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31

Using $a^2-b^2 = (a+b)(a-b)$ \begin{align*} 2^{96}-3^{16} ={} & (2^{48}+3^8)(2^{48}-3^8) \\ ={} & (2^{48}+3^8)(2^{24}+3^4)(2^{24}-3^4) \\ ={} & (2^{48}+3^8)(2^{24}+3^4)(2^{12}+3^2)(2^{12}-3^2) \\ ={} & (2^{48}+3^8)(2^{24}+3^4)(2^{12}+3^2)(2^6+3)(2^6-3) \end{align*} and $2^6+3=67$, $2^6-3=61$, ...

9

Gauss sums Your sum is strongly related to the Gauss sum. The usual trick is to compute the modulus. This works particularly smoothly over $\mathbf{Z}/p\mathbf{Z}$ as with usual Gauss sums, but essentially it works here too: If $S = \sum_{k=0}^{n-1} e^{ik^2},$ then \begin{align} |S|^2 &= \sum_{k=0}^{n-1} \sum_{k'=0}^{n-1} e^{i(k'^2 - k^2)}\\ &= \...

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}

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

Hint: for any $x \in \Bbb R$ with $x \neq 1$, $$1 + x + x^2 + \cdots + x^{n} = \frac{1-x^{n+1}}{1-x}$$

7

Hint: you can write $\sum\limits_{n=4}^{n= \infty } \frac{2^n + 8^n}{10^n} = \sum\limits_{n=4}^{n= \infty } \frac{2^n }{10^n} + \sum\limits_{n=4}^{n= \infty } \frac{8^n }{10^n}$ because the two series on the right hand side converge. And then you have two geometric series, which you can probably solve.

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

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 ...

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*}$$in ... 5 The nth step is 2^n. If you want 2^n\geq A, then you want$$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.$$5 If x, y, z>1, then y^z, z^x, x^y \geq 2 so x^2y^2z^2 \mid x^{y^z}y^{z^x}z^{x^y}=5xyz so xyz \mid 5, impossible since x, y, z \geq 2. Thus at least one of x, y, z is 1. The equation is cyclic, so we may assume z=1, so the equation becomes x^yy=5xy so x^{y-1}=5. It is now clear that we must have x=5, y=2. Therefore all solutions are ... 5 Recall the geometric series (see Wikipedia): for any y with |y|<1,$$\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.$$Now let y=e^{-x} (though observe that we need x>0 to have e^{-x}<1). 5$$\sum_{k=0}^\infty\frac{x^{k+a}}{(k+a)!}~=~e^x~\bigg[1-\dfrac{\Gamma(a,x)}{\Gamma(a)}\bigg].$$5 I agree with Lucian and Alex, such a function can be expressed in terms of an incomplete Gamma function or a hypergeometric function \phantom{}_{1} F_1. I just wanted to add that for \alpha=\frac{1}{2} we have something that depends on the error function:$$ \sum_{k\geq 0}\frac{x^{k+1/2}}{(k+1/2)!} = e^x\operatorname{Erf}(\sqrt{x})=\frac{e^x}{\sqrt{\...

5

No, such expression is not known in the world of commonly used special functions. Its closest analog is the sum $\sum\nolimits_{n=0}^{\infty}e^{-n^2}$, which is already quite nontrivial: it is expressed in terms of elliptic theta functions. One way to convince yourself that the sum is "too exotic" is to consider a generalization $$f(z)=\sum\limits_{n=0}^{\... 5 I guess the answer ("it is hard") have been given. The numerical computation gives \sum_{n=0}^\infty{\mathrm e}^{-n^3}=1.368\dots I want to add that a natural first order approximation given by the Euler–Maclaurin formula, is$$\sum_{n=0}^\infty{\mathrm e}^{-n^a}\approx\frac{1}{2}+\Gamma\left(1+\frac{1}{a}\right)\approx 1.5-\frac{\gamma}{a},$$where ... 4 Write : L^2=L(L-1)+L and use derivative. For  L \geq 2 :$$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 \[ \frac{(n-... 4 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 ... 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 Note that,$$ \int_{0}^{x}f(t)\ \mathrm dt=\sum_{n=1}^{\infty}\frac{1}{n^4}\int_{0}^{x}\sin(nt) \ \mathrm dt=\sum_{n=1}^{\infty}\frac{1}{n^4}\left( \frac{1}{n}-\frac{\cos(nx)}{n} \right)  \implies \int_{0}^{x}f(t) \ \mathrm dt=\sum_{n=1}^{\infty}\frac{1}{n^5}-\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^5} \implies f(x)=\zeta(5)-\frac{1}{2}\sum_{n=1}^{\...

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

Here are two facts you might find useful: $$e^{2y+2} = e^2 e^{2y};$$ $$\frac{(2(y+1))^n}{n!} = \frac{2^n}{n!}(y+1)^n.$$

4

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: $$\frac{x_1\mathrm{exp}(w_1x_1+b_1)}{\mathrm{exp}(w_1 * x_1 + b_1) + \mathrm{... 4 By definition, \displaystyle\sum_{n=-\infty}^\infty r^{n^2}=\theta_3(0,r). See Jacobi elliptic \theta function. 4 Hint You may try with the Sophie Germain identity. As (10^4+324)=(10^4+4\times3^4) 4 In the binomial theorem,$$(x+y)^n={n \choose 0}x^{n-0}y^0+{n \choose 1}x^{n-1}y^1+{n \choose 2}x^{n-2}y^2+\cdots+{n \choose n}x^{n-n}y^n=\sum_{i=1}^{n} {n \choose i} x^{n-i}y^i Put $x=1$ and $y=2$.

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