Tag Info

7

Square each side of your equation, move constant terms to the right hand side, and square again. The answer will come out clearly.

7

Stirling's approximation gives a pretty tight bound for the natural logarithm of $n!$ (but not an exact value), and then the base-10 logarithm is only a matter of dividing by $\ln 10$.

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

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 $e^{-x}<1$).

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 $$\sum_{x \in \mathbb{Z}/n \mathbb{Z}} e\left(\frac{kx^2}{n} \right) = \left\{ \begin{array}{lcl} \varepsilon_n \left( \frac{k}{n} \right) \sqrt{n} & & n \equiv 1 \pmod{2}\\ 0 ... 4 Unfortunately your intuition is incorrect, for kind of an annoying reason: you can find smaller sets of elements, of cardinalities k and m-k, which individually sum to 0. For example, take u=30 and m=5. Then x_0+x_{15}=0 and x_1+x_{11}+x_{21}=0, and so the set \{x_0,x_1,x_{11},x_{15},x_{21}\} has the right sum but not the nice form you hoped ... 4 You have that$$0.3 + 0.03 + 0.003 + \cdots = \sum\limits_{n = 1}^\infty {\frac{3}{{{{10}^n}}}} $$Now use that$$\sum\limits_{n = 1}^\infty {{a^n}} = \frac{a}{{1 - a}}$$whenever |a|<1 4 Well if x,y\in \mathbb{R} then by definition $$\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$$ when at least one of the ... 4 Write : L^2=L(L-1)+L and use derivative. For  L \geq 2 :$$L^2 s^L = s^2L(L-1)s^{L-2}+ s Ls^{L-1}= s^2(s^L)'' + s (s^L)'$$We get :$$\sum_{L=0}^M L^2s^L=0^2+1^2 s + s^2 \left(\sum_{L=2}^{M} s^L \right)''+s \left(\sum_{L=2}^{M} s^L \right)'$$4 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 \[ ... 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 By definition, \displaystyle\sum_{n=-\infty}^\infty r^{n^2}=\theta_3(0,r). See Jacobi elliptic \theta function. 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$$\left\lceil \log_2A \right\rceil.$$4 Here are two facts you might find useful:$$e^{2y+2} = e^2 e^{2y};\frac{(2(y+1))^n}{n!} = \frac{2^n}{n!}(y+1)^n.3 Notice that: \begin{align*} 3^{x-5} + 3^{x-7} + 3^{x-9} &= 3^{4 + (x-9)} + 3^{2 + (x-9)} + 3^{x-9} \\ &= (3^4)3^{x-9} + (3^2)3^{x-9} + 3^{x-9} \\ &= (81)3^{x-9} + (9)3^{x-9} + (1)3^{x-9} \\ &= (81 + 9 + 1)3^{x-9} \\ &= (91)3^{x-9} \\ \end{align*} 3 The following approach does not use the Mellin transform, but it is worth mentioning. It is quite easy to prove that:\sum_{n=1}^{+\infty}\frac{\cos(\pi n)}{n} e^{-nk\pi} = -\log\left(1+e^{-k\pi}\right)$$hence the original sum equals:$$-\log\prod_{k=1}^{+\infty}\frac{1+e^{-(2k-1)\pi}}{1+e^{-2k\pi}}.$$The last product is clearly related with the Jacobi ... 3 Using$$\tanh\left(x\right)=1+2\underset{n=1}{\overset{\infty}{\sum}}\frac{\left(-1\right)^{n}}{e^{2nx}}$$we have$$\underset{n\geq1}{\sum}\frac{\left(-1\right)^{n}}{n\left(e^{\pi n}+1\right)}=\frac{1}{2}\underset{n\geq1}{\sum}\frac{\left(-1\right)^{n}\left(1-\tanh\left(\frac{\pi ...

3

Consider, instead, the series: \begin{align} \sum_{n=0}^\infty nx^n &= x\sum_{n=0}^\infty nx^{n-1} \\ &= x\sum_{n=0}^\infty\frac{d}{dx}x^n \\ &= x\frac{d}{dx}\sum_{n=0}^\infty x^n \\ &= x\frac{d}{dx}\left(\frac{1}{1-x}\right) \\ &= x\left(\frac{1}{(1-x)^2}\right) \\ &= \frac{x}{(1-x)^2} \end{align} This converges whenever ...

3

Using the series definition of exponential: $$e^{iG\lambda}A e^{-iG\lambda} = \sum_{p=0}^\infty\frac{(iG\lambda)^p}{p!}A\sum_{q=0}^\infty\frac{(-iG\lambda)^q}{q!} = \sum_{p=0}^\infty\sum_{q=0}^\infty(-)^q\frac{(i\lambda)^{p+q}}{p!q!}G^pAG^q=\\ \sum_{s=0}^\infty\sum_{d=0}^s(-)^d\frac{(i\lambda)^s}{d!(s-d)!}G^{s-d}AG^d=\\ ... 3 We have,$$\sum_{\mathbf{x}\in\mathcal{S}}\exp\left[f(n)\sum_{i=1}^nx_i\right] = \sum_{\mathbf{x}\in\mathcal{S}} \prod_{(x_i)_{i=1}^{n} = \mathbf{x}} e^{f(n)x_i}=\left(e^{f(n)}+e^{-f(n)}\right)^n$$The identity is a consequence of: \displaystyle \left(t^{1}+t^{-1}\right)^{n} = \sum\limits_{\substack{1 \le i \le n\\ \epsilon_i = \pm1}} ... 3 We have that there are n\choose r vectors in S with exactly r negative components, and that the sum of the components of any one of these vectors is n-2r, so your sum is$$A_{f(n)}(n)=\frac{1}{2^n}\sum_{\mathbf{x}\in\mathcal{S}}\exp\left[f(n)\sum_{i=1}^nx_i\right] = \frac{1}{2^n}\sum_{r=0}^n {n \choose r} \exp(f(n)(n-2r))\le ...

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

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 $$1\le ... 3 This is a partial sum of a geometric series:$$\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|}$$with q:=e^{(2 \pi)^2 i h}. If your question is whether there is an upper bound for this independent of h, the ... 3 This is just an application of the chain rule$$ \frac{\partial\mathrm{log}(\mathrm{exp}(w_1 * x_1 + b_1) + \mathrm{exp}(w_2 * x_2 + b_2))}{\partial w_1} = \frac{\partial}{\partial w_1}(log(f(w_1)) = \frac{\frac{\partial f(w_1)}{\partial w_1}}{f(w_1)} $$So your final answer is:$$ \frac{x_1\mathrm{exp}(w_1x_1+b_1)}{\mathrm{exp}(w_1 * x_1 + b_1) + ...

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)\\ & ...

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