Math videos covering a wide variety of topics from theory to application. Calculus, Differential Equations, Number Theory, Proofs, unique problems and much more. I have a second channel that has full course material for higher level math called MathMajor which is geared toward more structured videos that support real classes I teach. My goal is to spread math education, knowledge, and understanding to as many people as possible.
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2:1494 views · today · 2:14
Around 440 BCE, Hippocrates of Chios found a curved shape with exactly the area of a triangle. Build a semicircle on the chord of a quarter circle, and the crescent outside the circle is his lune. Its area is a semicircle, minus a sector, plus a triangle.
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1:344K views · 1d ago · 1:34
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1:132K views · 1d ago · 1:13
Ruler and compass can't construct pi, so what extra tool would? Hippias of Elis found a curve around 420 BCE: slide one edge of a square down at a steady speed while another swings down like a clock hand, and trace where they cross. That curve is the quadratrix, the square-maker.
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30:4615K views · 1d ago · 30:46
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2:3012K views · 2d ago · 2:30
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1:292K views · 3d ago · 1:29
Clairaut's theorem says mixed partial derivatives commute. On R×Z one of them is a forward difference, and the proof is short: the x-derivative passes straight through f(x,n+1) − f(x,n), so ∂x Δn = Δn ∂x. Full video
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2:233K views · 4d ago · 2:23
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2:003K views · 4d ago · 2:00
Add zero in the right form and the forward difference of a product splits in two: Δ(fg) = (Δf) g(n+1) + f(n) Δg. It looks like the usual product rule, except one factor is shifted. Full video
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2:5011K views · 5d ago · 2:50
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1:514K views · 5d ago · 1:51
Sum the forward differences of f from c to d − 1 and almost everything cancels: each term is wiped out by its neighbour. Only f(x,d) − f(x,c) survives. That is the discrete fundamental theorem of calculus. Full video
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2:372K views · 6d ago · 2:37
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1:302K views · 6d ago · 1:30
Take f(x,n) = x2 2n. The x-derivative is 2n+1 x, and the forward difference in n gives back x2 2n, the function we started with. In discrete calculus 2n is its own derivative, the way ex is in ordinary calculus. Full video
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2:072K views · 7d ago · 2:07
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2:203K views · 7d ago · 2:20
A line integral on R×Z: slide along a floor, hop between floors. Two routes from (0,0) to (2,3) for n2 dx + xn dn give 6 and 18, so this field is not path independent. Full video
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2:106K views · 8d ago · 2:10
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2:099K views · 8d ago · 2:09
In discrete calculus the derivative is the forward difference: Δn f(n) = f(n+1) − f(n). Apply it to n2 and you get 2n + 1, not 2n. The power that does behave like xk is the falling power n to the k underline. Full video
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47:1816K views · 8d ago · 47:18
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2:1210K views · 10d ago · 2:12
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2:354K views · 11d ago · 2:35
For y' = x−2 + y2, substitute y = v/x. The equation becomes x v' = v2 + v + 1, which is separable; complete the square and integrate to get (2/√3) arctan((1 + 2v)/√3) = ln x + C. So n = −2 has a formula too, by a trick unlike the others.
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2:349K views · 12d ago · 2:34
y' = 1 + y2 is separable: write it as dy/(1 + y2) = dx and take antiderivatives to get arctan y = x + C. So y = tan(x + C), and the tangent addition formula rewrites it as (tan x + A)/(1 − A tan x) with A = tan C. Full video
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2:487K views · 13d ago · 2:48
Set ν = 1/(n + 2). Every exponent the two tricks reach from n = 0 has ν in 1⁄2 + Z, and those are the only equations y' = xn + y2 solvable by formulas. Jacob Bernoulli's n = 2 gives ν = 1⁄4, and the tricks only bounce it around ±1⁄4 + Z, always a quarter away from the nearest half integer.
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2:503K views · 14d ago · 2:50
Two tricks turn a solution of y' = xn + y2 into one for a new exponent: n ↦ −n − 4 and n ↦ −n/(n + 1). Starting from n = 0 they hop through −4, −4/3, −8/3, −8/5, ..., every number of the form −4k/(2k + 1), crowding toward −2. Jacob Bernoulli's n = 2 is never reached.
