All started with the problem $9$ (books.google.ro/… that I studied 2 years ago, but I kept in my mind some specific values and I recognized those values again when I worked on this integral. Then I realized that something amazing is about to happen....
Nice exercise: suppose $f:\Bbb R^n\to\Bbb R$ is $\mathscr C^1$ with $n\geq 2$. Suppose there is $p$ such that $f(p)=0$ and $\nabla f(p)\neq 0$. Prove $f(x)=0$ for infinitely many $x\in\Bbb R^n$.
It what way? It is dominated convergence:$$ \begin{align} \frac1{2^{k+1}}\frac{n}{n-k} &=\frac1{2^{k+1}}+\frac1{2^{k+1}}\frac{k}{n-k}\\ &\le\frac1{2^{k+1}}+\frac{k}{2^{k+1}} \end{align} $$
With my idea, it boils down to showing $$\mathop {\lim }\limits_{\alpha \to \infty } \alpha \int\limits_0^\alpha {{e^{ - \frac{{{x^2}}}{2}}}\sin \alpha xdx} = 1$$
This essentially becomes a Diophantine equation:
$$9x\equiv{24}\pmod{21} \\
9x-24=21y \\
9x-21y=24 \\
3x-7y=8$$
Solve the Diophantine equation: $3u-7v=1$ using the Euclidean algorithm:
$$3(2)+1=7\\
3(-2)-7(-1)=1\\
u=-2;\,v=-1$$
Now multiply out by $8$ to get the original Diophantine equation...
Does anyone know why my answer is different than everyone else's?
@Ataraxia Solving an equation of the form $ax\equiv b\pmod n$ is the same as solving $[a]x=[b]$ in the ring of equivalence classes $\mathbb Z/n\mathbb Z$.
It is also a rather broad question, which easily permits "dumb" (non-specifically insightful) answers. In fact, let $r(a)$ be any function. For every $a$, let $c_1(a),c_2(a),\cdots$ be any sequence of nonzero numbers converging to $r(a)$. Then $\prod_{i=0}^\infty q_i(a)=r(a)$, where $q_0(a):=c_1(a)$ and $q_i(a):=c_{i+1}(a)/c_i(a)$.
note that the answer wasn't approached in anything but the telescopy perspective (and used the recursion definition for a distinguished telescoping product)
like I said, the question is "broad" and the answer format is "dumb" in the sense that the construction process used in the answer has really nothing to do with the particulars of the question. as I explained in detail above, the same process works for literally any function r(a), with nested radicals or otherwise.
Okay, this is really banal and you've all certainly seen this before, but can someone give a hint for $\int_0^{\frac{\pi}{2}}\frac{dx}{1+a^2\tan^2(x)}$?
@user1 well if the question is, can you start with any integer $n$ and end up with any 5th power of any other integer $m$, then it's obviously false, just let $n>m^5$
This essentially becomes a Diophantine equation:
$$9x\equiv{24}\pmod{21} \\
9x-24=21y \\
9x-21y=24 \\
3x-7y=8$$
Solve the Diophantine equation: $3u-7v=1$ using the Euclidean algorithm:
$$3(2)+1=7\\
3(-2)-7(-1)=1\\
u=-2;\,v=-1$$
Now multiply out by $8$ to get the original Diophantine equation...
