Maclaurin series for ln(exponential expressions)

AQA FP3 2013 January Q6

6

It is given that $y = \ln \left( \mathrm { e } ^ { 3 x } \cos x \right)$.
1. Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 - \tan x$.
2. Find $\frac { \mathrm { d } ^ { 4 } y } { \mathrm {~d} x ^ { 4 } }$.
Hence use Maclaurin's theorem to show that the first three non-zero terms in the expansion, in ascending powers of $x$, of $\ln \left( \mathrm { e } ^ { 3 x } \cos x \right)$ are $3 x - \frac { 1 } { 2 } x ^ { 2 } - \frac { 1 } { 12 } x ^ { 4 }$.
(3 marks)
Write down the expansion of $\ln ( 1 + p x )$, where $p$ is a constant, in ascending powers of $x$ up to and including the term in $x ^ { 2 }$.
1. Find the value of $p$ for which $\lim _ { x \rightarrow 0 } \left[ \frac { 1 } { x ^ { 2 } } \ln \left( \frac { \mathrm { e } ^ { 3 x } \cos x } { 1 + p x } \right) \right]$ exists.
2. Hence find the value of $\lim _ { x \rightarrow 0 } \left[ \frac { 1 } { x ^ { 2 } } \ln \left( \frac { \mathrm { e } ^ { 3 x } \cos x } { 1 + p x } \right) \right]$ when $p$ takes the value found in part (d)(i).

AQA FP3 2010 June Q5

5

Write down the expansion of $\cos 4 x$ in ascending powers of $x$ up to and including the term in $x ^ { 4 }$. Give your answer in its simplest form.
1. Given that $y = \ln \left( 2 - \mathrm { e } ^ { x } \right)$, find $\frac { \mathrm { d } y } { \mathrm {~d} x } , \frac { \mathrm {~d} ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ and $\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }$.
  (You may leave your expression for $\frac { \mathrm { d } ^ { 3 } y } { \mathrm {~d} x ^ { 3 } }$ unsimplified.)
2. Hence, by using Maclaurin's theorem, show that the first three non-zero terms in the expansion, in ascending powers of $x$, of $\ln \left( 2 - \mathrm { e } ^ { x } \right)$ are $$- x - x ^ { 2 } - x ^ { 3 }$$
Find $$\lim _ { x \rightarrow 0 } \left[ \frac { x \ln \left( 2 - \mathrm { e } ^ { x } \right) } { 1 - \cos 4 x } \right]$$

AQA FP3 2007 June Q6

6

The function f is defined by $$\mathrm { f } ( x ) = \ln \left( 1 + \mathrm { e } ^ { x } \right)$$ Use Maclaurin's theorem to show that when $\mathrm { f } ( x )$ is expanded in ascending powers of $x$ :
1. the first three terms are $$\ln 2 + \frac { 1 } { 2 } x + \frac { 1 } { 8 } x ^ { 2 }$$
2. the coefficient of $x ^ { 3 }$ is zero.
Hence write down the first two non-zero terms in the expansion, in ascending powers of $x$, of $\ln \left( \frac { 1 + \mathrm { e } ^ { x } } { 2 } \right)$.
Use the series expansion $$\ln ( 1 + x ) = x - \frac { 1 } { 2 } x ^ { 2 } + \frac { 1 } { 3 } x ^ { 3 } - \ldots$$ to write down the first three terms in the expansion, in ascending powers of $x$, of $\ln \left( 1 - \frac { x } { 2 } \right)$.
Use your answers to parts (b) and (c) to find $$\lim _ { x \rightarrow 0 } \left[ \frac { \ln \left( \frac { 1 + \mathrm { e } ^ { x } } { 2 } \right) + \ln \left( 1 - \frac { x } { 2 } \right) } { x - \sin x } \right]$$

AQA Further AS Paper 1 2021 June Q7

7 Show that the Maclaurin series for $\ln ( \mathrm { e } + 2 \mathrm { e } x )$ is $$1 + 2 x - 2 x ^ { 2 } + a x ^ { 3 } - \ldots$$ where $a$ is to be determined.