I was looking at algebraic rules that certain functions follow. For example, exponential functions have the rule:
$e^a e^b = e^{a+b}$
And I was wondering if this is the only function that obeys this rule. So I set up the equation:
$z(x+y)=z(x)z(y)$
and tried taking partial derivatives on both sides. First with respect to x:
$\frac{\partial}{\partial x}f(x+y)=[\frac{d}{du}f(u)][\frac{\partial}{\partial x}(x+y)]=\frac{df}{du}=\frac{\partial}{\partial x}[z(x)z(y)]=z(y)\frac{dz}{dx}$
(Where $u=x+y$.) I did the same with y:
$\frac{\partial}{\partial y}f(x+y)=[\frac{d}{du}f(u)][\frac{\partial}{\partial y}(x+y)]=\frac{df}{du}=\frac{\partial}{\partial y}[z(x)z(y)]=z(x)\frac{dz}{dy}$
This would imply:
$z(x)\frac{dz}{dy}=z(y)\frac{dz}{dx}$
Or:
$\frac{1}{z(x)}\frac{dz}{dx}=\frac{1}{z(y)}\frac{dz}{dy}$
Now, if we take a partial derivative with respect to either x or y, we see that:
$\frac{\partial}{\partial x} \frac{1}{z(x)}\frac{dz}{dx}=\frac{\partial}{\partial x} \frac{1}{z(y)}\frac{dz}{dy}=\frac{\partial}{\partial y} \frac{1}{z(x)}\frac{dz}{dx}=\frac{\partial}{\partial y} \frac{1}{z(y)}\frac{dz}{dy} = 0$.
And hence:
$\frac{1}{z(x)}\frac{dz}{dx}=k$
$\frac{1}{z(y)}\frac{dz}{dy}=k$
We can now solve both of these differential equations and get:
$z(x)=Ae^{kx}$
$z(y)=Be^{ky}$
But we want $z(x)=z(y)$ if $x=y$. So therefore $A=B$. What's more is that we want $z(x+y)=z(x)$ when $y=0$ So, sice:
$z(x)z(y)=A^2e^{k(x+y)}=Ae^{k(x+y)}$,
$A=1$ ($A=0$ would be boring). So all of this seems to imply, that:
$z(x)=e^{kx}$
Is the only solution to the rule:
$z(x+y)=z(x)z(y)$.
I also tried this with logarithms:
$z(xy)=z(x)+z(y)$
Again, taking some partial derivatives, we end up with:
$\frac{\partial}{\partial x}f(xy)=[\frac{d}{du}f(u)][\frac{\partial}{\partial x}(xy)]=y\frac{df}{du}=\frac{\partial}{\partial x}[z(x)+z(y)]=\frac{dz}{dx}$
$\frac{\partial}{\partial y}f(xy)=[\frac{d}{du}f(u)][\frac{\partial}{\partial y}(xy)]=x\frac{df}{du}=\frac{\partial}{\partial y}[z(x)+z(y)]=\frac{dz}{dy}$
Rearranging some things, we get:
$\frac{df}{du}=\frac{1}{x}\frac{dz}{dy}=\frac{1}{y}\frac{dz}{dx}$
$x\frac{dz}{dx}=y\frac{dz}{dy}$
Using the same argument from before, by taking partial derivatives with respect to either x or y on both sides we get:
$y\frac{dz}{dy}=k$
$x\frac{dz}{dx}=k$
Solving these, we get:
$z(x)=klnx + A$
$z(y)=klny + B$
After similar arguments from before, we get that:
$z(x)=klnx$
is the only function that obeys the rule:
$z(xy)=z(x)+z(y)$.
Since I was curious, I wanted to investigate other rules. So I set up the equation:
$[z(x)]^{z(y)}=z(x+y)$
Now looking at this, we can see that it doesn't seem like any function obeys this rule. Since $z(x+y)=z(y+x)$, this would imply:
$[z(x)]^{z(y)}=[z(y)]^{z(x)}$
And it doesn't seem like any function could really obey this equation, because, for example, $2^3 \ne 3^2$. But I still wanted to see what the math would lead me to. So, again I took some derivatives and blah, blah, blah. I ended up with the equation:
$\frac{1}{z(y)}\frac{dz}{dy}=\frac{1}{z(x)ln(z(x))}\frac{dz}{dx}=k$
Which would imply that the functions $z(x)$ and $z(y)$ cannot have the same solution. Therefore, we can already say that there is no function that obeys the equation:
$[z(x)]^{z(y)}=z(x+y)$,
if x and y are to be independent variables.
So my question is "Is this a way you can figure out what function obeys a specific algebraic equation?". Because if it is a valid way of doing this and if I really didn't make any dumb mistakes then obviously all mathematicians know about this. I'm still a high-schooler, so I probably wouldn't be adding anything new to the table, since I'm sure all mathematicians are way smarter than me. But I'm asking about this because I kinda just thought of this on my own and I've never had anyone teach me about this before either. And if this whole process is valid and if someone could tell me it is, I'd be really happy.
