9 problems found
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*}
Given that \(y = \cos(m \arcsin x)\), for \(\vert x \vert <1\), prove that \[ (1-x^2) \frac {\d^2 y}{\d x^2} -x \frac {\d y}{\d x} +m^2y=0\,. \] Obtain a similar equation relating \(\dfrac{\d^3y}{\d x^3}\,\), \(\dfrac{\d^2y}{\d x^2}\, \) and \(\, \dfrac{\d y}{\d x}\,\), and a similar equation relating \(\dfrac{\d^4y}{\d x^4}\,\), \(\dfrac{\d^3y}{\d x^3}\,\) and \(\,\dfrac{\d^2 y}{\d x^2}\,\). Conjecture and prove a relation between \(\dfrac{\d^{n+2}y}{\d x^{n+2}}\,\), \(\dfrac{\d^{n+1}y}{\d x^{n+1}}\;\) and \(\;\dfrac{\d^n y}{\d x^n}\,\). Obtain the first three non-zero terms of the Maclaurin series for \(y\). Show that, if \(m\) is an even integer, \(\cos m\theta\) may be written as a polynomial in \(\sin\theta\) beginning \[ 1 - \frac{m^2\sin^2\theta}{2!}+ \frac{m^2(m^2-2^2)\sin^4\theta}{4!} -\cdots \,. \, \tag{\(\vert\theta\vert < \tfrac12 \pi\)} \] State the degree of the polynomial.
Solution: \begin{align*} && y &= \cos(m \arcsin x) \\ && y' &= -m \sin (m \arcsin x) \cdot (1-x^2)^{-\frac12} \\ && y'' &= -m^2 \cos(m \arcsin x) \cdot (1-x^2)^{-1} -m \sin(m \arcsin x) \cdot (1-x^2)^{-\frac32} \cdot (-x) \\ &&&= -m^2 y (1-x^2)^{-1} + x(1-x^2)^{-1} y' \\ \Rightarrow && 0 &= (1-x^2)y'' - x y' + m^2y \\ \\ && 0 &= (1-x^2)y^{(3)} -2xy'' - xy''-y' + m^2y' \\ &&&= (1-x^2)y^{(3)} - 3xy'' + (m^2-1)y' \\ \\ && 0 &= (1-x^2)y^{(4)} - 2xy^{(3)} - 3xy^{(3)} - 3y^{(2)} + (m^2-1)y^{(2)} \\ &&&= (1-x^2)y^{(4)}- 5xy^{(3)} - (m^2-4)y^{(2)} \end{align*} Claim: \(0 = (1-x^2)y^{(n+2)} - (2n+1)y^{(n+1)} + (m^2-n^2)y^{(n)}\) Proof: (By induction) Clearly the first few base cases are true. Suppose it is true for some \(n\), then \begin{align*} && 0 &= (1-x^2)y^{(n+2)} - (2n+1)xy^{(n+1)} + (m^2-n^2)y^{(n)} \\ \Rightarrow && 0 &= (1-x^2)y^{(n+3)} - 2xy^{(n+2)} - (2n+1)xy^{(n+2)} - (2n+1)y^{(n+1)} + (m^2-n^2)y^{(n+1)} \\ &&&= (1-x^2)y^{(n+3)} - (2n+3)xy^{(n+2)} + (m^2-n^2-2n-1)y^{(n+1)} \\ &&&= (1-x^2)y^{(n+1+2)} - (2(n+1)+1)xy^{(n+1+1)} +(m^2-(n+1)^2)y^{(n)} \end{align*} And so we can conclude the result by induction. Notice that \begin{align*} && y(0) &= \cos(m 0) = 1 \\ && y'(0) &= -m\sin(m 0) = 0 \\ && y''(0) &= -m^2 y(0) = -m^2\\ \end{align*} Notice that \(y^{(n+2)}(0) + (m^2-n^2)y^{(n)} = 0\) so in particular all the odd terms will be \(0\) and the even terms will be \(1, -m^2, m^2(m^2-2^2), \cdots\), therefore \begin{align*} && \cos (m \arcsin x) &= 1 -\frac{m^2}{2!} x^2 + \frac{m^2(m^2-2^2)}{4!}x^4 - \cdots \\ \Rightarrow && \cos(m \theta) &= 1 - \frac{m^2}{2!} \sin^2 \theta + \frac{m^2(m^2-2^2)}{4!} \sin^4 \theta \end{align*} Notice that if \(m\) is even, then at some point we will have \(m^2-m^2\) appearing in our expansion and all remaining terms will be zero. Therefore we will end up with a polynomial series, of degree \(m\) in \(\sin \theta\)
Given that \(y = \ln ( x + \sqrt{x^2 + 1})\), show that \( \displaystyle \frac{\d y}{\d x} = \frac1 {\sqrt{x^2 + 1} }\;\). Prove by induction that, for \(n \ge 0\,\), \[ \l x^2 + 1 \r y^{\l n + 2 \r} + \l 2n + 1 \r x y^{\l n + 1 \r} + n^2 y^{\l n \r} = 0\;, \] where \(\displaystyle y^{(n)} = \frac{\d^n y}{\d x^n}\) and \(y^{(0)} =y\,\). Using this result in the case \(x = 0\,\), or otherwise, show that the Maclaurin series for \(y\) begins \[ x - {x^3 \over 6} +{3 x^5 \over 40} \] and find the next non-zero term.
