Question
The function f is defined by f (x) = ex sinx , where x ∈ R.
(a) Find the Maclaurin series for f (x) up to and including the x3 term.
(b) Hence, find an approximate value for \(\int_{0}^{1}e^{x^{2}} sin(x^{2})dx.\)
The function g is defined by g (x) = ex cos x , where x ∈ R.
(c) (i) Show that g(x) satisfies the equation g ″(x) = 2(g′(x) – g(x)).
(ii) Hence, deduce that g(4) (x) = 2(g″′(x) – g ″(x)) .
(d) Using the result from part (c), find the Maclaurin series for g(x) up to and including the x4 term.
(e) Hence, or otherwise, determine the value of \(\lim_{x\rightarrow 0}\frac{e^{x}cos x – 1 – x}{x^{3}}.\)
▶️Answer/Explanation
Ans:
(a) METHOD 1
recognition of both known series
attempt to multiply the two series up to and including x3 term
Note: Condone absence of limits up to this stage.
Note: Accept working with each side separately to obtain -2ex sin x .
Note: Accept working with each side separately to obtain -4ex cos x .
Note: Do not award any marks for approaches that do not use the part (c) result.
Note: Condone the omission of +… in their working.
Question
Let f (x) = \(\sqrt{1+x}\) for x > -1 .
(a) Show that f (x)′′ = − \(\frac{1}{\sqrt[4]{(1+x)}^{3}}\) [3]
(b) Use mathematical induction to prove that f(n) (x) = \((-\frac{1}{4})^{n-1}\frac{(2n-3)!}{(n-2)!}(1+x)^{\frac{1}{2}-n}\)
for \(n\in \mathbb{Z}\;,n\geqslant 2\) [9]
Let g (x) = emx , m \(\in \mathbb{Q}\)
Consider the function h defined by h (x) = f (x) x g (x) for x > -1 .
It is given that the x2 term in the Maclaurin series for h (x) has a coefficient of \(\frac{7}{4}\)
(c) Find the possible values of m . [8]
▶️Answer/Explanation
Ans
(a)
attempt to use the chain rule
\(f'(x)= \frac{1}{2}(1+x)^{\frac{-1}{2}}\)
f”(x)= – \(\frac{1}{\sqrt[4]{(1+x)^3}}\)
Question
a.By evaluating successive derivatives at \(x = 0\) , find the Maclaurin series for \(\ln \cos x\) up to and including the term in \({x^4}\) .[8]
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.[5]
▶️Answer/Explanation
Markscheme
attempt at repeated differentiation (at least 2) M1
a.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 \(\mu \).
a.i.Show that \({f^{(4)}}x = f(x)\);[4]
b.i.Write down a series in terms of \(\mu \) for the probability \(p = {\text{P}}[X \equiv 0(\bmod 4)]\).[2]
b.ii.Show that \(p = {{\text{e}}^{ – \mu }}f(\mu )\).[1]
b.iii. Determine the numerical value of \(p\) when \(\mu = 3\). [2]
▶️Answer/Explanation
Markscheme
a.i.\(f’(x) = \frac{{{{\text{e}}^x} – {{\text{e}}^{ – x}} – 2\sin x}}{4}\) (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 \(f(x) = {{\rm{e}}^x}\cos x\) .
