^r=(1-x)^{-n-1}$. I know that your sum starts with $n_1,n_2,\ldots,n_k\geq 1$, but for the sake of simplicity, I will calculate the sum starting with $n_1,n_2,\ldots,n_k\geq 0$. You could probably easily modify my answer to get the sum that starts with $n_1,n_2,\ldots,n_k\geq 1$. ...


1

Let $x \in \mathbb{R}$ be such that $\lvert x \rvert > 6$. If $n$ is a positive integer such that $\lvert \sin(n) \rvert \ge 1/2$, we have $$ \left\lvert \frac{n^3 sin(n) x^n}{6^n} \right\rvert \ge \lvert\sin(n)\rvert\left\lvert\frac{x}{6}\right\rvert^n > \frac{1}{2} $$ Since there are infinitely many such $n$, this inequality shows that $\frac{k^3 ...


2

Here is a simple method that works well to find the first few terms in the power-series. We first expand the denominator in a power-series around $z=0$: $$\frac{z}{\sin(z)} = \frac{1}{1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \ldots}$$ This has the form of a geometrical series $\frac{1}{1-x} = 1+x + x^2 + \ldots$ for $x = \frac{z^2}{3!} - \frac{z^4}{5!} + ...


0

Note that $f$ is odd. Thus, it suffices to look at the interval $[0, \infty)$. Lemma: For $x$ > $\pi$, we have $$\sin(x) > \frac{(\pi^2-x^2)x}{\pi^2+x^2}.$$ Proof: See here. Thus, for $x > \pi$, we have $$ \frac{1}x - \frac{\sin(x)}{x^2} < \frac{1}x + \frac{\pi^2-x^2}{x(\pi^2+x^2)} := g(x).$$ For $x > \pi$ (or even $x > 0$), the RHS ...


4

Using $f(x)=\frac{x-\sin(x)}{x^2}$, the inequality is equivalent to: $$ \frac{x-\sin(x)}{x^2}≤\frac{1}{\pi}\iff x\left(1-\frac{x}{\pi}\right)≤\sin(x) $$ Thus, define $g(x):=x\left(1-\frac{x}{\pi}\right)$. Note that $g(x)=g(\pi-x)$ and $\sin(x)=\sin(\pi-x)$. Therefore it suffices (by symmetry) to prove the inequality for ...


-1

That is o.k. for obtaining which solution get max and which min use second derivative test: if $f^"(a)<0$ then $a$ get max and if $f^"(a)>0$ then $a$ get min.


0

Here is an "almost" solution for integer coefficients. The zeros of $x^4+12x^3+14x^2-12x+1$ are algebraic integers. So are their reciprocals. Now in the series $$ (1-xa)^{-1/4} = 1+\frac{a}{4} x+\frac{5 a^2}{32} x^2 +\frac{15a^3}{128} x^3 + \cdots $$ where $a$ is an alebraic integer, the coefficients are alebraic integers (except for power-of-2 ...


0

Wolfram Alpha can find the inverse in terms of $c$. The denominators of each term are easy to express in closed form. The polynomials in the numerators seem to follow Euler's triangle.


1

Consider the power series \begin{align} \sum_{n=1}^{\infty}{a_n (x-c)^n}. \end{align} Define \begin{align} R &= \frac{1}{\limsup_{n \to \infty}{|a_n|^{1/n}}}. \end{align} So, \begin{align} \limsup_{n \to \infty}{|a_n|^{1/n}} &= \frac{1}{R}. \end{align} We note \begin{align} \limsup_{n \to \infty}(|a_n (x - c)^n|)^{1/n} &= |x-c| \limsup_{n ...


2

The radius of convergence of $f(x) =\sum_{n=0}^{\infty} a_n x^n $ is defined as the value $r$ such that $f(x)$ converges for $|x| < r$ and diverges for $|x| > r$. As zhw said in a comment, it is a theorem that follows from the definition that, if $\lim_{n \to \infty} |a_n|^{1/n} = 1/R$, then $R$ is the radius of convergence.


1

Let $f(x)=\sum_{n\ge 1}a_n x^n$ be a power series, and set $R$ as above. Then, if $|x|>R$ it implies that $f(x)$ diverges. Conversely, if $|x|<R$ then it converges. Notice that if $|x|=R$ then you don't know the behaviour of $f(x)$.


1

The first proof needs some improvement. Note that $|z| \le R_2$ does not imply the convergence of $P(z)$. Only if $|z| < R_2$ you can infer the convergence of the power series [and this needs to be proven!]. For the second part you first need to show that the sequence stays bounded for any $z$ with $|z| < R_1$. This is easy to prove, but not trivial. ...


0

In differentiating you have forgeted a $\frac{1}{2}.$ The derivative of \begin{align*} \sum_{n=0}^{\infty} \frac{1}{2} \bigg( \frac{x}{2} \bigg)^{n} \end{align*} is \begin{align*} \sum_{n=1}^{\infty} \frac{n}{2} \bigg( \frac{x}{2} \bigg)^{n-1}.\frac{1}{2} \end{align*} Which you forgot the last $ \frac{1}{2}$ and so the answer is is \begin{align*} ...


0

Both series are correct. The one from the lecture is the series expansion around $x=0$, while the one derived in the posted question is the series expansion around $x=2$. And one could choose other arbitrary points around which to expand the function. Using a straightforward approach we see that for $f(x)=\log(3-x)$, we have for $n>0$ ...


0

Hint. Your route is OK, but you should rather start with $$ \ln(3-x) = -\int_0^x { \frac{1}{3-t} dt}+\ln 3 $$ then follow the same path to obtain the right answer.


