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7

Here is a slightly more general strategy that can be adapted here: If $|r| < 1$, we have $$\sum\limits_{x = 0}^{\infty} r^x = \frac{1}{1 - r}$$ Taking a derivative on both sides leads to $$\sum\limits_{x = 1}^{\infty} x r^{x - 1} = \frac{1}{(1 - r)^2}$$ or by a change of indices, $$\sum\limits_{x = 0}^{\infty} (x + 1) r^{x} = \frac{1}{(1 - r)^2}$$ ...

7

We have \begin{align} \sum_{i=1}^n\sum_{j=1}^i(2^j-1) &=\sum_{i=1}^n\sum_{j=1}^i2^j-\sum_{i=1}^n\sum_{j=1}^i1\\ &=\sum_{i=1}^n(2^{i+1}-2)-\sum_{i=1}^ni\\ &=2\sum_{i=1}^n2^{i}-\sum_{i=1}^n2-\sum_{i=1}^ni\\ &=2(2^{n+1}-2)-2n-\frac12n(n+1)\\ &=2^{n+2}-4-\frac52n-\frac12n^2\\ \end{align}

6

For good ballpark estimates, use the fact that $2^{10}=1024\approx 1000$. So $10$ doublings is about the same as multiplying by $1000$. Since $10^{82}=10\times 1000^{27}$, we can see that $270$ doublings will get us past $10^{81}$. There is some slack, and another $3$ doublings get us over.

5

They do not provide a derivation, but this is actually written up in Wikipedia. I use the standard notation $e(x) = \exp(2 \pi i x)$. Assuming $\gcd(k,n) = 1$, we have $$\sum_{x \in \mathbb{Z}/n \mathbb{Z}} e\left(\frac{kx^2}{n} \right) = \left\{ \begin{array}{lcl} \varepsilon_n \left( \frac{k}{n} \right) \sqrt{n} & & n \equiv 1 \pmod{2}\\ 0 ... 5 Clearly,$$\begin{align*} \int_{0}^{\infty} \sin mx \sin nx \; dx &= \frac{1}{2}\int_{-\infty}^{\infty} \sin mx \sin nx \; dx \\ &= \frac{1}{4}\int_{-\infty}^{\infty} (\cos (m-n)x - \cos(m+n)x ) \; dx \\ &= \frac{1}{4}\int_{-\infty}^{\infty} (e^{i(m-n)x} - e^{i(m+n)x}) \; dx \\ &= \frac{\pi}{2}(\delta(m-n) - \delta(m+n)), \end{align*}$$in ... 5 Recall the geometric series (see Wikipedia): for any y with |y|<1,$$\frac{1}{1-y}=1+y+y^2+\cdots=\sum_{k=0}^\infty y^k.$$Therefore, for any such y, we also have$$\frac{y}{1-y}=y+y^2+y^3+\cdots=\sum_{k=1}^\infty y^k.$$Now let y=e^{-x} (though observe that we need x>0 to have e^{-x}<1). 4 If x, y, z>1, then y^z, z^x, x^y \geq 2 so x^2y^2z^2 \mid x^{y^z}y^{z^x}z^{x^y}=5xyz so xyz \mid 5, impossible since x, y, z \geq 2. Thus at least one of x, y, z is 1. The equation is cyclic, so we may assume z=1, so the equation becomes x^yy=5xy so x^{y-1}=5. It is now clear that we must have x=5, y=2. Therefore all solutions are ... 4 The nth step is 2^n. If you want 2^n\geq A, then you want$$n = \log_2(2^n) \geq \log_2(A) = \frac{\ln(A)}{\ln(2)} = \frac{\log_{10}A}{\log_{10}(2)}.$$So the first n at which 2^n\geq A will be the least positive integer greater than or equal to \log_2(A), which is denoted$$\left\lceil \log_2A \right\rceil.$$3 This is a partial sum of a geometric series:$$\left| \frac{1}{n} \sum_{m=1}^n e^{2 \pi i h (2 \pi m)} \right| =\left| \frac{1}{n} \sum_{m=1}^n q^m \right| =\left| \frac{q}{n} \frac{1-q^n}{1-q}\right| \le \frac{2}{n\left|1-q\right|}$$with q:=e^{(2 \pi)^2 i h}. If your question is whether there is an upper bound for this independent of h, the ... 3 Try to make the inner expression look like a derivative:$$ \begin{align} \sum_{L=0}^M\left(Ls^{L-1}\right)sL & =s\sum_{L=0}^M\left(\partial_ss^L\right)L\\ & =s\partial_s\sum_{L=0}^Ms^LL\\ & =s\partial_s\sum_{L=0}^M\left(Ls^{L-1}\right)s\\ & =s\partial_s\left(s\sum_{L=0}^M\left(Ls^{L-1}\right)\right)\\ & ...