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1:4174K views · 15d ago · 1:41
y' = 1 + y2 is solved by the tangent function. Change it to y' = x2 + y2 and the solution curves look almost the same, but no formula for them exists. Jacob Bernoulli fought it in 1697; in 1724 Riccati asked which powers xn allow a formula, and Daniel Bernoulli answered with an anagram.
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41:4369K views · 15d ago · 41:43
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2:453K views · 17d ago · 2:45
Ramanujan's third congruence says 11 divides p(11n+6), and the first case is p(6) = 11. Sort those 11 partitions by rank mod 11 and it falls apart: slots 1 and 10 take two each, slots 4 and 7 come out empty. Eleven partitions, nine piles. Dyson said some other statistic must work for 11 and called it the crank.
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2:232K views · 18d ago · 2:23
The Durfee square is the biggest square sitting in the top-left of a Ferrers diagram. What is left is a piece to the right and a piece below, and the rank only sees those two: columns on the right minus rows below. Tag each right-hand column with z and each bottom row with z inverse, and the power of z is exactly the rank.
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9:5713K views · 18d ago · 9:57
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2:233K views · 19d ago · 2:23
The 5 partitions of 4 have ranks 3, 1, 0, -1, -3 — one in each class mod 5. The 7 partitions of 5 give one in each class mod 7. Sort the 30 partitions of 9 by rank mod 5 and you get six in every pile, and the 135 partitions of 14 split into five piles of 27.
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2:373K views · 20d ago · 2:37
The rank of a partition is its largest part minus its number of parts. Draw it as a Ferrers diagram and reflect across the diagonal: the top row and the first column swap, so the rank turns into minus itself. That one fact pairs up the piles and does real work later on.
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1:204K views · 21d ago · 1:20
Euler's product has one factor for each part size: the first chooses how many 1s, the second how many 2s, and so on. Multiply them out and the coefficient of q^n counts the partitions of n. Throw away the even factors and it counts partitions into odd parts instead — change the factors, change what you count.
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1:566K views · 21d ago · 1:56
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32:4234K views · 21d ago · 32:42
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1:584K views · 22d ago · 1:58
p(n) counts the ways to write n as a sum of positive whole numbers. Line up p(4), p(9), p(14), p(19), p(24) and every one of them is a multiple of 5 — and it never stops. Ramanujan proved it in 1919, along with the same pattern mod 7 and mod 11.
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2:163K views · 22d ago · 2:16
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2:473K views · 23d ago · 2:47
sin(ax)/(ax) is the average of cos(xy) over y from −a to a. That one observation, plus a product-to-sum formula, computes ∫0^∞ sinc(x)·sinc(x/3) dx by hand: both signs come out +1, and the answer is exactly π/2. Full video
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1:269K views · 23d ago · 1:26
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1:223K views · 24d ago · 1:22
Proving the Dirichlet integral, part 2. From Im ∫0^∞ 1/(ic − y) dy: multiply through to get c/(c2 + y2), integrate to arctan(y/c), and let y → ∞. That gives π/2 when c is positive and −π/2 when c is negative. Full video
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2:019K views · 25d ago · 2:01
Proving the Dirichlet integral without contour integration, part 1. Write 1/x as ∫0^∞ e^(−xy) dy, turn sine into the imaginary part of an exponential, swap the order of integration, and the x-integral collapses to Im ∫0^∞ 1/(ic − y) dy. Part 2 finishes it.
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1:227K views · 25d ago · 1:22
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2:1811K views · 26d ago · 2:18
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1:535K views · 27d ago · 1:53
1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/13 ≈ 0.955, just under 1, so every sign in the Borwein integral stays +1 and the answer is exactly π/2. Add 1/15 and the sum becomes 46027/45045 ≈ 1.022. It spills over 1, some signs flip, and the integral falls just short of π/2.
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1:295K views · 28d ago · 1:29
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