Solution: \begin{align*} && y & = \ln ( x + \sqrt{x^2+1}) \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{1}{x+\sqrt{x^2+1}} \cdot \frac{\d }{\d x} \left ( x + \sqrt{x^2+1} \right) \\ &&&= \frac{1}{x+\sqrt{x^2+1}} \left (1 + \frac12 \frac{2x}{\sqrt{x^2+1}} \right) \\ &&&= \frac{1}{x+\sqrt{x^2+1}} \left ( \frac{\sqrt{x^2+1} + x}{\sqrt{x^2+1}}\right) \\ &&&= \frac{1}{\sqrt{x^2+1}} \end{align*} Note that \(\displaystyle y^{(2)} = - \frac12 \frac{2x}{(x^2+1)^{3/2}} = - \frac{x}{(x^2+1)^{3/2}}\), and in particular \((x^2+1)y^{(2)} + xy^{(1)} = 0\). Now applying Leibnitz formula: \begin{align*} 0 &= \left ( (x^2+1)y^{(2)} + xy^{(1)} \right )^{(n)} \\ &= \left ( (x^2+1)y^{(2)}\right )^{(n)} + \left (xy^{(1)} \right )^{(n)} \\ &= (x^2+1)y^{(n+2)} +n2xy^{(n+1)} + \binom{n}{2}2y^{(n)} + xy^{(n+1)} + n y^{(n)} \\ &= (x^2+1)y^{(n+2)} + (2n+1)xy^{(n+1)} + (n^2-n+n)y^{(n)} \\ &= (x^2+1)y^{(n+2)} + (2n+1)xy^{(n+1)} + n^2y^{(n)} \end{align*} as required. When \(x = 0\): \begin{align*} && y(0) &= \ln(0 + \sqrt{0^2+1}) \\ &&&= \ln 1 = 0 \\ && y'(0) &= \frac{1}{\sqrt{0^2+1}} = 1 \\ && y^{(n+2)} &= -n^2 y^{(n)} \\ && y^{(2k)} &= 0 \\ && y^{(3)} &= -1 \\ && y^{(5)} &= 3^2 \\ && y^{(7)} &= - 5^2 \cdot 3^2 \\ \end{align*} Therefore the Maclaurin series about \(x = 0\) is \begin{align*} y &= x - \frac{1}{3!} x^3 + \frac{3^2}{5!} x^5 - \frac{3^2 \cdot 5^2}{7!} x^7 + \cdots \\ &= x - \frac{1}{6} x^3 + \frac{3}{1 \cdot 2 \cdot 4 \cdot 5} x^5 - \frac{5}{1 \cdot 2 \cdot 4 \cdot 2 \cdot 7} x^7 + \cdots \\ &= x - \frac{1}{6}x^3 + \frac{3}{40} x^5 - \frac{5}{56} x^7 + \cdots \end{align*}
Let $$ {\rm S}_n(x)=\mathrm{e}^{x^3}{{\d^n}\over{\d x^n}}{(\mathrm{e}^{-x^3})}.$$ Show that \({\rm S}_2(x)=9x^4-6x\) and find \({\rm S}_3(x)\). Prove by induction on \(n\) that \({\rm S}_n(x)\) is a polynomial. By means of your induction argument, determine the order of this polynomial and the coefficient of the highest power of \(x\). Show also that if \(\displaystyle \frac{\d S_n}{\d x}=0\) for some value \(a\) of \(x\), then \(S_n(a)S_{n+1}(a)\le0\).
