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I am looking for a function $f(n)$ that satisfies the following two conditions at the same time $$ \frac{f(n-1)}{f(n)}=(-1)^n\quad ,\quad \frac{f(n+1)}{f(n)}=(+1)^n\equiv 1,\quad \forall n\in\mathbb{N}\ . $$ As I am not even sure if such a function exists, I'd appreciate any help or comment. Thank you very much in advance!

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Doesn't the second condition imply this function is constant? – Julian Rosen Jul 24 '12 at 16:23
@PinkElephants: That depends on the domain the function is defined on. If only defined on $\mathbb{N}$, then yes, the second condition implies the function is constant. If the function is defined on any larger domain $D$, then it does not restrict the function for any non-natural number (e.g. on $\mathbb{R}$ the second condition would be fulfilled for $f(x)=\cos(2\pi x)$). Also note that the first condition (if we ignore that it contradicts the second) implies that $f$ is defined for at least one $x\notin\mathbb{N}$ (because it contains $f(n-1)$ but shall hold for all $n\in\mathbb{N}$). – celtschk Jul 24 '12 at 16:36
up vote 9 down vote accepted

No such function exists. For the requirements to be well-defined, we need $f(n)\neq 0$, so in particular $f(1)=k\neq 0$. Then $f(2)=f(1+1)=(+1)^1f(1)=k$, but this gives $$k=f(1)=f(2-1)=(-1)^nf(2)=-k$$ hence $k=0$, a contradiction.

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Please accept my apologies if I am incorrect, but did your logic just go: f(2) = f(1+1) = f(1)*f(1) = (+1)^1 * f(1) = k? But that logic isn't correct. Using this logic, g(0) = g(1 + -1) = g(1) * g(-1), but if g(x) = x^2 then we get a contradiction (g(0) would be 0, g(1) * g(-1) = 1) – pelson Jul 24 '12 at 21:58
@pelson The logic is correct assuming the function $f$ satisfies the properties mentioned in the question. The function $x^2$ does not satisfy these properties, so my deduction would not be valid for $x^2$. – Alex Becker Jul 24 '12 at 22:13

Another way of recognizing the impossibility without doing any calculation (and one that shows why a lot of similar problems are inherently impossible) is by juggling indices. Your second condition, $\displaystyle\frac{f(n+1)}{f(n)}=1$, can also be written (since the right side is always invertible) as $\displaystyle\frac{f(n)}{f(n+1)}=1$, and then by setting $m=n+1$, as $\displaystyle\frac{f(m-1)}{f(m)}=1$ - and this is obviously incompatible with the first condition. In general, the two expressions you've written are dependent (as sequences) in that one is just a shifted inverse of the other, so for exactly the same reason you couldn't have e.g. $\displaystyle\frac{f(n-1)}{f(n)}=2$ and $\displaystyle\frac{f(n+1)}{f(n)}=\frac{1}{3}$ for all $n$ in some suitable domain.

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