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phanthanhtom

Member

Registered: 2012-06-22

Posts: 290

Determining a polynomial

Determine all polynomials (of all degrees) satisfying

[P(2x)]^2 = 4P(x^2) for all x.

So far I have determined the following polynomials:

Undefined degree: P(x) = 0

Degree 0: P(x) = 4

Degree 1: P(x) = x

Degree 2: P(x) = (1/4)*x^2

I believe that such P(x) exists for all degrees, and it equals m*x^n for all degrees n not smaller than 1. However, I have  been able to prove neither the formula (P(x)=m * x^n) nor any formula for the coefficient m.

Help! Hilfe! Hjælpe!

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Determining a polynomial

hi phanthanhtom

That suggestion looks good to me.

Then

and

For these to be the same

That shows that your function works and gives the value of m in terms of n.

When n = 1 and n = 2 it gives the earlier functions you have suggested.  Here's a check with n = 3

and

To finish this question off it would be good to show that no other function will also have the property.  I'm working on that. 

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

phanthanhtom

Member

Registered: 2012-06-22

Posts: 290

Re: Determining a polynomial

So the formula for m can be written as m = (1/4)^(n-1).

It gets really tricky because it can be of any degree. I'm still trying to figure out a catch-all strategy.

phanthanhtom

Member

Registered: 2012-06-22

Posts: 290

Re: Determining a polynomial

Okay I'll have a try.

For constant polynomials we easily get P(x) = 0 and P(x) = 4.

Suppose the polynomial is written as

(

should not be 0)

We will expand both sides of the original equation and consider the coefficients of the two polynomials we receive, degree by degree they should be equal.

Degree 2n:

LHS:

RHS:

Thus

Degree 2n-1:

LHS:

RHS: 0

Thus

and from now we ignore this coefficient.

Degree 2n - 2:

LHS:

RHS: 0

Thus

We can then prove all coefficients except for the highest is equal to 0.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Determining a polynomial

hi phanthanhtom

Some of your [/math] tags had got reversed to [\math].  I've tidied these up for you.

I started to try a similar approach, but starting with a0, a1 a2 etc.  I am cutting a long hedge so I didn't get very far with a proof before I needed to return to the cutting.

I agree with a0 = 4 or 0 and with your degree 2n analysis.

But I couldn't follow the next bit:

Shouldn't that be

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

phanthanhtom

Member

Registered: 2012-06-22

Posts: 290

Re: Determining a polynomial

Well no.

is the coefficient of degree n-1, which when multiplied with

and a constant gives us the coefficient of degree n + (n-1) = 2n-1.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Determining a polynomial

hi phanthanhtom

My apologies.  In my version (incomplete) I was looking at degree n, not degree 2n, so I got confused.  I'll look at it properly leter.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob