Exponential Function

Defining Differential Equation

The solution to \frac{dx}{dt} = v with v=x and x(0)=1 , that is the solution x(t) to the IVP

  • \dot x(t) = x(t)\quad\mbox{for all } t,\quad x(0)=1,

is the exponential function

  • \exp(t) =\int_0^t\exp(s) ds,\quad \exp(0)=1.

The exponential functions \exp(x) and \exp(-x) for x\ge 0.

Computing The Exponential Function

Updating dx=xdt gives

  • x^{n+1} = x^n + x^ndt^n=(1+dt)x^n,\quad \mbox{for} n=0,1,2,...,

and after summation, assuming x^0=1,

  • x^{n} = (1+dt)^n.

With t=ndt, we thus have

  • \exp(t)\approx (1+\frac{t}{n})^n .

Compound Capital

The above formula states that \exp(t) is the compund capital after n years of interest at a yearly rate of \frac{t}{n}=dt  with a starting capital of 1.

Varying the Time Step

We shall see below that as n increases the approximation \exp(t)\approx (1+\frac{t}{n})^n improves and can be made as small as we like. Decreasing the time step \frac{t}{n} in the formula

  • \exp(t)\approx (1+\frac{t}{n})^n

by increasing n, we thus obtain for t=1 and n=1,2,3,4,5,7,10,20,100,1000,10000,... the following values for \exp(1)\approx (1+\frac{1}{n})^n:

  • 2
  • 2.25
  • 2.37
  • 2.4414
  • 2.4883
  • 2.5216
  • 2.5465
  • 2.5937
  • 2.6533
  • 2.7048
  • 2.7169
  • 2.7181

Increasing n we can this way compute any number of decimals of the number e\equiv\exp(1),
or more generally of \exp(t) for any t>0.

Properties of the Exponential Function

Properties of the exponential function \exp(t) can be derived from the defining differential equation
\dot u(t)=u(t). For example, the basic property of the exponential function

  • \exp(t+s)=\exp(t)\exp(s)

follows by noting that u(t)=\exp(t+s) viewed as a a function of t, satisfies

  • \dot u(t)=u(t)\quad\mbox{for }t>0, u(0)=\exp(s),

which means that u(t)=\exp(s)\exp(t)=\exp(t)\exp(s), which proves (\ref{expst}).

The exponential \exp(-t) is the solution of the differential equation \dot u=-u with u(0)=1.

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