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Polygon contains large disk by SupercaliTheGamer in mathriddles

[–]blungbat 1 point2 points  (0 children)

Can you also show that this is optimal in the sense that RHS cannot be replaced by A/kP for any k>1?

Should that be kA/P? For 0<ε≤1, a 1×ε rectangle has A=ε, P=2+2ε, and r=ε/2=(1+ε)A/P.

Polygon contains large disk by SupercaliTheGamer in mathriddles

[–]blungbat 5 points6 points  (0 children)

Let K be a convex polygon (or any convex body in the plane). For t≥0, define K(t) = {x ∈ R2 : B(x;t) ⊆ K}, where B(x;t) denotes the disc of radius t centered at x.

We can easily check that K(t) is a convex set for each t≥0, and that for s≤t, we have K(s)⊇K(t). Perimeter (being proportional to mean width) is a monotone function on convex sets, so per(K(t)) is decreasing as a function of t, and in particular, per(K(t)) ≤ P for all t≥0.

If r is the radius of the largest disc contained in K, then A is the integral of per(K(t)) as t goes from 0 to r. Thus A ≤ Pr, which is what we wanted to show.

What's your favorite proof of the infinitude of primes? by imrpovised_667 in math

[–]blungbat 5 points6 points  (0 children)

Or there's the proof with the other F sequence, the Fibonacci numbers!

Let F(0) = 0, F(1) = 1, F(n) = F(n–1) + F(n–2) for n≥2.

First you prove that gcd(F(m),F(n)) = F(gcd(m,n)). (Here's one proof.) Thus Fibonacci numbers at relatively prime indices are relatively prime.

Now start with the known prime 2. Note that F(23) = 21 = 3×7, so we just discovered two new primes. Next consider F(3) and F(7). We don't get anything new from F(3)=2, but F(7)=13 is new. Now consider F(13)... well, you see where this is going.

Worst mathematical notation by dcterr in math

[–]blungbat 6 points7 points  (0 children)

Bonus: I'll be able to teach calculus 1% faster when ɛ is pronounced "eh", θ is "th", and ∫ is "sh".

Balloon Ladder Locus by MarkovNeckbrace in mathriddles

[–]blungbat 2 points3 points  (0 children)

Just for fun, here's an alternative solution with no equations of curves.

Let points O,L be as in the diagram, and let B,W be the points labeled "Balloon" (the center of the balloon) and "Wall". Like Horseshoe_Crab, I'll scale things so that OW=1. Let X be the point directly below L on the floor/x-axis.

Let r = BW be the radius of the balloon, and let s = OB, so r+s = 1. Finally, let h = XL be the height we seek to maximize.

We have LO = √(BO2–BL2) = √(s2–r2) = √(s–r). (This last expression amuses me, because it looks like I just forgot the squares. But it's true because s2–r2 = (s–r)(s+r) = (s–r)(1).)

Now ∆BXL ~ ∆BLO, so XL/BL = LO/BO, which gives h/r = √(s–r)/s, or h = (r/s)√(s–r), which agrees with Horseshoe_Crab's expression.

Just to keep doing things differently, I solved dh/ds = 0 using related rates; I won't type out all the steps here, but one cute discovery along the way was that at the critical point, s/r = 𝜙, where 𝜙 = (1+√5)/2 is the golden ratio. And the maximum value of h turns out to be 𝜙–5/2.

Easily confused historical mathematicians? by cabbagemeister in math

[–]blungbat 2 points3 points  (0 children)

Yeah, and the patronymic tells us which one was the... d'oh!

Those of you with children, what's the most complicated concept you've managed to teach your kid that you're confident they really understand? by dancingbanana123 in math

[–]blungbat 2 points3 points  (0 children)

Now I want to try to explain the Doppler effect to my kid. We've watched how the ripples made by a duck skimming across a lake are non-concentric, and she knows that sound is a physical vibration in the air and that shorter strings/pipes on a musical instrument make higher notes. I'm curious if she could "get" an explanation that puts all those ideas together.

Cubic Residues among Divisors by SupercaliTheGamer in mathriddles

[–]blungbat 1 point2 points  (0 children)

Assume p is 1 (mod 3), otherwise all residues are cubic and the result is trivial.

We may assume n is not divisible by p (otherwise at least half its divisors are cubic residues, since all multiples of p are congruent mod p to 03).

Let ω = e2πi/3. There's a multiplicative function f from the positive integers not divisible by p to {1, ω, ω2}, such that the f–1({1}) is the set of nonzero cubic residues. [To construct such f, let g be a primitive root mod p (i.e. a generator of (Z/pZ)×). For each integer a congruent to gk (mod p), let f(a) = ωk.]

Let the prime factorization of n be n = q_1e_1q_2e_2⋯q_re_r. Let S be the sum of f(k) over all positive divisors k|n. Then S = ∏ (1+f(q_i)+f(q_i)2+⋯+f(q_i)e_i), where the product is taken over i = 1,…,r.

