# Interesting math problem involving a weirdly-defined polynomial function **URL:** \ **Category:** Factual Questions\ **Created:** [February 20, 2006, 3:23am UTC](https://boards.straightdope.com/t/interesting-math-problem-involving-a-weirdly-defined-polynomial-function/345283 "2006-02-20T03:23:26Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![iwakura43](https://avatars.discourse-cdn.com/v4/letter/i/9de0a6/32.png) [@iwakura43](https://boards.straightdope.com/u/iwakura43)\ **Post date:** [February 20, 2006, 3:23am UTC](https://boards.straightdope.com/t/interesting-math-problem-involving-a-weirdly-defined-polynomial-function/345283/1 "2006-02-20T03:23:26Z")
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? ---
**Author:** ![Cabbage](https://avatars.discourse-cdn.com/v4/letter/c/f07891/32.png) [@Cabbage](https://boards.straightdope.com/u/Cabbage)\ **Post date:** [February 20, 2006, 3:45am UTC](https://boards.straightdope.com/t/interesting-math-problem-involving-a-weirdly-defined-polynomial-function/345283/2 "2006-02-20T03:45:35Z")
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). ---
**Author:** ![Cabbage](https://avatars.discourse-cdn.com/v4/letter/c/f07891/32.png) [@Cabbage](https://boards.straightdope.com/u/Cabbage)\ **Post date:** [February 20, 2006, 3:47am UTC](https://boards.straightdope.com/t/interesting-math-problem-involving-a-weirdly-defined-polynomial-function/345283/3 "2006-02-20T03:47:53Z")
Of course, I forgot the constant factor for P: P(x) = x(x + 6)(x + 2)/21 (since P(1) = 1). ---
**Author:** ![bitwise](https://avatars.discourse-cdn.com/v4/letter/b/ea5d25/32.png) [@bitwise](https://boards.straightdope.com/u/bitwise)\ **Post date:** [February 20, 2006, 5:07am UTC](https://boards.straightdope.com/t/interesting-math-problem-involving-a-weirdly-defined-polynomial-function/345283/4 "2006-02-20T05:07:54Z")
> [@Cabbage](#): > > Of course, I forgot the constant factor for P: > > P(x) = x(x + 6)(x + 2)/21 (since P(1) = 1). 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). ---
**Author:** ![Saint\_Cad](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/saint_cad/32/18907_2.png) [@Saint\_Cad](https://boards.straightdope.com/u/Saint_Cad)\ **Post date:** [February 20, 2006, 5:16am UTC](https://boards.straightdope.com/t/interesting-math-problem-involving-a-weirdly-defined-polynomial-function/345283/5 "2006-02-20T05:16:11Z")
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. ---
**Author:** ![bitwise](https://avatars.discourse-cdn.com/v4/letter/b/ea5d25/32.png) [@bitwise](https://boards.straightdope.com/u/bitwise)\ **Post date:** [February 20, 2006, 5:30am UTC](https://boards.straightdope.com/t/interesting-math-problem-involving-a-weirdly-defined-polynomial-function/345283/6 "2006-02-20T05:30:56Z")
Let D denote the derivative operator. 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. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [February 20, 2006, 5:54am UTC](https://boards.straightdope.com/t/interesting-math-problem-involving-a-weirdly-defined-polynomial-function/345283/7 "2006-02-20T05:54:53Z")
> [@bitwise](#): > > Thus, we must conclude that g is a constant function. Since g(1) = 1, it follows that g(x) = 1 for all x. Except that g(0) = 0. ---
