Math Is Fun Forum

tony123

Member

Registered: 2007-08-03

Posts: 229

f(x)

Determine all polynomials

satisfying 

for all

.

Last edited by tony123 (2009-06-24 10:17:20)

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: f(x)

When x=1, the RHS is 0 and so the LHS must be 0 too.
Hence f(1/3)=0.
But then the RHS is 0 when x=1/3, so we also need f(1/9)=0.

Continuing, we get that f(1/3^n) = 0 for any n.

This is a polynomial with infinitely many zeroes and so the only solution is f(x)=0.

---

Why did the vector cross the road?
It wanted to be normal.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: f(x)

When x=1, the RHS is 0 and so the LHS must be 0 too.
Hence f(1/3)=0.
But then the RHS is 0 when x=1/3, so we also need f(1/9)=0.

Continuing, we get that f(1/3^n) = 0 for any n.

This is a polynomial with infinitely many zeroes and so the only solution is f(x)=0.

First of all, it is

.

Secoind, there are not infinitely many real roots. IIt is true that

for

but it is possible that

, so the polynomial can have just 5 real roots.

You can check that the polynomial

[align=center]

[/align]

satisfies the given condition.

Last edited by JaneFairfax (2009-06-24 11:55:52)

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: f(x)

Aha, that makes a lot more sense. I thought there should have been more to the question that what I had.

---

Why did the vector cross the road?
It wanted to be normal.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: f(x)

I think I’ve got hte answer.

[align=center]

[/align]

where

is a real constant.

[align=center]http://www.mathhelpforum.com/math-help/ … 1-f-x.html[/align]

Last edited by JaneFairfax (2009-06-25 02:51:57)