I am an Vietnamese grade 12 student. We were studying about exponential functions until I read the part that writes:
For the function $f(x) = a^{x}$, we always assume a is an positive real number that is not 1. In other words, $0< a \neq 1$.
What would happen if we set an $a$ that violates those properties? $a = 0$ and $a = 1$ yields pretty boring results (we would have $$f(x) = \begin{cases} 0 & x\subset \mathbb{R} \setminus \left \{ 0 \right \}\\undefined & x = 0\end{cases}$$ and $f(x) = 1$). AFter all, for complex numbers we defined the factorial and Fibonacci numbers, which are definitely much more difficult to define.
What if we let $a$ be some negative number instead? Due to the rule $a^{x} b^{x} = (ab)^{x}$ (if we assume it holds anyway), we only need to define exponentiation for $-1$, since if we somehow made a function that is $(-1)^{x}$, we can define $a^{b}$ for all $a$ and $b$ in $\mathbb{R}$ (except $a = 0$ and $b = 0$) by writing $$a^{b} =\left | a \right |^{b} sgn(a)^{b} $$
Now onto the actual topic. Using GeoGebra, I plotted a few points in space. Due to the nature of this function, it is impossible to plot a consistent line. Points on GeoGebra.
I noticed that the function $f(x) = (-1)^{x}$ also has the following properties:
- It is periodic with period 2 since : $(-1)^{x + 2} = (-1)^{x} 1^{2} = (-1)^{x}$
- It takes values at $-1$ for odd integer inputs and $1$ for even integer inputs. These are also its maximum and minimum values over $\mathbb{N}$
The generalization I suggest will have these properties:
- Continuous over $\mathbb{R}$
- Has maximum value $1$ at even integer inputs and minimum value at $-1$ at odd integer inputs.
Therefore I suggest one such generalization is $f(x) = (-1)^{x} =\cos (\pi x)$. This would also allow negative exponentiation to expand indefinitely to the complex numbers, since $\cos (\pi x)$ is well-defined over $\mathbb{C}$. Is there a hole in my argument? Please point it out to me, thanks for reading!
One of the problems you immediately encounter will be losing property $a^{x + y} = a^x \cdot a^y$. If you take $a = -1$, $x = y = 1/2$, then you need to have $(a^{1/2})^2 = (-1)^1 = -1$, but even with your definition equation $t^2 = -1$ has no real solutions. Given that linking multiplication and addition is one of the most important (if not the most important) property of exponentiation, I doubt such generalisation can be useful. But your formula satisfies properties you required, good work!
The standard way to define power of any (real or complex) number is by saying $a^z = \exp(z \cdot \operatorname{Log} a)$, where $\operatorname{Log}$ is complex logarithm, and $\exp$ is complex exponentiation (defined by Euler formula or by series). This allows to save the identity $a^{x + y} = a^x \cdot a^y$, but violates, for example, $(a^x)^y = a^{xy}$ (and there is actually no good way to save it).