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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.

Definition 8.5.1.

The exponential function \(\exp\) is the unique solution to the differential equation \(f'=f\) which has \(f(0)=1\text{.}\)

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.

Find a solution to the differential equation \(f''=f+f'\) which satisfies \(f(0)=1, f'(0)=2\text{.}\)

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 \(\cos x\text{.}\)

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 \(\cosh x\text{.}\)