One of the coolest functional equations I have seen!

[@andreben6224]

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)

[@goodplacetostop2973]

15:31

[@HagenvonEitzen]

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

[@DonkoXI]

Yes. This. I see it all the time. Nobody remembers the n=0 base case because it's too easy.

[@chobes1827]

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.

[@M4DA.]

Yeah, but not only shows that this is homomorphism but actually shows that there is only one such homomorphism from Z[x] into Z

[@chobes1827]

@@M4DA. I think you might be a little bit confused. There are infinitely many different homomorphisms from Z[x] to Z, all of them are fully characterized by the image of x (which can be any integer, so that all evaluation maps are homomorphisms). You need to use the fact that a map from Z[x] to Z is a homomorphism if and only if it is evaluation at some integer in order to conclude that a map Z[x] to Z is a homomorphism if and only if it satisfies the two conditions given in the video.

[@DonkoXI]

In general, the set of ring homomorphism from Z[x] to R corresponds to the elements of R via the evaluation map. The ring Z[x] plays the same role among rings that the one element set plays among sets (maps from Z[x] correspond to elements of the target). A direct way to see this connection is via the free-forgetful adjunction between rings and sets. More generally. Fix a (commutative?) ring k, and let [R, S] denote the k-linear ring homomorphisms R -> S, where R and S are k-algebras (note that being a ring is equivalent to being a Z-algebra). Then there is an isomorphism [k[x], R] = R given by the evaluation map.

[@DonkoXI]

As for the recursive and divisibility aspect, this unfortunately doesn't generalize as well. Z is special here because it can be generated in a recursive way. To get a similar statement with another ring, you'll need a similar recursive definition of that ring.

[@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)

[@m2a2x2000]

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.

[@m2a2x2000]

oooh I see the answer in previous comments. Question is gone. Thanks

[@givrally]

Small challenge : Find all pairs of integers a,b

[@gregoryk_lite]

Take two perfect squares to be a and b , and then multiply both by the same product of primes (where each prime appears at most once), then check that they are less than n. As to the algorithm, I think you can precompute all such products of primes

[@ritwinnarra]

Related to USAJMO 2010 Problem 1

[@briemann4124]

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

[@telnobynoyator_6183]

Love the font in the thumbnail btw

[@potawatomi100]

You’re the best, Michael.

[@abrahammekonnen]

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

[@aweebthatlovesmath4220]

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

[@ΕχιΜιμζ]

nice name

[@aweebthatlovesmath4220]

@@ΕχιΜιμζ thank you!

[@NicholasPellegrino]

Thanks for another great video!

[@agamanbanerjee9048]

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]?

[@brunojani7968]

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.

[@schweinmachtbree1013]

I will add to Angel's great explanation that when division is also well-defined (division by _non-zero_ elements, of course) and multiplication is commutative, we call this kind of structure a "field" (if you have division but multiplication is not commutative then this is called a skew-field) - so for example, the rational numbers, the real numbers, and the complex numbers are fields, and an example of a skew-field is the quaternions (however the next step up the "complex ladder", which are called the octonions, are not a skew-field). And I will also note that there is another property of general rings besides possibly-non-commutative multiplication which is unintuitive: the notion of "zero-divisors". A zero-divisor is a non-zero element _c_ which "divides zero" in the sense that there exists another non-zero element _d_ such that _cd_ = 0. Simply: in a general ring the product of non-zero elements can be zero - for example this happens with matrices, or a slightly easier example if you know what modular arithmetic is is that 2=/=0 mod 6 and 3=/=0 mod 6 but 2*3=0 mod 6. A commutative ring in which this kind of thing doesn't happen (in which there are no zero-divisors) is called an "integral domain"* - for example the integers are an integral domain, as are polynomials with real coefficients. *or without the commutativity requirement, a "domain" (but then one can to make a distinction between left zero-divisors and right-zero-divisors) - for example polynomials whose coefficients are quaternions form a domain.

[@eric3813]

Wow, what a fun functional equation! :D

[@AnarchoAmericium]

Ah yes, secretly the Yoneda lemma.

[@cd-zw2tt]

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

[@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

[@jussari7960]

What part are you talking about?

