18

This is not an answer. Interesting is $$\int_1^{\infty} \frac{dx}{(x-\log x)^2}=\sum _{n=1}^{\infty }\frac {n!} {n^n}$$ which does converge very fast. Consider the partial sums $$S_p=\sum _{n=1}^{p }\frac {n!} {n^n}$$ For $50$ significant figures, some numbers $$\left( \begin{array}{cc} p & S_p \\ 10 & \color{red} {1.879}...


16

The video solution is really clever, but you can also solve the problem by just slogging along. As the presenter said at the beginning, we have to choose $2k$ east-west moves, $k$ of which will be west, and then we have to choose $1009-k$ north moves. The remaining $1009-k$ moves will be south. This gives $$ n=\sum_{k=0}^{1009}{2018\choose k}{2018 -k\...


12

A more general relationship : $$\int_1^\infty \frac{dx}{(x-\ln(x))^p}=\sum_{k=0}^\infty \frac{(p+k-1)!}{(p-1)!\:(k+p-1)^{k+1}}\qquad p> 1 \:,\: p \text{ integer.}$$ This could be extended to real $p>1$ thanks to the function $\Gamma$. Or, on an equivalent form : $$\int_1^\infty \frac{dx}{(x-\ln(x))^p}=\frac{1}{(p-1)!}\sum_{n=p-1}^\infty \frac{n!}{n^{...


6

Let's try to simplify the expression $\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$. Multiply through by $a + b + c$ and you obtain the equivalent expression $$\frac{a+b+c}{a+c} + \frac{a + b + c}{b + c} = 3$$ This is equivalent to $$\frac{b}{a+c} + \frac{a}{b + c} = 1$$ Multiplying through by $(a+c)(b+c)$ this is the same as $$b(b+c) + a(a + c) = (a + c)(b+...


6

You can just count how many possible paths the ant can take by assuming it makes $k$ steps to the east and $k$ to the west, then deciding the order of the east-west paths, similarly those of the north-south ($1009-k$) and finally deciding how to mix both sequences. Everything can hence be expressed in terms of binomial coefficients: $$\sum_{k=0}^{N}C_k^{2k}...


5

This problem is solved by using the Zero Factor Property from algebra; i.e., $ab = 0$ iff $a = 0$ or $b=0$. Hence, we have $x - 1 = 0$ or $x^{2} - 4 = (x+2)(x-2) = 0$; giving $$ x = 1; x = \pm 2 $$ Remark1: $x=1$ causes the second term to be imaginary; yet, as a commentator points out, is a real solution. Therefore, the solution set is $$ x = 1; \pm 2.$...


5

I just tried to solve the problem in my head, and thought of the change-of-coordinate trick within a minute or two (at which point I stopped, knowing that it is only computation starting from there). To answer your question of "how does one come up with it", let me try to retrace the intuitive process in my mind. I thought roughly as follows: "OK, so a ...


5

$p(2013)-p(1515)$ is divisible by $2013-1515=498,$ but $1789-1830=-41$ is not. It follows from the following reasoning. Let $p(x)=a_0x^n+a_1x^{n-1}+...+a_n,$ where $a_i\in\mathbb Z$. Thus, $$p(m)-p(k)=a_0(m^n-k^n)+a_1(m^{n-1}-k^{n-1})+...+a_{n-1}(m-k)=$$ $$=(m-k)(a_0(m^{n-1}+m^{n-2}k+...+k^{n-1})+a_1(m^{n-2}+...+k^{n-2})+...+a_{n-1}).$$


4

$$ x y z = \frac{1}{24}\left( (x+y+z)^3 - (x+y-z)^3 -(x-y+z)^3 -(-x+y+z)^3 \right) \text{.} $$


4

Let $G$ be a finite group with $|G| \equiv 0$ mod $8$ and assume that the number of conjugacy classes is at most $4$. This is equivalent to the statement that the number of irreducible complex characters of $G$ is at most $4$. Since $|G|$ is the sum of the squares of its irreducible character degrees we get $|G|=1+a^2+b^2+c^2$, where the $1$ comes from the ...


