Let \(f(x) = \frac{6 + 7x}{(2-x)(1+x^2)}\).
(i) Express \(f(x)\) in partial fractions.
(ii) Show that, when \(x\) is sufficiently small for \(x^4\) and higher powers to be neglected,
\(f(x) = 3 + 5x - \frac{1}{2}x^2 - \frac{15}{4}x^3\).
Solution
(i) Express \(f(x)\) in partial fractions:
Assume \(f(x) = \frac{A}{2-x} + \frac{Bx+C}{x^2+1}\).
Equating coefficients, solve for \(A, B,\) and \(C\):
\(A = 4, \ B = 4, \ C = 1\).
Thus, \(f(x) = \frac{4}{2-x} + \frac{4x+1}{x^2+1}\).
(ii) Show the expansion:
Use the binomial expansion for \((1-x)^{-1}\) and \((1+x^2)^{-1}\):
\((1-x)^{-1} = 1 + x + x^2 + x^3 + \ldots\)
\((1+x^2)^{-1} = 1 - x^2 + x^4 - \ldots\)
Multiply the expansions: \((4x+1)(1-x^2)^{-1}\) and simplify.
Combine terms to obtain: \(f(x) = 3 + 5x - \frac{1}{2}x^2 - \frac{15}{4}x^3\).
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