Problems

Filters
Clear Filters

1 problem found

2006 Paper 3 Q3
D: 1700.0 B: 1500.0
Taylor series trigonometric power series cot tan cosec trigonometric identities series coefficients

  1. Let \[ \tan x = \sum\limits_{n=0}^\infty a_n x^n \text{ and } \cot x = \dfrac 1 x +\sum\limits_{n=0}^\infty b_nx^n \] for \(0< x < \frac12\pi\,\). Explain why \(a_n=0\) for even \(n\). Prove the identity \[ \cot x - \tan x \equiv 2 \cot 2x\, \] and show that \[a_{n} = (1-2^{n+1})b_n\,.\]
  2. Let $ \displaystyle {\rm cosec}\, x = \frac1x +\sum\limits _{n=0}^\infty c_n x^n\,$ for \(0< x < \frac12\pi\,\). By considering \(\cot x + \tan x\), or otherwise, show that \[ c_n = (2^{-n} -1)b_n \,. \]
  3. Show that \[ \left(1+x{ \sum\limits_{n=0}^\infty} b_n x^n \right)^2 +x^2 = \left(1+x{ \sum\limits_{n=0} ^\infty} c_n x^n \right)^2\,. \] Deduce from this and the previous results that \(a_1=1\), and find \(a_3\).

Solution:

  1. Since \(\tan (-x) = -\tan x\), \(\tan\) is an odd function, and in particular all it's even coefficients are zero. \begin{align*} && 2 \cot 2x &\equiv \frac{2 cos 2x}{\sin 2 x} \\ &&&\equiv \frac{2(\cos^2 x- \sin^2 x)}{2 \sin x \cos x} \\ &&&\equiv \frac{\cos x}{\sin x} - \frac{\sin x}{ \cos x} \\ &&&\equiv \cot x - \tan x \end{align*} Therefore \begin{align*} && \underbrace{\frac1x + \sum_{n=0}^\infty b_nx^n}_{\cot x} - \underbrace{\sum_{n=0}^\infty a_n x^n}_{\tan x} &= 2 \left (\underbrace{\frac{1}{2x} + \sum_{n=0}^\infty b_n(2x)^n}_{\cot 2x} \right) \\ \Rightarrow && \sum_{n=0}^\infty a_n x^n &= \sum_{n=0}^\infty b_nx^n - 2\sum_{n=0}^\infty b_n(2x)^n \\ &&&= \sum_{n=0}^{\infty}b_n(1-2^{n+1})x^n \\ [x^n]: && a_n &= (1-2^{n+1})b_n \end{align*}
  2. \(\,\) \begin{align*} && \cot x + \tan x &= \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} \\ &&&= \frac{1}{\sin x \cos x} \\ &&&=2\cosec 2x \\ \\ \Rightarrow && \underbrace{\frac1x + \sum_{n=0}^\infty b_nx^n}_{\cot x} + \underbrace{\sum_{n=0}^\infty a_n x^n}_{\tan x} &= 2\left (\underbrace{ \frac1{2x} +\sum\limits _{n=0}^\infty c_n (2x)^n}_{\cosec 2x} \right) \\ \Rightarrow && \sum_{n=0}^\infty 2^{n+1}c_n x^n &= \sum_{n=0}^{\infty}(a_n+b_n)x^n \\ &&&= \sum_{n=0}^{\infty}\left((1-2^{n+1})b_n+ b_n\right)x^n \\ &&&= \sum_{n=0}^{\infty}\left(2-2^{n+1}\right)b_nx^n \\ [x^n]: && c_n &= (2^{-n}-1)b_n \end{align*}
  3. \(\,\) \begin{align*} && \cot^2 x + 1 &= \cosec^2 x \\ \Rightarrow && x^2 \cot^2 x + x^2 &= x^2 \cosec^2 x \\ \Rightarrow && x^2 \left ( \underbrace{\frac1x + \sum_{n=0}^\infty b_nx^n}_{\cot x} \right)^2 + x^2 &= x^2 \left (\underbrace{ \frac1{x} +\sum\limits _{n=0}^\infty c_n x^n}_{\cosec x} \right)^2 \\ \Rightarrow && \left ( 1 + x\sum_{n=0}^\infty b_nx^{n} \right)^2 + x^2 &= \left ( 1 +x\sum\limits _{n=0}^\infty c_n x^{n} \right)^2 \\ \\ \Rightarrow && \left ( 1 + x(b_1x + b_3 x^3 + \cdots) \right)^2 + x^2 &= \left ( 1 + x(c_1x + c_3 x^3 + \cdots) \right)^2 \\ \Rightarrow && 1 + (1+2b_1)x^2+(2b_3+b_1^2)x^4 + \cdots &= 1 + 2c_1x^2 + (2c_3+c_1^2)x^4 + \cdots \\ \Rightarrow && 1 + 2b_1 &= 2(2^{-1}-1)b_1 \\ \Rightarrow && b_1 &= -\frac13 \\ \Rightarrow && a_1 &= (1-2^{2})(-\tfrac13) = 1 \\ && c_1 &= \frac16\\ \Rightarrow && 2b_3+\frac19&= 2c_3+\frac1{36} \\ \Rightarrow && 2b_3 -2(2^{-3}-1)b_3 &= -\frac{1}{12} \\ \Rightarrow && \frac{15}{4}b_3 &= -\frac{1}{12} \\ \Rightarrow && b_3 &= -\frac{1}{45} \\ \Rightarrow && a_3 &= -(1-2^4)\frac{1}{45} = \frac13 \end{align*}

View