Solving a Nice Polynomial Equation

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#ChallengingMathProblems #PolynomialEquations

PLAYLISTS 🎵 :

[yoav]

The problem is simple, but the method of solution and the explenation are great!😀💯

[GlorifiedTruth]

Love it! I started off on the right track but quickly got lost. More like this one, please!

[rex_yourbud]

You're just brilliant brother.
Love from an IIT Aspirant who is a Math freak !!

[hassanalihusseini]

This time I am so proud of me. 🙂 Before I started watching I guessed the solution correctly.

[reggaetyro]

1 with 2D vision on a 3D plane with a hidden <> 4D memory.
You have to be careful with knowledge and wisdom the former comes before the latter. The latter comes before the former for truth

[rasmusa]

But in the initial equations there is p(1/x) terms which means that x can't be zero, so I wonder if the solution can be something like (x^4+x)/x which basically X^3+1 when x isn't zero.

[emanuellandeholm]

I reasoned like this: if x=1 then p(x) = p(1/x), and you get a quadratic like: p(1)^2 = 2 p(1), which is solved for p(1) by 0 or 2. So my second solution for p(1) agrees with your solution at p(1), because 1^3 + 1 = 2. But what about the first solution p(1) = 0? Is that just a trivial solution?

Also, since p(x) = p(1/x) / (p(1/x) - 1), for x such that p(1/x) != 1, we have that p(0) = lim (1/x)-> 0 = lim y->oo p(y) / (p(y) - 1) = 1, which agrees with your solution.

[Chrisoikmath_]

How did you guess that you can replace P(x) by x^n+1?

[thierryvieville]

Such a nice problem again, as every day, huge thanks. I may niggle, i.e., split hairs (``cut the hairs in four´´ we use to say in French) but …

Is it that obvious, beyond common sense reasoning, that
Q(x) Q(1/x) = 1 => Q(x) = +- x^n
indeed ?

I may have missed the argument and need on my side to consider a general polynomial
Q(x) = sum(c_n x^n, n = 0 .. d) for degree d, thus with c_d != 0
in order to show by induction that all c_n, n < d are 0.

Let us write:
1 = Q(x) Q(1/x)
= sum(c_n c_m x^(n-m), n = 0.. d, m = 0 ..d) # expanding
= sum( sum(c_m c_(m+r), m = max(0, r) .. min(d, d - r)) x^r, r = -d .. d) # writing r = n -m and grouping terms, and in particular for r = d we get
1 = c_0 c_d x^d + terms of degree below d
so that c_0 c_d = 0 because the term of degree d must vanish
so that c_0 = 0 because c_d != 0 by definition.

If you dislike the previous heavy formula you may also write something like
1 = Q(x) Q(1/x)
= (c_d x^d + terms of degree below d) (c_0 + terms of degree negative)
= c_d x^d c_0 + (terms of degree below d) c_0 + c_d x^d (terms of degree negative)) # so that all other terms are below d
= c_0 c_d x^d + terms of degree below d
and obtain c_0 = 0.

But if c_0 = 0 we may write Q(x) = x R(x) where R(x) is of degree d-1, i.e.:
R(x) = c_1 + ... + c_d x^(d-1)
with
R(x) R(1/x) = 1
and the same derivation yields c_1 = 0
And so on ... until finally, we are left with Q(x) = c_d x^d, with now
c_d x^d c_d 1/x^d = 1 => c_d^2 = 1
yielding the result.

Is there a cleverer simpler way to give an evidence?

Bye bye and take care :)

[necro]

f(x)(1/x) = f(x) + f(1/x), only 1+x^n and 1-x^n (n € Z) satisfies this relation.

P(x) = 1+x³

[hassanawdi]

How did u know that Q(x) only has x^n not multiple?

[janibashashaik]

Good explanation sir.

JANI basha sir , INDIA

[leickrobinson]

Attempting to solve just in my head again, but I think it’s [SPOILER]...





X^3 + 1?
I noticed that if you define G(x) = P(x) - 1, then
G(x)G(1/x) = 1
This leads to G(x) = x^n, and from the initial condition we get n = 3. :-D