Solution: \begin{align*} && S_2(x) &= e^{x^3} \frac{d^2}{\d x^2} \left [e^{-x^3} \right] \\ &&&= e^{x^3} \frac{d}{\d x} \left [e^{-x^3}(-3x^2) \right] \\ &&&= e^{x^3} \left [e^{-x^3}(9x^4-6x) \right] \\ &&&=9x^4-6x \\ \\ && S_3(x) &= e^{x^3} \frac{\d^3}{\d x^3} \left [ e^{-x^3} \right]\\ &&&= e^{x^3} \frac{\d}{\d x} \left [ e^{-x^3}(9x^4-6x) \right ] \\ &&&= e^{x^3} e^{-x^3}\left [ (-3x^2)(9x^4-6x)+(36x^3-6) \right ] \\ &&&= -27x^6 +54x^3-6 \end{align*} Claim: \(S_n\) is a polynomial of degree \(2n\) with leading coefficient \((-3)^n\). Proof: Clearly this is true for \(n = 1, 2, 3\) by demonstration. Suppose it is true for some \(n = k\), then \begin{align*} && S_k(x) &= e^{x^2} \frac{\d^k}{\d x^k} \left [ e^{x^3}\right] \\ && (-3)^kx^{2k} +\cdots &= e^{x^3} \frac{\d^k}{\d x^k} \left [ e^{x^3}\right] \\ \Rightarrow && \frac{\d^k}{\d x^k} \left [ e^{x^3}\right] &= e^{-x^3} \left ( (-3)^kx^{2k} +\cdots\right) \\ \Rightarrow && \frac{\d^k}{\d x^k}\left [ e^{x^3}\right] &= e^{-x^3} (-3x^2)\left ( (-3)^kx^{2k} +\cdots\right) + e^{-x^3} S_k'(x) \\ &&&= e^{-x^3} \left (\underbrace{ (-3)^{k+1}x^{2k+2} + \cdots + S_k'(x)}_{\deg =2k+2}\right) \\ \Rightarrow && S_{k+1}(x) &= (-3)^{k+1}x^{2k+2} + \cdots + S_k'(x) \end{align*} And therefore \(S_{k+1}\) is a polynomial degree \(2(k+1)\) with leading coefficient \((-3)^{k+1}\) so by induction it's true for all \(n\). If \(S'_n(a) = 0\) then \(S_{n+1}(a) = (-3a^2)S_n(a) + S_n'(a) \Rightarrow S_{n+1}(a)S_n(a) = -3 (aS_n(a))^2 \leq 0\)
Show that \(y=\sin^{2}(m\sin^{-1}x)\) satisfies the differential equation \[ (1-x^{2})y^{(2)}=xy^{(1)}+2m^{2}(1-2y), \] and deduce that, for all \(n\geqslant1,\) \[ (1-x^{2})y^{(n+2)}=(2n+1)xy^{(n+1)}+(n^{2}-4m^{2})y^{(n)}, \] where \(y^{(n)}\) denotes the \(n\)th derivative of \(y\). Derive the Maclaurin series for \(y\), making it clear what the general term is.
