# Find the period of $g(x) = f(2^x,0)$ for $f(x,y) = f(2x + 2y, 2y - 2x)$

Let $f(x,y)$ be a function satisfying the condition $$f(x,y) = f(2x + 2y, 2y - 2x)$$ with $x, y \in \mathbb{R}$. Now we define a function $g(x) := f(2^x, 0)$.

Decide if $g(x)$ is periodic or not. If it is, find its period.

Attempt: I didn't get the approach. I didn't understand the function. I found $g(x) = f(2^{x+1}, -2^{x+1})$. But got confused as to what to do next.

For all $a$, we have \begin{align*} f(a,0)&=f(2a,-2a)\\[4pt] &=f(0,-2^3a)\\[4pt] &=f(-2^4a,2^4a)\\[4pt] &=f(-2^6a,0)\\[4pt] &=f(b,0)\;\;\;\text{[letting $b=-2^6a$]}\\[4pt] &=f(-2^6b,0)\\[4pt] &=f(-2^6(-2^6a),0)\\[4pt] &=f(2^{12}a,0)\\[4pt] \end{align*} hence, letting $a=2^x$, we get $$g(x)=f(2^x,0)=f(a,0)=f(2^{12}a,0)=f(2^{12}2^x,0)=f(2^{x+12},0)=g(x+12)$$ so $g$ is periodic.

• Why do I do that until I get $2^{12}$? How do I know where to stop? Do we apply it again and again until we get $2^x$ form? Jun 17 '18 at 13:06
• Some initial experimentation can hopefully reveal a pattern. The fact that powers of $2$ appear as coefficients lends support to the possibility that iteratively applying the specified rule to $g(x)$ will lead to an expression of the same form as the initial one. Jun 17 '18 at 16:44
• Also note: Iteration on $f(a,0)$ is less cumbersome than iteration on $f(2^x,0)$. Jun 17 '18 at 16:47
• In my latest edit, the fact that line $4$ yields $f(a,0)=f(-2^6a,0)$ suggests iterating that identity (since the zero part will stay fixed). Jun 17 '18 at 17:16

Define $h\colon \Bbb C\to \Bbb R$ as $h(z)=f(\operatorname{Re}z,\operatorname{Im}z)$, i.e., $h(x+iy)=f(x,y)$. From $(2x+2y)+(2y-2x)i=2(x+iy)(1-i)$, we see that $h(z)=h(wz)$ with $w:=2(1-i)$. by induction, $h(z)=h(w^nz)$ for all $n\in\Bbb N$. Now from $(1-i)^2=-2i$, we see $\bigl(2(1-i)\bigr)^8=2^8\cdot (-2)^4i^4=2^{12}$. Hence $h(z)=h(2^{12}z)$, which translates to $f(x,y)=f(2^{12}x,2^{12}y)$ and $g(x)=g(x+12)$. Hence $12$ is a period of $g$.

Note that it is possible that $g$ also has smaller periods than $12$, for example, $f,h,g$ might be constant. In fact, the only functions with $h(z)=h(wz)$ for all $z$ that are continuous at $z=0$ are the constants. But if we do not require $f$ to be continuous, we can define $g$ arbitrarily (hence certainly without any smaller period) on, say, $[0,12[$, then obtain from that a partial definition of $h$ on $[1,2^{12}[$, can define $h$ arbitrarily in the remainder of the quadrangle with vertices $1,2^{12}, 2-2i, 2^{13}-2^{13}i$ (except that the values at the edges must match), can use $h(wz)=h(z)$ to extend $h$ to $\Bbb C\setminus\{0\}$, and define $h(0)$ arbitrarily. This will give us an $f$ with $f(x,y)=f(2x+2y,2y-2x)$ such that the $g$ derived from it assumes the values on $[0,12[$ that we started with.

• I didn't understand this back in last June. But, now I do and it is brilliant. Thank you. Mar 29 '19 at 7:03

We have $$f(x,y)=f(2x+2y,2y-2x)=f(2(2x+2y+2y-2x),2(2y-2x-2y-2x))=f(8y,-8x)$$

Again applying it we get $$f(x,y)=f(8(-8x),-8(8y))=f(-2^6x,-2^6y)=f(-2^6(-2^6x),-2^6(-2^6y))=f(2^{12}x,2^{12}y)$$

Now we have $g(x)=f(2^x,0)=f(2^{12}2^x,0)=g(x+12)$. Thus $g$ is periodic. One possible period being $12$.

• Do we apply it again to get $2^x$ form? Jun 17 '18 at 13:07
• yeah after deducing $f(x,y)=f(2^{12}x,2^{12}y)$ set $x=2^x$ and $y=0$
– user428700
Jun 18 '18 at 2:29