4

A simple answer can be based on the fact that a $2$-group has a non-trivial center (the same applies to any $p$-group). The assumption about the order $n=8k$ of the group $G$ implies by basic Sylow theory that there is a subgroup $P\le G$ of order $8$. Let $z\neq1_G$ be an element of $Z(P)$. It follows that the centralizer of $z$ has order $8\ell$, $\ell\ge1$...


3

You have to be careful when dividing with $-3d$, because as you said it can be either positive or negative. For some reason your last formula has $8$ instead of $168$ in it. But even after you correct that mistake, your formula $(4)$ is only valid for $d<0$, and for $d>0$ you need to flip the inequality sign.


3

$$a+b+c+d=(a+b+c+d)\sum_{cyc}\frac{a}{b+c+d}=$$ $$=\sum_{cyc}\frac{a^2+a(b+c+d)}{b+c+d}=a+b+c+d+\sum_{cyc}\frac{a^2}{b+c+d},$$ which says $$\sum_{cyc}\frac{a^2}{b+c+d}=0.$$ In the full writing the last equality it's: $$\frac{a^2}{b+c+d}+\frac{b^2}{c+d+a}+\frac{c^2}{d+a+b}+\frac{d^2}{a+b+c}=0.$$


2

If you divide by -3d you first of all have to rule out $d=0$. [Edit: I just see, that $d,z\neq 0$] Also you have to seperate cases, because if $-3d<0$ the 'direction' of the inequality changes. It is $a\leq b\Leftrightarrow -a\geq -b$


2

Brute force is an entirely acceptable way to solve this problem. If 17 is opposite 56, then 16 is opposite 57, 15 is opposite 58, and so on. Keep going. Which house is then opposite 1?


2

Writing your equation in the form $$\sqrt{1+x^2}=3\sqrt{3}-x$$ and by squaring we get $$1+x^2=27+x^2-6\sqrt{3}x$$ so we get $$6\sqrt{3}x=26$$ Can you finish?


2

In the case you are dealing with, the three vectors are $(1,0)$, $(-1/2,\sqrt3/2)$ and $(-1/2,-\sqrt3/2)$. The sum of the squares of the $y$-coordinates is $$0^2+\left(\frac{\sqrt3}2\right)^2+\left(-\frac{\sqrt3}2\right)^2=\frac32.$$ The general solution though, has vectors $(\cos\alpha,\sin\alpha)$, $(\cos(\alpha+2\pi/3),\sin(\alpha+2\pi/3))$ $(\cos(\...


2

Premise This is another non-answer, I find myself in agreement with what is written by Claude Leibovici. Numerical experiment Writing in Wolfram Mathematica: int1 = NIntegrate[1/(x - Log[x])^2, {x, 0, ∞}, WorkingPrecision -> 100] int2 = (WolframAlpha[ToString[int1], {{"PossibleClosedForm", 12}, "FormulaData"}, PodStates -> {"...


2

We approach this from the perspective of colouring the graph $Z$ where each vertex represents an integer, the colours correspond to the desired partitions and $n,n+p,n+q$ are given different colours. If $\gcd(p,q)=d>1$, $Z$ is the disjoint union of $d$ copies of itself. Hence we may assume $\gcd(p,q)=1$, in which case $Z$ is a triangular lattice as shown ...


2

The maximum value of $\int_0^1 f(x)^5 \mathrm{d}x$ is $\frac{1}{16}$. We first prove that $\frac{1}{16}$ is an upper bound. Note that if $t\le 1$ then $t^5\le \frac{5}{4} t^3 - \frac{5}{16} t + \frac{1}{16}$. Indeed, $$t^5 - \left(\frac{5}{4} t^3 - \frac{5}{16} t + \frac{1}{16}\right) = (t-1)\left(t^2+\frac t2 - \frac 14 \right)^2 \le 0.$$ Putting $t=f(x)$...


