Forum index » GR8767 #20

Here is the question:

Suppose that f(1+x) = f(x) for all real x. If f is a polynomial and f(5) =11, then f(15/2) is ___?

A) -11
B) 0
C) 11
D) 33/2
E) not uniquely determined by the information given

They say the correct answer is C. In my opinion this assumes the polynomial is finite, which is not necessary from the language of the question. The only way a finite polynomial could be 11 at every integer is if it was the constant function f(x) = 11. But if we are to consider infinite polynomials, (without going into every detail of the construction) one could transform the sine function so that it's hitting 11 at every integer, and cycling through whatever amplitude range you want in between them. Then, of course, any transformation of the sine function can be represented as an infinite polynomial.

Okay fine I will go into the details of it. Easyest sine transformation I see that will do the trick is

f(x) = 11sin(2pi*(x + 1/4))
=11[sum from n=0 to n=infinity of (((-1)^n)*(2pi*(x + 1/4))^(2n+1))/((2n+1)!)]

There's one polynomial satisfying the criteria that is -11 at 15/2.

f(x) = 11 is another polynomial satisfying the criteria and it's 11 at 15/2.

=> E.

A polynomial is by definition an element of the ring R[X] where R is a commutative ring. There are no such things as infinite polynomials; these are called power series and the ring in question is denoted by R[[X]].

I think this question was solved by many ways on this forum
From my understanding, we do not have some thing like "infinite polynomials", just like eof said we have power series.

yep, a power series is not a polynomial.

if p(x) is not a constant polynomial, then |p(x)| --> infty for |x| --> infty. in the problem, this cannot happen, so p(x) must be constant.