a.Show that \(f”(x) = – 2{{\rm{e}}^x}\sin x\) .[2]
b.Determine the Maclaurin series for \(f(x)\) up to and including the term in \({x^4}\) .[5]
c.By differentiating your series, determine the Maclaurin series for \({{\rm{e}}^x}\sin x\) up to the term in \({x^3}\) .[4]
▶️Answer/Explanation
Markscheme
a.\(f'(x) = {{\rm{e}}^x}\cos x – {{\rm{e}}^x}\sin x\) 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]
MAA HL 5.20 MACLAURIN SERIES – EXTENSION OF BINOMIAL THEORM – Neha
Question
Find the Maclaurin series of the function \(f(x)=e^{2x}\) up to and including the term in \(x^{4}\)
(a) by using the formula of the Maclaurin series \(f(x)=f(0)+f'(0)+\frac{{f}^”(0)}{2!}x^{2}+…\)
(b) by using the Maclaurin series of \(e^{x}=1+x+\frac{x^{2}}{2!}+…\)
▶️Answer/Explanation
Ans
(a) \(f(x)=e^{2x}\) \(f(0)=1\)
\(f'(x)=2e^{2x}\) \(f'(0)=2\)
\(f{}”(x)=4e^{2x}\) \(f{}”(0)=4\)
\({f}”'(x)=8e^{2x}\) \({f}”'(0)=8\)
\(f^{(4)}=16e^{x}\) \(f^{(4)}(0)=16\)
\(f(x)=1+2x+\frac{4}{2!}x^{2}+\frac{8}{3!}x^{3}+\frac{16}{4!}x^{4}…=1+2x+2x^{2}+\frac{4}{3!}x^{3}+\frac{2}{3!}x^{4}…\)
(b) \(e^{2x}=1+2x+\frac{(2x)^{2}}{2!}x^{2}+\frac{(2x)^{3}}{3!}x^{3}+\frac{(2x)^{4}}{4!}x^{4}…=1+2x+2x^{2}+\frac{4}{3!}x^{3}+\frac{2}{3!}x^{4}…\)
Question
Find the Maclaurin series of the function \(f(x)=(2+x)^{3}\)
(a) by using the formula of the Maclaurin series \(f(x)=f(0)+f'(0)x+\frac{{f}^”(0)}{2!}x^{2}+…\)
(b) by using the expansion of the binomial theorem.
▶️Answer/Explanation
Ans
(a) \(f(x)=(2+x)^{3}\) \(f(0)=8\)
\(f'(x)=3(2+x)^{2}\) \(f'(0)=12\)
\({f}”(x)=6(2+x)\) \({f}”(0)=12\)
\({f}”'(x)=6\) \({f}”'(0)=6\)
\(f(x)=8+12x+\frac{12}{2!}x^{2}+\frac{6}{3!}x^{3}+0+…=8+12x+6x^{2}+x^{3}\)
(b) \((2+x)^{3}=2^{3}+3\times 2^{2}\times x+3\times 2\times x^{2}+x^{3}=8+12x+6x^{2}+x^{3}\)
Question
Find the Maclaurin series of the function \(f(x)=\sqrt{2+x}=(2+x)^{\frac{1}{2}}\) , up to \(x^{2}\)
(a) by using the formula of the Maclaurin series \(f(x)=f(0)+f'(0)x+\frac{{f}^”(0)}{2!}x^{2}+…\)
(b) by using the extended version of the binomial theorem.
▶️Answer/Explanation
Ans
(a) \(f(x)=\sqrt{2+x}f(0)=\sqrt{2}\)
\(f'(x)=\frac{1}{\sqrt{2+x}}=\frac{1}{2}(2+x)^{\frac{1}{2}}f'(0)=\frac{1}{2\sqrt{2}}\)
\({f}”(x)=-\frac{1}{4}(2+x)^{-\frac{3}{2}}=-\frac{1}{4\sqrt{(2+x)^{3}}}\)
\({f}”(0)=-\frac{1}{8\sqrt{2}}\)
The Maclaurin series is
\(f(x)=\sqrt{2}+\frac{1}{2\sqrt{2}}x-\frac{1}{16\sqrt{2}}x^{2}+…=\sqrt{2}+\frac{\sqrt{2}}{4}x-\frac{\sqrt{2}}{32}x^{2}+…\)
(b) \((2+x)^{\frac{1}{2}}=2^{\frac{1}{2}}=\left ( 1+\frac{x}{2} \right )^{\frac{1}{2}}=\sqrt{2}\left ( 1+\frac{1}{2}\frac{x}{2}+\frac{\frac{1}{2}(-\frac{1}{2})}{2}\left ( \frac{x}{2} \right )^{2}+… \right )\)
\(=\sqrt{2}\left ( 1+\frac{x}{4}+\frac{x^{2}}{32}+…\right )=\sqrt{2}+\frac{\sqrt{2}}{4}x-\frac{\sqrt{2}}{32}x^{2}+…\)
Question
(a) Find the Maclaurin series of the function \(f(x)=\sqrt{x^{2+1}}\), up to and including the term \(x^{6}\) in by using the extended version of the binomial theorem.