3

Let $$S^{(n)}(\lambda)=\sum_{k=0}^\infty\frac{\lambda^{nk}}{(k!)^n}.$$ Then $$\lim_{n\to\infty}S^{(n)}(\lambda)=\begin{cases} 0 & \text{if }0\le\lambda<1,\\ 2 & \text{if }0\le\lambda=1,\\ \infty &\text{if }\lambda>1. \end{cases}$$ Let's prove it. If $0\le\lambda<1$ then $$1\le ... 3 Say you have a random variable Y=aX, where X is a random variable and a is a scalar. Then$$M_Y(t)=M_{aX}(t)=E[e^{t(aX)}]=E[e^{(ta)X}]=E[e^{(at)X}]=M_X(at)$$Since you're taking an average, then your new random variable is the sum of your Y_i's, each multiplied by the scalar \frac1{n}. By properties of the moment generating function, ... 3 1) The moment generating function of a sum of independent random variables is the product of the individual moment generating functions. 2) If W=aV, where a is a constant, then the moment generating functions M_V(t) and M_W(t) are related by the equation$$M_W(t)=M_V(at).$$If your exponentials Y_i have parameter \lambda, then each has moment ... 3 Using the series definition of exponential:$$ e^{iG\lambda}A e^{-iG\lambda} = \sum_{p=0}^\infty\frac{(iG\lambda)^p}{p!}A\sum_{q=0}^\infty\frac{(-iG\lambda)^q}{q!} = \sum_{p=0}^\infty\sum_{q=0}^\infty(-)^q\frac{(i\lambda)^{p+q}}{p!q!}G^pAG^q=\\ \sum_{s=0}^\infty\sum_{d=0}^s(-)^d\frac{(i\lambda)^s}{d!(s-d)!}G^{s-d}AG^d=\\ ...

3

Let's note that $$(2^1 - 1) + (2^2 - 1) + \cdots + (2^k - 1) = 2^{k+1} - 2-k$$ where we have used the geometric series. Thus, the desired sum is actually $$\sum_{k=1}^n{2^{k+1}-2-k}$$. As this is a finite sum, we can evaluate each of the terms separately. We get the sum is $$2\left(\frac{2^{n+1}-1}{2-1}-1\right) - 2n- \frac{n(n+1)}{2} = 2^{n+2}-4 - ... 3 Since we only have linear powers of x (that is, powers of the form ax+b, where a,b constants with a\ne 0), then one approach we can take (that puts off logarithms until the very end) is this:$$7^x=5^{x-4}\\7^x=\frac{5^x}{5^4}\\7^x=\frac{5^x}{625}\\7^x\cdot625=5^x\\625=\frac{5^x}{7^x}\\625=\left(\frac57\right)^x$$At this point, we can take a ... 3 Well if x,y\in \mathbb{R} then by definition $$\exp(x)\exp(y)=(\sum_{k=0}^{\infty}\frac{x^k}{k!})(\sum_{k=0}^{\infty}\frac{y^k}{k!})$$ The Cauchy's Multiplication Theorem tells as that $$\sum_{k=0}^{\infty}\sum_{k=0}^{n}a_kb_{n-k}=\sum_{k=0}^{\infty}a_k\sum_{k=0}^{\infty}b_k$$ when at least one of the ... 3 Every positive integer n is either even, i.e., of the form n=2k for k \geq 1, or odd, i.e., of the form n=2k-1 for k \geq 1. More formally, if f:\mathbb{N}\rightarrow\mathbb{R}, then$$\sum_{n=1}^\infty f(n)=\sum_{\text{odd } n=1}^\infty f(n)+\sum_{\text{even } n=1}^\infty f(n)$$since addition of real numbers is commutative and hence ... 3 Unfortunately your intuition is incorrect, for kind of an annoying reason: you can find smaller sets of elements, of cardinalities k and m-k, which individually sum to 0. For example, take u=30 and m=5. Then x_0+x_{15}=0 and x_1+x_{11}+x_{21}=0, and so the set \{x_0,x_1,x_{11},x_{15},x_{21}\} has the right sum but not the nice form you hoped ... 2 The proof using the generating function:$$\frac{t}{e^t-1}$$Can be found here. In essence, the proof follows from noting that:$$e^{kt}=\sum k^m \frac{t^m}{m!}$$So that the sum \ \sum k^m is related to the sum of the geometric series which is in turn related to the generating function. 2 For a random variable uniform on [0,1] the density function is 1 on [0,1] and 0 elsewhere. Median should be easy to pick up. The density function attains a maximum everywhere on our interval. Lotsa modes! Let X have exponential distribution parameter \lambda. Then X has density function f_X(x)=\lambda e^{-\lambda x} for x\ge 0, and 0 ... 2 Others have given the correct answer; here’s how you could have simplified your incorrect expression.$$\begin{align*} n(2^1) + (n-1)\cdot2^2 + (n-2)\cdot2^3 +\ldots&=\sum_{k=1}^n(n-k+1)2^k\\ &=(n+1)\sum_{k=1}^n2^k-\sum_{k=1}^nk2^k\\ &=(n+1)\left(2^{n+1}-2\right)-\sum_{k=1}^n\sum_{i=1}^k2^k\\ ...

