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Question
By evaluating successive derivatives at \(x = 0\) , find the Maclaurin series for \(\ln \cos x\) up to and including the term in \({x^4}\) .
Consider \(\mathop {\lim }\limits_{x \to 0} \frac{{\ln \cos x}}{{{x^n}}}\) , where \(n \in \mathbb{R}\) .
Using your result from (a), determine the set of values of \(n\) for which
(i) the limit does not exist;
(ii) the limit is zero;
(iii) the limit is finite and non-zero, giving its value in this case.
Answer/Explanation
Markscheme
attempt at repeated differentiation (at least 2) M1
let \(f(x) = \ln \cos x\) , \(f(0) = 0\) A1
\(f'(x) = – \tan x\) , \(f'(0) = 0\) A1
\(f”(x) = – {\sec ^2}x\) , \(f”(0) = – 1\) A1
\(f”'(x) = – 2{\sec ^2}x\tan x\) , \(f”'(0) = 0\) A1
\({f^{iv}}(x) = – 2se{c^4}x – 4se{c^2}xta{n^2}x\) , \({f^{iv}}(0) = – 2\) A1
the Maclaurin series is
\(\ln \cos x = – \frac{{{x^2}}}{2} – \frac{{{x^4}}}{{12}} + \ldots \) M1A1
Note: Allow follow-through on final A1.
[8 marks]
\(\frac{{\ln \cos x}}{{{x^n}}} = – \frac{{{x^{2 – n}}}}{2} – \frac{{{x^{4 – n}}}}{{12}} + \ldots \) (M1)
(i) the limit does not exist if \(n > 2\) A1
(ii) the limit is zero if \(n < 2\) A1
(iii) if \(n = 2\) , the limit is \( – \frac{1}{2}\) A1A1
[5 marks]
Question
(a) Assuming the Maclaurin series for \({{\text{e}}^x}\), determine the first three non-zero terms in the Maclaurin expansion of \(\frac{{{{\text{e}}^x} – {{\text{e}}^{ – x}}}}{2}\).
(b) The random variable \(X\) has a Poisson distribution with mean \(\mu \). Show that \({\text{P}}\left( {X \equiv 1(\bmod 2)} \right) = a + b{{\text{e}}^{c\mu }}\) where \(a\), \(b\) and \(c\) are constants whose values are to be found.
Answer/Explanation
Markscheme
(a) \({{\text{e}}^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} + \frac{{{x^5}}}{{5!}} + \ldots \)
\({{\text{e}}^{ – x}} = 1 – x + \frac{{{x^2}}}{{2!}} – \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} – \frac{{{x^5}}}{{5!}} + \ldots \) A1
\(\frac{{{{\text{e}}^x} – {{\text{e}}^{ – x}}}}{2} = x + \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} + \ldots \) (M1)A1
Note: Accept any valid (otherwise) method.
[3 marks]
(b) \({\text{P}}\left( {X \equiv 1(\bmod 2)} \right) = {\text{P}}(X = 1,{\text{ }}3,{\text{ }}5,{\text{ }} \ldots )\) (M1)
\( = {{\text{e}}^{ – \mu }}\left( {\mu + \frac{{{\mu ^3}}}{{3!}} + \frac{{{\mu ^5}}}{{5!}} + \ldots } \right)\) A1
\( = \frac{{{{\text{e}}^{ – \mu }}({{\text{e}}^\mu } – {{\text{e}}^{ – \mu }})}}{2}\) A1
\( = \frac{1}{2} – \frac{1}{2}{{\text{e}}^{ – 2\mu }}\) A1
\(\left( {a = \frac{1}{2},{\text{ }}b = – \frac{1}{2},{\text{ }}c = – 2} \right)\)
[4 marks]
Question
The function \(f\) is defined by
\[f(x) = \frac{{{{\text{e}}^x} + {{\text{e}}^{ – x}} + 2\cos x}}{4},{\text{ }}x \in \mathbb{R}.\]
The random variable \(X\) has a Poisson distribution with mean \(\mu \).
Show that \({f^{(4)}}x = f(x)\);
By considering derivatives of \(f\), determine the first three non-zero terms of the Maclaurin series for \(f(x)\).
Write down a series in terms of \(\mu \) for the probability \(p = {\text{P}}[X \equiv 0(\bmod 4)]\).
Show that \(p = {{\text{e}}^{ – \mu }}f(\mu )\).
