All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Real Polynomial Puzzle (Posted on 2015-06-27)
Given that x is a real number, determine all polynomials with real coefficients satisfying this relationship:
P(x2) = P(x)*P(x+2)

 No Solution Yet Submitted by K Sengupta No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Many, if not all. Comment 1 of 1
Among constant functions P(x)=0 and P(x)=1 are clearly the only possibilities.

Among linear functions P(x)=ax+b
a little algebra shows a=1, b=-1 is the only possibility.
So P(x)=x-1.

a lot of algebra shows a=1, b=-2, c=1 is the only possibility.
So P(x)=x²-2x+1=(x-1)²

From which appears a nice generalization:
P(x)=(x-1)^n for all non-negative integers n.

Proof:
(x^2 -1)^n = (x-1)^n*(x+1)^n

The bold above are the solutions.

I did not check for other higher order polynomials.  It seems these are the only ways to make the zeros to work out right.

 Posted by Jer on 2015-06-27 16:40:37

 Search: Search body:
Forums (0)