Determine all polynomials f(x) such that, for some positive integer n:?

Question

f(x^n) -( x^3)f(x)= 2[(x^3) - 1]

Answer 1

First let's peg down what f(x^n) and x^3*f(x) look like. Using a rather generic representation of f(x), we have:
f(x) = a + bx + cx^2 + dx^3 + ... + gx^k
x^3*f(x) = ax^3 + bx^4 + cx^5 + dx^6 + ... gx^(k+3)
f(x^n) = a + bx^n + cx^2n + dx^3n + ... + gx^kn

For these formulas, we assume that g is not 0 (since otherwise we could have stopped one term earlier in the description of the function).

Now consider the RHS, as suggested by the previous answerer. We need to have 2x^3 - 2. Focus first on the -2. Only f(x^n) from the LHS can have a constant term, so a=-2. Now, since we know a=-2, we can determine the coefficient on the x^3 term. On the RHS, the coefficient is 2, and on the LHS we have, from the -x^3*f(x) term, a 2. Then the f(x^n) must have 0 as its coefficient of any x^3 term. Note that this means either n=1 and d=0, or n=3 and b=0, or n is a positive integer other than 1 or 3.

Now then, since there are no other terms on the RHS, any other terms must cancel out. Consider the degrees of x^3*f(x) and f(x^n). If one is greater than the other, the subtraction would leave that highest degree term, so that the greatest degree can only be 3 (since it's okay to have this term leftover). Otherwise, if the degrees are the same, then we have that kn=k+3, so that k(n-1)=3. Since k and n must be positive integers, this can only be (k,n)=(1,4) or (3,2).

Case (k,n)=(1,4): Then f(x) is just linear, or
f(x^n) - x^3*f(x) = bx^4 - 2 - bx^4 + 2x^3
= 2x^3-2, so that b can be any real number (sweet!)

Case (k,n)=(3,2): Then f(x) is a cubic, so
f(x^n) - x^3*f(x) = dx^6 + cx^4 + bx^2 - 2 - (dx^6 + cx^5 + bx^4 - 2x^3)
= -cx^5 + (c-b)x^4 + 2x^3 + bx^2 - 2,
so that c=0, b=0 (and here d can be any real number).

Therefore f(x) can be
ax-2 or ax^3-2, where a is any real number (we take n as 4 or 2, respectively).

That seems rather complex, you could probably whittle the argument down a bit; I just typed this as I thought it up. Also, would anybody care to double-check these? I'm tired and won't be up for it for a while :)

EDIT: Looking again at the first answerer's post, I realize that a degree zero polynomial does indeed work, something I didn't bother to check. Of course, this is taken care of in both given answers, taking a=0, but I could have easily missed out on it. Good eye back2nature...and a negative degree makes a function non-polynomial, so no there is no such thing.

Answer 2

I didn't learn this. So not sure if it is complete.

Since RHS is a polynomial of degree 3, f(x) - (x^3)f(x) must not be a polynomial of degree more than 3, thus f(x) may be of degree 0, or -3 (if there is such thing)

For degree 0, A case is f(x) = -2, and n = 1.
For degree -3, f(x) = 2x^(-3), and n = -1.

I doubt there are any other satisfying polynomials.

Answer 3

i think the respond is n = 40. i'm unsure a thank you to remedy this appropriate yet, yet using a working laptop or computing gadget Algebra device, I got here up with n = 40. right here is what I did: define a "determining" function d(N, i) whose cost is i if N % i = 0, in any different case 0. D(n) turns into the sum from i=a million to n of (d(n, i))^2. Assuming it incredibly is right, except I ignored some thing, n=40 is the smallest n such that D(n) is a minimum of 1998 (D(40) = 2210), and the sum seems to be incredible for smaller numbers that i ought to truly make certain. apart from, diagnosis of the above sum famous that D(n) isn't an incredibly superb function. It jumps around lots in accordance with if n is fundamental, a small distinctive of a substantial, et cetera. There may be a thank you to squeeze what n must be to a sufficiently small band to truly discover it. **edit** It looks like Quadrillerator have been given it.