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Prove that any solution $f: \mathbb{R} \to \mathbb{R}$ of the functional equation

$$ f(x + 1)f(x) + f(x + 1) + 1 = 0 $$

cannot have range $\mathbb{R}$.


I transformed it into

$$ f(x) = \frac {-1} {f(x + 1)} - 1 = \frac {-1 - f(x + 1)} {f(x + 1)} $$ I tried to evaluate $f(x + 1)$ and $f(x + 2)$ and put them into the transformed equation:

1) after $f(x + 1)$

$$ f(x) = \frac {1} {-1 - f(x + 2)}$$

2) after $f(x + 2)$

$$ f(x) = \frac {1} {\frac {-1}{f(x + 3)}} = -f(x + 3) $$

What am I supposed to do next?

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  • $\begingroup$ What do you mean by 'can't have real solutions of $f(x)$ on the whole number line'? $\endgroup$ – Workaholic Nov 29 '15 at 16:05
  • $\begingroup$ @user236182 That will always be the case, as, if $f(x+1) = 0$ for any $x$ then $0 f(x) + 0 + 1 = 0$, a contradiction. $\endgroup$ – James Nov 29 '15 at 16:08
  • $\begingroup$ @Workaholic I meant that $\exists$ real number $y$ so that $y$ isn't a solution for the original exuation. $\endgroup$ – Denis Nov 29 '15 at 16:09
  • $\begingroup$ @Denis If you mean you want to show there are no real solutions to $f(x)=0$, then James's comment answers it. $\endgroup$ – Peter Woolfitt Nov 29 '15 at 16:10
  • $\begingroup$ @PeterWoolfitt Clearly he simply wants to show the functional equation $f(x+1)f(x)+f(x+1)+1=0$ has no solutions, where $f:\Bbb R\to \Bbb R$. I.e. there are no functions $f:\Bbb R\to \Bbb R$ such that $f(x+1)f(x)+f(x+1)+1=0$ for all $x\in\Bbb R$. $\endgroup$ – user236182 Nov 29 '15 at 16:13
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You should always be careful when dividing, because you'll run into trouble if you try and divide by 0.

We have that for all $x$, $$f(x+1)(f(x) + 1) = -1$$

Notice that $f$ can never be zero, then, because that would mean $0$ on the left-hand side was $-1$ on the right-hand side. This justifies your division step, and proves that $0$ is never in the range of $f$; that is all you need.


As an appendix, we prove that there are, in fact, solutions to the functional equation.

If we fix $x=0$, obtain what is effectively the recurrence $$x_{n+1} = \frac{-1}{x_n + 1}$$ where $x_n = f(n)$. This certainly does have solutions (for example, if $x_0 = 1$, then it cycles $1, -\frac{1}{2}, -2, 1, -\frac{1}{2}, -2, \dots$), so the recurrence does have a solution over the reals.

One such solution is therefore $$f(x) = \begin{cases} 1 & \lfloor x \rfloor \equiv 0 \pmod{3} \\ -\frac{1}{2} & \lfloor x \rfloor \equiv 1 \pmod{3} \\ -2 & \lfloor x \rfloor \equiv 2 \pmod{3} \end{cases} $$

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  • $\begingroup$ I wanted to say that range of values of $f(x)$ doesn't belong to $\Bbb R$ $\endgroup$ – Denis Nov 29 '15 at 16:24
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    $\begingroup$ But it can. $f(x) = 1$ if $\lfloor x \rfloor \equiv 0 \pmod{3}$; $f(x) = -\frac{1}{2}$ if $\lfloor x \rfloor \equiv 1 \pmod{3}$; $f(x) = -2$ otherwise. $\endgroup$ – Patrick Stevens Nov 29 '15 at 16:27
  • $\begingroup$ ...doesn't belong to full range of $\Bbb R$. $\endgroup$ – Denis Nov 29 '15 at 16:28
  • $\begingroup$ I mean it can't be any number from $\Bbb R$ $\endgroup$ – Denis Nov 29 '15 at 16:29
  • $\begingroup$ Is your question "Show that $\text{range}(f) \not = \mathbb{R}$"? $\endgroup$ – Patrick Stevens Nov 29 '15 at 16:30
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Assume that the range of $f$ is all of $\Bbb R$. Then there are $a,b\in\Bbb R$ with $f(a)=0$ and $f(b)=-1$. But then $$ 1=f(a)f(a-1)+f(a)+1=0$$ and $$ 1=f(b+1)f(b)+f(b+1)+1=0,$$ both of which are absurd.

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