**Author:** ![Larry\_Borgia](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/larry_borgia/32/156_2.png) [@Larry\_Borgia](https://boards.straightdope.com/u/Larry_Borgia)\ **Post date:** [February 20, 2006, 8:08am UTC](https://boards.straightdope.com/t/interesting-math-problem-involving-a-weirdly-defined-polynomial-function/345283/8 "2006-02-20T08:08:32Z")
Just out of curiosity, how can you find the zeros at -6 and -2? ---
**Author:** ![RM\_Mentock](https://avatars.discourse-cdn.com/v4/letter/r/e274bd/32.png) [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)\ **Post date:** [February 20, 2006, 8:17am UTC](https://boards.straightdope.com/t/interesting-math-problem-involving-a-weirdly-defined-polynomial-function/345283/9 "2006-02-20T08:17:34Z")
> [@ultrafilter](#): > > Except that g(0) = 0. P(0)=0, g(0)=1 ---
**Author:** ![RM\_Mentock](https://avatars.discourse-cdn.com/v4/letter/r/e274bd/32.png) [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)\ **Post date:** [February 20, 2006, 8:21am UTC](https://boards.straightdope.com/t/interesting-math-problem-involving-a-weirdly-defined-polynomial-function/345283/10 "2006-02-20T08:21:12Z")
> [@Larry Borgia](#): > > Just out of curiosity, how can you find the zeros at -6 and -2? Substitute X=-7 and X=-1 in the equation (X+7)\*P(2X) = 8X \* P(X+1), and use the previously known information that P(0)=0 ---
**Author:** ![RM\_Mentock](https://avatars.discourse-cdn.com/v4/letter/r/e274bd/32.png) [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)\ **Post date:** [February 20, 2006, 8:27am UTC](https://boards.straightdope.com/t/interesting-math-problem-involving-a-weirdly-defined-polynomial-function/345283/11 "2006-02-20T08:27:32Z")
And so the answer is -5/21? ---
**Author:** ![Saint\_Cad](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/saint_cad/32/18907_2.png) [@Saint\_Cad](https://boards.straightdope.com/u/Saint_Cad)\ **Post date:** [February 20, 2006, 5:34pm UTC](https://boards.straightdope.com/t/interesting-math-problem-involving-a-weirdly-defined-polynomial-function/345283/12 "2006-02-20T17:34:50Z")
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. ---
**Author:** ![Larry\_Borgia](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/larry_borgia/32/156_2.png) [@Larry\_Borgia](https://boards.straightdope.com/u/Larry_Borgia)\ **Post date:** [February 20, 2006, 6:19pm UTC](https://boards.straightdope.com/t/interesting-math-problem-involving-a-weirdly-defined-polynomial-function/345283/13 "2006-02-20T18:19:47Z")
> [@RM Mentock](#): > > Substitute X=-7 and X=-1 in the equation (X+7)\*P(2X) = 8X \* P(X+1), and use the previously known information that P(0)=0 Damn, that’s embarrassing. Oh well, it was 3:00 am. Thanks **RM**! ---
**Author:** ![yohanbaby](https://avatars.discourse-cdn.com/v4/letter/y/91b2a8/32.png) [@yohanbaby](https://boards.straightdope.com/u/yohanbaby)\ **Post date:** [February 20, 2006, 6:26pm UTC](https://boards.straightdope.com/t/interesting-math-problem-involving-a-weirdly-defined-polynomial-function/345283/14 "2006-02-20T18:26:41Z")
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. ---
**Author:** ![Omphaloskeptic](https://avatars.discourse-cdn.com/v4/letter/o/bcef8e/32.png) [@Omphaloskeptic](https://boards.straightdope.com/u/Omphaloskeptic)\ **Post date:** [February 20, 2006, 7:22pm UTC](https://boards.straightdope.com/t/interesting-math-problem-involving-a-weirdly-defined-polynomial-function/345283/15 "2006-02-20T19:22:24Z")
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). ---
