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For any quadratic function, prove that:

$$ f(x+3)-f(x) = 3[f(x+2)-f(x+1)] $$

This was a fairly straightforward problem, I solved it by assuming $f(x)=ax^2 + bx + c$ and by simplifying $g(x)=f(x+3)-3f(x+2)+3f(x+1)-f(x)=0$. Is there any way of looking at this problem geometrically, as perhaps a consequence of some property of parabolas?

I also tried to generalize it to a cubic function, but there is a slight problem;

Let the cubic be $h(x) = Ax^3 + f(x)$ where $f(x)$ is some quadratic. So I just needed to simplify the cube since I had already proved it for any quadratic;

$$ g(x) = A[(x+3)^3 - 3(x+2)^3 + 3(x+1)^3 - x^3]$$

$$=> g(x) = A [ x^3+9x^2+27x+27-3(x^3+6x^2+12x+8)+3(x^3+3x^2+3x+1)-x^3]$$

$$=>g(x) = 6A$$

So, nearly everything cancels out but we are left with a constant term. How should I change the expression of g(x) to make it 0 instead of 6A? And how will this generalize to higher polynomial expressions? Is there any theorem for these kinds of functional polynomial expressions? I have a feeling that this has to do with the binomial coefficients which are generated when this expression expands, but that is only an intuition and I have no way of concretizing it.

Edit: Now trying the cubic with $f(x)=Ax^3+Bx^2+Cx+D$

$g(x)=f(x+4)-4f(x+3)+6f(x+2)-4f(x+1)+f(x)$

First, looking at the first term of expansion;

$A[(x+4)^3-4(x+3)^3+6(x+2)^3-4(x+1)^3+x^3]$ This evaluates out to 0.

Next quadratic term;

$B[(x+4)^2-4(x+3)^2+6(x+2)^2-4(x+1)^2+x^2]$

$=>B[x^2+8x+16-4(x^2+6x+9)+6(x^2+4x+4)-4(x^2+2x+1)+x^2]=16-36+24-4=0$

Linear term: $C[(x+4)-4(x+3)+6(x+2)-4(x+1)+x]=0$

Constant term: $D-D+D-D+D=0$

So, this can now be generalized as for an (n)th degree polynomial,

$$f(x+n+1)=(-1)^n\left[\sum_{k=0}^n (-1)^k \binom{n+1}{k}f(x+k)\right]$$

This can be proved easily by summing up individual powers and using the result established in;

prove that a polynom is zero

Cognoscenti
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  • Your feeling about binomial coefficients is right. See https://math.stackexchange.com/q/918552/42969 for the general case. – Martin R Jan 12 '24 at 05:48
  • Um, I can't understand how to use that result for (x-k) to prove it for a general polynomial of the mth degree. – Cognoscenti Jan 12 '24 at 05:57
  • Are you saying that for a cubic function the identity would be f(x+4)-4f(x+3)+6f(x+2)-4f(x+1)+f(x)=0? – Cognoscenti Jan 12 '24 at 06:01
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    Yes, exactly... – Martin R Jan 12 '24 at 06:01
  • Oh, never mind I got it oof, took me long enough – Cognoscenti Jan 12 '24 at 06:50
  • I would'nt say geometric its more closely related to binomial coeffs considering ur edits you noticed this too. You can also draw relations to the pascal triangle since every linear also solves the quadratic identity and so on so the remove the constant term in ur first attempt you add another equation of the same kind by shifting x to x+1 this is analogous to the entries in pascal triangle – RandomGuy Jan 12 '24 at 07:09
  • That is an interesting way to think about it, I just noticed that the combinations of binomial coefficients for any nth degree polynomial works for lower degree polynomials too. – Cognoscenti Jan 12 '24 at 07:11
  • You are re-discovering the calculus of finite differences : this remark isn't at all critical : kudos for remarking the necessary presence of binomial coefficients. – Jean Marie Jan 12 '24 at 07:42
  • Oh shoot, I've seen that topic before in 1220 olympiad problems. I didn't realise it referred to this. Thank you so much for directing me to the right topic @Jean Marie! – Cognoscenti Jan 12 '24 at 12:59

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