(b) Given that \(f'(x)=\frac{x}{\sqrt{x^{2}+1}}\) find the Maclaurin series of \(g(x)=\frac{x}{\sqrt{x^{2}+1}}\) up to and including the term in \(x^{5}\).
▶️Answer/Explanation
Ans
(a) \(f(x)=\sqrt{x^{2}+1}=(1+x^{2})^{\frac{1}{2}}=1+\frac{1}{2}x^{2}+\frac{\frac{1}{2}(-\frac{1}{2})}{2}x^{4}+\frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})}{3!}x^{6}+…\)
\(=1+\frac{1}{2}x^{2}-\frac{1}{8}x^{4}+\frac{1}{16}x^{6}+…\)
(b) The Maclaurin series of \(f'(x)=\frac{x}{\sqrt{x^{2}+1}}\)
\(\left ( 1+\frac{1}{2}x^{2}-\frac{1}{8}x^{4}+\frac{1}{16}x^{6}+…\right )’=1+x-\frac{1}{2}x^{3}+\frac{3}{8}x^{5}+…\)
Question
By using the Maclaurin series find the Maclaurin series \(e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+…\), find the Maclaurin series of
(a) \(f(x)=e^{-x}\) up to and including the term in \(x^{3}\)
(b) \(f(x)=e^{-x^{2}}\) up to and including the term in \(x^{6}\)
(c) \(f(x)=xe^{x}\) up to and including the term in \(x^{4}\)
(d) \(f(x)=x^{2}e^{-x}\) up to and including the term in \(x^{5}\)
(e) \(f(x)=e^{x}-e^{-x}\) up to and including the term in \(x^{5}\)
(f) \(f(x)=(x+1)e^{4x}\) up to and including the term in \(x^{3}\)
▶️Answer/Explanation
Ans
By appropriate substitutions of \(x\) on \(e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+…\) we obtain
(a) \(e^{-x}=1-x+\frac{x^{2}}{2!}-\frac{x^{3}}{3!}+…=1-x+\frac{x^{2}}{2}-\frac{x^{3}}{6}+…\)
(b) \(e^{-x^{2}}=1-x^{2}+\frac{x^{4}}{2!}-\frac{x^{6}}{3!}+…=1-x^{2}+\frac{x^{4}}{2}-\frac{x^{6}}{6}+…\)
(c) \(xe^{x}=x\left ( 1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+… \right )=x+x^{2}+\frac{x^{3}}{2!}+\frac{x^{4}}{3!}+…=x+x^{2}+\frac{x^{3}}{2}+\frac{x^{4}}{6}+…\)
(d) \(x^{2}e^{-x}=x^{2}\left ( 1-x+\frac{x^{2}}{2!}-\frac{x^{3}}{3!}+… \right )=x^{2}-x^{3}+\frac{x^{4}}{2!}-\frac{x^{5}}{3!}+…=x^{2}-x^{3}+\frac{x^{4}}{2}-\frac{x^{5}}{6}+…\)
(e) \(e^{x}-e^{-x}=\left ( 1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+… \right )-\left ( 1-x+\frac{x^{2}}{2!}-\frac{x^{3}}{3!}+… \right )=2x+2\frac{x^{3}}{3!}+2\frac{x^{5}}{5!}+…
=2x+\frac{x^{3}}{3}+\frac{x^{5}}{60}+…\)
(f) \((x+1)e^{4x}=(x+1)\left ( 1+4x+\frac{(4x)^{2}}{2!}+\frac{(4x)^{3}}{3!}+…\right )\)
\(=x\left ( 1+4x+8x^{2}+\frac{32x^{3}}{3}+… \right )\)
\(=1+5x+12x^{2}+\frac{56x^{3}}{3}+…\)
Question
(a) Find the first three terms of the Maclaurin series for \(\ln (1+e^{x})\).