Solution: \begin{align*} && y &= \sin^2 (m \sin^{-1} x) \\ \Rightarrow && y' &= 2 \sin (m \sin^{-1} x) \cdot \cos (m \sin^{-1} x) \cdot m \cdot \frac1{\sqrt{1-x^2}} \\ \Rightarrow && y'' &= 2 \cos^2(m \sin^{-1} x) \cdot m^2 \cdot \frac{1}{1-x^2} + \\ &&&\quad\quad-2\sin^2(m \sin^{-1} x) m^2 \frac{1}{1-x^2} + \\ &&&\quad\quad\quad-\sin(m \sin^{-1} x) \cdot \cos(m \sin^{-1} x) \cdot m \cdot (1-x^2)^{-\frac32} \cdot (-2x) \\ \Rightarrow && (1-x^2)y^{(2)} &= 2m^2-4m^2y+xy' \\ &&&= xy^{(1)} + 2m^2(1-2y) \\ \\ \Rightarrow && (1-x^2)y^{(n+2)}-2nxy^{(n+1)}-2\binom{n}{2}y^{(n)} &= xy^{(n+1)}+ny^{(n)} -4m^2y^{(n)} \\ \Rightarrow && (1-x^2)y^{(n+2)} &= (2n+1)xy^{(n+1)}+(n(n-1)+n-4m^2)y^{(n)} \\ &&&= (2n+1)xy^{(n+1)}+(n^2-4m^2)y^{(n)} \\ \end{align*} \begin{align*} && y(0) &= \sin^2(m \sin^{-1} 0) \\ &&&= \sin^2 0 = 0 \\ \\ && y'(0) &= 0 \\ && (1-0^2)y^{(2)}(0) &= 2m^2(1-2y(0)) \\ \Rightarrow && y^{(2)}(0) &= 2m^2 \\ \\ && y^{(n+2)} (0) &= (2n+1) \cdot 0 \cdot y^{(n+1)} +(n^2-4m^2)y^{(n)}(0) \\ &&&= (n^2-4m^2)y^{(n)}(0) \\ \\ && y^{(2)}(0) &= 2m^2 \\ && y^{(4)}(0) &= (4-4m^2) \cdot 2m^2 \\ &&&= -8m(m+1)m(m-1) \\ && y^{(6)}(0) &= 32m(m+2)(m+1)m(m-1)(m-2) \\ && y^{(2k)}(0) &= (-1)^{k+1}2^{2k-1}m (m+k)\cdots(m-k) \text{ if }k < m \\ \\ && y &= m^2x^2 -2m\binom{m+1}{3} x^4 + \frac{16}{3}m\binom{m+2}{5}x^6 - \cdots \\ &&&+ (-1)^{k}\frac{2^{2k}}{k+1} m \binom{m+k}{2k+1}x^{2k+2}+\cdots \\ &&&= mx^2\sum_{k=0}^{m-1} \frac{(-1)^k2^{2k}}{k+1}\binom{m+k}{2k+1}x^{2k} \end{align*}
If \(\mathrm{Q}\) is a polynomial, \(m\) is an integer, \(m\geqslant1\) and \(\mathrm{P}(x)=(x-a)^{m}\mathrm{Q}(x),\) show that \[ \mathrm{P}'(x)=(x-a)^{m-1}\mathrm{R}(x) \] where \(\mathrm{R}\) is a polynomial. Explain why \(\mathrm{P}^{(r)}(a)=0\) whenever \(1\leqslant r\leqslant m-1\). (\(\mathrm{P}^{(r)}\) is the \(r\)th derivative of \(\mathrm{P}.\)) If \[ \mathrm{P}_{n}(x)=\frac{\mathrm{d}^{n}}{\mathrm{d}x^{n}}(x^{2}-1)^{n} \] for \(n\geqslant1\) show that \(\mathrm{P}_{n}\) is a polynomial of degree \(n\). By repeated integration by parts, or otherwise, show that, if \(n-1\geqslant m\geqslant0,\) \[ \int_{-1}^{1}x^{m}\mathrm{P}_{n}(x)\,\mathrm{d}x=0 \] and find the value of \[ \int_{-1}^{1}x^{n}\mathrm{P}_{n}(x)\,\mathrm{d}x. \] {[}Hint. \textit{You may use the formula \[ \int_{0}^{\frac{\pi}{2}}\cos^{2n+1}t\,\mathrm{d}t=\frac{(2^{2n})(n!)^{2}}{(2n+1)!} \] without proof if you need it. However some ways of doing this question do not use this formula.}{]}
