Problem: Let $\Omega\subset\mathbb{R}^n$ be a domain and $u\in C^2(\Omega)\cap C(\overline{\Omega})$ a solution of $$\Delta u=u^3\text{ in } \Omega.$$ Then $u$ attains no nonnegative (local) maximum and no nonpositive (local) minimum in $\Omega$, if $u\neq 0$ holds. Attempt: Assume that $u$ has a maximum at $x_0\in\Omega$ so that $u(x_0)>0$. Since $x_0$ is a maximum $\Delta u(x_0)\leq 0$ holds. But $\Delta u(x_0)=(u(x_0))^3>0$ is a contradiction. Assume that $u$ has a minimum at $x_0\in\Omega$ so that $u(x_0)<0$. Since $x_0$ is a minimum $\Delta u(x_0)\geq 0$ holds. But $\Delta u(x_0 ..read more

For my profession I have to calculate the weighted average number of days it takes a business to review contracts. The count of each contract types reviewed is the weight, and the sum of days it takes to review is the number. The weighted average for this below data set is 35.6 days, however the raw data indicates that not a single contract took more than 25 days to review. Should my days be calculated as the average of days instead of sum? If that's the case, my weighted average is 8 days, which seems more accurate. I don't know why the weighted average using the Sum of Days is so much higher ..read more

Given the equations: $$\sin x + \sin 3x + \sin 5x = 1,$$ $$\cos x + \cos 3x + \cos 5x = \sqrt{2} + 1,$$ we are asked to find the value of $\tan 3x$. Firstly, let's use the sum-to-product formulas for sine and cosine: For sine: $$\sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right),$$ For cosine: $$\cos A + \cos B = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right).$$ Applying these to our equations, we get: For the first equation: $$\sin x + \sin 5x = 2\sin\left(\frac{x+5x}{2}\right)\cos\left(\frac{x-5x}{2}\right) = 2\sin(3x)\cos(-2x) = 2\sin(3x)\cos(2x ..read more

The sequence OEIS A333646: $1,6,15,28,30,33,40,42,51,66,69,84,91,95,102,105,117,120,135,138,140,141,145,159,165,182,186,190,210,213,\cdots$ Contains numbers that are divisible by the largest prime factor of the sum of their own divisors. For instance the sum of divisors of $15$ is $24$, the biggest prime divisor of $24$ is $3$, and $3$ divides $15$. If $G$ is a group and $|G|>6$ is featured in the sequence above, it seems guaranteed there will be at least one element of $G$ that has an order which is not a power of a prime. I've verified this up to and including $213$, and the pattern feels ..read more

I am given the differential equation: $$ xy \dot{y} = y - 1 $$ I have tried to solve it by separating the variables, but I have only come this far: $$ \int 1 + \frac{1}{y-1} dy = \int \frac{1}{x} dx = y + \log|y-1| = \log{|x|} + C $$ I am struggeling on how to isolate $ y(x) $. Can somebody please give me a hint on how to continue ..read more

The following question is a Math Olympiad Problem: Find $x>0$ that solves $x\sqrt{x\sqrt x}=2$ The answer is $\sqrt[7]{16}$ but I got $2\sqrt[3]{2}$ by: $$\begin{align}x\sqrt{x\sqrt x}=2&\to \sqrt{x\sqrt{x}}=\frac2x\\&\to x\sqrt{x}=\frac{4}{x^{2}}\\&\to \sqrt{x}=\frac{4}{x}\to x=\frac{16}{x^{2}}\\&\to x^{3}=16\\&\to x=\sqrt[3]{16}\\&\to x=2\sqrt[3]2\end{align}$$ What have I done wrong ..read more

Apologies for the unclear title, I have no idea if the property I'm looking for has a better name. I'm wondering if there exists a pair of functions $f, g : \mathbb{R} \rightarrow \mathbb{R}$ such that : $g$ is a bijection and is nowhere continuous (for an example, see this answer). $f$ is continuous and not constant. $f \circ g$ is continuous ..read more

Evaluate the integral $$\int_{-\infty}^{\infty}\binom{n}{x}dx$$ This question came in Cambridge Integration Bee and I have no clue what to do in this. I rewrote $\binom{n}{x}$ as $\frac{n!}{{x!}{(n-x)!}}$ but I don't know how to integrate those factorials. Also what to do if $n$ isn't an integer as no information regarding it is given$?$ Any help is greatly appreciated ..read more

I got a expression that is related to the combinatorics, and it looks like this: $$\sum_{k=1}^n \frac{2^{2k-1}}{k}\binom{2n-2k}{n-k}\cdot \frac{1}{\binom{2n}{n}}$$ it is from a question I've been studying, and with another approach, I got the answer: $$\sum_{k=1}^n\frac{1}{2k-1}$$ so I think the two should be the same, which means that $$\sum_{k=1}^n \frac{2^{2k-1}}{k}\binom{2n-2k}{n-k}\cdot \frac{1}{\binom{2n}{n}}=\sum_{k=1}^n\frac{1}{2k-1}$$ and to verify this, I wrote a python program to justify. And the outcome is that the two are very likely to be the same. (I tried when $n=1$ to $n=50 ..read more

Let $f$ be a entire function and $P$ be a polynomial of degree at least $2$. If $$f(z)=f(P(z)),\quad \forall z\in\mathbb C,$$ Is the entire function $f$ constant function? My thought: If $f$ is not constant, then $f$ must be transcendental entire function. If the sequence $\{z_n=P^n(z)\}$ iterated by $P$ has a limit point $w_0$ in $\mathbb C$, then $$f(z)=f(z_1)=\cdots=f(z_n)=\cdots,\quad \lim_{n\to\infty}f(z_n)=f(w_0),$$ then $f(z)\equiv w_0$ by Identity theorem. But this is not always ture. For enample, if $P(z)=z^2+1$, take $z=1$, then $z_n\to\infty$. Also if $$f(z)=f(e^z),\quad \fora ..read more