A Fun Polynomial Equation

[@RomanOrekhov]

You promised a wolframalpha output...

[@SyberMath]

I forgot. Check the community post

[@fahrenheit2101]

beautiful stuff.

[@SyberMath]

Thank you! 😍

[@vaggelissmyrniotis2194]

Pretty cool method!Unfortunately i missed the P(x)=(x-2)^n although i was kinda close but i got the P(x)=0 or 1.

[@scottleung9587]

Nice!

[@giuseppemalaguti435]

P(x) =x^2-4x+4

[@dan-florinchereches4892]

I think if y=x+3 then x=y-3 we substitute in and get P(y^2-6y+9+2y-6+3)=(P(y))^2 P(y^2-4y+6)=(P(y))^2 If P(y)=a(y-y1)(y-y2)... With y1,y2... real or complex Thenit follows the leading term from lhs and rhs are equal so a=a^2 => a=1 or a=0. We can verify that P(x)=0 is a solution but not a very exciting one Checking some values P(x)=c => c=c^2 so P(x)=1 is a new solution If P((y-2)^2+2)=(P(y))^2 and P(y)=y-y1 then (y-2)^2+2-y1=(y-y1)^2 for any y y^2-4y+6-y1=y^2-2y*y1+y1^2 y1=2 and y1^2=6-y1 so P(y)=y-2 is a solution If P(y)=(y-a)(y-b) then (y^2-4y+6-a)(y^2-4y+6-b)=(y-a)^2(y-b)^2 So we need a and b to be double roots for lhs polynomial This means that either we have y^2-4y+6-a=(y-a)^2 which has solution a=2 from previous point and also b=2 also Or y^2-4y+6-a=(y-a)(y-b) but this implies a=b and forces out the same solution So P(y)=(y-2)^2 Generalising P(y)=(y-2)^k are solutions So it results that we have the following solutions: P(x)=0; P(x)=1; P(x)=(x-2)^k with k >=1 PS to question 0 is a natural number in Romania, you can write N+ for the set of nautrals without 0

[@dansimpson6844]

You do “u”

[@mtaur4113]

I have seen a book take the convention that the zero polynomial has degree minus infinity, so that the deg(PQ)=deg(P)+deg(Q) is always true with no further comment.

[@mtaur4113]

Also, I don't remember how we defined them in high school, but I think starting at 1 was "whole", and starting at 0 was "natural"? In grad school, N and Z^+ for shorthand. As we see here, we need constant polys and zero if we want polys to be stable under addition and subtraction. So degree zero deserves to be included by default. But giving the zero poly "degree -infty" means it has to have a special consideration here, as you can no longer declare the degree to be a natural number, as there is no smallest power to go with the empty set of coefficients. In general, we want to be able to intersect two finite sets and have the cardinality be a natural number, so 0 being natural is convenient.

[@fahrenheit2101]

Ohhh that's why we did it that way. Our lecturer just used this convention without telling us why...

[@mtaur4113]

@@fahrenheit2101 There might be other ways to justify or explain it, but yeah. You don't have to believe in debt or negatives, but you can still say how many apples you have no matter what.

[@anshult]

kzbin.infoQql-4XAAfyQ?si=ShpTEQ2fx2Z6Fu3F

[@promixinc.8434]

Nice video. By the way Most of materials in my country assume zero is not a element in Nature numbers.

[@SyberMath]

Thanks for letting us know. Where are you from?

[@promixinc.8434]

@SyberMath From Sri Lanka

[@Don-Ensley]

problem P(x² +2x+3) = [ P(x+3) ]² Let P(x) = ax + b + c , where a, b and c are real undetermined coefficients. Substitute. P(x+3) = ax + 3 a + b + c [P(x+3)]² = a² x² + 2a(3a +b+c) x + 9 a² + 6 a (b+c) + (b+c)² P(x² +2x+3) = a x² + 2a x + 3a + b+c Set coefficients of the same powers of x equal to each other in the last 2 equations. Solve for a and b+c. a = a² a = 1 (a = 0 will make everything 0. The trivial solution of P(x) = 0). 2 = 6 + 2(b + c) 3 + b+c = 9 + 6 (b+c) + (b+c)² b+c = -2 b+c = 6 + 6 (b+c) + (b+c)² Plugging -2 from the first into the second for b+c: -2 = 6 -12 + 4 -2 = -2 true. Therefore P(x) = x - 2 is one solution for P. Another was found by this same method but with P(x) = ax² + bx + c + d to be P(x) = (x-2)² Could it be that P(x) = (x-2)ⁿ is the answer? Let's see. The hypothesis is that P(x) = (x-2)ⁿ, for n integer ≥ 0. [P(x+3)]² = (x+1)²ⁿ P(x² +2x+3) = [(x+1)²]ⁿ By the rules of exponents, P(x² +2x+3) = [(x+1)²]ⁿ = (x+1)²ⁿ = [P(x+3)]² answer P(x) ∈ { 0, (x-2)ⁿ , ( n ≥ 0, n ∈ ℤ) }

[@SyberMath]

Nice

[@Don-Ensley]

@@SyberMathexcellent! ❤