1994 Paper 3 Q5

Year: 1994
Paper: 3
Question Number: 5

Course: UFM Pure
Section: Taylor series

Difficulty: 1700.0 Banger: 1486.8

Taylor series Maclaurin series inverse trigonometric functions arcsin Leibniz rule differential equation logarithmic function inequality coefficient comparison proof by induction

Problem

The function \(\mathrm{f}\) is given by \(\mathrm{f}(x)=\sin^{-1}x\) for \(-1 < x < 1.\) Prove that \[ (1-x^{2})\mathrm{f}''(x)-x\mathrm{f}'(x)=0. \] Prove also that \[ (1-x^{2})\mathrm{f}^{(n+2)}(x)-(2n+1)x\mathrm{f}^{(n+1)}(x)-n^{2}\mathrm{f}^{(n)}(x)=0, \] for all \(n>0\), where \(\mathrm{f}^{(n)}\) denotes the \(n\)th derivative of \(\mathrm{f}\). Hence express \(\mathrm{f}(x)\) as a Maclaurin series. The function \(\mathrm{g}\) is given by \[ \mathrm{g}(x)=\ln\sqrt{\frac{1+x}{1-x}}, \] for \(-1 < x < 1.\) Write down a power series expression for \(\mathrm{g}(x),\) and show that the coefficient of \(x^{2n+1}\) is greater than that in the expansion of \(\mathrm{f},\) for each \(n > 0\).

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Problem source

The function $\mathrm{f}$ is given by $\mathrm{f}(x)=\sin^{-1}x$
for $-1 < x < 1.$ Prove that 
\[
(1-x^{2})\mathrm{f}''(x)-x\mathrm{f}'(x)=0.
\]
Prove also that 
\[
(1-x^{2})\mathrm{f}^{(n+2)}(x)-(2n+1)x\mathrm{f}^{(n+1)}(x)-n^{2}\mathrm{f}^{(n)}(x)=0,
\]
for all $n>0$, where $\mathrm{f}^{(n)}$ denotes the $n$th derivative
of $\mathrm{f}$. Hence express $\mathrm{f}(x)$ as a Maclaurin series. 
The function $\mathrm{g}$ is given by 
\[
\mathrm{g}(x)=\ln\sqrt{\frac{1+x}{1-x}},
\]
for $-1 < x < 1.$ Write down a power series expression for $\mathrm{g}(x),$
and show that the coefficient of $x^{2n+1}$ is greater than that
in the expansion of $\mathrm{f},$ for each $n > 0$.

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