(b) Hence, or otherwise, determine the value of \(\lim_{x\rightarrow 0}\frac{2\ln (1+e^{x})-x-\ln 4}{x^{2}}\).
▶️Answer/Explanation
Ans
(a) METHOD 1
\(f(x)=\ln (1+e^{x});f(0)=\ln 2\)
\(f'(x)=\frac{e^{x}}{1+e^{x}};f'(0)=\frac{1}{2}\)
\({f}”(x)=\frac{e^{x}({1+e^{x}})-2e^{2x}}{(1+e^{x})^{2}};{f}”(0)=\frac{1}{4}\)
\(\ln (1+e^{x})=\ln 2+\frac{1}{2}x+\frac{1}{8}x^{2}+…\)
METHOD 2
\(\ln (1+e^{x})=\ln (1+1+x+\frac{1}{2}x^{2}+…)\)
\(=\ln 2+\ln(1+\frac{1}{2}x+\frac{1}{4}x^{2}+…)\)
\(=\ln 2+\left ( \frac{1}{2}x+\frac{1}{4}x^{2} +…\right )-\frac{1}{2}\left ( \frac{1}{2}x+\frac{1}{4}x^{2} +… \right )^{2}+…\)
\(=\ln 2+\frac{1}{2}x+\frac{1}{4}x^{2}-\frac{1}{8}x^{2}+…\)
=\(\ln 2+\frac{1}{2}x+\frac{1}{8}x^{2}+…)\)
(b) METHOD 1
\(\lim_{x\rightarrow 0}\frac{2\ln (1+e^{x})-x-\ln 4}{x^{2}}=\lim_{x\rightarrow 0}\frac{\left ( 2\ln 2+x+\frac{x^{2}}{4}+… \right )-x-\ln 4}{x^{2}}\)
\(=\lim_{x\rightarrow 0}\left ( \frac{1}{4}+ powers\ of\ x \right )=\frac{1}{4}\)
METHOD 2
Using 1’Hôpital’s Rule
\(\lim_{x\rightarrow 0}\frac{2\ln (1+e^{x})-x-\ln 4}{x^{2}}=\lim_{x\rightarrow 0}\frac{2e^{x}\div (1+e^{x})-1}{2x}=\lim_{x\rightarrow 0}\frac{2e^{x}\div (1+e^{x})^{2}}{2}=\frac{1}{4}\)
Question
The function \(f\) is defined by \(f(x)=\ln \left ( \frac{1}{1-x} \right )\).
(a) Write down the value of the constant term in the Maclaurin series for \(f(x)\).
(b) Find the first three derivatives of \(f (x)\) and hence show that the Maclaurin series for \(f (x)\) up to and including the \(x^{3}\) term is \(x+\frac{x^{2}}{2}+\frac{x^{3}}{3}\).
(c) Use this series to find an approximate value for \(\ln 2\).
▶️Answer/Explanation
Ans
(a) Constant term=0
(b) \(f'(x)=\frac{1}{1-x}\)
\({f}”(x)=\frac{1}{(1-x)^{2}}\)
\({f}”'(x)=\frac{2}{(1-x)^{3}}\)
\(f'(0)=1;{f}”(0)=1;{f}”'(0)=2\)
\(f(x)=0+\frac{1\times x}{1!}+\frac{1\times x^{2}}{2!}+\frac{2\times x^{3}}{3!}+…=x+\frac{x^{2}}{2}+\frac{x^{3}}{2}\)
(c) \(\frac{1}{1-x}=2\Rightarrow x=\frac{1}{2}\)
\(\ln 2\approx \frac{1}{2}+\frac{1}{8}+\frac{1}{24}=\frac{2}{3}(0.667)\)
Question
Let \(f(x)=\ln (1+\sin x)\)
(a) Show that \({f}^”(x)=\frac{-1}{1+\sin x}\)
(b) Find the third and the fourth derivatives of \(f\).
