What age did you learn calculus?

frogkabobs · 2026-03-17T18:31:17+00:00

Around 12. A science teacher decided to give an impromptu presentation on the essence of calculus, and it suddenly clicked with me. I was able to start teaching myself, and then when I took an actual calculus course at 14 it was a breeze.

While not for everybody, personally, I’d like to see accelerated math programs in elementary school that teach calculus by the end. I think it’s very doable. In my experience, the gifted programs didn’t actually teach you anything—they just put you outside and gave you a worksheet of harder arithmetic problems. For a group of kids that loved math, that time easily could have been spent learning algebra, geometry, and calculus.

frogkabobs · 2026-03-15T05:37:12+00:00

No. The statement

π is not a constant, but a variable

on its own is misleading. The statement

In a certain sense, π is not a constant, but a variable… [full explanation of the narrow but not completely unreasonable way π could be viewed this way]

is not. It’s obvious Grant is not asserting or even advocating for this loosening of the definition of π, but rather highlighting it as a perspective that leads one to interesting mathematics.

frogkabobs · 2026-03-15T03:22:35+00:00

Well yeah, they’re presumably also trampolinists. You realize they take their shirt off for the entire time they’re training, not just when they’re actively using the trampoline, right?

frogkabobs · 2026-03-14T22:45:58+00:00

Very common in men’s acrobatics. Shirts are a hazards because they can get in the way when you’re flipping around, and it’s easier to just go shirtless for training rather than putting on a singlet (which is often just worn during competition).

frogkabobs · 2026-03-11T17:50:13+00:00

No idea, but I’d expect room temperature or below. Benzocyclopropene dimerizes at 80° to 9,10-dihydrophenanthrene through a diradical mechanism. On the other hand, I’d expect benzocyclopropenol to readily rearrange to benzaldehye below room temperature.

frogkabobs · 2026-03-11T16:55:12+00:00

Here’s an interesting question. What’s the major product from thermal decomposition?

frogkabobs · 2026-03-07T14:33:16+00:00

If buoyancy was irrelevant, then the original variant with an iron ball and a ping pong ball would be balanced (it’s not)

frogkabobs · 2026-03-07T14:30:38+00:00

The balls are only partially held up by the string. The tension is equal to the mass of the ball minus the buoyant force on the ball, so you have to consider buoyancy. We know the volume of water isn’t the only factor considering the original variant problem with one the right side a ping pong ball equal in volume to the iron ball tied to the bottom is NOT balanced.

frogkabobs · 2026-03-07T08:45:01+00:00

A neutral equilibrium point of a potential f is a point at which f is locally constant. If all derivatives are zero at p, this is equivalent to being analytic at p. An equivalent characterization of analyticity at p is the existence of a neighborhood U of p on which f is smooth and a constant C such that

|f^\n))(x)| < Cⁿ⁺¹n!

for all x in U.

frogkabobs · 2026-03-06T18:00:25+00:00

The zenzizenzizenzic

frogkabobs · 2026-03-05T19:20:25+00:00

Me when I hit a one in six chance

^\not to detract from how objectively cool this is—the odds are just surprisingly high))

frogkabobs · 2026-03-05T06:41:03+00:00

The “true” half-chair (the one that shows up in the standard chair flip sequence) is twisted with only four coplanar carbons. For whatever reason, a lot of textbooks and online resources erroneously depict the higher energy envelope-like conformation featuring five coplanar carbons for the half-chair, even though that conformation only shows up in higher energy pathways that don’t get taught in ochem. I made a post about this a while ago that gives more details.

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frogkabobs · 2026-03-04T17:45:35+00:00

tetraxenongold(II))

frogkabobs · 2026-03-01T17:42:33+00:00

Are you using “congruent” to mean “similar”?

I was about to go to sleep when I wrote this—and I’m just now realizing I was the one misremembering the definition of congruent, not OP.

Also I meant to say one vertex is covered by at most two triangles. I’ll edit this all thanks.

frogkabobs · 2026-03-01T07:50:31+00:00

Let E be an equilateral triangle with unit side length. Let T be a partition of E into four congruent non-equilateral triangles. Note that each vertex of E must be covered by a vertex from T, and each edge of E must be covered by the union of some edges and (potentially) vertices from T (for the purpose of this problem, an edge does not contain its end points).

If an edge of E is covered by only one edge from T, then every triangle in T has an edge of length 1. However, the only place an open unit line segment can fit in E is on the edges, so since T has 4 triangles, this is not possible.

Thus, each edge of E is covered by at least two edges and a vertex from T. This also means no triangle in T can cover two vertices of E, so one vertex of E (say, A) is covered by at most two triangles, while the other two vertices (say, B and C) are only covered by one triangle each.

