A calculus of the absurd

11.12 Introduction

Some functions can be written as an "infinite power series", which is a sum in the form

\begin{equation*} a + bx + cx^2 + dx^3 + ... \end{equation*}

How would we find the values of \(a, b, c, d, ...\) for a specific function?

Differentiation! Let’s take \(\sin (x)\):

\begin{align} \sin (x) &= c_1 + c_2 x + c_3 x^2 + c_4 x^3 + ... \\ \cos (x) &= c_2 + 2c_3 x + 3 c_4 x^2 + ... \tag {By differentiating both sides} \\ -\sin (x) &= 2c_3 + 3 \cdot 2 c_4 x + 4 \cdot 3 c_5 x^2 ... \tag {By differentiating both sides again} \\ -\cos (x) &= 3 \cdot 2 \cdot 1 c_4 + 4 \cdot 3 \cdot 2 c_5 x \tag {By differentiating both sides once more}\\ \sin (x) &= 4 \cdot 3 \cdot 2 \cdot 1 c_5 x \tag {Differentating again} \\ ... \end{align}

How do we find the value of a coefficient? Another algebra trick - plug in a specific value for \(x\), in this case \(0\). This is because every term in the power series, except the constant one, depends on \(x\) and thus is zero when \(x\) is zero. In order to find the value of the next constant, we can just differentiate, which brings all the powers down by one.

Therefore

\begin{align*} c_1 &= \sin (0) \\ c_2 &= \cos (0) \\ 2c_3 &= -\sin (0) \\ 3 \cdot 2 \cdot 1 c_4 &= -\cos (x) \\ 4 \cdot 3 \cdot 2 \cdot 1 c_5 &= \sin (x) \\ ... \end{align*}

This leads directly to the general case (i.e. the value of \(c_n\))

\begin{equation} c_n = \frac {\sin ^{(n)}(0)}{n!} \end{equation}

Where \(sin^{(n)}(0)\) means the value of the nth derivative at the point \(x=0\). Therefore, overall, we can write the power series of any function8484 Note that are some additional conditions - the function must be infinitely differentiable (we can keep differentiating forever) and each derivative must be defined at the point \(x=0\). as

\begin{equation} \label {maclaurin series in general} f(x) = \sum _{s=0}^{\infty } \left [\frac {f^{(s)}(0)}{s!}x^s\right ] \end{equation}

Maclaurin series of common functions

This is just a list, which is also given in the formula booklet.

11.12.1 Derivation of the sum of an infinite geometric series
  • Example 18 Prove the formula for a sum of an infinite geometric series using Maclaurin series.

We know8585 If not, see Section 3.4.2 that

\begin{equation} \frac {1}{1 - z} = 1 + z + z^2 + z^3 + ... \end{equation}

This can be derived by finding an expression for the \(n\)th derivative of \(\frac {1}{1-z}\).

Finding the first few derivatives,

\begin{equation} \frac {d}{dz} \left [ \frac {1}{1-z} \right ] = \frac {d}{dz} \left [(1-z)^{-1}\right ] \end{equation}

By the chain rule, this gives that the first derivative is equal to

\begin{equation} (-1) \cdot (-1) (1-z)^{-2} = (1-z)^{-2} \end{equation}

The second derivative is equal to

\begin{equation} 2 (1-z)^{-3} \end{equation}

The third is

\begin{equation} 1 \cdot 2 \cdot 3 (1-z)^{-4} \end{equation}

And so on (this can be proved by induction). Therefore, the Maclaurin expansion for the function is

\begin{align} \frac {1}{1-z} &= \sum _{k \geqq 0} \left [ \frac {\text {kth derivative at 0}}{k!} z^{k} \right ] \\ &= \sum _{k \geqq 0} \left [ \frac {k! (1-0)^{-k-1}}{k!} z^k \right ] \\ &= \sum _{k \geqq 0} \left [ z^k \right ] \\ &= 1 + z + z^2 + z^3 + ... \end{align}