2018 MJC Promo Q8

Given that \(y=\cos \left( {{\mathrm{e}}^{2x}}-1 \right)\), show that \(\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}-2\frac{\mathrm{d}y}{\mathrm{d}x}+4y{{\mathrm{e}}^{4x}}=0\). By further differentiation of this result, find the Maclaurin series of the function \(\cos \left( {{\mathrm{e}}^{2x}}-1 \right)\) up to and including the term in \({{x}^{3}}\).[7]

Verify the correctness of your answer in part (ai) by using the standard series for \({{\mathrm{e}}^{x}}\) and \(\cos x\).[2]

Explain why it is not possible to obtain a Maclaurin series for \(\ln x\) .[1]

A Taylor series is an expansion of a real function \(\mathrm{f}\left( x \right)\) about a point \(x=a\) and it is defined by
\[\mathrm{f}\left( x \right)=\mathrm{f}\left( a \right)+\mathrm{f}'\left( a \right)\left( x-a \right)+\frac{\mathrm{f}''\left( a \right)}{2!}{{\left( x-a \right)}^{2}}+...+\frac{{\mathrm{f}^{\left( n \right)}}\left( a \right)}{n!}{{\left( x-a \right)}^{n}}+... \]

where \({{\mathrm{f}}^{\left( n \right)}}\left( a \right)\) is the value of the \(n\)th derivative of \(\mathrm{f}\left( x \right)\) when \(x=a\).

Given that \(\mathrm{f}\left( x \right)=\ln x\), find \(\mathrm{f}\left( 1 \right)\), \(\mathrm{f}'\left( 1 \right)\) and \(\mathrm{f}''\mathrm{ }\left( 1 \right)\). Hence write down the Taylor series for \(\mathrm{f}\left( x \right)\) about the point \(x=1\), giving your answer in the form of \(\ln x=b+c\left( x-a \right)+d{{\left( x-a \right)}^{2}}+...\), where \(a\), \(b\), \(c\) and \(d\) are real constants.[3]