Compute approximately: 1° (1.004)/(0.996); 2° (0.998)^5

Compute approximately: 1° $\dfrac{1.004}{0.996}$; 2° $\bigl(0.998\bigr)^{5}$; 3° $\bigl(1.002\bigr)^{15}$.

Step 1: The rule

\[ \bigl(1+\alpha\bigr)^{n} \approx 1+n\alpha \qquad (|\alpha|\ll1). \tag{$*$} \]

Step 2: Part 1°: a quotient of two perturbations

Write both numbers as deviations from $1$ and apply $(*)$ with $n = -1$ to the denominator: \[\frac{1.004}{0.996} = \bigl(1+0.004\bigr)\bigl(1-0.004\bigr)^{-1} \approx \bigl(1+0.004\bigr)\bigl(1+0.004\bigr).\]

Multiplying out and dropping the term in $\bigl(0.004\bigr)^{2}$: $\frac{1.004}{0.996} \approx 1+2\times0.004 = 1.008$.

Step 3: Part 2°: $\bigl(0.998\bigr)^{5}$

\[ \bigl(1-0.002\bigr)^{5} \approx 1+5\times\bigl(-0.002\bigr) = 1-0.010 = 0.990. \]

Step 4: Part 3°: $\bigl(1.002\bigr)^{15}$

\[ \bigl(1+0.002\bigr)^{15} \approx 1+15\times0.002 = 1.030. \]

Step 5: How accurate are they?

The discarded second-order term is $\dbinom{n}{2}\alpha^{2} = \dfrac{n(n-1)}{2}\alpha^{2}$, which grows quadratically with the exponent — it is not simply $\alpha^{2}$. The three cases therefore behave quite differently:

\[\begin{array}{c|c|c|c|c} \text{case} & \text{estimate} & \text{exact} & \text{error} & \text{predicted } \tbinom{n}{2}\alpha^{2} \\ \hline 1.004/0.996 & 1.008 & 1.0080321 & 3.2\cdot10^{-5} & 2\alpha^{2} = 3.2\cdot10^{-5} \\ \bigl(0.998\bigr)^{5} & 0.990 & 0.9900399 & 4.0\cdot10^{-5} & 10\alpha^{2} = 4.0\cdot10^{-5} \\ \bigl(1.002\bigr)^{15} & 1.030 & 1.0304237 & 4.2\cdot10^{-4} & 105\alpha^{2} = 4.2\cdot10^{-4} \end{array}\]

\[\text{The predicted and actual errors agree in every row} \;\;\checkmark\]

\[\text{Cases 1}^{\circ}\text{ and 2}^{\circ}\text{ are good to four decimals; case 3}^{\circ}\text{ only to three} \;\;\checkmark\]