# Tag Info

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.

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

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

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

$$\sum_{k=0}^\infty\frac{x^{k+a}}{(k+a)!}~=~e^x~\bigg[1-\dfrac{\Gamma(a,x)}{\Gamma(a)}\bigg].$$

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: $$\sum_{k\geq 0}\frac{x^{k+1/2}}{(k+1/2)!} = ... 5 The nth step is 2^n. If you want 2^n\geq A, then you want$$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.$$5 Clearly,$$\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.$$Now let y=e^{-x} (though observe that we need x>0 to have e^{-x}<1). 5 Well if x,y\in \mathbb{R} then by definition $$\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

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 ...

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

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

Hint You may try with the Sophie Germain identity. As $(10^4+324)=(10^4+4\times3^4)$

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

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

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: $$\frac{x_1\mathrm{exp}(w_1x_1+b_1)}{\mathrm{exp}(w_1 * x_1 + b_1) + ... 4 By definition, \displaystyle\sum_{n=-\infty}^\infty r^{n^2}=\theta_3(0,r). See Jacobi elliptic \theta function. 4 In the binomial theorem,$$(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$$Put x=1 and y=2. 4 Here are two facts you might find useful:$$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 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

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

It's, unfortunately, not a particularly well-defined problem, as infinite power towers aren't always well defined. However, if we want to apply algebraic techniques anyhow, notice that we can write it as $$2x^{\left(2x^{2x^{2x\ldots}}\right)}=4$$ where the inner expression on the left is equal to four for a solution, giving $$2x^4=4$$ which is easier to ...

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