\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
Since \(y = \cos (m \arcsin x)\), we have \[ \DiffFrac {y}{x} = - \frac {m \sin (m \arcsin x)}{\sqrt {1 - x^2}}, \] and \begin {align*} \NdiffFrac {2}{y}{x} & = - \frac {m^2 \cos (m \arcsin x) \cdot \frac {1}{\sqrt {1 - x^2}} \cdot \sqrt {1 - x^2} - m \sin (m \arcsin x) \cdot (-x) \cdot \frac {1}{\sqrt {1 - x^2}}}{1 - x^2} \\ & = -\frac {m}{1 - x^2} \left (m \cos (m \arcsin x) + x \sin (m \arcsin x) \cdot \frac {1}{\sqrt {1 - x^2}}\right ). \end {align*}
Hence, the left-hand side of the differential equation reduces to \begin {align*} & \phantom {=} (1 - x^2) \DiffFrac {y}{x} - x \DiffFrac {y}{x} + m^2 y \\ & = -m \cdot \left (m \cos (m \arcsin x) + x \sin (m \arcsin x) \cdot \frac {1}{\sqrt {1 - x^2}}\right ) \\ & \phantom {=} + \frac {mx \sin (m \arcsin x)}{\sqrt {1 - x^2}} + m^2 \cos (m \arcsin x) \\ & = -m^2 \cos (m \arcsin x) + m^2 \cos (m \arcsin x) \\ & \phantom {=} - \frac {mx \sin (m \arcsin x)}{\sqrt {1 - x^2}} + \frac {mx \sin (m \arcsin x)}{\sqrt {1 - x^2}} \\ & = 0, \end {align*}
as desired.
Differentiating both sides of this equation with respect to \(x\), we get \[ (-2x) \NdiffFrac {2}{y}{x} + (1 - x^2) \NdiffFrac {3}{y}{x} - \DiffFrac {y}{x} - x \NdiffFrac {2}{y}{x} + m^2 \DiffFrac {y}{x} = 0, \] which reduces to \[ (1 - x^2) \NdiffFrac {3}{y}{x} -3x \NdiffFrac {2}{y}{x} + (m^2 - 1) \DiffFrac {y}{x} = 0. \]
Differentiating both sides with respect to \(x\) again, we get \[ (-2x) \NdiffFrac {3}{y}{x} + (1 - x^2) \NdiffFrac {4}{y}{x} - 3 \NdiffFrac {2}{y}{x} - 3x \NdiffFrac {3}{y}{x} + (m^2 - 1) \NdiffFrac {2}{y}{x} = 0, \] which reduces to \[ (1 - x^2) \NdiffFrac {4}{y}{x} - 5x \NdiffFrac {3}{y}{x} + (m^2 - 4) \NdiffFrac {2}{y}{x} = 0. \]
The conjecture is for all \(n \geq 0\), \[ (1 - x^2) \NdiffFrac {n + 2}{y}{x} - (2n + 1) \NdiffFrac {n + 1}{y}{x} + (m^2 - n^2) \NdiffFrac {n}{y}{x} = 0. \]
The base case where \(n = 0\) is already shown. We show the inductive step. Assume this statement is true for some \(n = k\), i.e. \[ (1 - x^2) \NdiffFrac {k + 2}{y}{x} - (2k + 1)x \NdiffFrac {k + 1}{y}{x} + (m^2 - k^2) \NdiffFrac {k}{y}{x} = 0. \]
Differentiating both sides with respect to \(x\) gives \[ (-2x) \NdiffFrac {k + 2}{y}{x} + (1 - x^2) \NdiffFrac {k + 3}{y}{x} - (2k + 1) \NdiffFrac {k + 1}{y}{x} - (2k + 1)x \NdiffFrac {k + 2}{y}{x} + (m^2 - k^2) \NdiffFrac {k + 1}{y}{x} = 0, \] which reduces to \[ (1 - x^2) \NdiffFrac {k + 3}{y}{x} - (2k + 3)x \NdiffFrac {k + 2}{y}{x} + (m^2 - (k + 1)^2) \NdiffFrac {k + 1}{y}{x} = 0. \]
This is precisely the statement for when \(n = k + 1\).