Solution: \begin{align*} && P(x) &= (x-a)^mQ(x) \\ \Rightarrow && P'(x) &= m(x-a)^{m-1}Q(x) + (x-a)^mQ'(x) \\ &&&= (x-a)^{m-1}(\underbrace{mQ(x) + (x-a)Q'(x)}_{\text{a polynomial}}) \\ &&&= (x-a)^{m-1}R(x) \end{align*} Therefore \(P^{(r)}(a) = 0\) for \(1 \leq r \leq m-1\) since each time we differentiate we will have a factor of \((x-a)^{m-r}\) which is zero when we evaluate at \(x = a\). If \(P_n(x) = \frac{\d^n}{\d x^n}(x^2-1)^n\) then we are differentiating a degree \(2n\) polynomial \(n\) times. Each time we differentiate we reduce the degree by \(1\), therefore the degree of \(P_n\) is \(n\). \begin{align*} && \int_{-1}^1 x^mP_n(x) \d x &= \left [x^m \underbrace{\frac{\d^{n-1}}{\d x^{n-1}}\left ( (x-1)^{n} (x+1)^{n} \right)}_{\text{has a factor of }x-1\text{ and }x+1}\right]_{-1}^1 - \int_{-1}^1 mx^{m-1}\frac{\d^{n-1}}{\d x^{n-1}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&&= 0 - \int_{-1}^1 mx^{m-1}\frac{\d^{n-1}}{\d x^{n-1}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&&= -\left [mx^{m-1} \underbrace{\frac{\d^{n-2}}{\d x^{n-2}}\left ( (x-1)^{n} (x+1)^{n} \right)}_{\text{has a factor of }x-1\text{ and }x+1}\right]_{-1}^1+ \int_{-1}^1 m(m-1)x^{m-2}\frac{\d^{n-2}}{\d x^{n-2}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&&= m(m-1)\int_{-1}^1 x^{m-2}\frac{\d^{n-2}}{\d x^{n-2}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&& \cdots \\ &&&= (-1)^m m!\int_{-1}^1 \frac{\d^{n-m}}{\d x^{n-m}} \left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&&= 0 \end{align*} If \(n = m\), we have \begin{align*} && \int_{-1}^1 x^n P_n(x) \d x&= (-1)^nn! \int_{-1}^1 (x^2-1)^n \d x \\ && &= (-1)^{2n}n! \cdot 2\int_{0}^1 (1-x^2)^n \d x \\ x = \sin \theta, \d x = \cos \theta \d \theta: &&&= 2 \cdot n!\int_{0}^{\pi/2} \cos^{2n} \theta \cdot \cos \theta \d \theta \\ &&&= 2 \cdot n!\int_{0}^{\pi/2} \cos^{2n+1} \theta \d \theta \\ &&&= 2 \cdot n!\frac{(2^{2n})(n!)^{2}}{(2n+1)!} \\ &&&= \frac{(2^{2n+1})(n!)^{3}}{(2n+1)!} \\ \end{align*}
The function \(\mathrm{f}\) is given by \(\mathrm{f}(x)=\sin^{-1}x\) for \(-1 < x < 1.\) Prove that \[ (1-x^{2})\mathrm{f}''(x)-x\mathrm{f}'(x)=0. \] Prove also that \[ (1-x^{2})\mathrm{f}^{(n+2)}(x)-(2n+1)x\mathrm{f}^{(n+1)}(x)-n^{2}\mathrm{f}^{(n)}(x)=0, \] for all \(n>0\), where \(\mathrm{f}^{(n)}\) denotes the \(n\)th derivative of \(\mathrm{f}\). Hence express \(\mathrm{f}(x)\) as a Maclaurin series. The function \(\mathrm{g}\) is given by \[ \mathrm{g}(x)=\ln\sqrt{\frac{1+x}{1-x}}, \] for \(-1 < x < 1.\) Write down a power series expression for \(\mathrm{g}(x),\) and show that the coefficient of \(x^{2n+1}\) is greater than that in the expansion of \(\mathrm{f},\) for each \(n > 0\).
