Vrftsjtryk

I can name at least 4 different ways of representing $exp$ function:

1. Taylor series: For $x in mathbb{R}, exp(x) = sum_{k=0}^{infty} frac{x^k}{k!}$.


2. Differential equation: $f: mathbb{R} to mathbb{R}$ differentiable with $f'(x) = f(x)$ and $f(0)=1$.


3. Inverse function of $ln(x) = int_1^x frac{dt}{t}$ for $x>0$.


4. Exponent: The number $e$ (defined e.g. as $sum_{k=0}^{infty} frac{1}{k!}$) raised to the power $x$, for $x in mathbb{R}$
   

I managed to proved equivalence among the first $3$ but I am a bit puzzled by $4.$.

An easy way would be to look at $a^b = exp(b ln(a))$. But I am not sure that is meaningful.

Is there any other way of defining $a^b$ without involving $exp$ that would give a more meaningful answer? Or how would you approach proving that 4. is equivalent to 1-3?

You could define $exp :mathbb{R}to mathbb{R}$ to be a continuous function that satisfies
$$
expleft(frac{p}{q}right) = sqrt[q]{e^{p}}
$$
for all $p$, $qin mathbb{Z}$, then show that such a function exists and is unique.

If you're interested, here are are two other ways of defining $exp$:

1.$$
exp (x) = lim_{nto infty} left( 1 +frac{x}{n}right)^n
$$

2.A continuous function $exp : mathbb{R} to mathbb{R}$ that satisfies
$$
exp '(0) = 1
$$
$$
exp(x+y) = exp(x)exp{y}
$$
for all $x,yin mathbb{R}$. Of course you'll have to prove existance and uniqueness.

The only problem in defining $a^b$ is when $b$ is irrational. This can be handled as follows.

Let $b_n$ be a sequence of rationals tending to $b$ and we can define $a^b=lim_{ntoinfty} a^{b_n} $. This approach is slightly difficult and presented here.