Hence, by the principle of mathematical induction, the conjecture holds for all integers \(n \geq 0\).
Now, we evaluate this at \(x = 0\), and we have \[ \LEvalAt {\NdiffFrac {n + 2}{y}{x}}{x = 0} + (m^2 - n^2) \LEvalAt {\NdiffFrac {n}{y}{x}}{x = 0} = 0 \] for all \(n \geq 0\), which rearranged gives \[ \LEvalAt {\NdiffFrac {n + 2}{y}{x}}{x = 0} = (n^2 - m^2) \LEvalAt {\NdiffFrac {n}{y}{x}}{x = 0}. \]
Notice that \[ \LEvalAt {y}{x = 0} = \cos (m \arcsin 0) = 1, \] and \[ \LEvalAt {\DiffFrac {y}{x}}{x = 0} = - \frac {m \sin (m \arcsin 0)}{\sqrt {1 - 0^2}} = 0 \]
Hence, \[ \LEvalAt {\NdiffFrac {2}{y}{x}}{x = 0} = (0^2 - m^2) \LEvalAt {y}{x = 0} = -m^2, \] and \[ \LEvalAt {\NdiffFrac {3}{y}{x}}{x = 0} = (1^2 - m^2) \LEvalAt {\DiffFrac {y}{x}}{x = 0} = 0, \] and \[ \LEvalAt {\NdiffFrac {4}{y}{x}}{x = 0} = (2^2 - m^2) \LEvalAt {\NdiffFrac {2}{y}{x}}{x = 0} = -m^2 (2^2 - m^2), \]
In general, we have \[ \LEvalAt {\NdiffFrac {2n + 1}{y}{x}}{x = 0} = 0, \] and \[ \LEvalAt {\NdiffFrac {2n}{y}{x}}{x = 0} = \prod _{k = 0}^{n - 1} (4k^2 - m^2) = (-1)^n \prod _{k = 0}^{n - 1} (m^2 - 4k^2) \] for all integers \(n \geq 0\).
Hence, the Maclaurin series for \(y\) satisfy that \begin {align*} y & = \sum _{n = 0}^{\infty } \frac {\LEvalAt {\NdiffFrac {n}{y}{x}}{x = 0} x^n}{n!} \\ & = \sum _{n = 0}^{\infty } \frac {(-1)^n \prod _{k = 0}^{n - 1} (m^2 - 4k^2) x^{2n}}{(2n)!} \\ & = 1 - \frac {m^2 x^2}{2!} + \frac {m^2 (m^2 - 2^2) x^4}{4!} - \cdots . \end {align*}
In the case where \(m\) is even, notice that when \(m = 2k\), \(m^2 - 4k^2 = 0\), and so for all \(n \geq \frac {m}{2} + 1\), \[ \prod _{k = 0}^{n - 1} (m^2 - 4k^2) x^{2n} = 0, \] and hence this infinite sum becomes finite: \begin {align*} y & = \sum _{n = 0}^{\infty } \frac {(-1)^n \prod _{k = 0}^{n - 1} (m^2 - k^2) x^{2n}}{(2n)!} \\ & = \sum _{n = 0}^{\frac {m}{2}} \frac {(-1)^n \prod _{k = 0}^{n - 1} (m^2 - k^2) x^{2n}}{(2n)!}. \end {align*}
Now, let \(x = \sin \theta \), we have \(\theta = \arcsin x\) since \(\abs *{\theta } < \frac {1}{2}\pi \), and \(y = \cos m\theta \). Hence, \[ \cos m\theta = 1 - \frac {m^2 \sin ^2\theta }{2!} + \frac {m^2 (m^2 - 2^2) \sin ^4\theta }{4!} - \cdots , \] where the sum is finite (and hence a polynomial), and the degree of this polynomial is \(m\);