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Hopeful mathematics student with some competition level problem solving experience. Studying algebraic topology and algebraic geometry at the moment, but I feel like anyone who know anything know more than I do.


22h
comment Formula to calculate a length to a point on hypotenuse according to given angle
Not as you've written it (note the $x$ on both sides of your equals sign: they will cancel and you're left with $1 = \dfrac{\frac{\tan\alpha}{1 + \tan\alpha}}{\sin 45^\circ}$ which is not true).
22h
comment Digits of $n$ factorial
It is the most probable thing to happen. The digits (apart from the trailing zeros) are in some sense psuedo-random, so you would expect there to be about as many of each one of them. I would consider it far more worthy of research was this not the case. Also note that the first digit is nowhere near uniformly distributed; $1$ is as likely as $5, 6, 7, 8$ and $9$ combined as a consequence of Benford's law.
22h
answered Percentage operation gives different results with two scientific calculator
22h
answered Formula to calculate a length to a point on hypotenuse according to given angle
2d
comment How can a simple closed curve not look locally like the rotated graph of a continuous function?
"Nice enough" basically means "not fractal in nature". If the curve is fractal-like, then it is not obvious that there is any interior big enough to contain a square..
Apr
20
comment L'Hopital quicky
Well, in the case of $\lim_{x\to\infty}\frac{n}{n^3}$ this obviously doesn't work, since $\frac{\frac{f(x)}{g(x)} }{ \frac{f'(x)}{g'(x)}} = \frac{1/n^2}{1/3n^2}= 3\neq 1$.
Apr
17
comment Proving a relation is transitive
@JohnAnderson That would be (3, 3) in the {(1, 3),(3, 1)}-example. Because you can read it the "other way", that is {(3, 1), (1, 3)}, so you need (3, 3) as well. In your big example, you just have to check that for every pair of pairs (a, b), (b, c) you also have (a, c). You have (1, 3), (3, 5) therefore you need (1, 5) and so on.
Apr
17
comment Why is the method to finding the order of a torsion subgroup different than finding the maximum order of a given element of a direct product?
@Frumpy Yes, that is the gist of it. That is the reason why when decomposing finite abelian groups into sums of $\Bbb Z_n$'s, you can split any $n$ according to its maximal prime power divisors, but you cannot split the prime powers themselves, e.g. you can split $\Bbb Z_{36}$ into $\Bbb Z_4\times \Bbb Z_9$, but you cannot split the $4$ or $9$ any further (that would change the $\operatorname{lcm}$)
Apr
17
comment Proving a relation is transitive
To answer your particular question: Yes, transitivity plus symmetry of a relation implies reflexivity. So you would need (1, 1) if you had (1, 3), (3, 1) (you would also need (3, 3), just so you know). In the case of a list of ordered pairs like yours, you actually have to check every single pair of pairs, see if they make a "demand" of transitivity (i.e. one of them ends with the same number the other begins with), and check whether that demanded pair is actually an element of the relation. If your relation is rather given as a general description, that'd take a different strategy.
Apr
14
comment Prove the inequality $({1+\frac{a}b})^n$ + $(1+\frac{b}a)^n$ $\geq$ $2^{n+1}$
Binomial theorem plus repeated application of $x + 1/x\geq2$ might do the trick.
Apr
13
answered GNU Octave draw figure of 2 planes
Apr
10
comment What exactly is a dimension?
There are the concepts of Hausdorff dimension and topological dimension. They might be what you need.
Apr
9
comment Flipping a coin 1000 times.
The maximum likelihood estimation is one way. There you look at what value of $p$ yields the largest value of $P(\text{Your result}\mid\text {Bias}=p)$. In the case of a coin flip, though, it is more accurate the closer to $50-50$ the result is so it might not be the best approach.
Apr
9
comment Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $
WolframAlpha says $\sqrt{3}$.
Apr
8
revised Are there arbitrarily large arithmetic progressions of primes for some fixed progression width?
added 213 characters in body
Apr
8
answered Are there arbitrarily large arithmetic progressions of primes for some fixed progression width?
Apr
8
revised Arithmetic progression of primes question
added 209 characters in body
Apr
8
comment Arithmetic progression of primes question
@TonyK I am aware that the asker starts counting at $1$ and I start at $0$, but I do not see the big difference as long as the number of terms are correct (technically $a = -23, k = 6$ is the example I've shown, yes, I know). Let's say I use $b$ instead, just to make you happy.
Apr
8
answered Arithmetic progression of primes question
Apr
8
answered Riddle about amusement park profit in relation to ticket's price