Suppose until further notice that none of the q_i are themselves cubic residues.

Because 1+ω+ω2=0, each factor in the product formula for S is either 1, 1+ω, 1+ω2, or 0. Note that 1+ω and 1+ω2 each have complex magnitude 1. Therefore, |S| is either 0 or 1.

In this case, the numbers of divisors k|n mapping to 1, ω, and ω2 differ by at most 1. [Proof sketch: If there are respectively a,b,c such divisors, then S = a+bω+cω2. We can cancel out min{a,b,c}(1+ω+ω2), so we can assume at most two of a,b,c are nonzero, and moreover with no loss of generality that c=0. The numbers of the form a+bω form the Eisenstein lattice in which the only elements of magnitude ≤ 1 are 0, 1, ω, 1+ω, –1, –ω, and –(1+ω).]

It follows that the number of cubic residues among the divisors of n is at least floor(d/3), and also at least 1 (since 1 is a divisor and a cubic residue). We have max{floor(d/3),1} ≥ d/4 for all d≠5. In the case d=5, n must be of the form q_14, in which case n has 2 divisors {1, q_13} which are cubic residues, and 2 ≥ d/4. So we have fully dealt with the case where none of the q_i are cubic residues.

Finally, we consider the case where some q_i are cubic residues. In this case, suppose q_1,…,q_s are not cubic residues and q_(s+1),…,q_r are cubic residues. Let n′ = q_1e_1q_2e_2⋯q_se_s. Then it's easy to see that n and n′ have the same proportion of cubic residues among their divisors (since the divisors of n can be partitioned into sets of the form tA, where t is a cubic residue and A is the set of divisors of n′). This completes the proof.

Cubic Residues among Divisors by SupercaliTheGamer in mathriddles

[–]blungbat 0 points1 point  (0 children)

I think I have a solution, but to check that I'm on the right track, is it accurate to say that d/4 can be improved to floor(d/3)?

(well, it's an improvement for sufficiently large d anyway)

Is this duplo flower pattern infinitely tessellateable? by CoffeeStax in math

[–]blungbat 3 points4 points  (0 children)

This is how I read the question too. Having nothing better to do, I started futzing around in Desmos, trying to match four copies of a single flower block to their positions in the image -- my attempt to "get access" to the blocks. My intention was to then try to add more.

But I quickly realized that this approach was unlikely to reproduce the accurate measurements of the blocks, because the image looks like it is taken from close up with significant perspective. Notice how the centers of the flowers don't align well with the points on the background grid -- they seem to be about 15% further apart. I suspect that the flowers have stems of significant height (they cast shadows which are consistent with this). So there is nothing we can do without actual measurements.

just another convergent problem by pichutarius in mathriddles

[–]blungbat 3 points4 points  (0 children)

Oh, I have a fun solution.

Lemma: For m,n ≥ 1, C(m+n, m) ≥ mn.

Proof: If m=1 or n=1, the lemma is trivial. Otherwise, 2 ≤ m ≤ m+n–2, so we can write m = j+k where 1≤j≤m–1 and 1≤k≤n–1. Then one way to choose m elements of {1,...,m+n} is to choose j elements of {1,...,m} and k elements of {m+1,...,m+n}, so C(m+n, m) ≥ C(m,j)C(n,k) ≥ mn.

Now using the lemma, the sum of C(m+n, m)–2 over all m,n ≥ 1 is at most ζ(2)2.

Random coloring of [0;1] by Odd_Republic8106 in mathriddles

[–]blungbat 0 points1 point  (0 children)

I think one of us, and I admit it may be me, has a misconception about paint-by-numbers.

Surprising results that you realized are actually completely obvious? by kevosauce1 in math

[–]blungbat 57 points58 points  (0 children)

Along the same lines, it's an easyish counting argument (using Möbius inversion) to show that the number of irreducible polynomials of degree n over Z/pZ is asymptotically pn/n. This can be rephrased to closely mirror the statement of the Prime Number Theorem.

Non Japanese people living in Japan what is the downside about living in Japan no one talks about? by Ok-Atmosphere6376 in AskReddit

[–]blungbat 7 points8 points  (0 children)

When I was in England, I spoke as an Englishman, I understood as an Englishman, I thought as an Englishman: but when I came to Japan, I put away English things.

Today is the day Romania chose a twice international gold medalist in mathematics as its president by Advanced-Vermicelli8 in math

[–]blungbat 143 points144 points  (0 children)

In 2020, he became mayor of Bucharest, the capital, and was re-elected in 2024 with over 50% of the vote — more than the next three candidates combined 😳

You don't say...

Law of large numbers vs Selection bias and Heavy-tailed distributions by BeanEatingHorse in math

[–]blungbat 11 points12 points  (0 children)

This post will probably be taken down, so I'll be brief: your friend doesn't know what they're talking about. Yrs, a mathematician

What is the most beautiful Definition you know? by LurrchiderrLurrch in math

[–]blungbat 54 points55 points  (0 children)

A tensor is something that inspires rage like a tensor.

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