Calculus without limits

Exponential Function

We have seen that $f'(x)=Df(x) = (f(x+h)-f(x))/h$ satisfies $D[x]^n = n [x]^{n-1}$.
We will often leave the constant $h$ out of the notation and use terminology like $f'(x) = Df(x)$ for the “derivative”. It makes sense not to simplify $[x]^n$ to $x^n$ since the algebra structure is different.

Define the exponential function as
$exp(x) = \sum_{k=0}^{\infty} [x]^k/k!$. It solves the equation $Df=f$. Because each of the approximating polynomials $exp_n(x) = \sum_{k=0}^{n} [x]^k/k!$ is monotone and positive also $exp(x)$ is monotone and positive for all $x$. The fixed point equation $Df=f$ reads $f(x+h) = f(x) + h f(x) = (1+h) f(x)$ so that for $h=1/n$ we have $f(x+1) = f(x+n h) = (1+h)^n f(x) = e_n f(x)$
where $e_n \to e$. Because $n \to e_n$ is monotone, we see that the exponential function $\exp(x)$ depends in a monotone manner on h and that for $h \to 0$ the graphs of $\exp(x)$ converge to the graph of $\exp(x)$ as $h \to 0$.

Since the just defined exponential function is monotone, it can be inverted on the positive real axes. Its inverse is called $\log(x)$. We can also define trigonometric functions by separating real and imaginary part of $\exp(i x) = \cos(x) + i \sin(x)$. Since $D\exp=\exp$, these functions satisfy $D\cos(x) = - \sin(x)$ and $D\sin(x) = \cos(x)$ and are so both solutions to $D^2 f = -f$.