some very nice values of cosine

[@scipython]

Alright ... but when is a good place to stop?

[@DeJay7]

It's never a good place to stop. Always keep going with mathematics!

[@JordanBiserkov]

360 degrees is a good place to stop, because that's the period of cosine.

[@HoSza1]

...and no backflip either 😢

[@MarcinSzyniszewski]

No good place to stop this time, we have to keep going!

[@sbares]

If one knows a little bit of algebraic number theory, then this can be proven in a very elegant way: For any rational number p/q, e^(2pi i p/q) and e^(-2pi i p/q) are algebraic integers (they are roots of the monic integer polynomial x^q - 1), thus so is e^(2pi i p/q) + e^(-2pi i p/q) = 2 cos(2pi p/q). If 2 cos(2pi p/q) is also rational, this implies that it is an integer.

[@zygoloid]

That's an extremely neat proof, thanks for sharing! Prof Penn's approach also shows that 2 cos (2pi p/q) is a root of a monic integer polynomial of degree q; can we get the same result this way? (Does addition of algebraic integers of equal degree always produce an algebraic integer of that degree?)

[@sbares]

​@@zygoloid One can produce an explicit polynomial satisfied by a+b from polynomials satisfied by a and b, for example one may take the product of (X - a' - b') over all the roots a' and b' of the two polynomials (one can show this still has coefficients in the same ring as the original polynomials using the theory of symmetric polynomials, or using Galois theory if one wants to be fancy). The degrees will not generally match in the way you describe though (e.g. in the previous example we get the product of the degrees), and that certainly won't generally be the the case for the degrees of the minimal polynomials (=degrees of the numbers) either. For an example relevant to this video e^(2pi i / 6) and e^(-2 pi i / 6) are of degree 2 (note: not 6, as their minimal polynomial is the 6th cyclotomic polynomial x^2 - x + 1, not x^6 - 1), while their sum 2*cos(2 pi i / 6) = 1 is of degree 1.

[@amos9274]

Did I understand this correctly: basically because 2*cos with rational angle can be represented by an addition of two algebraic integers, the result if rational must be an integer. I don't know why that last part is. Is that a theorem?

[@zygoloid]

@@amos9274 Yes, any algebraic integer that is rational is by definition a rational root of x^n + a_1 x^(n-1) + ... + a_n, and by the rational root theorem must be an integer (and moreover a factor of a_n).

[@amos9274]

@@zygoloid Oh right, now I remember that theorem from algebra lectures in my electrical engineering course. Though we only used it to derive horners method I believe :) I would have never made that connection, thanks!

[@yashmehta9299]

11:24 bless you!

[@gael8828]

Chebychev polynomials: Hello there

[@quqgaming]

I'm pretty sure it's spelled Chebyshov

[@holyshit922]

Although I cant speak Russian (I was forced to learn English) in my opinion your transcription of name of this guy is poor and should be written as Chebyshov

[@bjornfeuerbacher5514]

@@holyshit922 Chebychev is a common transcription used in Germany. Other transcriptions used here are Tschebyscheff, Tschebycheff, Tschebyschew, Tschebyschev, Chebyshev. I've never seen Chebyshov seen used in German, maybe that's more common in English.

[@holyshit922]

@@bjornfeuerbacher5514 In my opinon your transcription is poor because it leads to misreading In my opinion transcription Chebyshov is as close as possible to the original language In my opinion for German transctiption the closest to the original would be Tschebyschow, because of these two dots over last e in cyrilic which is alphabet of his native language The source of this problem is that Russians usually dont write these two dots over e but it does not imply that they are not present there You have your umlauts and imagine that someone missed these two dots Will it change way of reading the word ?

[@bjornfeuerbacher5514]

@@holyshit922 Again: That is not _my_ transcription, that are the _usual_ transcriptions in Germany. I don't know who came up with those, but you have to take it up with them, not with me.

[@holyshit922]

