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2013 Paper 3 Q2
D: 1700.0 B: 1516.0
Taylor series Maclaurin series Leibniz rule differential equation arcsin recurrence relation infinite series evaluation repeated differentiation

In this question, you may ignore questions of convergence. Let \(y= \dfrac {\arcsin x}{\sqrt{1-x^2}}\,\). Show that \[ (1-x^2)\frac {\d y}{\d x} -xy -1 =0 \] and prove that, for any positive integer \(n\), \[ (1-x^2) \frac{\d^{n+2}y}{\d x^{n+2}} - (2n+3)x \frac{\d ^{n+1}y}{\d x ^{n+1}} -(n+1)^2 \frac{\d^ny}{\d x^n}=0\, . \] Hence obtain the Maclaurin series for \( \dfrac {\arcsin x}{\sqrt{1-x^2}}\,\), giving the general term for odd and for even powers of \(x\). Evaluate the infinite sum \[ 1 + \frac 1 {3!} + \frac{2^2}{5!} + \frac {2^2\times 3^2}{7!}+\cdots + \frac {2^2\times 3^2\times \cdots \times n^2}{(2n+1)!} + \cdots\,. \]

Solution: \begin{align*} && y &= \frac{\arcsin x}{\sqrt{1-x^2}} \\ && \frac{\d y}{\d x} &= \frac{(1-x^2)^{-1/2} \cdot (1-x^2)^{1/2}-\arcsin x \cdot (-x)(1-x^2)^{-1/2}}{1-x^2} \\ &&&= \frac{1+ xy}{1-x^2} \\ \Rightarrow && 0 &= (1-x^2) \frac{\d y}{\d x} -xy-1\\ \\ \frac{\d^n}{\d x^{n+1}}: && 0 &= \left ( (1-x^2) y' \right)^{(n+1)} - (xy)^{(n+1)} \\ \Rightarrow && 0 &= (1-x^2)y^{(n+2)} + \binom{n+1}{1}(1-x^2)^{(1)}y^{(n+1)}+\binom{n+1}{2} (1-x^2)^{(2)}y^{(n)} - (xy^{(n+1)} +\binom{n+1}{1} y^{(n)} ) \\ &&&= (1-x^2)y^{(n+2)}+\left ( (n+1)\cdot(-2x)-x \right)y^{(n+1)} + \left ( \frac{(n+1)n}{2} \cdot (-2)-(n+1) \right)y^{(n)} \\ &&&= (1-x^2)y^{(n+2)}-\left ( 2n+3 \right)xy^{(n+1)} - \left ( (n+1)n+(n+1)\right)y^{(n)} \\ &&&= (1-x^2)y^{(n+2)}-\left ( 2n+3 \right)xy^{(n+1)} - \left ( n+1\right)^2y^{(n)} \\ \end{align*} Since \(y(0) = 0, y'(0) = 1\) we can look at the recursion: \(y^{(n+2)} - (n+1)^2y^{(n)}\) for larger terms, ie \(y^{(2k)}(0) = 0\) \(y^{(1)}(0) = 1, y^{(3)}(0) = (1+1)^2 \cdot 1 = 2^2, y^{(5)}(0) = (3+1)^2 y^{(3)} = 4^2 \cdot 2^2\) and \(y^{(2k+1)}(0) = (2k)^2 \cdot (2k-2)^2 \cdots 2^2 \cdot 1^2 = 2^{2k} \cdot (k!)^2\). Therefore \begin{align*} && \frac{\arcsin x}{\sqrt{1-x^2}} &= \sum_{k=0}^{\infty} \frac{2^{2k} \cdot (k!)^2}{(2k+1)!} x^{2k+1} \\ \\ \Rightarrow && \frac{\arcsin \frac12}{\sqrt{1-\left (\frac12 \right)^2}} &= \sum_{k=0}^{\infty} \frac{2^{2k} \cdot (k!)^2}{(2k+1)!} 2^{-2k-1}\\ &&&= \frac12 \sum_{k=0}^{\infty} \frac{ (k!)^2}{(2k+1)!} \\ &&&= \frac12 \left ( 1 + \frac1{3!} + \frac{2^2}{5!} + \cdots+ \right) \\ \Rightarrow&& S &= 2 \frac{2\frac{\pi}{6}}{\sqrt{3}} = \frac{2\pi}{3\sqrt{6}} \end{align*}

