Sangchul Lee
Reputation
93/100 score
 20h comment Are mappings $f: \mathbb{R} \to \mathbb{R}$ with $|f'(x)| < 1$ contractions? @AnanthRaviKumar, That is exactly what I am claiming: there exists a differentiable function $f : \Bbb{R} \to \Bbb{R}$ with $|f'| < 1$ which is not a contraction. 1d answered Are mappings $f: \mathbb{R} \to \mathbb{R}$ with $|f'(x)| < 1$ contractions? 1d answered On the limit of $f(n)$, specifically having to do with integration of an iterated $\arctan$ 2d awarded Nice Answer Nov 21 comment Kolmogorov's 0-1 Law related : proving that $\mathbb{P} \left(\lim_{n \to \infty} \frac{X_1+…+X_n}{n}=1 \right) = 1$ Hint: As long as each $X_i$ are a.s.-finite, then for any $m$, we have $$\limsup_{n\to\infty} \frac{X_1 + \cdots + X_n}{n} = \limsup_{n\to\infty} \frac{X_{m} + \cdots + X_n}{n}$$ and likewise for liminf. Thus we can ignore finitely many terms without affecting the convergence. Nov 21 answered Evaluation of a cauchy Singular Integral Nov 20 answered Indefinite integral: chance of analytic solution? Nov 19 comment Limit with fractional part $\lim \limits_{n\to \infty}\{(2+\sqrt{3})^n\}$ Then what bothers you to jump into the conclusion? You can simply take $n$ and $\delta$ as $(2+\sqrt{3})^n + (2-\sqrt{3})^n \in \Bbb{Z}$ and $(2-\sqrt{3})^n \in (0, 1)$. Nov 19 comment Limit with fractional part $\lim \limits_{n\to \infty}\{(2+\sqrt{3})^n\}$ Let me simplify the situation: If $n$ is an integer and $0 < \delta < 1$, then can you check that $\{n-\delta\} = 1-\delta$? Nov 19 comment Limit with fractional part $\lim \limits_{n\to \infty}\{(2+\sqrt{3})^n\}$ Also notice that $0 < (2-\sqrt{3})^n < 1$ and consequently $\{ (2+\sqrt{3})^n \} = 1 - (2-\sqrt{3})^n$. Nov 19 comment Limit with fractional part $\lim \limits_{n\to \infty}\{(2+\sqrt{3})^n\}$ Hint. Can you check that $(2+\sqrt{3})^n + (2-\sqrt{3})^n$ is always an integer? Nov 16 accepted Extending the result $\int_{0}^{\infty} \left( ( 1 - 2C(x))^{2} + (1-2S(x))^{2} \right) \, dx = \frac{4}{\pi}$ Nov 14 comment IS the following series convergent or divergent? I don't think so. We may have one for $k=1$ in view of the Glaisher-Kinkelin constant together with the functional equation for the Riemann zeta function. This is a disguise, however, since the derivative is just absorbed into that magical constant. Nov 14 comment IS the following series convergent or divergent? Note that the logarithm $\log n$ grows slower than any positive power of $n$. So you can apply p-series test. Also the exact value of the series is $\zeta^{(k)}(2)$. Nov 14 comment Inner product equals for all vectors means @Hurkly, Initially I only considered complex inner product spaces, and then thought that the proof also works for real inner product spaces. Now I found that it doesn't. So you're right, in real inner product spaces we have a counterexample. Nov 14 comment Inner product equals for all vectors means You're welcome! I am pretty sure you already know this, but I just wanted to assure myself of the fact by leaving a comment. :) Nov 14 comment Inner product equals for all vectors means To complement this answer, recall the fact that an operator $T$ is orthogonal projection if and only if $T = T^*$ and $T^2 = T$. Then it is clear that this implies the given condition. Conversely, from polarization argument we can check that the given condition implies $T = T^* T$, which then shows that $T = T^*$ and $T = T^2$. Nov 12 comment Show the map from $0-1$ sequences to the corresponding binary numbers is continuous @user58902, Exactly. That will be one of the fastest way to guarantee the continuity of $f$. Nov 12 comment Show the map from $0-1$ sequences to the corresponding binary numbers is continuous $f$ is the sum of functions $f_n(a_1, a_2, \cdots) = 2^{-n}a_n$, and these functions are continuous since they are scalar multiple of projection maps. Now you may apply the Weierstrass M-test. Nov 10 revised Prove that a recursively given sequence converges from any starting number from interval $(2,\infty)$ LaTeXification