Problem: Find the first few terms of the Maclaurin series of

g(x) = \frac{1}{5 − 3x} .

Method 1: Apply Taylor’s formula directly: Calculate g(0), g′(0), g′′(0), … and construct

g(x) ≈ g(0) + g′(0)x + \frac{1}{2}g′′(0)x^{2} + · · · .

(This is probably the worst way to do the problem — certainly it’s the most boring.)

Method 2: Treat this as a long division problem (see Stewart p. 662):

\frac{1}{5} + \frac{3}{25}x + ⋯

5 − 3x)\overline{1 + 0x + 0x^{2} + ⋯}

(\underline{1 − \frac{3}{5}x}) / (\frac{3}{5}x)

\underline{\frac{3}{5}x − \frac{9}{25}x^{2}}

\frac{9}{25}x^{2} ⋯

Method 3: Write \frac{1}{5−3x} as \frac{1}{5} \frac{1}{1− \frac{3}{5} x} . Now apply the geometric series

\frac{1}{1 − z} = 1 + z + z^{2} + z^{3} + ⋯

with z = \frac{3}{5}x.

Method 4 (the method marketed by our sponsor): We are seeking a solution for the equation

(5 − 3x)g(x) = 1.

Assume that

g(x) = a_{0} + a_{1}x + a_{2}x^{2} + · · · .

Then

\begin{aligned}1 & = (5 − 3x)(a_{0} + a_{1}x + a_{2}x^{2} + ⋯) \\ & = 5a_{0} + 5a_{1}x − 3x_{0}x + 5a_{2}x^{2} − 3a_{1}x^{2} + · · · . \end{aligned}

After we combine terms, the total coefficient of each power x^{j} must match the corresponding coefficient on the other side of the equation. This principle gives a sequence of recursion relations,

1 = 5a_{0} ,  (x^{0})

0 = 5a_{1} − 3a_{0} ,  (x^{1})
0 = 5a_{2} − 3a_{1} ,  (x^{2})

= ….

These can be solved in succession for the coefficients:

a_{0} = \frac{1}{5},
a_{1} = \frac{3}{5}a_{0} = \frac{3}{25},
a_{2} = \frac{3}{5}a_{1} = \frac{9}{125},

= ….

It’s easy to see that the general solution is

a_{j+1} = \frac{3^{j+1}}{5^{j+2}} .