Problem Statement
(a) Find the first three non-zero terms in the Maclaurin series expansion of $(1 - 2x)^{-1/2}$ in ascending powers of $x$, stating the validity range. [4 Marks]
(b) By substituting $x = \frac{1}{50}$, estimate the value of $\frac{1}{\sqrt{0.96}}$ as a fraction in simplest form. [3 Marks]
Verified Solution & Marking Scheme
Part (a): Apply Generalised Binomial Theorem
$(1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \dots$
Here $u = -2x$ and $n = -\frac{1}{2}$:
$(1 - 2x)^{-1/2} = 1 + \left(-\frac{1}{2}\right)(-2x) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}(-2x)^2 + \dots$
$= 1 + x + \frac{\frac{3}{4}}{2}(4x^2) = 1 + x + \frac{3}{2}x^2 + \dots$
State Domain of Convergence
The expansion is valid for $|-2x| < 1 \iff |x| < \frac{1}{2}$ or $-\frac{1}{2} < x < \frac{1}{2}$.
Part (b): Substitute x = 1/50
For $x = \frac{1}{50} = 0.02$:
$1 - 2x = 1 - \frac{2}{50} = 1 - 0.04 = 0.96$
Hence $(1 - 2x)^{-1/2} = (0.96)^{-1/2} = \frac{1}{\sqrt{0.96}}$.
Using the first three terms:
$\approx 1 + \frac{1}{50} + \frac{3}{2}\left(\frac{1}{50}\right)^2 = 1 + \frac{1}{50} + \frac{3}{2(2500)} = 1 + \frac{1}{50} + \frac{3}{5000}$
$= \frac{5000 + 100 + 3}{5000} = \frac{5103}{5000} = 1.0206$