This lemma is in fact recurrence relation for Chebyshov polynomials (Name of this guy in original is Пафнутий Львович Чебышёв) because tese two dots over last e there cannot be e but o You can find formula for coefficients two ways by solving recurrence relation via exponettial generating function then differentiate it, by solving ordinary differential equation via power series Exponential generating fuction is E(x,t) = exp(xt)cos(sqrt(1-x^2)t) (Here wikipedia gives hyperbolic cosine but trigonometric suits better because avoids complex argument) I derive this recurrence in following way cos((n+1)t) = cos(nt)cos(t) - sin(nt)sin(t) cos((n-1)t) = cos(nt)cos(t) + sin(nt)sin(t) I add them up cos((n+1)t) + cos((n-1)t) = 2cos(nt)cos(t) Rearrange and here it is cos((n+1)t) = 2cos(nt)cos(t) - cos((n-1)t) To find coefficients of these polynomials you will need one of following two sums depending on what method you want to use If you try to find it by solving recursion via exponential generating function you will need following sum If you decide to use complex numbers you will get the same sum to calculate but i suggest to solve recurrence because later for Legendre polynomials which produces similar calculations (standarization is the same , ode is slightly different) complex numbers trick doesn't work Let S(n,m) = \sum_{k=m}^{\lfloor\frac{n}{2} floor} {n \choose 2k}\cdot {k \choose m} then T_{n}(x) = \sum_{m=0}^{\lfloor\frac{n}{2} floor}(-1)^m*S(n,m)*x^{n-2m} If you want to solve ordinary differential equation to get coefficients you need to calculate following sum \sum_{k=0}^{\lfloor\frac{n}{2} floor}\frac{(-1)^k}{4^k}\cdot\frac{n}{n - k} \cdot{n - k \choose k} Let S(n) = \sum_{k=0}^{\lfloor\frac{n}{2} floor}\frac{(-1)^k}{4^k}\cdot\frac{n}{n - k} \cdot{n - k \choose k} then T_{n}(x) = \frac{1}{S}\cdot \sum_{k=0}^{\lfloor\frac{n}{2} floor} \frac{(-1)^k}{2^n}\cdot\frac{n}{n - k}\cdot }{n - k \choose k}\cdot (2x)^{n-2k} but this ode solution is valid only for n > 0

[@Tabu11211]

loved the video! Had a giggle when you said "Lets expand this out a bit" and it was to me already pretty expanded.

[@Wielorybkek]

wow that's actually pretty crazy there is just a handful of them

[@japanada11]

This proof was unfortuntely a lot messier and sketchier than it needed to be - there's a much shorter and cleaner proof! Write 2cos(theta) = a/b in simplest form. We'll prove by induction that 2cos(n theta) is a rational number which, when written in simplest form, has denominator exactly b^n. As base cases, for n=0 we have 2cos(0 theta) = 2/1 with denominator b^0, and 2cos(1 theta) has denominator b^1 by definition. Now for n at least 2, we have the trig identity Michael proved (I just multiplied it by 2): 2cos(n theta) = 4cos((n-1)theta)cos(theta) - 2cos((n-2)theta). We have 2cos(theta) = a/b, and by induction hypothesis, we can write 2cos((n-2)theta) = c/b^{n-2} and 2cos((n-1)theta) = d/b^{n-1} for some integers c,d relatively prime to b. Thus 2cos(n theta) = (a/b)(d/b^{n-1}) - c/b^{n-2} = (ad-bc)/b^n. Now ad is relatively prime to b, so ad-bc is relatively prime to b as well. Therefore this fraction is in simplest form, proving the inductive step. But if theta is a rational number of degrees, then for some positive value of n, n theta is a whole number of full rotations, so 2cos(n theta) = 2/1. We showed above that the denominator must equal b^n, so we can conclude that b=1. So cos(theta)=a/2 is an integer multiple of 1/2.

[@Keithfert490]

Why restrict the lemma to n >= 2? It follows for all n

[@ed.puckett]

Great problem and solution!

[@wilderuhl3450]

I’m awaiting part 2 as we were never brought to a good place to stop

[@Alan-zf2tt]

Whooosh! Look out Michael is on fire again! Yet there were some stunning revelations for me and the most important is so so obvious. That irrational π is not just irrational - in degree-land π is rational! (sound of hand slapping its own forehead) And as for construction of polynomials - well it seems they get everywhere them polynomial. They do - they really get every where. And it got me wondering how things in one land can be irrational and in another rational

[@allanjmcpherson]

Shouldn't the constant term in the polynomial be 2a_0 - 2 instead of a_0 - 2? Edit: Or has he absorbed the factor of 2 into a_0?

[@iooooooo1]

I think you're right. Fortunately I don't think that affects the rest of the argument since all he uses about that constant term is that it is an integer; he restricts the factors it can possibly have via the possible values of the cosine function, not any properties of the specific integer that constant term might be.

[@iooooooo1]

​@@edoardodaurelio4102 I think you may be slightly misremembering the rational root theorem. The theorem says that if x=p/q in lowest terms is a rational root of a polynomial with integer coefficients, then p divides the constant term and q divides the leading term. Even if the constant term is even, 1 and -1 would still be valid factors of that number. The divisibility-by-2 restriction would only be true if the divisibility relationship worked the other way (i.e. if the constant term were a factor of p, rather than the other way around).

[@edoardodaurelio4102]

@@iooooooo1 you’re right, mb

[@josemanuelbarrenadevalenci653]

Where can I buy a T-shirt like yours.

[@ojas3464]

Serge Lang in his volume on Transcendental numbers establishes that for nonzero x, both x and sin x cannot be algebraic. Here sin x could be replaced by cos x, or exp (x), and the whole thing could be then taken as a Lemma pdf version of Analytical Trigonometry contains a chapter on Expansion of cosⁿ θ and sinⁿ θ Would like a review whether these help towards simplification.

[@AnkhArcRod]

That is super neat!

