Tag Info

Hot answers tagged

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\\ ...


6

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} ...


5

Split the integral up at $x=3$ to get $$\int_2^3 dx \frac{\sqrt{\log{(9-x)}}}{\sqrt{\log{(9-x)}}+\sqrt{\log{(3+x)}}}+\int_3^4 dx \frac{\sqrt{\log{(9-x)}}}{\sqrt{\log{(9-x)}}+\sqrt{\log{(3+x)}}}$$ In the second integral, sub $x=6-y$. Then add the 2 integrals together. The answer is $1$.


5

Split the integrand: $$\frac{e^{\alpha x}-e^{\beta x}}{(e^{\alpha x}+1)(e^{\beta x}+1)}= \frac{1}{(e^{\beta x}+1)}-\frac{1}{(e^{\alpha x}+1)}$$ Then recognize this as an integral of another function, evaluated at the limits $\alpha$ and $\beta$. Which function? Well, take the derivative over $\beta$ to find that out: ...


5

Incomplete answer: $$ \begin{align*} I&=\int^{\infty}_{0}\frac{\log(1+x)\log(1+x^{-2})}{x}dx\\ &=\int^{1}_{0}\frac{\log(1+x)\log(1+x^{-2})}{x}dx+\int^{\infty}_{1}\frac{\log(1+x)\log(1+x^{-2})}{x}dx\\ &=\int^{1}_{0}\frac{\log(1+x)\log(1+x^{-2})}{x}dx+\int^{1}_{0}\frac{\log(1+x^{-1})\log(1+x^2)}{x}dx\\ ...


4

Nice proof! This kind of proof, an "infinite" construction based on an assumption that is false (and that we want to prove false), can be misunderstood. So it is useful to make the steps fully explicit. We give three ways of doing so, the last of which is descent. 1: Fermat's Method of Infinite Descent is, essentially, induction, in the form of what is ...


4

The $\nu_2$ (the exponent of the greatest power of $2$ dividing) of the RHS is at least $4010$, so there must be an element $\xi\in X=\{x+1,\ldots,x+2014\}$ such that $\nu_2(\xi)\geq 1996$. This gives that, if a solution $(x,y)$ exists, it must be huge, with $x>2^{1995}$ and $y$ around $\sqrt{x}$. For the same reason (with $\nu_3$ in place of $\nu_2$), if ...


4

For any integer $m$, we have $\text{gcd}(a_m, a_{2m}) = \text{gcd} (2m, m) = m$, and so $m$ divides $a_m$. Then, it follows that for any other integer $n$, $m$ divides $a_n$ if and only if it divides $\text{gcd}(a_m, a_n) = \text{gcd}(m, n)$. Thus $a_n$ has exactly the same divisors as $n$. Hence it must equal $n$, for all $n$. References: R. Gelca and T. ...


4

This is an algebraic approach which shows the locus of the center lies on a circle centered at centroid. The algebra is not as horrible as I originally thought. I hope this can inspire someone to construct a more geometrical proof for this interesting problem.` WOLOG, we will choose a coordinate system where the centroid $C$ of $\triangle A_0A_1A_2$ is the ...


4

Here is a proof assuming that $D={\mathbb R}_{+}$ means $(0,+\infty)$ (and so does not include $0$). Step 1. $f(x)\neq x$ for every $x\in D$. Indeed, if $f(x)=x$, then $f(a+f(x))=f(a+x)+f(x)$ becomes $f(x)=0$, so $x=0$ which is impossible. Step 2. $f(x) > x$ for every $x\in D$. Suppose by contradiction that $f(x)<x$ for some $x\in D$. Let ...


4

We have $$\ln(\sinh(x)) = \ln(e^x-e^{-x}) - \ln(2) = x - \ln(2) + \ln(1-e^{-2x}) = x - \ln(2) - \sum_{k=1}^{\infty} \dfrac{e^{-2kx}}{k}$$ Hence, $$\int \ln(\sinh(x)) dx = \dfrac{x^2}2 - x\ln(2) + \sum_{k=1}^{\infty} \dfrac{e^{-2kx}}{2k^2} = \dfrac{x^2}2 - x\ln(2) + \dfrac{\text{Li}_2(e^{-2x})}2+C$$ Similarly, we have $$\ln(\cosh(x)) = \ln(e^x+e^{-x}) - ...


3

For a fixed $n$, let $ A_r $ be the number of sequences of $n$ terms, whose sum is $r \pmod{11}$. Then, we want the value of $ \sum A_r ^2 $. As pointed out by ABC, the generating function $f(x) = \left ( 1 + x + x^2 + \ldots + x^9 \right) ^n $ gives us the number of sequences of $n$ terms, with a total sum of $r$. We find the sum of those coefficients with ...


3

We need to say a bit more after Erik's answer. We have $$(b+c)(b+d)(c+d)=\pm311,\,\pm312,\,\pm313\ .$$ Now the sum of the three numbers $b+c$, $b+d$, $c+d$ is even and so at least one of the numbers must be even. Therefore the only possibility is that $$(b+c)(b+d)(c+d)=\pm312$$ and we have $$M=312^2\ .$$ Note: we cannot straight away rule out the ...


3

Consider coordinates where $0 \le x, y \le 3$. This has area $9$. The points where $y \ge x+1$ make up the part of this square on or above the line $y = x+1$. This is a right triangle with sides 2 and 2, so its area is 2. The probability wanted is the ratio of these or 2/9. More generally, suppose you want to find the probability that $y \ge x+c$ where $0 ...


3

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 $x=15$.


3

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 ...


2

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 ...


2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} ...


2

Elegant proof Let's start with a generic cyclic quadrilateral, and define \begin{align*} \angle ACB = \angle ADB &= \alpha \\ \angle CAD = \angle CBD &= \beta \\ \angle DBA = \angle DCA &= \gamma \\ \angle BAC = \angle BDC &= 180°-\alpha-\beta-\gamma \end{align*} This answer is based on a key observation: $K$ is the center of the circle ...


2

We can let $a=0$ and see that: $$f(f(b)) = f(b) + f(b)$$ $$f(f(b)) = 2.f(b)$$ This function $f$ has a domain that is the same as its range. Substituting $f(b)$ for $x$: $$f(x) = 2x $$ Which gives: $$f:Range(f) \rightarrow Range(f)$$ $$f: x \mapsto 2x$$ This allows us to then define the range arbitrarily (so long as the range is closed under multiplication ...


2

$$\int_0^{\pi/2}dx \ln \sinh x$$ by partial integration $$x \ln \sinh x - \int dx x \frac{1}{\sinh x} \cosh x$$ where $\cosh x$ is the term from the inner differentiation. Then underivative of the second term then contributes $$\int \frac{x dx}{\tanh x} = -\frac{x^2}{2}+x \ln[1-e^x] +Li_2(e^x)+x\ln[1+e^x]+Li_2(-e^x)$$ where $Li_2(y)\equiv \int_1^y \frac{\ln ...



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