Definition 8.5.1.
The exponential function \(\exp\) is the unique solution to the differential equation \(f'=f\) which has \(f(0)=1\text{.}\)
Section 8.5 The Transcendental Functions
Theorem 8.4.8 enables a use of power series that ends up being pretty handy: solving differential equations. We’ll use this approach to define the transcendental functions we all know and love: the exponential, sine, and cosine functions.
This definition is nice, but it doesn’t give too much to work with. We’d like to get a better handle, so acting on faith (or, in fancier parlance, making an ansatz), we’ll just go ahead and assume that \(\displaystyle\exp(x)=\sum_{k=0}^\infty a_k x^k\) can be expressed as a power series centered at \(x=0\text{.}\) Then our job is to figure out what the \(a_k\) are.
By Theorem 8.4.8, \(\displaystyle \frac{d}{dx}\exp x=\sum_{k=0}^\infty (k+1) a_{k+1} x^k\text{,}\) so the defining differential equation becomes
\begin{equation*}
\displaystyle \sum_{k=0}^\infty (k+1) a_{k+1} x^k=\sum_{k=0}^\infty a_k x^k
\end{equation*}
which can only happen if \(a_k=(k+1)a_{k+1}\text{,}\) or in other words, if \(a_{k+1}=\frac{1}{k+1}a_k\text{.}\) We further know that \(\exp(0)=1\text{,}\) so \(a_0=1\text{.}\) Thus we obtain:
\begin{align*}
a_0&=1\\
a_1&=\frac{1}{1}a_0=1\\
a_2&=\frac{1}{2}a_1=\frac{1}{2}\\
a_3&=\frac{1}{3}a_2=\frac{1}{3\cdot 2}
\end{align*}
and so on. Thus, if \(\exp\) is a power series, that power series can only be
\begin{equation*}
\exp(x)=\sum_{k=0}^\infty \frac{1}{k!}x^k\ \ \ .
\end{equation*}
We should then check that the claimed series actually does converge.
Checkpoint 8.5.2.
Find a solution to the differential equation \(f''=\frac{1}{2}f-2f'\) which satisfies \(f(0)=1, f'(0)=2\text{.}\)
Checkpoint 8.5.3.
If we define \(\sin x\) and \(\cos x\) to be functions which satisfy:
\begin{align*}
\frac{d}{dx}\sin x&=\cos x\\
\frac{d}{dx}\cos x&= -\sin x\\
\sin(0)&=0\\
\cos(0)&=1
\end{align*}
find a power series representation, centered at \(x=0\text{,}\) for \(\sin x\) and \(\cos x\text{.}\)
Checkpoint 8.5.4.
If we define \(\sinh x\) and \(\cosh x\) to be functions which satisfy:
\begin{align*}
\frac{d}{dx}\sinh x&=\cosh x\\
\frac{d}{dx}\cosh x&= \sinh x\\
\sinh(0)&=0\\
\cosh(0)&=1
\end{align*}
find a power series representation, centered at \(x=0\text{,}\) for \(\sinh x\) and \(\cosh x\text{.}\)