(c) Hence, find the Maclaurin series, up to the term in \(x^{4}\), for \(f(x)\)
▶️Answer/Explanation
Ans
(a) \(f'(x)=\frac{\cos x}{1+\sin x}\)
\(f”(x)=\frac{-\sin x(1+\sin x)-cos^{2}x}{(1+\sin x)^{2}}=\frac{-\sin x-1}{(1+\sin x)^{2}}=\frac{-1}{1+\sin x}\)
(b) \(f”'(x)=\frac{\cos x}{(1+\sin x)^{2}}\)
\(f^{iv}(x)=\frac{-\sin x(1+\sin x)^{2}-2(1+\sin x)cos^{2}x}{(1+\sin x)^{4}}\)
(c) \(f(0)=0,f'(0)=1,f”(0)=-1,f”'(0)=1,f^{iv}(0)=-2\)
\(\ln (1+\sin x)=x-\frac{1}{2}x^{2}+\frac{1}{6}x^{3}-\frac{1}{12}x^{4}+…\)
Question
The variables \(x\) and \(y\) are related by \(\frac{\mathrm{d} y}{\mathrm{d} x}-y\tan x=\cos x\).
Find the Maclaurin series for \(y\) up to and including the term in \(x^{2}\) given that \(y=-\frac{\pi }{2}\) when \(x=0\).
▶️Answer/Explanation
Ans
from \(\frac{\mathrm{d} y}{\mathrm{d} x}=y\ tan x+\cos x,f'(0)=1\)
now \(\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}}=y\sec ^{2}x+\frac{\mathrm{d} y}{\mathrm{d} x}\tan x-\sin x\)
\(\Rightarrow f”(0)=-\frac{\pi }{2}\)
\(\Rightarrow y=-\frac{\pi }{2}+x–\frac{\pi x^{2} }{4}\)
Question
(a) Given that \(y=\ln \cos x\), show that the first two non-zero terms of the Maclaurin series for \(y\) are \(-\frac{x^{2}}{2}-\frac{x^{4}}{12}\).
(b) Use this series to find an approximation in terms of \(\pi \) for \(ln 2\).
▶️Answer/Explanation
Ans
\(f(x)=\ln \cos x\)
\(f'(x)=\frac{-\sin x}{\cos x}=-\tan x\)
\(f”(x)=-\sec^{2}x\)
\(f”'(x)=-2\sec x\sec x\tan x\)
\(f^{iv}(x)=-2sec^{2}x(sec^{2}x)-2\tan x(2\sec ^{x}\tan x)\)
\(=-2sec^{4}x-4sec^{2}x\tan^{2}x\)
\(f(x)=f(0)+xf'(0)+\frac{x^{2}}{2!}f”(0)+\frac{x^{3}}{3!}f”'(0)+\frac{x^{4}}{4!}f^{iv}(0)+…\)
\(f(0)=0\),
\(f'(0)=0\),
\(f”(0)=-1\),
\(f”'(0)=0\),
\(f^{iv}(0)=-2\),
\(\ln(\cos x)\approx -\frac{x^{2}}{2!}-\frac{2x^{4}}{4!}=-\frac{x^{2}}{2}-\frac{x^{4}}{12}\)
(b) EITHER Taking \(x=\frac{\pi }{3}\)
\(\ln \frac{1}{2}\approx \frac{\left ( \frac{\pi }{3} \right )^{2}}{2!}-\frac{2\left ( \frac{\pi }{3} \right )^{4}}{4!}\Rightarrow -\ln 2\approx -\frac{\frac{\pi ^{2}}{9}}{2!}-\frac{2\frac{\pi ^{4}}{81}}{4!}\)
\(\ln 2\approx \frac{\pi ^{2}}{18}+\frac{\pi ^{4}}{972}=\frac{\pi ^{2}}{9}\left ( \frac{1}{2}+\frac{\pi ^{2}}{108} \right )\)
OR Taking \(x=\frac{\pi }{4}\)
\(\ln \frac{1}{\sqrt{2}}\approx -\frac{\left ( \frac{\pi }{4} \right )^{2}}{2!}-\frac{2\left ( \frac{\pi }{4} \right )^{4}}{4!}\)