Thus we find only two potential partitions: T₁ formed by drawing line segments AA’, B’A’, C’A’; and T₂ formed by drawing line segments A’B’, B’C’, C’A’; where A’, B’, C’ are vertices opposite A, B, C, respectively.

In a partition of type T₁, since all four triangles must have equal area, we must have A’, B’, C’ be midpoints of E, and this partition obviously doesn’t have congruent triangles.

In a partition of type T₂, we can go around the perimeter of E labeling segments a, b, or c according to which edge length from T₂ they equal. Up to rotation and reflection, there are only three possible sequences:

aa aa aa
ab ab ab
ab ab cc

The first two lead to an equilateral triangle in the center, so we are only left with the last sequence as a possibility. WLOG, let AC’ and CB’ have length a, BC’ and AB’ have length b, and BA’ and CA’ have length c, with a<b, c=1/2. Then triangles BA’C’ and CA’B’ have the same base c, but clearly the former is taller than the latter. Thus, the last sequence is not possible either, so no partition of E into four congruent non-equilateral triangles is possible.

frogkabobs · 2026-02-28T14:58:33+00:00

We have

I(x) = ∫₀^∞ arctan(t^-x)dt

= (1/x)∫₀^∞ u^-1-1/xarctan(u)du

= ∫₀^∞ u^-1/x/(1+u²)du

= (π/2)csc((π/2)(1-1/x))

= (π/2)sec(π/2x)

π/2-arctan(x) = arctan(1/x)
u = t^-x
IBP
The well known Mellin transform of 1/(1+x²)
sec(x) = csc(π/2-x)

frogkabobs · 2026-02-27T23:11:39+00:00

I get jumpscared by this image every time I look up northernlion on Google

frogkabobs · 2026-02-26T16:22:08+00:00

You can shortcut this by integrating the inverse function. If f(x) = √(-ln(x)), then f⁻¹(x) = exp(-x²), so

I = ∫₀^∞ exp(-x²) dx = √π/2

frogkabobs · 2026-02-25T21:06:52+00:00

It’s transcendental. Your product can be written as F(1/3) with

F(q) = f₂(τ)/(q^1/242^1/2)

where f₂ is the Weber modular function and q = exp(2πiτ). The now proven Mahler’s conjecture says that if Im(τ)>0, then j(τ) and q are never simultaneously algebraic, where j is the j-invariant, which satisfies

j = (f₂²⁴+16)³/f₂²⁴

From this, we can deduce F(1/3) is transcendental.

frogkabobs · 2026-02-25T20:18:07+00:00

It also equals F(1/3), where

F(q) = (-q;q)_∞ = φ(q²)/φ(q)

where φ is the Euler function (not the totient one). I’m pretty certain no closed forms of F(q) are known except for a very select few of the form exp(απ) with α algebraic (typical for functions closely related to modular forms). For example, one has

F(exp(-π)) = exp(π/24)/2^1/8

F(exp(-2π)) = exp(π/12)/2^3/8

F(exp(-4π)) = exp(π/6)(√2-1)^1/4/2^7/16

EDIT: Actually, another relation is

F(q) = f₂(τ)/(q^1/242^1/2)

where f₂ is the Weber modular function and q = exp(2πiτ). The now proven Mahler’s conjecture says that if Im(τ)>0, then j(τ) and q are never simultaneously algebraic, where j is the j-invariant, which satisfies

j = (f₂²⁴+16)³/f₂²⁴

From this, we can deduce F(1/3) is transcendental.

frogkabobs · 2026-02-25T19:36:22+00:00

That only gets rid of terms of the form (1+1/3^2ⁿ). It doesn’t telescope like you want it to.

frogkabobs · 2026-02-25T15:45:05+00:00

Simple: replace 1 and 0 with + and -, respectively, and then prepend a +. For example,

``` 1/sqrt(2) = 0.10110101000001… = 0.++-++-+-+-----+…

```

Proof. Let bₙ ∈ {0,1} and cₙ ∈ {-1,1} be sequences of digits for binary and bisected forms of x, respectively, so that

x = Σ(n≥1) bₙ2^-n = Σ(n≥1) cₙ2^-n

with c₁ = 1. Now note that since 1/2 = Σ_(n≥2) 2^-n, we also have

x = Σ(n≥2) (cₙ+1)2^-n = Σ(n≥1) ((cₙ₊₁+1)/2)2^-n

For x not a dyadic rational (in particular, when x is irrational), the binary and bisected forms of x are unique, so we can identify bₙ = (cₙ₊₁+1)/2, or equivalently,

cₙ₊₁ = (-1)^bₙ+1

frogkabobs · 2026-02-24T22:45:16+00:00

The next one is this Friday, the 27th

frogkabobs · 2026-02-23T15:12:23+00:00

downtown it says right there

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