2018 MJC Promo Q8
(a)
(i) Given that $y=\cos \left( {{\text{e}}^{2x}}-1 \right)$, show that $\frac{{{\text{d}}^{2}}y}{\text{d}{{x}^{2}}}-2\frac{\text{d}y}{\text{d}x}+4y{{\text{e}}^{4x}}=0$. By further differentiation of this result, find the Maclaurin series of the function $\cos \left( {{\text{e}}^{2x}}-1 \right)$ up to and including the term in ${{x}^{3}}$.
[7]
(ii) Verify the correctness of your answer in part (ai) by using the standard series for ${{\text{e}}^{x}}$ and $\cos x$.
[2]
(b)
(i) Explain why it is not possible to obtain a Maclaurin series for $\ln x$ .
[1]
(ii) A Taylor series is an expansion of a real function $\text{f}\left( x \right)$ about a point $x=a$ and it is defined by
$\text{f}\left( x \right)=\text{f}\left( a \right)+\text{f}\,\text{ }\!\!’\!\!\text{ }\left( a \right)\left( x-a \right)+\frac{\text{f}\,”\left( a \right)}{2!}{{\left( x-a \right)}^{2}}+…+\frac{{{\text{f}}^{\left( n \right)}}\left( a \right)}{n!}{{\left( x-a \right)}^{n}}+…$
where ${{\text{f}}^{\left( n \right)}}\left( a \right)$ is the value of the $n$th derivative of $\text{f}\left( x \right)$ when $x=a$.
Given that $\text{f}\left( x \right)=\ln x$, find $\text{f}\left( 1 \right)$, $\text{f}\,\text{ }\!\!’\!\!\text{ }\left( 1 \right)$ and $\text{f}\,\text{ }\!\!’\!\!\text{ }\!\!’\!\!\text{ }\left( 1 \right)$. Hence write down the Taylor series for $\text{f}\left( x \right)$ 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]
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