One of the coolest functional equations I have seen!

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[andreben]

Wow this is pretty cool!
Starting at 11:41 you could have concluded a bit faster:
Since a-b divides both \theta(f(x))-f(b) and f(a)-f(b), a-b must also divide their difference so
a-b will divide \theta(f(x))-f(a).
What's cool about this is that a-b varies with b whereas \theta(f(x))-f(a) does not, thus it is divisible by all non-zero integers, but the only integer that can be divisible by all non-zero integers is 0, so
\theta(f(x)) = f(a)

[HagenvonEitzen]

4:43 Well, that's precisely the reason why I *would* use the n=0 case as base case. :)

[chobes]

This video was super cool. It's really interesting that the given functional equation and divisibility condition are equivalent to the definition of a ring homomorphism from Z[x] to Z. I'm curious if a generalization holds for any other commutative rings.

[HagenvonEitzen]

13:48 Or, to avoid that interval stuff, you could go indirect: assume theta(f(x)) != f(a), explicitly take b = a + 2f(a)-2theta(f(x)) and arrive at 1/2 in Z, contradiction - hence theta(f(x))=f(a)

[briemann]

I’m excited for another video, Michael. Thanks for the upload!

[max]

I have a stupid question about roots in Z[x]
at 9:02 Michael said that x=b being a root of f(x) - f(b) means that it can be written as (x-b)g(x) for g(x) in Z[x]
my question: why there exists such g(x) in Z[x] ?
sorry If I have wasted someone's time.

[givrally]

Small challenge : Find all pairs of integers a, b <= 100 such that √a and √b aren't integers, but their product is. In other words, a and b aren't perfect squares but their product is.
A bit harder : Construct an efficient algorithm to list all such pairs less than n.
(It's not a very hard problem but it's a pretty fun exercise)

[aweebthatlovesmath]

5:34 i for some reason really enjoyed that proof.

[brunojani]

Never really understood the notion of rings, but this example helps me understand it, even thought I have never seen this branch of math before.

[abrahammekonnen]

Really nice problem. I think I'll definitely be rewatching this video again to absorb it properly.

[eric]

Wow, what a fun functional equation! :D

[agamanbanerjee]

So am I right in saying that the only solutions are the ring homomorphisms from Z[x]-> Z which is constant on Z and is determined by the evaluation of the element x in Z[x]?

[cd-zwtt]

1:05 it better be or my train of thought has been for nought!!!

[frentz]

In the instructions .. that means "for all q(x), , " I guess?

[lucachiesura]

Very good solution. The image of trasformation of polinomial must be Z, Principal Ideal Domain

[cd-zwtt]

3:44 can you expand for a non-math major why this "is" a ring? please correct me in formal math -- I do not understand the translation from english to math :D

[matthieumoussiegt]

I am troubled by the conclusion of the argument involving polynomial g. Is it sure that g cannot be in Q[X]. I have not found a counter example so that must be true and in fact it would be a good things to show a quick proof

[francocosta]

What is the domain of the polinomial? Is Z also?I think that but its not written.

[ChefSalad]

I'm not sure that function is the right word to describe Θ(x). I think it's a functional, since it inputs a polynomial, which is a type of function, and outputs a number. It seems weird to think of elements of ℤ[x] as either functions or numbers (even though they're kind of both here) but it seems even weirder to think of objects like Θ as functions instead of as functionals. Maybe this is one of those areas where the terminology gets kind of arbitrary, since the objects are so abstract.

[howwitty]

Hmmm... intriguing function. Would this fall under the umbrella of real analysis or algebra?