Evaluating $\frac{f(2016)}{f(2015)}$

Question

Let $f(x)$ be a polynomial such that $xf(x)=(x+1)f(x-1)$. Find $\frac{f(2016)}{f(2015)}$.

Evaluating at $x=0$:

$$f(-1)=0$$

so we have that $x+1$ is a root of $f(x)$.

We can now write $f(x) = (x+1) \cdot g(x)$, where $g(x)$ is some other polynomial.

From here we have that $x(x+1)g(x)=x(x+1)g(x-1)$. Here is where I can't really go forward. I could divide the equation by $x(x+1)$ and get that $g(x)=g(x-1)$, but that doesn't tell me anything. What should I do?

You overcomplicated this. Plug in $x=2016$ into the given formula. — Rushabh Mehta, May 09 '20 at 21:22
Actually, $g(x) = g(x-1)$ tells you that if $g$ has any roots, say $a$, it has infinitely many, $a\pm1$, $a \pm 2$, ..., so is not a polynomial. Therefore, $g$ is a constant. None of this is on-point to solve the problem. — Eric Towers, May 09 '20 at 21:26

score 2 · Answer 1 · answered May 09 '20 at 21:22

2

This is simpler than you're making it. Divide both sides by $f(x-1)$ and by $x$, on the premise they're nonzero. Then you have

$$\frac{f(x)}{f(x-1)} = \frac{x+1}{x}$$

Let $x=2016$.

answered May 09 '20 at 21:22

PrincessEev

43,815

score 1 · Answer 2 · answered May 09 '20 at 21:22

1

From $$ xf(x)=(x+1)f(x-1) $$ we have $$ \frac{f(x)}{f(x-1)}=\frac{x+1}{x} \text{.} $$ Specialize to $x = 2016$.

answered May 09 '20 at 21:22

Eric Towers

67,037

score 0 · Answer 3 · answered May 09 '20 at 23:18

Potentially, $f(x) = 0$ wich would satisfy the premise that $f(x)$ is a polynomal and $xf(x)= (x+1) f(x)$ but then $\frac {f(2016)}{f(2015)}$ is not defined.

Anyway following the logic from the OP.

$f(x) = (x+1)g(x)\\ f(x-1) = xg(x-1)\\ x(x+1)g(x) = (x+1)(x)g(x-1)\\ g(x) = g(x-1)$

$g(x)$ is a periodic function. If $g(x)$ is a polynomial (of fininte degree) then $g(x)$ must be constant.

Nonetheless, as others have pointed out, this bit of functional analysis is not necessary to answer the question that has been asked.

Evaluating $\frac{f(2016)}{f(2015)}$

3 Answers3