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2005 Paper 2 Q6
D: 1600.0 B: 1500.0
generalised binomial theorem power series binomial expansion summation negative exponent differentiation of series infinite series evaluation

  1. Write down the general term in the expansion in powers of \(x\) of \((1-x)^{-1}\), \((1-x)^{-2}\) and \((1-x)^{-3}\), where \(|x| <1\). Evaluate \(\displaystyle \sum_{n=1}^\infty n 2^{-n}\) and \(\displaystyle \sum_{n=1}^\infty n^22^{-n}\).
  2. Show that $\displaystyle (1-x)^{-\frac12} = \sum_{n=0}^\infty \frac{(2n)!}{(n!)^2} \frac{x^n}{2^{2n}}\( , for \)|x|<1$. Evaluate \(\displaystyle \sum_{n=0}^\infty \frac{(2n)!} {(n!)^2 2^{2n}3^{n}} \) and \(\displaystyle \sum_{n=1}^\infty \frac{n(2n)!} {(n!)^2 2^{2n}3^{n}}\).

Solution:

  1. \(\displaystyle (1-x)^{-1} = \sum_{n=0}^\infty x^n\), \(\displaystyle (1-x)^{-2} = \sum_{n=0}^\infty (n+1)x^n\), \(\displaystyle (1-x)^{-3} = \sum_{n=0}^\infty \frac{(n+2)(n+1)}{2}x^n\) \begin{align*} && \sum_{n=1}^{\infty} n2^{-n} &= \frac12\sum_{n=0}^{\infty}(n+1)2^{-n} \\ &&&= \frac12 (1-\tfrac12)^{-2} = 2 \\ \\ && \sum_{n=1}^{\infty} nx^n&= x(1-x)^{-2} \\ \Rightarrow && \sum_{n=1}^{\infty} n^2x^{n-1}&= (1-x)^{-2}+2x(1-x)^{-3} \\ \Rightarrow && \sum_{n=1}^{\infty} n^22^{-n} &= \frac12 \left ( (1-\tfrac12)^{-2}+2\cdot \tfrac12 \cdot (1-\tfrac12)^{-3} \right) \\ &&&= \frac12 \left ( 4 +8\right) = 6 \end{align*}
  2. By the generalised binomial theorem, \begin{align*} && (1-x)^{-\frac12} &= 1 + \sum_{n=1}^{\infty} \frac{(-\tfrac12)\cdot(-\tfrac32)\cdots(-\tfrac12-n+1)}{n!}(-x)^n \\ &&&= 1 + \sum_{n=1}^{\infty} \frac{(-1)^n(\tfrac12)\cdot(\tfrac32)\cdots(\tfrac{2n-1}2)}{n!}(-x)^n \\ &&&= 1 + \sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2^nn!}x^n \\ &&&= 1 + \sum_{n=1}^{\infty} \frac{(2n)!}{2^nn! \cdot 2^n n!}x^n \\ &&&= 1 + \sum_{n=1}^{\infty} \frac{(2n)!}{2^{2n}(n!)^2}x^n \\ &&&= \sum_{n=0}^{\infty} \frac{(2n)!}{2^{2n}(n!)^2}x^n \\ \end{align*} \begin{align*} && \sum_{n=0}^\infty \frac{(2n)!} {(n!)^2 2^{2n}3^{n}} &= (1-\tfrac13)^{-\frac12} \\ &&&= \sqrt{\frac32} \\ \\ && (1-x)^{-\frac12} &= \sum_{n=0}^{\infty} \frac{(2n)!}{2^{2n}(n!)^2}x^n \\ \Rightarrow && \tfrac12(1-x)^{-\frac32} &= \sum_{n=0}^{\infty} \frac{n(2n)!}{2^{2n}(n!)^2}x^{n-1} \\ \Rightarrow && \sum_{n=1}^\infty \frac{n(2n)!} {(n!)^2 2^{2n}3^{n}} &= \frac16(1-\tfrac13)^{-3/2} \\ &&&= \frac16 \sqrt{\frac{27}{8}} = \frac14\sqrt{\frac{3}2} \end{align*}

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