# The Unapologetic Mathematician

## The Exponential Differential Equation

So we long ago defined the exponential function $\exp$ to be the inverse of the logarithm, and we showed that it satisfied the exponential property. Now we’ve got another definition, using a power series, which is its Taylor series at ${0}$. And we’ve shown that this definition also satisfies the exponential property.

But what really makes the exponential function what it is? It’s the fact that the larger the function’s value gets, the faster it grows. That is, the exponential function satisfies the equation $f(x)=f'(x)$. We already knew this about $\exp$, but there we ultimately had to use the fact that we defined the logarithm $\ln$ to have a specified derivative. Here we use this property itself as a definition.

This is our first “differential equation”, which relates a function to its derivative(s). And because differentiation works so nicely for power series, we can use them to solve differential equations.

So let’s take our equation as a case in point. First off, any function $f$ that satisfies this equation must by definition be differentiable. And then, since it’s equal to its own derivative, this derivative must itself be differentiable, and so on. So at the very least our function must be infinitely differentiable. Let’s go one step further and just assume that it’s analytic at ${0}$. Since it’s analytic, we can expand it as a power series.

So we have some function defined by a power series around ${0}$:

$\displaystyle f(x)=\sum\limits_{k=0}^\infty a_kx^k$

We can easily take the derivative

$\displaystyle f'(x)=\sum\limits_{k=0}^\infty (k+1)a_{k+1}x^k$

Setting these two power series equal, we find that $a_0=1a_1$, $a_1=2a_2$, $a_2=3a_3$, and so on. In general:

$\displaystyle a_k=\frac{1}{k}a_{k-1}=\frac{1}{k}\frac{1}{k-1}a_{k-2}=...=\frac{1}{k}\frac{1}{k-1}...\frac{1}{3}\frac{1}{2}\frac{1}{1}a_0=\frac{a_0}{k!}$

And we have no restriction on $a_0$. Thus we come up with our series solution

$\displaystyle f(x)=\sum\limits_{k=0}^\infty\frac{a_0}{k!}x^k=a_0\sum\limits_{k=0}^\infty\frac{x^k}{k!}$

which is just $a_0=f(0)$ times the series definition of our exponential function $\exp$! If we set the initial value $f(0)=a_0=1$, then the unique solution to our equation is the function

$\displaystyle\exp(x)=\sum\limits_{k=0}^\infty\frac{x^k}{k!}$

which is our new definition of the exponential function. The differential equation motivates the series, and the series gives us everything else we need.

October 10, 2008 -

## 3 Comments »

1. [...] Homogeneous Linear Equations with Constant Coefficients Now that we solved one differential equation, let’s try a wider class: “first-degree homogeneous linear equations with constant [...]

Pingback by First-Degree Homogeneous Linear Equations with Constant Coefficients « The Unapologetic Mathematician | October 10, 2008 | Reply

2. [...] we can, and we’ll use the same techniques as we did before. We again find that any solution must be infinitely differentiable, and so we will assume that [...]

Pingback by Sine and Cosine « The Unapologetic Mathematician | October 13, 2008 | Reply

3. Now, I try to verify your statement …
“That is, the exponential function satisfies the equation f(x)=f’(x). We already knew this about exp, but there we ultimately had to use the fact that we defined the logarithm ln to have a specified derivative” to prove that:

d(e^x)/dx = e^x, (1)

Let differentiate the both sides of eq.(1) respect to x, and suppose d(e^x)/dx = u, hence the result can be written as following

du/dx = u, (2)

Next, separating variables gives

du/u = dx, (3)

and integrating the both sides of eq.(3) gives ln(u) = x. Here I put the integrating constant c=0, because at x=1, ln(1)=0. From ln(u) = x , I find u = e^x. After u is written into initial form, then I can prove that d(e^x)/dx = e^x.

Okay, your statement is right.
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Comment by rohedi | November 30, 2008 | Reply