[@matthieumoussiegt]

@@jussari7960 it is the theorem that says that if we have a P € Z[X] polynomials with integer roots r there exist Q € Z[X] s.t P=(X-r)Q

[@jussari7960]

@@matthieumoussiegt Take any p ∈ Z[X] (of degree ≥1) with a root at r ∈ Z and let q(x) be the polynomial that satisfies p(x) = (x-r)q(x). Write p and q as p(x) = a_n*x^n + ... + a_1*x + a_0 and q(x) = b_(n-1)*x^(n-1) + ... + b_1*x + b_0 and note that a_i are all integers. Then from p(x) = (x-r)q(x) we get a_i = b_(i-1) + r*b_i for all 0≤i

[@matthieumoussiegt]

@@jussari7960 the problem is not that it is not in Q[X] the problem that it is specified that q is in Z[X]

[@jussari7960]

@@matthieumoussiegt Oh, I see what you mean. i think that is true too, (though not immediately obvious). Take the notation from my previous comment. Then we have b_(n-1) = a_n is an integer, and inductively b_(i-1) = a_i - r*b_i is an integer for i≥0. (I initially messed up the formulas for a_i but they should hopefully be correct now)

[@cd-zw2tt]

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

[@lucachiesura5191]

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

[@francocosta1]

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

[@chobes1827]

In this context, the polynomials in Z[x] are being treated as formal expressions as opposed to functions. However, in the solution it gets established that theta must be equal to the map sending each polynomial to its evaluation at some integer. Theta cannot be evaluation for any domain other than the integers. If it were, then it would map the polynomial p(x) = x to some object that is not an integer, which would contradict the assumption that the codomain of theta is the integers.

[@frentz7]

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

[@pushpasharma1412]

7:25 OK

[@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.

[@mathboy8188]

When doing algebra, for a R a ring, you should view R[x] as being a RING, not as a SET of functions. Here's why: Suppose R = Z/(3Z) = ring of equivalence classes of integers modulo 3. For any a in R, a^3 = a (Check: [0]^3 = [0], [1]^3 = [1], [2]^3 = [8] = [2+3(2)] = [2].) Let p in R[x] be p(x) = x^3 - x. Let q in R[x] be q(x) = 0 (which is of course 0 in R, so probably should be written [0]). Then if you think of p as a FUNCTION from R to R, then p is the zero function... p maps every element of R = { [0], [1], [2] } to the value [0] in R. Thus thought of AS FUNCTIONS, p and q are the same... they both map every element of R to [0]. But they aren't the same AS ELEMENTS of R[x]... p is a degree 3 polynomial, while q is a constant polynomial, p has 2 terms while q has only 1, and so on. When it becomes useful to consider them as functions, what you're actually doing is applying an evaluation homomorphism to them. If f in R[x], a in R, then E(a) : R[x] -> R is a ring homomorphism defined by E(a)(f) = f(a) in R, meaning what you'd expect: replace x everywhere by a in f, and then compute the result.

[@chaparral82]

don't understand the Q expansion. Q is a Lebesgue Zero Set, so how should Integration work on that.

[@endersteph]

(Disclaimer: I am not knowledgeable in math and am not sure of what I'm talking about) The polynomial in ℚ[x] has coefficients in ℚ, and integrating over it from a to b just means integrating it over [a ; b], because x is totally allowed to be any number, there's not restriction on that. For example p₀(x) ≔ 3x² - x/8 ∈ ℚ[x], and since p₀(x) is defined for all x ∈ [0 ; 1], you could define the function: Θ : ℚ[x] ⟶ ℚ p(x) ⟼ ∫ p(x)dx with bounds 0 and 1 and then get that Θ(p₀(x)) = ∫ p₀(x)dx, bounds 0 & 1 = ∫ (3x² - x/8)dx, bounds 0 & 1 = 15/16 (∈ ℚ) Hope this answers your question!

[@chaparral82]

@@endersteph if you extend the polynomial function on R that is where the integration normally defined and you define the function that it is p(x) on rational points and 0 on irrational points and you integrate it then you receive always the value 0 for the integral, because Q is a Lebesgue zero set in R.

[@endersteph]

@chaparral82 I know but this doesn't apply here. The function you are describing, 𝟙_ℚ × p, is not a polynomial. If the coefficients of p are in ℚ, then p is in ℚ[X], and integrating it over real numbers yields a rational number. Therefore, the function which takes a polynomial with rational coefficients and outputs its integration over say [a ; b] for reals a and b - which is what Michael was talking about - is a function from polynomials with rational coefficients (ℚ[X]) to ℚ. Again what I want to get at is that polynomials are integrable over real numbers, so there is no problem in doing this. For reminder: the coefficients of x ↦ a_n · x^n + a_n-1 · x^n-1 + ... + a_2 · x^2 + a_1 · x + a_0 are the constants a_0 through a_n. If A is a set that contains all of those coefficients, then this polynomial is in A[X]. This does not change the fact that the polynomial is defined over the reals, and that you can integrate it over [a ; b].

[@howwitty]

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

[@andersen9044]

i believe this problem is more algebraic than analytic what about you ?

[@NicholasPellegrino]

Definitely algebra

[@smolboi9659]

I think it's in the realm of analytic number theory?

[@howwitty]

@@andersen9044 I agree, but he did mention open intervals...

[@jussari7960]

There's nothing analytic about this other than the examples of derivative and integral evaluation in the beginning

[@pmcate2]

So in summary we proved that if the theta function is an evaluation function for polynomials of degree one with no coefficients, then it must be an evaluation function for all polynomials? What if we had made a different assumption?

