I was tutoring today and a student brought me a problem from an MIT competition, as follows:
P is a polynomial function. We are given P(1) = 1.
P(2X) / P(X+1) = 8 - 56/(x + 7) , defined wherever possible
Find P(-1)
We got so far as to find that P(0) = 0, P(-6) = 0, and P(-2) = 0. Those would work even if P was discrete, rather than polynomial. We also decided that the function was probably cubic, based on rewriting the equation as
(X+7)*P(2X) = 8X * P(X+1). But then we got stuck, and our session ended. Any ideas?
I don’t know if I’ll have any time to actually play around with it, but I did just want to mention one basic thing (which you may already be aware of). If you suspect that P is a cubic, then it’s clear what P would have to be: P(x) = x(x + 6)(x + 2) (since you’ve already found those 3 zeroes of P).
From there it would be trivial to check if this guess for P is correct (which I have not done).
p(x) = (1/21)x(x+2)(x+6) does satisfy the original equation. However, I don’t know how you can prove the polynomial is cubic. So far, we know that p(x) has the form p(x) = (1/21)x(x+2)(x+6)g(x) where g(1) = 1 and g(2x) = g(x+1).
Too drunk to look at this right now, but typically these problems have a trick to them. I remember that on the AP Calculus BC test there was a question about f(-h). We knew f(h), f(a+b), and f(0) so the trick was a=h and b=-h.
Let n denote the degree of g, and assume n >= 1. Let a_n denote the nth coefficient of g. Since g(2x) = g(x+1), we have
(2^n)(D^n)g(2x) = (D^n)g(x+1)
However, we must have (D^n)g(x) = n! a_n for all x. Therefore,
(2^n) n! a_n = a_n
which is true only if a_n = 0. This is a contradiction, since we assumed n is the degree of g and n >= 1. Thus, we must conclude that g is a constant function. Since g(1) = 1, it follows that g(x) = 1 for all x.
I doubt it’s cubic. If P(-6)=0 then what is P(-14)? P(-30)? In fact, by your analysis (I suspect something similar to If x/0 = real then x=0) then there is an infinite number of zeros. Note the phrase “defined whenever possible” would not apply to indeterminate forms.
I believe it’s cubic. Because if x is really large, the right side of the equation is approximately 8. And the left side is approximately p(2x) / P (x). A cubic is the only polynomial I can think of that would have this property.
Here’s another solution method. Rewrite the given equation as
P(2(x-1))=(8(x-1)/(x+6)) * P(x).
Now x+6 must divide P(x) (since otherwise P(2(x-1)) is not a polynomial), so we can write P(x)=(x+6)Q(x), where Q(x) is another polynomial. Plug this expression for P in, lather, rinse, repeat. Eventually we find
S(2(x-1))=S(x)
for polynomial S, which means S(x) is constant. Combining all the factors gives the cubic found above (evaluating at 1 finds the constant term).
Except when x=7 we can rewrite the original relation as:
P(2x)(x+7) = 8xP(x+1)
Now P is stated to be a polynomial, so both the LHS and RHS above are. Let the highest order term be ax^n, then the highest order terms on each side of the equation are
a(2x)^n x = 8x ax^n => 2^n = 8 or n = 3. So P(x) is a third order polynomial.
P(x) = ax^3+bx^2+cx+d
Now the RHS clearly has no constant term so the LHS can’t either. If P(x) had a constant term (i.e., d not equal to 0), then P(2x)(x+7) would as well, so P(x) must not have a constant term and P(x) =xQ(x) where Q(x) =ax^2+bx+c is a 2nd-order polynomial. Therefore,
I’m not sure how this could be a cubic to begin with because take any polynomial of degree n. (i.e. ax^n + bx^(n-1) + … + k) and then replace the variable x with either (x+1) or 2x and you still have a polynomial of the same degree, n. Take those two equations and divide them. How do you end up with a polynomial of degree 1 or -1?
I would like to believe this is a simple polynomial, but I don’t see how it could be.