Prove that, for any integers \(n\) and \(r\), with \(1\leqslant r\leqslant n,\) \[ \binom{n}{r}+\binom{n}{r-1}=\binom{n+1}{r}. \] Hence or otherwise, prove that \[ (uv)^{(n)}=u^{(n)}v+\binom{n}{1}u^{(n-1)}v^{(1)}+\binom{n}{2}u^{(n-2)}v^{(2)}+\cdots+uv^{(n)}, \] where \(u\) and \(v\) are functions of \(x\) and \(z^{(r)}\) means \(\dfrac{\mathrm{d}^{r}z}{\mathrm{d}x^{r}}\). Prove that, if \(y=\sin^{-1}x,\) then \((1-x^{2})y^{(n+2)}-(2n+1)xy^{(n+1)}-n^{2}y^{(n)}=0.\)
Solution: \begin{align*} \binom{n}{r} + \binom{n}{r-1} &= \frac{n!}{r!(n-r)!} + \frac{n!}{(r-1)!(n-r+1)!} \\ &= \frac{n!}{(r-1)!(n-r)!} \left ( \frac{1}{r} + \frac{1}{n-r+1} \right) \\ &= \frac{n!}{(r-1)!(n-r)!} \frac{(n-r+1)+r}{r(n-r+1)} \\ &= \frac{n! (n+1)}{r! (n-r+1)!} \\ &= \frac{(n+1)!}{r!(n+1-r)!} \\ &= \binom{n+1}{r} \end{align*} Claim: \(\displaystyle (uv)^{(n)} = \sum_{r=0}^n \binom{n}{r} u^{(n-r)} v^{(r)}\) Proof: (By induction on \(n\)). Base case: \(n = 0\) is clear. Inductive step: Suppose it is true for \(n = k\), then consider \begin{align*} (uv)^{(k+1)} &= \left ( (uv)^{(k)} \right)' \\ &= \left ( \sum_{r=0}^k \binom{k}{r} u^{(k-r)} v^{(r)} \right)' \tag{by assumption} \\ &=\sum_{r=0}^k \binom{k}{r} \left ( u^{(k-r)} v^{(r)}\right)' \tag{linearity} \\ &=\sum_{r=0}^k \binom{k}{r} \left ( u^{(k-r+1)} v^{(r)} + u^{(k-r)}v^{(r+1)}\right) \\ &= \sum_{r=0}^{k} \binom{k}{r} u^{(k-r+1)} v^{(r)} + \sum_{r=0}^{k} \binom{k}{r} u^{(k-r)}v^{(r+1)} \\ &= \sum_{r=0}^{k} \binom{k}{r} u^{(k-r+1)} v^{(r)} + \sum_{r=1}^{k+1} \binom{k}{r-1} u^{(k-r+1)}v^{(r)} \\ &= u^{(k+1)}v + \sum_{r=1}^k \left (\binom{k}{r} + \binom{k}{r-1} \right)u^{(k-r+1)}v^{(r)} + u v^{(k+1)}\\ &= u^{(k+1)}v + \sum_{r=1}^k \binom{k+1}{r} u^{(k-r+1)}v^{(r)} + u v^{(k+1)}\\ &= \sum_{r=0}^{k+1} \binom{k+1}{r} u^{(k-r+1)}v^{(r)}\\ \end{align*} Therefore if our statement is true for \(n = k\) it is true for \(n = k+1\). Since it is true for \(n = 0\) by the principle of mathematical induction it is true for all integer \(n \geq 0\) Suppose \( y = \sin^{-1} x\), then \(y' = \frac{1}{\sqrt{1-x^2}}\), \(y'' = \frac{x}{(1-x^2)^{3/2}}\). Not that this means that \((1-x^2)y'' - xy' = 0\) (which is our formula when \(n = 0\)). Now apply Leibniz's formula to this. \begin{align*} 0 &= \left ( (1-x^2)y'' - xy' \right)^{(n)} \\ &= \left ( (1-x^2)y'' \right)^{(n)} -\left ( xy' \right)^{(n)} \\ &= \left ( (1-x^2)y^{(n+2)} - n\cdot 2x \cdot y^{(n+1)}-\binom{n}{2} \cdot 2 \cdot y^{(n)} \right )- \left (xy^{(n+1)}+ny^{(n)} \right) \\ &= (1-x^2)y^{(n+2)} - (2n+1)y^{(n+1)} - \left ( n(n-1)+n \right)y^{(n)} \\ &= (1-x^2)y^{(n+2)} - (2n+1)y^{(n+1)} - n^2y^{(n)} \\ \end{align*} as required