2

First, supposing $a_{n+1} = 0$ for some $n$, then $a_n = 0$, and if $n > 1$, we get $a_{n-1} = 0$. Repeatedly, we have $a_n = 0$, so $a_n \neq 0$ whenever $n > 0$. $\frac{n(n+2)a_n}{n+1} - \frac{(n-1)(n+1)a_{n+1}}{n} = n(n+1)a_na_{n+1}$ is equivalent to $\frac{n(n+2)}{(n+1)a_{n+1}} - \frac{(n-1)(n+1)}{na_n} = n(n+1)$ We obtain that \[ ...

2

Rewrite $L^2 = L(L-1)+L.$ Then, \begin{align} \sum_{L=0}^{M} { L^2 s^L } &= \sum_{L=0}^{M} {L(L-1)s^L} + \sum_{L=0}^{M} {Ls^L}\\ &=s^2 \cdot \frac{\partial^2}{\partial s^2} \sum_{L=0}^{M}{s^L} + s \cdot \frac{\partial}{\partial s}\sum_{L=0}^{M} {s^L}\\ \end{align} You can find formulas for the summations with detailed descriptions of their ...

2

For the first one, recall that $$\frac{1}{1 - x} = \sum\limits_{n = 0}^{\infty} x^n = 1 + x + x^2 + ...$$ Differentiating both sides leads to $$\frac{1}{(1 - x)^2} = \sum\limits_{n = 1}^{\infty} = 1 + 2x + 3x^2 + ... = \sum\limits_{n = 1}^{\infty} nx^{n - 1}$$ This is valid within the disk of convergence, which has radius $1$ around $x = 0$.

2

Hint: For the first one if $\sum_{i = 3}^{\infty} x^{i}=f(x)$ then $\sum_{i = 3}^{\infty} i \times x^{i-1}=f'(x)$. For the second one consider that: $$\sum_{k = 2}^{i-1} a^{i-k} b^{k-2}=a^{i-2}\frac{1-\frac{b^{i-2}}{a^{i-2}}}{1-\frac{b}{a}}= \frac{a^{i-2}-{b^{i-2}}}{1-\frac{b}{a}}$$ Then you can decompose the series into two one and use the previous ...

2

It's a geometric series, with common ratio $r = e^{-1/\mu}$. It converges if and only if $|r|<1$ (this includes the case where $r$ is complex). If it converges, the easiest way to remember its value is $$\frac{\mbox{[First Term In Series]}}{1-\mbox{[Common Ratio]}}.$$ That way you don't have to fuss about whether the series ...

2

You won't find a nice algebraic solution, so to get an accurate answer you will need to use a numeric root finder. To get started, you can approximate the sum with an integral. $$\int_0^{N-1}x^n\; dx \le \sum_{n=1}^N x^n\le \int_1^N x^n\; dx\\ \frac {(N-1)^{n+1}}{n+1}\le \sum_{n=1}^N x^n\le \frac {N^{n+1}-1}{n+1}\\$$ so you can start from $n \approx \frac ... 1 a convergent power series of the form $$f(x)=\sum_{k=0}^{\infty}\frac{g(k)}{k!}x^k$$ represents the zero function iff$g(k)=0$for all$k$(here$g(k)=f^{(k)}(0)$). this can be zero for given values of$x$as noted in the comments e.g. $$\sin(\pi/2)=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}(\pi/2)^{(2k+1)}$$ corresponding to a not-identically-zero$g(k)$... 1 If you want each one to be smaller than the last by a ratio$r$, the sum of$1+r+r^2+\dots+r^{n-1}$($n$terms) is$\frac{1-r^n}{1-r}$so you could give$100\frac{r^{i-1}(1-r)}{1-r^n}$to iterm$i$, where$i$ranges from$1$to$n$. So for$r=\frac{2}{3}, n=4$, the sum is$\frac{65}{27}$and the points are about$41.53, 27.69, 18.46, 12.30.\$ Does this work ...

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