We now construct the exponential function \exp(t)=e^t as the solution to the differential equation or initial value problem

  • \frac{dx}{dt} =x for  t>0
  • x(0)=1

which we compute by time stepping  x = x +x*dt (expressing dx =x*dt).

Note that the initial value problem \frac{dx}{dt} =x for  t>0 with x(0)=C, where C is any number, is solved by C*\exp(t), as an expression of the linearity of the differential equation \frac{dx}{dt}=x allowing multiplication by any number, because \frac{dC*x}{dt}=C*\frac{dx}{dt}, or computationally C*x = C*x + C*x*dt if x=x+x*dt.

We can extend \exp(t) to t < 0 by time stepping backwards in time.

The exponential function is the unique function which is equal to the derivative of itself as a form of complete self satisfaction.

To do:

  • Play with the code changing e g into dx =alpha*x*dt with alpha a given number.
  • Show that by construction \exp(t+s) =\exp(t)*\exp(s) by using the linearity of the differential equation (not the solution) discussed above. Hint: Show that x(t)=\exp(t+s) satisfies \frac{dx}{dt}=x for t > 0 with x(0)=\exp(s), and so is equal to the multiple \exp(s)*\exp(t) of \exp(t).