Determine the numerical value of \(p\) when \(\mu = 3\).
Answer/Explanation
Markscheme
\(f’(x) = \frac{{{{\text{e}}^x} – {{\text{e}}^{ – x}} – 2\sin x}}{4}\) (A1)
\(f’’(x) = \frac{{{{\text{e}}^x} + {{\text{e}}^{ – x}} – 2\cos x}}{4}\) (A1)
\(f’’’(x) = \frac{{{{\text{e}}^x} – {{\text{e}}^{ – x}} + 2\sin x}}{4}\) (A1)
\({f^{(4)}}(x) = \frac{{{{\text{e}}^x} + {{\text{e}}^{ – x}} + 2\cos x}}{4} = f(x)\) AG
[4 marks]
therefore,
\(f(0) = 1\) and \({f^{(4)}}(0) = 1\) (A1)
\(f’(0) = f”(0) = f”'(0) = 0\) (A1)
the sequence of derivatives repeats itself so the next non-zero derivative is \({f^{(8)}}(0) = 1\) (A1)
the MacLaurin series is \(1 + \frac{{{x^4}}}{{4!}} + \frac{{{x^8}}}{{8!}}( + \ldots )\) (M1)A1
[4 marks]
\(p = {\text{P}}(X = 0) + {\text{P}}(X = 4) + {\text{P}}(X = 8) + \ldots \) (M1)
\( = \frac{{{{\text{e}}^{ – \mu }}{\mu ^0}}}{{0!}} + \frac{{{{\text{e}}^{ – \mu }}{\mu ^4}}}{{4!}} + \frac{{{{\text{e}}^{ – \mu }}{\mu ^8}}}{{8!}} + \ldots \) A1
[??? marks]
\(p = {{\text{e}}^{ – \mu }}\left( {1 + \frac{{{\mu ^4}}}{{4!}} + \frac{{{\mu ^8}}}{{8!}} + \ldots } \right)\) A1
\( = {{\text{e}}^{ – \mu }}f(\mu )\) AG
[??? marks]
\(p = {{\text{e}}^{ – 3}}\left( {\frac{{{{\text{e}}^3} + {{\text{e}}^{ – 3}} + 2\cos 3}}{4}} \right)\) (M1)
\( = 0.226\) A1
[??? marks]
Question
The function \(f\) is defined by \(f(x) = {{\rm{e}}^x}\cos x\) .
Show that \(f”(x) = – 2{{\rm{e}}^x}\sin x\) .
Determine the Maclaurin series for \(f(x)\) up to and including the term in \({x^4}\) .
By differentiating your series, determine the Maclaurin series for \({{\rm{e}}^x}\sin x\) up to the term in \({x^3}\) .
Answer/Explanation
Markscheme
\(f'(x) = {{\rm{e}}^x}\cos x – {{\rm{e}}^x}\sin x\) A1
\(f”(x) = {{\rm{e}}^x}\cos x – {{\rm{e}}^x}\sin x – {{\rm{e}}^x}\sin x – {{\rm{e}}^x}\cos x\) A1
\( = – 2{{\rm{e}}^x}\sin x\) AG
[2 marks]
\(f”'(x) = – 2{{\rm{e}}^x}\sin x – 2{{\rm{e}}^x}\cos x\) A1
\({f^{IV}}(x) = – 4{{\rm{e}}^x}\cos x\) A1
\(f(0) = 1\), \(f'(0) = 1\) ,\(f”(0) = 0\), \(f”'(0) = – 2\), \({f^{IV}}(0) = – 4\) (A1)
the Maclaurin series is
\({{\rm{e}}^x}\cos x = 1 + x – \frac{{{x^3}}}{3} – \frac{{{x^4}}}{6} + \ldots \) M1A1
Note: Accept multiplication of series method.
[5 marks]
differentiating,
\({{\rm{e}}^x}\cos x – {{\rm{e}}^x}\sin x = 1 – {x^2} – \frac{{2{x^3}}}{3} + \ldots \) M1A1
\({{\rm{e}}^x}\sin x = 1 + {x^{}} – \frac{{{x^3}}}{3} + \ldots – 1 + {x^2} + \frac{{2{x^3}}}{3} + \ldots \) M1
\( = x + {x^2} + \frac{{{x^3}}}{3} + \ldots \) A1
[4 marks]