[@ktostamktos]

from 8:11 we assume f is an arbitrary polynomial, so the proof works for all polynomials

[@jussari7960]

We never made the assumption, we just defined a to be the value of θ(x), (which is some integer) and then from that proved that for any function it must be evaluation at a.

[@smolboi9659]

When you say Z adjoint X for the polynomial ring Z[X] and Q adjoint root 2 for the extended field Q(root2) in the video released just after this, Are the 'adjoints' mathematically related? Or are they seperate concepts entirely?

[@schweinmachtbree1013]

yes they are. *Z* [X] is the smallest ring which contains *Z* and the indeterminate X, while *Q* (root 2) is the smallest field that contains *Q* and root 2. We use square brackets for "smallest ring containing" and round brackets for "smallest field containing". In the case of root 2, the smallest ring containing *Q* and root 2 turns out to be a field, so we have *Q* [root 2] = *Q* (root 2). (note: there shouldn't be gaps between *Z* and [X] and between *Q* and (root 2) but KZbin won't let me format it without a space)

[@smolboi9659]

@@schweinmachtbree1013 Very informative. Thanks.

[@mathboy8188]

The reply from schweinmachtbree is about the difference in meaning between the square bracket [] and parentheses () notation. If that was the extent of your question, then what follows might be a waste, ut I'm not sure exactly what you were asking. If you were asking if there's some relationship between polynomials (adjoining the indeterminate x) and these special rings/fields (adjoining a specific value like sqrt(2), then the answer is yes... there's a BIG TIME connection between the two ideas: Let s = square root of 2 (just to make writing it easier). First notice that Q(s) = Q[s]: If a,b in Q, a,b not both 0, then it's impossible that a^2 = 2b^2. That's because, if assume a^2 = 2b^2, then if one is 0, then both are, so neither is 0... and since neither is 0, a^2 = 2b^2 => (a/b)^2 = 2, and a/n in Q, and so 2 has a rational square root. Contradiction. If a,b in Q, a,b not both 0, then (a + b s) ( [a/(a^2-2b^2)] - [b/(a^2-2b^2)] s ) = 1. Thus every non-zero element of Q[s] has an inverse, making Q[s] a field. Since Q and s are in the field Q[s], have that Q(s) is a subset of Q[s]. Since Q and s are in the ring Q(s) (it's a field, hence it's a ring), have that Q[s] subset of Q(s). Therefore Q(s) = Q[s]. Now let J = < x^2 - 2 > = the ideal generated by the polynomial x^2 - 2 in Q[x]. If f in Q[x], then f + J in Q[x]/J (which is a field, since J is a maximal, since Q[x] a PID... but that's not important yet). By the Remainder Theorem, there exists unique q, r in Q[x] s.t. f(x) = q(x) (x^2 - 2) + r(x) with deg(r)

[@smolboi9659]

@@mathboy8188 Well i don't know enough to know the full extent of what i was asking. I just wanted to know if adjoint being used in both cases are the same mathematical concept. Any information helps thanks.

[@smolboi9659]

@@mathboy8188 1) phi is a ring homomorphism It evaluates the polynomial at s which clearly respects addition and multiplication. It also evaluates 1 as 1. 2) The kernel is all polynomials with s as a root. Polynomials with x^2 -2 as a factor. The kernel is the ideal generated x^2 - 2. 3) phi is onto. For any rational a,b. a+bx evaluates to a+bs. By the ring isomorphism theorem we have Q[s] isomorphic to Q[x]/J

[@cd-zw2tt]

2:25 l'hopital be like

[@nicepajuju3900]

clever!!

[@singious]

Ah~ I have heard of the digital pulse function.

[@romajimamulo]

What happens when you have it be functions of rational numbers to a rational number? Well... From what I've done, seemingly a giant mess

[@chuckaway6580]

I suspect Gauss's Lemma is relevant here.

[@jaimeduncan6167]

The solution a*ln(x) is clearly more general than some log base b because a could be a negative number and it will satisfy the conditions, but clearly all the base b are solutions because for each one fo them there a proper a

[@tamirerez2547]

Can you please make a video about this problem? We have 2 numbers a and b The arithmetic mean of these numbers is c The geometric mean of them is d. Now we do the same with c and d Arithmetic mean of c and d is e, and geometric mean of c and d is f. We repeat this process with e and f to get two new numbers g and h so on again and again. The two numbers are finally coverage to one point. Question: How can we calculate the value of this final point, from the beginning. Without repeating the process. I will give you a and b, say 5.4 and 12.7 And you will tell me the average final point. Is there is such a formula or a function?

[@Tzizenorec]

I keep getting disappointed when I see the condition "this polynomial divides this other polynomial", then you point out that that you're talking about integer divisibility rather than about polynomial factorization.