# Tag Info

8

You are integrating over the part $0 \leqslant x \leqslant z \leqslant y \leqslant 1$ of the unit cube. The integrand is invariant under permutations of the coordinates, and the six permutations of $x,y,z$ cover the entire unit cube, so $$6\int_{x=0}^{x=1} \int_{y=x}^{y=1}\int_{z=x}^{z=y}f(x)f(y)f(z)\,dz\,dy\,dx = \int_0^1\int_0^1\int_0^1 ... 8 Separate into two parts, -\pi/2 to 0 and 0 to \pi/2. For the integral from -\pi/2 to 0, make the change of variable t=-x. We get$$\int_0^{\pi/2} \frac{2007^t}{2007^t+1} \frac{\sin^{2008}t}{\sin^{2008} t+\cos^{2008} t}\,dt.$$Change the variable back to x, and add to the integral from 0 to \pi/2.We get ... 7 The sum in Chen Wang's answer, that is  \displaystyle \sum_{n=1}^{\infty} \frac{H_{4n}}{n^{2}}, can be evaluated using contour integration by considering$$f(z) = \frac{ \big( \gamma + \psi(-4z) \big) \cot \pi z}{z^{2}} $$where \psi(z) is the digamma function. Now integrate around a square with vertices at \pm (N + \frac{1}{2}) \pm i (N ... 7 Because you are always evaluating the limit, this is an asymptotic expansion of the explicit expression for the solutions. Write$$x=2\pi n +\epsilon$$You get$$\sin \epsilon=\frac{1}{2\pi n +\epsilon}$$Your first limit in this notation is$$a=\lim_{n\to\infty}n\epsilon$$We are seeking the series for \epsilon expanded in inverse powers of n. ... 7 Using$$ \cos^2(x)=\frac{1+\cos(2x)}{2} $$we get that$$ \begin{align} &\int_0^1\int_0^1\cdots\int_0^1\cos^2\left(\frac{a\pi}{2n}(x_1+x_2+\dots+x_n)\right)\,\mathrm{d}x_1\,\mathrm{d}x_2\dots\,\mathrm{d}x_n\\ ...

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Hint $\$ Apply the following simple test. Parity Root Test $\$ A polynomial $\rm\,f(x)\,$ with integer coefficients has no integer roots when its constant coefficient and coefficient sum are both odd. Proof $\$ The test verifies that $\rm\ f(0) \equiv 1\equiv f(1)\ \ (mod\ 2),\$ i.e. that $\rm\:f(x)\:$ has no roots modulo $2$, hence no integer ...

3

This doesn't look so hard if you look at the sequence of complex numbers $z_n = a_n + i\cdot\frac{b_n}{n}$. Then the recurrence formula turns into $z_{n+1} = \frac{1}{2}\overline{z_n}^2$, which is very manageable. Alternatively, if you want to avoid complex numbers, you can look at the sequence $c_n = a_n^2 + \frac{b_n^2}{n^2}$. Note that the sequence in ...

3

My method is that we can ignore other terms except $a_{17}y^{17}=a_{17}(x+1)^{17}$, because this is the only term that is possible to product the term $-x^{17}$. $(x+1)^{17}=\sum_{n=0}^{17}C_{17}^nx^{17-n}$(Binomial theorem), then the coefficient of $x^{17}$ is $C_{17}^0=1$, and the coefficient of $x^{17}$ in the right side is $a_{17}C_{17}^0=a_{17}$. ...

3

I explain below an algorithmic construction which achieves the unique solution to the problem obtained when one adds the further constraint that all the consecutive $p$-sums be equal to $+1$ and all the consecutive $q$-sums equal to $-1$. The algorithm is simple enough (although the details are a little messy to write out), uses the Euclidean algorithm and ...

3

If you are attuned to the cyclic symmetry of things, you might notice it can be rewritten $$(x+y):(y+z):(z+x)=1:2:4\qquad\text{and}\qquad (x+y)+(y+z)+(z+x)=70$$ which gives $(x+y)+2(x+y)+4(x+y)=70$, or $x+y=10$. From this it follows that $y+z=2(x+y)=20$, and we can finish by invoking $x+y+z=35$, which gives $x=15$.

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You could start with Wikipedia A web search will turn up many references. It looks like you were applying it without knowing it. Here you are looking to solve $N \equiv 0 \pmod {2^4}, N\equiv -1 \pmod {5^4}$ or the other way around. Because $2^4,5^4$ are relatively prime, CRT says there will be exactly one solution $\pmod {2^4\cdot 5^4}$ Note that ...

2

The last $4$ digits of $n$ are $\,n\ {\rm mod}\ 10000,\,$ so if they are the same for both $\,n\,$ and $\,n^2\,$ then $\,10000\mid n^2-n,\,$ so $\,10^4 = 2^4 5^4\mid n(n\!-\!1).\,$ By $\,n,\,n\!-\!1\,$ coprime, either $\,2^4\mid n\,$ or $\,2^4\mid n\!-\!1,\,$ and $\,5^4\mid n\,$ or $\,5^4\mid n\!-\!1,\,$ leading to the following four possible cases ...

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The binomial theorem says \begin{align} 1-x+x^2-\dots-x^{17} &=\frac{1-x^{18}}{1+x}\\ &=\frac{1-(y-1)^{18}}{y}\\ &=\sum_{k=1}^{18}\binom{18}{k}(-1)^{k-1}y^{k-1}\\ &=\sum_{k=0}^{17}\binom{18}{k+1}(-1)^ky^k \end{align} Look at the $k=17$ term. The method in the question can be used to find any of the $a_k$, but it will involve solving ...

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