A Polynomial Functional System of Equations

If you need to post a picture of your solution or idea:
#ChallengingMathProblems #FunctionalEquations

EXPLORE 😎:

PLAYLISTS 🎵 :

[LavarBruh]

That’s really cool, never seen any problem like this before

[nitayweksler]

I learnt by looking in your vids how to do functional equations, and now i managed to solve it all alone :) thanks alot

[willyh.r.]

As a math lover, I really like your math puzzle:
1. Not a standard one and elegant; and
2. It always requires exploration, which is one of cognitive skills that US math education fails to address in math class.

[titassamanta]

When u told 'a polynomial composed with itself' I understood that P and Q are linear poly'als. But then, I did not have the patience to carry forward. So I watched the solution. Good problem.

[julienn]

The satisfaction of finding the answer before viewing the vid ^^
Great problem btw !

[yoav]

The day before system of eqations with integer solutions, now system of equations with polinomials with integers coeff solutions(at least one of the solutions).next, system of equations with matrices with integers entries solutions

[VSN]

Liked this question! I was stumped at first but realised how to solve it when you said the word “polynomial”. I was confused by your videos tumbnail at first lol

[igornikitin]

The other way is to use derivatives:
P'(x)*P'(x) = 4 =>
P(x) = 2x+c or P(x) = -2x+c. The rest is obvious

[honortruth]

From P(P(x)) we can also solve for P inverse, P^(-1)(y) = (y - 5/3)/2 and apply this to P(Q(x)) to get Q(x) = -3x + 4/3 for the first case, etc.

[math]

very well done bro, thanks for sharing

[swattikdas]

I can see the extra ordinary hardowrk you do for to get these brilliant question hats off to you 👏👏👏👌👌👌💖💖💖

[manojsurya]

It's a good n new problem sybermath, pls do some more new kinda problems in calculus too 😄

[leickrobinson]

What a nice problem, and a great solution! :-D

[azizulmostafa]

You solution is simply understand.thanks lot.

[jarikosonen]

Functional diophante's equation or deriving function from data set? Didn't see one yet in Youtube. Anyone to make this video soon?
5*f(x)+8*g(x)=33, f(x)&g(x) integers (or should it be g(y)?), and also x&y integers and f(x)<>x, g(y)<>y.
It looks like it could have a lot of solutions.

[rnite]

This sound pretty unique, Ive never seen such a thing before (Not making any silly jokes/memes anymore) I will watch this and bet this will be cool.
Edit 1:It was cool indeed

[aashsyed]

2nd again! coincidence! i am happy :) ! looks cool!

[angelmendez-rivera]

If deg(P) = n, then deg(P°P) = n·n. Since (P°P)(x) = 4·x + 7, deg(P°P) = 1, hence n·n = 1, thus n = 1. Therefore, P(x) = A·x + B. By a similar and analogous argument, Q(x) = C·x + D. (P°P)(x) = P[P(x)] = P(A·x + B) = A·(A·x + B) + B = A·A·x + A·B + B = 4·x + 7, while (P°Q)(x) = P[Q(x)] = P(C·x + D) = A·(C·x + D) + B = A·C·x + A·D + B = –6·x + 5. Therefore, A·A = 4, A·B + B = 7, A·C = –6, A·D + B = 5. A·A = 4 implies A = –2 or A = 2. Therefore, –2·B + B = 7, –2·C = –6, –2·D + B = 5, or 2·B + B = 7, 2·C = –6, 2·D + B = 5. This implies B = –7, C = 3, 2·D – 7 = 5, or B = 7/3, C = –2, 2·D + 7/3 = 5, which implies B = –7, C = 3, D = 6, or B = 7/3, C = –3, D = 4/3. Therefore, the solution set is {(–2·x – 7, 3·x + 6), (2·x + 7/3, –3·x + 4/3)}.

[ayaannarayan]

Bit confused at 2:14. If B(x)=x^3 then B= x^2 (divide both sides by x), so shouldn’t B(x^3)= x^2 * x^3= x^5?

[SamsungJ-kknr]

I don't now how show the outcome.