[@ocarinaknight]

The constant term should be (2a_0 -2) instead of (a_0-2) at the end

[@MichaelRothwell1]

Yep, I noticed that too. Fortunately it doesn't change the conclusion!

[@ocarinaknight]

@@MichaelRothwell1 Indeed. I only think it looks a bit confusing for the viewer, maybe it is worth adding an observation to the video.

[@GreenMeansGOF]

Does this mean that it is impossible for θ(in radians) to be rational and also have cos(θ) be rational?

[@TJStellmach]

It does not. Cos(0 radians)=1.

[@zygoloid]

I don't think this argument says anything about the case where θ (in radians) is rational.

[@GreenMeansGOF]

@@TJStellmach ok but is that the only example? If we have like 60,90,120, etc… those will all be irrational radian measures.

[@TrevorMag62]

Degrees are arbitrary - you've got to pick a number (integer) and divide a revolution into that many pieces. The fact that we've chosen 360 doesn't make it mathematically special. Fortunately, this video works for any chosen integer. In other words, for all arbitrary choices. But still arbitrary. Mathematicians have agreed that radians are more fundamental, less arbitrary. And that's what made the video title alluring...

[@AquaticWaters]

Great video!! 😊

[@pikkisir4664]

For what values of cosine(in radians) is equal to a rational value

[@Ninja20704]

Is it possible if we ask the same question but in radians instead? Since we don’t have anymore “trivial” solutions since pi/2 pi/3 etc are not rational anymore. Or how would we prove it is impossible?

[@3141minecraft]

cos(0)=1 I am not sure if there are more examples.

[@itsphoenixingtime]

Isn't it somewhat a short-lived question because radians is measured predominantly in pi so you can't really go anywhere with this because pi in the first place is irrational so other than 0 you'd get pretty much nowhere.

[@Ninja20704]

@@itsphoenixingtime Yes, but that alone doesn't prove that it is impossible for theta and cos(theta) to both be rational at the same time.

[@hhhhhh0175]

cos(x) is an algebraic function of e^ix, which is transcendental for x in Q\{0} by lindemann-weierstrass

[@tortoisebekkou]

It is called Niven's theorem.

[@aweebthatlovesmath4220]

engineers solution: sinx≈x Also cosx≈sinx so we have by Transitivity cox≈x so cosx=x thus whenever x is rational

[@bjornfeuerbacher5514]

"cosx≈sinx" Not even engineers use an approximation _that_ bad!

[@JordanBiserkov]

And the only "rational" values of X are 0, 60, 90, 120 and 180. So the engineer's solution is not worse than the mathematician's solution.

[@Kishblockpro]

@@bjornfeuerbacher5514as long as ur willing to have a variance of 1 😂

[@aweebthatlovesmath4220]

​​@@bjornfeuerbacher5514 bro i was just joking and also if |sinx-x| is a small error to you then |six-cosx| should be considered "small" too

[@bjornfeuerbacher5514]

@@aweebthatlovesmath4220 sinx≈x is only used for x close to zero, and in that case, the error indeed _is_ small. But for sinx = cosx, the error is in most cases large. Ok, when you say that x is approximately equal to pi/4, then I agree. ;)

[@Tabu11211]

some signs or sines? ;)

[@hellNo116]

if θ is in rads is there even another solution other than zero?

[@MarcoMate87]

Fantastic result, but at 13:01 it's not entirely obvious that a_(n-1) is an integer. Penn proved that the factor of (cos θ)^n is 2^n (after multiplying both sides by 2), but how about the coefficients of the lower degree terms?

[@VideoFusco]

That long initial demonstration is quite useless, just apply one of the prostapheresis formulas to get COS(n theta)+COS((n-2)theta)=2COS((n theta+(n-2)theta)/2)COS((n theta-(n-2)theta)/2)=2COS((n-1)theta)COS(theta) from which COS(n theta)=2COS((n-1)theta)COS(theta)-COS((n-2)theta) Alternatively, always much more quickly than in the initial demonstration, you can always start from the sum COS(n theta)+COS((n-2)theta) and rewrite it as COS((n-1)theta+theta)+COS((n-1)theta-theta) and then apply the cosine sum formula.

[@wesleydeng71]

A related problem: Is tan(1°) rational? And there is a neat solution to it.

[@vladthemagnificent9052]

I believe it is not rational, it is some root of high degree polynomial

[@gp-ht7ug]

No more “and this is a good place to stop”?

[@TheFinav]

And that's a good place to stop.

[@putruccu]

Hello Mr. Niven :) Perhaps a next step could be to treat the situation of rationally dependent roots of unity ;)

[@zeuspolancosalgado3385]

that was kind of anticlimactic :(

[@golddddus]

Theta degrees must also belong to rational numbers. It is easy to find a counter example. The "angle of life" is the angle of the sp3 hybridized C-C bond. and it is the angle between the heights of the tetrahedron. cos(θ°) = -1/3. 😎