Extreme workout with Polynomial

Algebra Level 4

Let P ( x ) P(x) be a polynomial with degree 5 5 and let P ( x ) P'(x) denotes the derivative of P ( x ) P(x) , if it exists such that P ( P ( x ) ) = P ( P ( x ) ) P'(P(x))=P(P'(x)) find the reciprocal of the leading coefficient of P ( x ) P(x) .

  • Bonus: What is the leading coefficient of P ( x ) P(x) if it has degree n 2 n\geq 2 ?


The answer is 625.

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1 solution

Kelvin Hong
Sep 9, 2018

Let P ( x ) = a 5 x 5 + a 4 x 4 + + a 0 P(x)=a_5x^5+a_4x^4+\cdots+a_0 , then P ( x ) = 5 a 5 x 4 + 4 a 4 x 3 + + a 1 P'(x)=5a_5x^4+4a_4x^3+\cdots+a_1 , calculate the composite function: P ( P ( x ) ) = 5 a 5 ( a 5 x 5 + ) 4 + P'(P(x)) = 5a_5(a_5x^5+\cdots )^4+\cdots P ( P ( x ) ) = a 5 ( 5 a 5 x 4 + ) 5 + P(P'(x))=a_5(5a_5x^4+\cdots)^5+\cdots

Since the leading term of these two is only 5 a 5 5 x 20 5a_5^5 x^{20} and 5 5 a 5 6 x 20 5^5 a_5^6x^{20} , they must be equal, so we ended up a 5 = 1 625 a_5=\dfrac1{625} , the desired reciprocal is 625 \boxed{625} .

  • This problem is inspired by the book Putnam and Beyond Question No.166.

  • Using similar process, any polynomial with degree n 2 n\geq 2 must have leading coefficient a n = 1 n n 1 a_n=\dfrac1{n^{n-1}} .

This is way over-determined. You've only got 6 unknowns, which are the coefficients of the 5th order equation. Yet, a total of 21 simultaneous equations must be satifisfied. There is no way there can be 6 coefficients that meets them all, there isn't sufficient degeneracy that would allow that. Such a polynomial P ( x ) P(x) does not exist.

It breaks down after solving about 7 or 8 of the necessary equations.

Michael Mendrin - 2 years, 9 months ago

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they could equal zero :)

Anton Curmanschiy - 2 years, 8 months ago

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It won't be 5 degree then :)

Praveen Kumar - 2 years, 8 months ago

Umm, P ( x ) = 1 625 x 5 P(x) = \frac{1}{625} x^5 works.

typical beam - 2 years, 8 months ago

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That works

Michael Mendrin - 2 years, 8 months ago

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