\(-\frac{1}{2}\ln 2\approx \frac{\frac{\pi ^{2}}{16}}{2!}-\frac{2\frac{\pi ^{4}}{256}}{4!}\)
\(\ln 2\approx \frac{\pi ^{2}}{16}+\frac{\pi ^{4}}{1536}=\frac{\pi ^{2}}{8}\left ( \frac{1}{2}+\frac{\pi ^{2}}{192} \right )\)
Question
It is given that the Maclaurin series of the function \(f(x)=\ln (1+\sin x)\), up to the term in \(x^{4}\) is
\(f(x)=x-\frac{1}{2}x^{2}+\frac{1}{6}x^{3}-\frac{1}{12}x^{4}+…\)
(a) Deduce the Maclaurin series, up to and including the term in \(x^{4}\), for
(i) \(y=\ln (1-\sin x);\)
(ii) \(y=\ln \cos x;\)
(iii) \(y=\ln \sec x;\)
(b) By differentiating the Maclaurin series of \(y=\ln \cos x\), deduce the Maclaurin series of \(y=\tan x\)
(c) Hence calculate the limits
(i) \(\lim_{x\rightarrow 0}\frac{\ln \sec x}{x\sqrt{x}}\). (ii) \(\lim_{x\rightarrow 0}\left ( \frac{\tan (x^{2})}{\ln \cos x} \right )\)
(d) By considering the difference of the two series of \(y=\ln (1+\sin x)\) and \(y=\ln (1-\sin x)\)
Deduce that \(\ln 3\approx \frac{\pi }{3}\left ( 1+\frac{\pi ^{2}}{216} \right )\).
▶️Answer/Explanation
Ans
(a) (i) \(\ln (1-\sin x)=\ln(1+\sin (-x))=-x-\frac{1}{2}x^{2}-\frac{1}{6}x^{3}-\frac{1}{12}x^{4}+…\)
(ii) \(\ln (1+\sin x)=\ln(1-\sin x)=\ln (1-\sin ^{2}x)=\ln \cos ^{2}x\)
So \(\ln \cos^{2}x=-x^{2}-\frac{1}{6}x^{4}+…\)
\(\ln \cos x=-\frac{1}{2}x^{2}-\frac{1}{12}x^{4}+…\)
(iii) \(\ln \sec x=-ln \cos x=\frac{x^{2}}{2}+\frac{x^{4}}{12}+…\)
(b) Differentiating, \(\frac{\mathrm{d} }{\mathrm{d} x}(\ln \cos x)=\frac{1}{\cos x}\times (-\sin x)=-\tan x\)
\(\tan x=x+\frac{1}{3}x^{3}+…\)
(c) (i) \(\frac{\ln \sec x}{x\sqrt{x}}=\frac{\sqrt{x}}{2}+\frac{x^{2}\sqrt{x}}{12}+…\)
Limit=0
(ii) \(\frac{tan(x^{2})}{\ln \cos x}=\frac{x^{2}+\frac{x^{3}}{3}+…}{-\frac{x^{2}}{2}-\frac{x^{4}}{12}+…}\)
\(\frac{1+\frac{x^{4}}{3}+…}{-\frac{1}{2}-\frac{x^{2}}{12}+…}\rightarrow -2\) as \(x\rightarrow 0\)
\(so \lim_{x\rightarrow 0}\left ( \frac{\tan(x^{2})}{\ln \cos x} \right )=-2\)
(d) \(\ln(1+\sin x)-\ln (1-\sin x)=\ln \left ( \frac{1+\sin x}{1-\sin x} \right )\approx 2x+\frac{x^{3}}{3}\)
let \(x=\frac{\pi }{6}\) then, \(\ln \left ( \frac{1+\frac{1}{2}}{1-\frac{1}{2}} \right )=\ln 3\approx 2\left ( \frac{\pi }{6} \right )+\frac{\left ( \frac{\pi }{6} \right )^{3}}{3}=\frac{\pi }{3}\left ( 1+\frac{\pi ^{2}}{216} \right )\)