# 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 $u(0)=1$.