# Exponential

We now construct the exponential function $\exp(t)=e^t$ as the solution $x(t)$ 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 $\exp(t)$.