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 Curious
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1d
answered Check the convergence of $a_{n+1}=\sqrt{a_n+\frac{4}{a_n}}$ where $a_1=4$
Apr
29
answered Guessing number in set 1-100 with weighted questions.
Feb
26
accepted What type of number is x?
Feb
26
accepted Prove that for any give sequence of digits, there is a perfect square starting with that sequence
Jan
29
answered how to compute the last 2 digits of 3^3^3^3 to n times?
Jan
6
answered Can a coin with an unknown bias be treated as fair?
Oct
22
comment It is easy to show that $S_m=\sum_{n=1}^\infty \frac{n}{2^n + m}$ converges for any natural$\ m$, but what is its value?
Not sure this is going to help (didn't try and don't have time for this), but did you try to find an asymptotic form? I typically start with $f(x)=\frac{x}{2^{x} + m}$ and $f(x)$ is descending from some $x$. Then $S_{m}=\sum_{n=1}^{\infty }f(n)\cdot (n+1 - n)\approx \int_{1}^{\infty }f(x)dx$. Then check this wolframalpha.com/input/?i=x+%2F+%282%5Ex+%2B+m%29
Sep
6
comment Is the sequence $(a_n)$ defined by $a_n=\tan{a_{n-1}}$, $a_0=1$, dense in $\Bbb{R}$?
Yes, it is. Consider $\tan(x)$ on $[ 0,\pi )$ (not $[-\frac{\pi }{2},\frac{\pi }{2}]$). Still, $\forall \alpha \in \mathbb{R}, \exists \beta \in[0, \pi): \tan(\beta)=\alpha$. If $\{b_{n}\}_{n=0}^{\infty}$ is dense in $[ 0,\pi )$, then $\exists b_{k}\approx \beta$ and $\alpha=\tan(\beta)\approx \tan(b_{k})=a_{k+1}$ (because $\tan(x)$ is continuous, except $\frac{\pi }{2}$, but it's easy to deal with it). And $\forall \alpha \in \mathbb{R}, \exists a_{k+1}: \alpha \approx a_{k+1}$, which means $\{a_{n}\}_{n=0}^{\infty}$ is dense in $\mathbb{R}$.
Sep
5
comment Is the sequence $(a_n)$ defined by $a_n=\tan{a_{n-1}}$, $a_0=1$, dense in $\Bbb{R}$?
Another trick that might turn to be useful, just narrowing down the problem. Consider $k_{n}\in \mathbb{Z}: k_{n}\cdot \pi \leq a_{n}<(k_{n}+1)\cdot \pi$ and the sequence $b_{n}=a_{n}-k_{n}\cdot \pi \in [0,\pi)$. We have $\tan(b_{n})=\tan(a_{n})=a_{n+1}$ or $\{\tan(b_{n})\}_{n=0}^{\infty}\bigcup \{1\}=\{ a_{n}\}_{n=0}^{\infty}$. If we "shift" $\{b_{n}\}_{n=0}^{\infty}$ by $t\cdot \pi, t\in \mathbb{Z}$ we get the same result. As a result, it's enough to prove that $\{b_{n}\}_{n=0}^{\infty}$ is dense in $[0,\pi)$
Aug
28
comment Is the sequence $(a_n)$ defined by $a_n=\tan{a_{n-1}}$, $a_0=1$, dense in $\Bbb{R}$?
Another result that might turn to be useful math.ust.hk/~majhu/Math203/Rudin/Homework15.pdf, point 4.4. Basically $\left \{ a_{n} \right \} = \left \{ \tan\left ( a_{n} \right ) \right \}\bigcup \left \{ 1 \right \}$
Aug
28
comment Is the sequence $(a_n)$ defined by $a_n=\tan{a_{n-1}}$, $a_0=1$, dense in $\Bbb{R}$?
Assume a function $f:A\rightarrow B$ that is surjective, continuous and periodic with period $T\in \mathbb{R}\setminus \mathbb{Q}$. Because $T$ is irrational, according to mathworld.wolfram.com/KroneckersApproximationTheorem.html, $\left \{ k\cdot T+n |n,k\in \mathbb{Z} \right \}$ is dense in $\mathbb{R}$, so $\forall \beta\in B,\exists \alpha \in A : \beta =f\left ( \alpha \right )\approx f\left ( k\cdot T+n \right )=f\left ( n \right )$ or $\left \{ f\left ( n \right ) |n\in \mathbb{Z}\right \}$ is dense in $B$, just a sketch ...
Aug
27
comment Is the sequence $(a_n)$ defined by $a_n=\tan{a_{n-1}}$, $a_0=1$, dense in $\Bbb{R}$?
I am wondering if the fact that $\left \{ \tan(n)| n\in \mathbb{Z} \right \}$ is dense on $\mathbb{R}$ could be of any use?
Jul
17
awarded  Curious
Jul
16
comment Prove that for any give sequence of digits, there is a perfect square starting with that sequence
From the educational perspective "so the solutions provided should include these tools (a bit of constraint to the problem)." ... I am sure there are other ways to prove it.
Jul
16
answered Prove that for any give sequence of digits, there is a perfect square starting with that sequence
Jul
16
asked Prove that for any give sequence of digits, there is a perfect square starting with that sequence
May
23
comment Given two concentric circles with radiuses r < R, can we estimate the number of chords in between the circles?
Another way, probably, to tackle the problem (and this is a brainstorming): $$n_{of chords}\sim \frac{\pi }{\sqrt{1-\alpha ^{2}}}=\pi \cdot \sqrt{1+\alpha ^{2}+\alpha ^{4}+...}< \pi \cdot \sqrt{n}\cdot \sqrt{1+\frac{1}{n}\sum_{i=n}^{\infty }\alpha ^{2\cdot i}}$$
May
23
comment Given two concentric circles with radiuses r < R, can we estimate the number of chords in between the circles?
Yes, sort of. For this particular case when $r=\sqrt{p_{n}}$ and $R=\sqrt{p_{n+1}}$, knowing that $\sqrt{p_{n+1}}<n$ from some n and $\sqrt{p_{n+1} - p_{n}} \geq \sqrt{2}$, $n_{of chords} < \frac{\pi }{\sqrt{2}}\cdot n$.
May
23
comment Given two concentric circles with radiuses r < R, can we estimate the number of chords in between the circles?
Yes, it goes to infinity, but how "quickly"?
May
23
comment Given two concentric circles with radiuses r < R, can we estimate the number of chords in between the circles?
yes, that's the ultimate goal ;)