Taylor Series (Edexcel A Level Further Maths: Further Pure 1): Exam Questions

$y=\text{ln}\left({\text{e}}^{2x}\text{cos}3x\right)-\frac{1}{2}<x<\frac{1}{2}$

Show that

$\frac{\text{d}y}{\text{d}x}=2-3\text{tan}3x$

Determine $\frac{{\text{d}}^{4}y}{\text{d}{x}^4}$

Hence determine the first 3 non-zero terms in ascending powers of $x$ of the Maclaurin series expansion of $\text{ln}\left({\text{e}}^{2x}\text{cos}3x\right)$, giving each coefficient in simplest form.

Use the Maclaurin series expansion for $\text{ln}(1+x)$ to write down the first 4 non-zero terms in ascending powers of $x$ of the Maclaurin series expansion of $\text{ln}(1+kx)$, where $k$ is a constant.

Hence determine the value of $k$ for which

$\underset{x\to 0}{\text{lim}}\left(\frac{1}{x^2}\text{ln}\frac{{\text{e}}^{2x}\text{ cos}3x}{1+kx}\right)$

exists.

The Taylor series expansion of $\text{f}\left(x\right)$ about $x=a$ is given by

$\text{f}\left(x\right)=\text{f}\left(a\right)+(x-a)\text{f}\text{'}\left(a\right)+\frac{(x-a{)}^2}{2!}\text{f}\text{'}\text{'}\left(a\right)+\dots +\frac{(x-a{)}^r}{r!}{\text{f}}^{\left(r\right)}\left(a\right)+\dots$

Use differentiation to determine the Taylor series expansion of $\text{ln}x$, in ascending powers of $(x-1)$, up to and including the term in $(x-1{)}^2$

Hence prove that

$\underset{x\to 1}{\text{lim}}\left(\frac{\text{ln}x}{x-1}\right)=1$

The Taylor series expansion of $\text{f}\left(x\right)$ about $x=a$ is given by

$\text{f}\left(x\right)=\text{f}\left(a\right)+(x-a)\text{f}\text{'}\left(a\right)+\frac{(x-a{)}^2}{2!}\text{f}\text{'}\text{'}\left(a\right)+\dots +\frac{(x-a{)}^r}{r!}{\text{f}}^{\left(r\right)}\left(a\right)+\dots$

Given that

$y=(1+\text{ln}x{)}^{2}x>0$

show that $\frac{{\text{d}}^{2}y}{\text{d}{x}^2}=-\frac{2\text{ln}x}{x^2}$

Hence find $\frac{{\text{d}}^{3}y}{\text{d}{x}^3}$

Determine the Taylor series expansion about $x=1$ of

$(1+\text{ln}x{)}^2$

in ascending powers of $(x-1)$, up to and including the term in $(x-1{)}^3$

Give each coefficient in simplest form.

Use this series expansion to evaluate

$\underset{x\to 1}{\lim }\frac{2x-1-{\left(1+\ln x\right)}^2}{{\left(x-1\right)}^3}$

explaining your reasoning clearly.

$\text{f}\left(x\right)=x^4\text{sin}\left(2x\right)$

Use Leibnitz's theorem to show that the coefficient of $(x-\pi {)}^8$ in the Taylor series expansion of $\text{f}\left(x\right)$ about $\pi$ is

$\frac{a\pi +b{\pi }^3}{315}$

where $a$ and $b$ are integers to be determined.

The Taylor series expansion of $\text{f}\left(x\right)$ about $x=k$ is given by

$\text{f}\left(x\right)=\text{f}\left(k\right)+(x-k)\text{f}\text{'}\left(k\right)+\frac{(x-k{)}^2}{2!}\text{f}\text{'}\text{'}\left(k\right)+\dots +\frac{(x-k{)}^r}{r!}{\text{f}}^{\left(r\right)}\left(k\right)+\dots$

$\frac{\text{d}y}{\text{d}x}=x-y^2\text{(I)}$

Show that

$\frac{{\text{d}}^{5}y}{\text{d}{x}^5}=ay\frac{{\text{d}}^{4}y}{\text{d}{x}^4}+b\frac{\text{d}y}{\text{d}x}\frac{{\text{d}}^{3}y}{\text{d}{x}^3}+c{\left(\frac{{\text{d}}^{2}y}{\text{d}{x}^2}\right)}^2$

where $a$, $b$ and $c$ are integers to be determined.

Hence find a series solution, in ascending powers of $x$ as far as the term in $x^5$, of the differential equation (I), given that $y=1$ at $x=0$

$\frac{{\text{d}}^{2}y}{\text{d}{x}^2}-2x\frac{\text{d}y}{\text{d}x}+y=0\text{(I)}$

Show that

$\frac{{\text{d}}^{5}y}{\text{d}{x}^5}=ax\frac{{\text{d}}^{4}y}{\text{d}{x}^4}+b\frac{{\text{d}}^{3}y}{\text{d}{x}^3}$

where $a$ and $b$ are integers to be found.

Hence find a series solution, in ascending powers of $x$, as far as the term in $x^5$, of the differential equation (I) where $y=0$ and $\frac{\text{d}y}{\text{d}x}=1$ at $x=0$

The Taylor series expansion of $\mathrm{f}\left(x\right)$ about $x=a$ is given by

$\mathrm{f}\left(x\right)=\mathrm{f}\left(a\right)+\mathrm{f}\text{'}\left(a\right)(x-a)+\frac{\mathrm{f}\text{'}\text{'}\left(a\right)}{2!}(x-a{)}^2+\dots +\frac{{\mathrm{f}}^{\left(r\right)}\left(a\right)}{r!}(x-a{)}^r+\dots$

The curve with equation $y=\mathrm{f}\left(x\right)$ satisfies the differential equation

$\frac{{\text{d}}^{2}y}{\text{d}{x}^2}+y\frac{\text{d}y}{\text{d}x}-y\text{sin}x=0$

Given that $\left(\frac{\pi }{6},2\right)$ is a stationary point of the curve,

determine the nature of this stationary point, giving a reason for your answer.

Show that $\frac{{\text{d}}^{3}y}{\text{d}{x}^3}=\sqrt{3}-2$ at this stationary point.

Hence determine a series solution for $y$, in ascending powers of $\left(x-\frac{\pi }{6}\right)$ up to and including the term in ${\left(x-\frac{\pi }{6}\right)}^3$, giving each coefficient in simplest form.