Understand the problem

Find all the real Polynomials P(x) such that it satisfies the functional equation:

P(2P(x)) = 2P(P(x)) + P(x)^{2} \forall real x .

Source of the problem

Unknown

Topic
Functional Equation, Polynomials
Difficulty Level
7/10
Suggested Book
Excursion in Mathematics

Challenges and Thrills in Pre College Mathematics

Start with hints

Do you really need a hint? Try it first!

Well, it is really good that the information polynomial is given! You should use that.

What is the first thing that you check in a Polynomial Identity?

Degree! Yes, check whether the degree of the Polynomial on both the LHS and RHS are the same or not. Yes, they are both the same n^2 . But did you observe something fishy?

 

Now rewrite the equation as P(2P(x)) - 2P(P(x)) = P(x)^{2}.

Do the Degree trick now… You see it right?

Yes, on the left it is n^2 and on the RHS it is 2n .

So, there are two cases now… Figure them out!

Case 1: 2n = n^2 … i.e. P(x) is either a quadratic or a constant function.

Case 2: P(2P(x)) - 2P(P(x)) has coefficient zero till x^2n.

We will study case 1 now.

Case 1: 2n = n^2… i.e. P(x) is either a quadratic or a constant function.

P(2P(x)) - 2P(P(x)) = P(x)^{2} = P(2y) - 2P(y) = y^{2} where y = P(x) .

Now, expand using P(x) = ax^2 + bx +c, it gives 2ay^2 -c = y^2 … Now find out all such polynomials satisfying this property.

For e.g. \frac{x^2}{2} is a solution.

If P(x) is constant, prove that P(x) = 0 / \frac{-1}{2} .

 

Case 2: P(2y) - 2P(y) = y^{2}.

Assume a general form of P(x) = $latex $and show that P(x) must be quadratic or lesser degree by comparing coefficients as you have a quadratic on RHS and n degree polynomial of the LHS.

Now, we have already solved it for quadratic or less degree.

 

  • Always Compare the Degree of Polynomials in identities like this. It provides a lot of information.
  • Compare the coefficients of Polynomials on both sides to equalize the coefficient on both sides.
  • Find all polynomials P(2P(x)) - 8P(P(x)) = P(x)^{2}.
  • Find all polynomials \( P(cP(x)) – d P(P(x)) = P(x)^{2} \) depending on the values of c and d.

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