Edexcel IAL October 2022 P4 (WMA14/01) Q4 Binomial Expansion & Approximation
\[g(x) = \frac{1}{\sqrt{4 – x^2}}\]
(a) Find, in ascending powers of \(x\), the first four non-zero terms of the binomial expansion of \(g(x)\). Give each coefficient in simplest form.
(5)
(b) State the range of values of \(x\) for which this expansion is valid.
(1)
(c) Use the expansion from part (a) to find a fully simplified rational approximation for \(\sqrt{3}\). Show your working and make your method clear.
(2)
Solution to Part a: Binomial Expansion of \(g(x)\) = \(\frac{1}{\sqrt{4-x^2}}\) and find the first four non-zero terms.
Key Concepts Used
- Algebraic Preparation: Factor out 4.
- General Binomial Expansion for \(n = -1/2\).
- Substitution: \(X = -\frac{1}{4}x^2\).
Step-by-Step Solution
- Prepare the expression
\[g(x) = (4 – x^2)^{-\frac{1}{2}} = \left[4\left(1 – \frac{1}{4}x^2\right)\right]^{-\frac{1}{2}} = 4^{\frac{-1}{2}}\left(1 – \frac{1}{4}x^2\right)^{-\frac{1}{2}} = \frac{1}{2}\left(1 + \left(-\frac{1}{4}x^2\right)\right)^{-\frac{1}{2}}\]
- Identify \(n\) and \(X\)
\(n = -\frac{1}{2}\) and \(X = -\frac{1}{4}x^2\).
- Apply the Binomial Formula
Term in \(x^0\):
\[x^0 = 1\]
Term in \(x^2\):
\[nX = \left(-\frac{1}{2}\right)\left(-\frac{1}{4}x^2\right) = \frac{1}{8}x^2\]
Term in \(x^4\):
\[\frac{n(n-1)}{2!}X^2 = \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2} \left(-\frac{1}{4}x^2\right)^2 = \frac{3}{8}\left(\frac{1}{16}x^4\right) = \frac{3}{128}x^4\]
Term in \(x^6\):
\[\frac{n(n-1)(n-2)}{3!} X^3 = \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{6} \left(-\frac{1}{4}x^2\right)^3 = \frac{5}{16} \left( \frac{1}{64} x^6 \right) = \frac{5}{1024} x^6\]
Multiply by \(\frac{1}{2}\)
\[\frac{1}{2} \left( 1 + \frac{1}{8} x^2 + \frac{3}{128} x^4 + \frac{5}{1024} x^6 + \cdots \right) = \frac{1}{2} + \frac{1}{16} x^2 + \frac{3}{256} x^4 + \frac{5}{2048} x^6 + \cdots\]
Final Answer
\[\frac{1}{2} + \frac{1}{16} x^2 + \frac{3}{256} x^4 + \frac{5}{2048} x^6 + \cdots\]
Solution to Part b: State the range of values of \(x\) for which this expansion is valid.
Key Concepts Used
- Range of Validity: The expansion is valid when \(|X| < 1\), where \(X = -\frac{1}{4} x^2\).
Step-by-Step Solution
- Set up the inequality
Valid for \(\left| -\frac{1}{4} x^2 \right| < 1\).
- Solve for \(x^2\)
\(\left|\frac{1}{4} x^2\right| < 1 \implies |x^2| < 4\)
- Solve for \(x\)
\(x^2 < 4 \implies -2 < x < 2\)
Final Answer
\(|x| < 2\)
Solution to Part c: Find a fully simplified rational approximation for \(\sqrt{3}\).
Key Concepts Used
- Approximation: Find the value of \(x^2\) that relates the function to \(\frac{1}{\sqrt{3}}\).
- Algebraic Manipulation: Solving for \(\sqrt{3}\).
Step-by-Step Solution
- Relate function to \(\sqrt{3}\)
We want the original function \(g(x) = \frac{1}{\sqrt{4-x^2}}\) to involve \(\sqrt{3}\).
Set the denominator equal to \(\sqrt{3}\)
\[\sqrt{4-x^2} = \sqrt{3} \Rightarrow 4-x^2 = 3 \Rightarrow x^2 = 1\] - Calculate the approximation for \(\frac{1}{\sqrt{3}}\)
Substitute \(x^2 = 1\) into the expansion from (a) (since only even powers of \(x\) appear)
\[\frac{1}{\sqrt{3}} \approx \frac{1}{2} + \frac{1}{16}(1) + \frac{3}{256}(1) + \frac{5}{2048}(1)\]
\[\frac{1}{\sqrt{3}} \approx \frac{1024 + 128 + 24 + 5}{2048} = \frac{1181}{2048}\] - Find the approximation for \(\sqrt{3}\)
Take the reciprocal of the fraction:
\[\sqrt{3} \approx \frac{2048}{1181}\]
Final Answer
\[\sqrt{3} \approx \frac{2048}{1181}\]