# Tag Info

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 $$\frac{(2n + 2)!}{(2n)!} \frac{(n!)^2}{(n + 1)!} = \frac{(2n + 2)(2n + 1)}{(n + 1)(n + 1)} = \frac{4n^2 + 6n + 2}{n^2 + 2n + 1} = \frac{4 + \frac{6}{n} + \frac{2}{n^2}}{1 + \frac{2}{n} + \frac{1}{n^2}}$$

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$5we 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 LetL$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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