0

I hope this works. The text around the sum is in danish so try to ignore that, but my professor rewrites the power series and I can't really see how he removes the $2n$.


1

Keep in mind that: $$\frac{\partial f(g(x))}{\partial x}=\frac{\partial f(g(x))}{\partial g(x)}\frac{\partial g(x)}{\partial x}$$ In you case $f(g(x))=(2x-2)^n$ where $g(x)=2x$, so $\dfrac{\partial g(x)}{\partial x}=2$ and $\dfrac{\partial f(g(x))}{\partial g(x)}=n(2x-2)^{n-1}$ etc...


1

HINT: refer to the chain rule. $$ \dfrac{d}{dx} \Big((2x -2)^n \Big) = n (2x-2)^{n-1} \dfrac{d}{dx}\Big(2x -2 \Big) = 2n(2x-2)^{n-1} $$ Therefore $$ \dfrac{df}{dx} = \dfrac{d}{dx} \left( \sum_{n \geq 1} \dfrac{(2x-2)^n}{n2^n+1}\right) = \sum_{n \geq 1}\left( \dfrac{1}{n2^n+1} \dfrac{d}{dx} \left((2x-2)^n\right)\right) = \sum_{n \geq ...


2

Let $f(x) = \sum_{n=1}^{\infty}a_nx^n $. What is needed is to get terms in the power series that are the terms in the recurrence. Those are $n a_n$, $(n-1)a_{n-1}$, and $a_{n-1}$. The operations that are typically used and differentiation, integration, and multiplying or dividing by a power of $x$. To get just $a_{n-1}$, multiply by $x$. Then $xf(x) ...


5

Setting $$ h(x)=a_0+\sum_{n=1}^\infty a_nx^n, $$ we have \begin{eqnarray} 3h'(x)&=&\sum_{n=1}^\infty 3na_nx^{n-1}=\sum_{n=1}^\infty[2a_{n-1}-3(n-1)a_{n-1}]x^{n-1}=2\sum_{n=1}^\infty a_{n-1}x^{n-1}-3\sum_{n=1}^\infty (n-1)a_{n-1}x^{n-1}\\ &=&2\sum_{n=0}^\infty a_nx^n-3\sum_{n=0}^\infty na_nx^n=2\sum_{n=0}^\infty a_nx^n-3x\sum_{n=1}^\infty ...


2

For the periodicity mod $5$, note that $$(1 - x^4)^4 \equiv (1 + 3 x + 4 x^2 + 2 x^3)^4 (1-12 x+14 x^2+12 x^3+x^4) \mod 5 $$ and thus in the field $\mathbb Z_5((x))$ of formal Laurent series in $x$ over the integers mod $5$ we have $$ \eqalign{&\dfrac{1}{1-12 x+14 x^2+12 x^3+x^4} = \left(\dfrac{1+3x+4x^2+2x^3}{1-x^4}\right)^4\cr &= ...


2

Here is a proof for positivitiy. Use the recurrence $$ (n+1)Q(n)+(12n+21)Q(n+1)+(14n+35)Q(n+2)+(-12n-39)Q(n+3)+(n+4)Q(n+4) = 0 $$ $Q(0)=1,Q(1)=3,Q(2)=19,Q(3)=147$. Prove by induction that $Q(n) \ge 3 Q(n-1)$ for $n \ge 1$. This is true for the first few terms. Assume true up to $Q(n+3)$, then prove it for $Q(n+4)$ as follows: $$ (n+4)Q(n+4) = ...


2

Let $L$ denote the matrix $$ L = \pmatrix{u \mathrm{I} + i S_3 && i S_- \\ i S_+ && u \mathrm{I} - i S_3} $$ My best guess is that whatever the author is getting at has something to do with the fact that $$ \operatorname{tr}(L^N) = 2I\,u^N + q_{2}u^{N-2} + \cdots + q_N $$ or, if $N$ is odd, $$ \operatorname{tr}(L^N) = 2I\,u^N + q_{2}u^{N-2} + ...


1

Another obstruction is that the proof has to work with $x$ replaced with $ax$ for positive $a$, which changes the $n$'th coefficient by a factor of $a^n$. A more accessible question is to find nice conditions on coefficients of a formal power series $1 + \sum a_n x^n$ so that the logarithm or the inverse of the series have negative coefficients.


0

Since $\|M^s B R^s\| \le \|M\|^s \|B\| \|R\|^s$ (using any sub-multiplicative matrix norm), the series converges if $\|M\| \|R\| < 1$ (this is a sufficient condition, not a necessary one). The sum $S$ will have to satisfy $$ S = B + MSR $$ Since the function $S \mapsto B + MSR$ is a strict contraction, this uniquely defines $S$. Note that if $v$ is ...


2

Let $A$ be the sum of the series, assuming it converges. Then $MAR = \sum_{s = 1}^\infty M^sBR^s = A - B$ and therefore $$ MAR - A = -B \, . $$ This matrix equation may be written as $$ -R^T \otimes M \cdot vec(A) + vec(A) = vec(B) $$ where $\otimes$ denotes the Kronecker product and $vec(A)$ is the vectorization of $A$ (stack all columns into a single ...


3

$$\dfrac{2x}{10+x}=\dfrac{2x}{10\left(1+\dfrac x{10}\right)}=\dfrac x5\left(1+\dfrac x{10}\right)^{-1}$$ For $\left|\dfrac x{10}\right|<1,$ $$\left(1+\dfrac x{10}\right)^{-1}=\sum_{r=0}^\infty\left(-\dfrac x{10}\right)^r$$



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