The exponential growth function exp(x) is a fundamental object of
calculus, which satisfies the defining differential equation
dy/dx = y. Here I will describe how applying Euler's Method (which
is normally understood to be approximate) to this differential equation can be
used to derive all of the important properties of the exponential growth
function.
Euler's Method approximately solves the differential equation
dy/dx = f(x, y) subject to the initial condition
y(0) = a, by taking the approximation
y(x+h)-y(x) ~= h*f(x, y(x)), giving the iterative method
y((n+1)*h) = y(n*h) + h * f(n*h, y(n*h)). By letting
h shrink to zero, while choosing n to be within a
finite distance of x/h, this method should converge to an
exact
value y(x), and then the derivative
(y(x+h) - y(x))/h should converge to the function
f(x, y) almost by construction. This method is simple and
(when presented more carefully) is easy to understand, and can approximate a
large variety of differential equations, but crucially, the larger the second
derivative of y, the less accurate this method will be for large
x, taking smaller and smaller h to converge. Thinking
about the limit that this method converges to, we can overcome this detail, but
it is important to remember when understanding the method in finite cases.
Applying this method to the exponential growth equation dy/dx = y
we get the recurrence relation y((n+1)*h) = (1 + h) * y(n*h),
which can be solved recursively to give y(n*h) = y(0) * (1 + h)^n.
Focussing on the specific case y(0) = 1, we can define the
exponential growth function exp(x) to be the limit of
(1 + h)^n as h goes to zero. Depending on how we
relate n and h, we can develop different formulas for
this same function, which should be equal, as long as the difference between
n and x/h is bounded.
By fixing h, and letting n grow, we can interpret
(1 + h)^n as a geometric sequence, and can observe, for example,
that
exp(a + b) ~= (1 + h)^(m+n) = (1 + h)^m(1 + h)^n ~= exp(a)exp(b),
i.e. that exp(a + b) ~= exp(a)exp(b). As we know, this equation is
exact, i.e. the difference between these two expressions converges to zero,
and we can show this by choosing m and n more
carefully. If a and b are both rational numbers, then
we can let k be an arbitrary multiple of their lowest common
denominator, so that m = ak and n = bk are both whole
numbers, and then fix h as 1/k, so that
m = a/h and n = b/k hold simultaneously. This gives
us a family of approximations of exp(a + b) and
exp(a)exp(b) that are exactly equal, and as k gets
arbitrarily large, these approximations should converge to both values
simultaneously, which demonstrates that they are equal to each other.
In a similar manner, this geometric interpretation also lets us show that
exp(ab) = exp(a)^b, this time picking k to be an
arbitrary multiple of the denominator of a times the denominator
of b. Both sides of the equation will be approximately equal to
(1 + 1/k)^(abk), and thus must be exactly equal to each other
since this approximation can be made arbitrarily accurate. In particular this
lets us define Euler's number e = exp(1), so that
exp(x) = e^x.
Alternatively, we can fix n and let h = x/n, in order
to approximate exp(x) as the polynomial (1 + x/n)^n,
which will be a continuous function in x that we can easily
compute as well as analyse.