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To try to understand this concretely, I chose function $\mathcal f$ to be defined by $\mathcal f(x) = x + 1$ and the piecewise function: $$g(x)= \begin{cases}\sin(x), & \text{0 $\le x\lt$ 1} \\x-1, & \text{$x\lt$ 0 or $x \ge$ 1}\end{cases}$$ for $\mathcal g$. But I don't seem to have gotten $\mathcal f \circ g$ = $\mathcal g \circ f$ for all $\mathcal x$ (perhaps I may have miscalculated). How do I prove this analytically, and$-$if this specific $\mathcal g(x)$ does not satisfy this condition$-$why not? My background in math is severely lacking, so try your best to be as simplistic as possible. Thank you.

Sahiba Arora
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Sourblob
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  • Why did you pick $g(x)=x-1$ outside the $[0,1)$ interval? What happens if you try to apply $g(x+1)=g(x)+1$ to the function you've written down? – A Simmons Jun 13 '17 at 17:26
  • Perhaps I'm not understanding what arbitrary means. I thought g can be anything between 0 and 1 so long as it fulfills the given condition for x $\lt$ 0 and x $\ge$ 1. – Sourblob Jun 13 '17 at 17:26
  • You can indeed take $g(x)$ to be anything you want on $[0,1)$ but then you have to extend it by the given functional equation.Thus if $g(x)=17$ for $x\in [0,1)$ we'd compute $g(\pi)$, say, by noting that $g(\pi-3)=17\implies g(\pi-2)=18\implies g(\pi-1)=19\implies g(\pi)=20$. – lulu Jun 13 '17 at 18:03

2 Answers2

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You appear to be misunderstanding what is being asserted here. The statement is that the functional equation $g(x) + 1 = g(x+1)$ has solutions where $g(x)$ is defined arbitrarily on the interval $0 \le x < 1$. On each other interval $n \le x < n+1$ for nonzero integer $n$, this choice determines what $g$ must be in order to make the functional equation true. If you want to take $g(x) = \sin(x)$ for $0 \le x \le 1$, then $g(x) = n + \sin(x-n)$ for $n \le x < n+1$.

Robert Israel
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  • i.e.: $g(x) = \left\lfloor x \right\rfloor + \sin \left( {x - \left\lfloor x \right\rfloor } \right) = x + \left( {\sin \left( {\left{ x \right}} \right) - \left{ x \right}} \right)$. Compare with my answer below with $\pi_{1}=\left( {\sin \left( {\left{ x \right}} \right) - \left{ x \right}} \right)$. – G Cab Jun 13 '17 at 20:45
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The equation $$ g(x + 1) = g(x) + 1\quad $$ is "standardly" understood as $$ \Delta _{\,x} \,g(x) = 1\quad \Leftrightarrow \quad \,g(x) = \Delta _{\,x} ^{\,( - 1)} 1 = \sum\nolimits_x 1 = x + c $$ where $\sum\nolimits_x {} $ indicates the Indefinite Sum and $\Delta _{\,x}$ the Forward Difference.

Now, the $c$ above is normally taken as a constant, but in the more general terms it can actually be any periodic function of $x$ with period (one of the periods) $= 1$, so the "general" solution is $$ g(x) = x + \pi _1 (x) $$

For instance, it can be taken as $$ \pi _1 (x) \in \left\{ {\sin (2\pi x),\;1 - \cos ^{\,2} (\pi x),\;\left\{ x \right\} = x - \left\lfloor x \right\rfloor ,\; \cdots } \right\} $$ or, as in the answer above, as ${\sin \left( {\left\{ x \right\}} \right) - \left\{ x \right\}}$.

So that is the indefinition /arbitrary choice you are wondering about.

G Cab
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