2

Not a full answer, but I hope to add more once I get to Mathematica in a few hours. Using an obvious substitution $x=e^y$, we can transform the integral to: $$I=\int_{-\infty}^\infty \frac{e^y dy}{(e^y-y)^2}=\int_0^\infty \frac{e^y dy}{(e^y-y)^2}+\int_0^\infty \frac{e^{-y} dy}{(e^{-y}+y)^2}$$ Some simple algebra gives us: $$I=2 \int_0^\infty \frac{(1+y^2)...


2

Since you're dividing by a quadratic polynomial, the remainder has to be a linear polynomial, because its degree must be strictly less than that of the divisor. So you can write your remainder as $R(x)=px+q$. Knowing that $R(2)=-6$ and $R(-3)=14$, you can set up and solve a system of two equations with two unknowns $p$ and $q$.


1

A good idea for whenever you don't know how to start a combinatorics problem that asks you to compute something for a very high integer (in this case, 2018) is to compute some small cases and look for a pattern. To get started on this one, note that the number of paths of length $n$ to any given point is the sum of paths of length $n-1$ to it's neighboring ...


1

The problem does not specify bad, good, or optimal functions: If $f_0$ is optimal then any "measure preserving horizontal rearrangement" of $f_0$ on the $x$-interval $[0,1]$ is admissible and gives the same objective integral as $f_0$. This means that we should not look for particular functions $f$ satisfying the constraints, but for measures $\mu$ on the ...


1

Sure: $$ \{ 2,4,9\} \quad \{3,5,7\} \quad \{1, 6, 8\} $$ or $$ \{2,6,7\} \quad \{ 1,5,9\} \quad \{ 3,4,8\} $$ The sum over elements is constant $15$. Fact: any such partition such that the sum of elements is constant will have this constant equal to $15$. Try to prove this for yourself.


1

There must be a minimal values as they are integers. If the numbers weren't all equal there would be one square with minimal value that was adjecent to a number that is not minimal. This would mean that the mean of the neighbouring squares would be larger than the value at the square.


1

Step by step: $$\frac{1}{(x^3+cx^{-3})^{1/3}}=\frac{1}{((x^{-3})(x^6+c))^{1/3}}=\frac{1}{(x^{-3})^{1/3}(x^6+c)^{1/3}}=\frac{1}{x^{-1}(x^6+c)^{1/3}} =\frac{x}{(x^6+c)^{1/3}}.$$


1

From Clarification regarding a question, $\alpha-\beta=\dfrac{2\pi}3+2\pi a,\beta-\gamma=\dfrac{2\pi}3+2\pi b$ $\implies\alpha-\gamma=\dfrac{4\pi}3+2\pi c=-\dfrac{2\pi}3+2\pi d$ If $S=\sin^2\alpha+\sin^2\beta+\sin^2\gamma$ $$2S=3-(\cos2\alpha+\cos2\beta+\cos2\gamma)$$ Method$\#1:$ $$\cos2\alpha+\cos2\beta=\cos2\left(\gamma-\dfrac{2\pi}3\right)+\cos2\...


1

In fact the cosine equation is not $$1-\cos\gamma-\cos\beta=0,$$ but $$\cos\beta\cos\gamma-\sin\beta\sin\gamma-\cos\gamma-\cos\beta=0,$$ by taking the dot-products between the vectors. The sine equation is obtained from the cross-products, $$-\sin\beta\cos\gamma-\cos\beta\sin\gamma-\sin\gamma+\sin\beta=0.$$ These are compatible with $\alpha+\beta+\...


1

One may get Mclaurin series solutions here. $$y(x)=\sum_{k=0}^{\infty} \frac{y^{(k)}(0)}{k!}x^k,~~~~(*)$$ $k$th derivatives of $y(x)$ at $x=0$ can be caklculated as by differentiation $$y'=x^2+y^2~~~(1)$$, repeatedly. we get $y'(0)=0$, then $$y''(x)=2x+2y(x)y'(x),~~~~(2)$$ gives $y''(0)=0$. Next, differentiating w.r.t $x$ $$y^{(3)}(x)=2+2y'^2(x)+2y(x)y''(x)...


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