**Author:** ![OldGuy](https://avatars.discourse-cdn.com/v4/letter/o/3bc359/32.png) [@OldGuy](https://boards.straightdope.com/u/OldGuy)\ **Post date:** [February 20, 2006, 10:35pm UTC](https://boards.straightdope.com/t/interesting-math-problem-involving-a-weirdly-defined-polynomial-function/345283/16 "2006-02-20T22:35:32Z")
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, 2xQ(2x)(x+7)=8x(x+1)Q(x+1) =\> Q(2x)(x+7)=4(x+1)Q(x+1) Now write Q(x) = xR(x)+c where R(x) is a polynomial, then 2xR(2x)+c=4(x+1)[(x+1)R(x+1)+c] =\> 7c = 4c or c = 0 so now Q(x) = xR(x) where R(x) = ax+b xR(2x)(x+7)=2(x+1)(x+1)R(x+1) Substituting ax+b in for R(x) gives x2ax+b=2(x+1)(x+1)[a(x+1)+b] Again the LHS has no constant term so the RHS cannot. Therefore b = 0 leaving us x(2ax)(x+7)=2a(x+1)^3 or 2ax^3+14ax^2 = 2a(x^3+3x^2+3x+1) which can’t hold for all x So I’m stumped. ---
**Author:** ![pikolo77](https://avatars.discourse-cdn.com/v4/letter/p/43a26b/32.png) [@pikolo77](https://boards.straightdope.com/u/pikolo77)\ **Post date:** [February 21, 2006, 6:37am UTC](https://boards.straightdope.com/t/interesting-math-problem-involving-a-weirdly-defined-polynomial-function/345283/17 "2006-02-21T06:37:53Z")
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. ---
**Author:** ![Cabbage](https://avatars.discourse-cdn.com/v4/letter/c/f07891/32.png) [@Cabbage](https://boards.straightdope.com/u/Cabbage)\ **Post date:** [February 21, 2006, 6:45am UTC](https://boards.straightdope.com/t/interesting-math-problem-involving-a-weirdly-defined-polynomial-function/345283/18 "2006-02-21T06:45:01Z")
> [@pikolo77](#): > > 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. For example: (x + 1) / x = 1 + (1/x) ---
**Author:** ![pikolo77](https://avatars.discourse-cdn.com/v4/letter/p/43a26b/32.png) [@pikolo77](https://boards.straightdope.com/u/pikolo77)\ **Post date:** [February 21, 2006, 2:18pm UTC](https://boards.straightdope.com/t/interesting-math-problem-involving-a-weirdly-defined-polynomial-function/345283/19 "2006-02-21T14:18:15Z")
> [@Cabbage](#): > > For example: > > (x + 1) / x = 1 + (1/x) Ah yes. And after retracing some earlier steps, the easiest solution is in fact the polynomial function given earlier where P(x) = x(x+6)(x+2). P(x) = x(x+6)(x+2) P(2x) = 2x(2x+6)(2x+2) P(x+1) = (x+1)(x+7)(x+3) P(2x)/P(x+1) = 8x(x+3)(x+1)/(x+1)(x+3)(x+7) which reduces to P(2x)/P(x+1) = 8x/(x+7) = 8 - 56/(x+7) Hardly a proof, but good enough for puzzle math. So, are we done? ---
**Author:** ![RM\_Mentock](https://avatars.discourse-cdn.com/v4/letter/r/e274bd/32.png) [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)\ **Post date:** [February 21, 2006, 5:34pm UTC](https://boards.straightdope.com/t/interesting-math-problem-involving-a-weirdly-defined-polynomial-function/345283/20 "2006-02-21T17:34:38Z")
> [@OldGuy](#): > > Now write Q(x) = xR(x)+c where R(x) is a polynomial, then > > 2xR(2x)+c=4(x+1)[(x+1)R(x+1)+c] =\> 7c = 4c or c = 0 > > ::snip:: > > which can’t hold for all x > > So I’m stumped. You cannot conclude that 7c = 4c, since R(x+1) could also have a constant term. > [@pikolo77](#): > > Ah yes. And after retracing some earlier steps, the easiest solution is in fact the polynomial function given earlier where P(x) = x(x+6)(x+2). It’s P(x) = x(x+6)(x+2)/21, since P(1)=1 [Next page](https://boards.straightdope.com/t/interesting-math-problem-involving-a-weirdly-defined-polynomial-function/345283.md?page=2)