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4h
comment Evaluate $\lim _{n\to \infty }\left(1-f\left(\frac{1}{\sqrt{n}}\right)\right)\cdot \sum _{k=1}^nf\left(\frac{k}{n}\right)$
Hint. $1 - f(n^{-1/2}) \sim 1/(2n)$.
5h
comment Show $\mathbb{E}(X \mid Y,Z) = \mathbb{E}(X \mid Y)$ if $Z$ is independent of $X$ and $Y$
A standard machinery in measure theory is the monotone class theorem. The rough idea is that you can first show the claim for indicator functions, then for simple functions, and then for any bounded measurable functions by approximation argument.
5h
answered Can I get better approximation of $\sum_{k=1}^{n} k^k$
1d
comment Wrong integral proof
Your setting actually leads to a solution, as long as you correctly figure out $x$. Indeed, using the arc-cosine function we get $u=\arccos\sin x = \frac{\pi}{2}-x$, which yields the desired formula through change of variable.
Apr
20
comment Limit property of a function: $\lim_{p \to 0} \frac{w(c p)}{w(p)} \in (0,\infty)$
Are you looking for regularly varying functions?
Apr
13
comment Proving associativity of symmetric set difference
Of if you want to do it in a more elementary way, the following interpretation will be useful: $$ A + B = \{ x \in X : x \text{ lies in exactly one of } A \text{ or } B\}, $$ and more generally, you can establish the following interpretation $$ A + (B + C) = \{ x \in X : x \text{ lies in exactly odd number of } A, B \text{ and } C\} $$. By establishing exactly the same interpretation for $(A+B)+C$ you can prove that they are equal.
Apr
12
comment Proving associativity of symmetric set difference
@AlexR, I agree that proving the correspondence requires a bit of work, but still the outcome may be worth it if OP wants more than just manipulating a jumble of set operations.
Apr
12
comment Proving associativity of symmetric set difference
Identify $A \subset X$ as a $\Bbb{Z}/2\Bbb{Z}$-valued indicator function $\mathbf{1}_A : X \to \Bbb{Z}/2\Bbb{Z}$ defined by $$ 1_{A}(x) = \begin{cases} 1, & x \in A \\ 0, & x \notin A \end{cases}. $$ Then you can easily prove that $\mathbf{1}_{A+B} = \mathbf{1}_A + \mathbf{1}_B$. This moves all the burden of set-operation to just algebraic operations, which makes proof almost trivial. For example, $$ \mathbf{1}_{A+(B+C)} = \mathbf{1}_A + \mathbf{1}_{B+C} = \mathbf{1}_A + \mathbf{1}_B + \mathbf{1}_C = \mathbf{1}_{A+B} + \mathbf{1}_C = \mathbf{1}_{(A+B)+C}. $$
Apr
12
comment Existence of complicated convex functions
Notice that $f'_{\pm}$ are increasing and differ exactly at points of jump discontinuity. Since an increasing function has at most countably many jumps, this implies that $D_f$ is at most countable.
Apr
12
comment How to solve $\int_0^{\frac{\pi}{2}}\frac{x^2\cdot\log\sin x}{\sin^2 x}dx$ using a very cute way?
(+1) I bow to your clever observation. I also finished my calculation and obtained exactly the same answer, so I guess your solution is among the shortest ones.
Apr
12
answered How to solve $\int_0^{\frac{\pi}{2}}\frac{x^2\cdot\log\sin x}{\sin^2 x}dx$ using a very cute way?
Apr
12
comment Can the hypergeometric function be extended analytically to the complex plane in the interval [1,$\infty$ )?
But definitely we should avoid $(1, \infty)$. (Even this condition is explicitly mentioned in the paper in the form $|\arg(-z)|<\pi$.) An easy way of checking the unavoidability of branch cut is to plug $(b, c) = (1, 2)$ and find an exact formula. (Of course, you may choose different branch cut that does not intersect $(1, \infty)$ but at least there is no hope of getting a meromorphic function on $\Bbb{C}$ for general parameters.)
Apr
12
comment Is it true that $f(x)=g(x)$ for all $x\in \mathbb R$
Hint. $\Bbb{Q}$ is dense in $\Bbb{R}$.
Apr
11
awarded  Enlightened
Apr
11
awarded  Nice Answer
Apr
10
answered Convergence of $\sum \frac{a_n}{1+a_n}$ when $\sum a_n$ and $\sum a_n^2$ converges.
Apr
10
answered $\lim_{n\to\infty} \sum\limits_{k=1}^n \frac{\ln k}{n} ( 1-\{\frac{n}{k} \} ) ( (1-\frac{k}{n} \{\frac{n}{k}\} )-\frac{1}{2})$
Apr
9
awarded  Yearling
Apr
7
comment Finding a combinatorial argument for an interesting identity.
For the sake of future reference, I would like to mention that both sides are indeed equal to $$ [x^m y^n] \frac{1}{1-x-y-xy}, $$ where the bracket notation extracts the coefficient of $x^m y^n$ from the Taylor series.
Apr
3
comment $\int_{0}^{\infty}e^{-st}h(t)dt=0 \Rightarrow h(t)=0.$
@robjohn, Thank you. I agree that Bernstein polynomial gives a short-cut proof. (Even some versions of Stone-Weierstrass proof exploit this nice family of polynomials, so my proof may be considered indirect.) By the way, depending on what type of integral OP is considering, the technicality I have introduced seems necessary. For example, if $h(x) = e^{x^2} \sin(e^{2x^2})$ then the Laplace transform exists only in improper sense when $s > 0$.