# 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 $$c_n$$)

$$c_n = \frac {\sin ^{(n)}(0)}{n!}$$

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

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

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

$$\frac {1}{1 - z} = 1 + z + z^2 + z^3 + ...$$

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

Finding the first few derivatives,

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

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

$$(-1) \cdot (-1) (1-z)^{-2} = (1-z)^{-2}$$

The second derivative is equal to

$$2 (1-z)^{-3}$$

The third is

$$1 \cdot 2 \cdot 3 (1-z)^{-4}$$

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}