What does $e^b$ means?

Question

What does it mean $a^b$ where $a$ and $b$ are real numbers not integers, not rationnals (I do not know the name of this set). Real numbers means $\mathbb{R}$ but $1\in \mathbb{R}$

So, what does it mean $a^b$ where $a$ and $b$ $\in\mathbb{R}\setminus\mathbb{Q}\cup\mathbb{N}$?

For example, how to interpret $e^e$?

I can suppose that $b$ is integer. How to interpret $a^b$ then? ($e^2$?)

I was reading about matrix exponentiation and I wanted to understand what does it mean exponential of a matrix $A$, i.e., $\exp A$. Suddenly, I saw that $\exp A=\sum\limits_{n=1}^{\infty}\dfrac{1}{n!}A^n$. So I wanted to explain what does $e^A$ means?

Any helps? Thank you very much for your time, help.

Answer 1

I'll answer your first question, as the second one requires something extra.

Originally, we learn that exponentiation is repeated multiplication. And this is true indeed, when the exponent is a natural number: $2^3 = 2 \times 2 \times 2$. Multiplication has an inverse, so we should also suspect that we can extend exponentiation to negative integers to "invert" the multiplications: $2^{-3} = \frac12 \times \frac12 \times \frac12$, $2^3 \times 2^{-3} = 1$. From this, we can derive lots of rules about multiplying exponents of common bases etc. I'm sure you know these.

Now, what if we want to invert with another exponentiation operation, not a multiplication? Well, using the rules we defined for integers, we can extend to rational exponents of the form $1/q$ quite easily. And more generally, to all rationals $p/q$. That is, $a^{p/q}$ is just $a^{1/q}$ multiplied $p$ times. And the rules we defined for integers are completely consistent with rationals.

But... irrationals are tricky. We can't decompose them into "sum/products of exponents" like we do with rationals. But, we can recognize one important fact: for any irrational number, there is a rational number above and below it with arbitrary precision. In other words, no matter how small a region you draw around an irrational number, you can find a rational on either side of that irrational.

Call our irrational number $\phi$. Let's consider the set of all nearby irrationals below $\phi$ and call it $B$, and let's also consider the set of all nearby irrationals above $\phi$, and call it $A$.

Then, we can take $x^a$ for every $a \in A$ and $x^b$ for every $b \in b$. Since $a,b$ are rational, we know how to work with them. And finally, note that $\phi$ is the least upper bound of $B$ and the greatest lower bound of $B$.

Therefore, $x^\phi$ is defined as the number that is simultaneously the LUB/GLB of $x^A$ and the GLB/LUB of $x^B$. Because we define it in terms of rational exponents in this way, then all the rules for irrational exponents hold as well: $x^\phi x^\psi = x^{\phi + \psi}$, etc.

Answer 2

For a positive real number $t$ and an arbitrary real number $x$, $t^x$ means, by definition, $\exp(x\log(t))$, where $\exp$ is the exponential function. If $x$ is an integer, this reduces by basic properties of $\log$ and $\exp$ to the usual thing. If $t$ is not positive, then there is difficulty in pinning down what $t^x$ should be (e.g. sometimes it doesn't make any sense as a real number). One is then led to consider branches of the logarithm in the complex plane, etc..