Proving Grant's Little Theorem | A Surprising Geometric Result

3blue1brown tucked a really neat geometric result in one of his videos. Here's my full proof and walkthrough of "Grant's Little Theorem".

Remember to post your solutions to the exercises in the comments!

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[zachstar]

Love seeing Jon help out small creators like 3b1b. Great video man!




Oh and first!

[upliftingcommunity]

What I love about this proof is how it branches across so many areas! It uses complex numbers, Euler’s formula, derivatives, limits, algebra, geometry. Super cool

[cinvhetin]

This is such an elegant proof, I love it. Also, I might go and check out this "two red three green" guy you mentioned!

[EpicMathTime]

Sorry for sounding like Chucky Finster in some parts, I got mega sick and had to wait a while to finish recording this. 😅

[chemistro]

I think you might literally be the most channel on youtube

[fiNitEarth]

I freaking LOVE this video! I'm gonna use this as a prime example when my class mates when they're starting to complain again about the "uselessness of maths". You first have to convert your problem into a arithmetic expression and THEN use those tools given to you..

[aokkishi]

Thats really cool, just a small detail: Instead of taking the limit you could have used the a^n - 1 formula and you'll get a sum and bim you finished haha

[camrouxbg]

I like how carefully you go through this. It is nice and methodical, showing each step. Great for high school students and beyond. Nice work.

[jonahdembe]

How am I just now finding this channel? This was so good.

[Jack_Callcott_AU]

This is one of the best maths videos I have seen. Bloody fantastic.

[GhizLob]

Great video. Just wanted to add that you can actually avoid having to deal with the modulus (up to a +- sign) as the product is already a real number: taking the complex conjugate will only invert the order of the roots of unity.

[MichaelRothwell]

A nice proof and a great video. You could avoid limits and l'Hopital by using (z^n-1)/(z-1) = 1+z+z^2+z^3+...+z^(n-1) [sum of a geometric series] and then setting z=1 on right gives n.

[KienPS]

1) The theorem is true for n=2 and n=3. For n=1, there will be only 1 point on the circle, so there isn't any length here (or the length would just be 0). Therefore, the theorem is valid for all n greater or equal to 2.


2) On my guess, if the radius of the circle is r (r >= 1), then the product of those lengths will be n*r^(n-1) (with n >= 2)
Proving this is similar to your proof. Drawing the circle on the complex plane, with the center is at the origin, the length between the center and any point on it is r*e^((2*k*pi*i)/n) for k runs from 1 to n (see 7:20). Then shift the circle by r, the product of the lengths will be |r*e^((2*k*pi*i)/n) - r| with k starts from 1 to n-1 (not counting the point sits on the origin), and we want this to be equal to n*r^(n-1). By factoring r out, we will get r^(n-1) appear on the left-hand side, and canceling that factor on both sides, we will eventually back to the formula that you've already proved in the video.


3) First, the length of any 2 points A, B sits on the circle (O is the center) is 2*sin(theta*pi), where theta is the angle between OA and OB, measured in radian. So if I have n points evenly-spaced on the circle, the length between the point you choose to any other points on it would be 2*sin(k*pi/n) (k runs from 1 to n-1), and this result is algebraic (You can find this fact on the internet) for any k and n.

[awabqureshi]

Thanks for shouting out a small math YouTuber!

[berserker]

Got to the final expression myself pretty easily, but I would never try using the limit without you mentioning it. Maybe because I haven't had a complex analysis course yet... That comes in a few weeks tho so hypeee!

[perappelgren]

Love that magic buzz at 12:53 when taking limits!

[edwardus]

I regret every moment of not knowing about this chanel, you're awesome <3

[toaj]

Verification for n=3. The points form an equilateral triangle. The centroid and circumcenter of this are the centre of the circle. The centroid divides the median/altitude (they are the same in an equilateral triangle) into a 2:1 ratio. The radius is thus 2/3 of the altitude. Therefore the altitude has length 3/2. The length of altitude=(sqrt(3)/2)*side. Therefore the length of each side is (3/2)*(2/sqrt(3))=sqrt(3). The product we want is the product of two of the sides=(sqrt(3))^2=3.

[ranjitsarkar]

3b1b. You know, one of those SMALL youtubers with potential

[Lance.]

Wow the step up in production is looking amazing lately