I am taking algebra two and questions like this popped into my head so I would like to learn how to solve them, I am currently working on composing functions.
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asked Jul 10, 2019 at 12:09
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$\begingroup$ I would start by thinking there might be a solution of the form $f(x) = x^a$ and $g(x) = x^b$ and trying to find $a$ and $b$. (I haven't tried to see if this works. You can.) $\endgroup$– Ethan BolkerJul 10, 2019 at 12:17
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1$\begingroup$ @EthanBolker No, this doesn't work because $f\circ g=g \circ f$ in this case. $\endgroup$– Kavi Rama MurthyJul 10, 2019 at 12:18
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1$\begingroup$ This is a very interesting question. I proved that no such functions exist here $\endgroup$– Maximilian JanischJul 10, 2019 at 12:19
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$\begingroup$ @KaviRamaMurthy True. That is where I'd have begun, and would quickly have found that it doesn't work. Nothing will, as per other comments. $\endgroup$– Ethan BolkerJul 10, 2019 at 12:22
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1$\begingroup$ you can take a look at this: math.stackexchange.com/questions/3065170/… is asked by @MaximilianJanisch $\endgroup$– KumarJul 10, 2019 at 12:23
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1 Answer
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We have $$ f(x^4) = f(g(f(x)) = (f(x))^3 $$ Let $F(y) := \log f(e^y)$. We have $$ F(4y) = \log f(e^{4y}) = \log \big((f(e^y))^3\big) = 3\log f(e^y) = 3F(y) $$ You can easily construct such functions. You can take any function $F : (-4,-1]\cup[1,4) \rightarrow \mathbb R$ and then define $F(4^k y) = 3^k F(y)$ for $k\in\mathbb Z$, which will define $F: (-\infty,0)\cup(0,\infty)\rightarrow \mathbb R$, and finally put $F(0)=0$.
You can take then $$ f(x) = \exp F(\log x)$$ which will define $f: (0,\infty) \rightarrow \mathbb (0,\infty)$. If this function turns out to be invertible (which happens if $F:\mathbb R\rightarrow\mathbb R$ is invertible), then you can define $g(x) = f^{-1}(x^3)$
Example: $$ F(y) = \left\{\begin{array}{ll} y^{\log_4 3} & \text{for }y>0 \\ 0 & \text{for }y=0 \\ -(-y)^{\log_4 3} &\text{for }y<0 \end{array}\right.$$ $$ f(x) = \left\{\begin{array}{ll} \exp\big((\log x)^{\log_4 3}\big) & \text{for }x>1 \\ 1 & \text{for }x=1 \\ \exp\big(-(-\log x)^{\log_4 3}\big) &\text{for }0<x<1 \end{array}\right.$$ $$ g(x) = \left\{\begin{array}{ll} \exp\big((3\log x)^{\log_3 4}\big) & \text{for }x>1 \\ 1 & \text{for }x=1 \\ \exp\big(-(-3\log x)^{\log_3 4}\big) &\text{for }0<x<1 \end{array}\right.$$
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$\begingroup$ +1 This is a nice construction; it is impossible to extend this to $\Bbb R$ though $\endgroup$ Jul 10, 2019 at 12:46