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A method to find the expansion of a function is to first differentiate the equation $y=f(x)$ twice with respect to $x$, and combine the results to form an equation in $y$, $y'$, and $y''$. Next, assume $y=A+Bx+Cx^2+\&\text{c}.$ Differentiate twice, and substitute the three values of $y$, $y'$, $y''$ so obtained in the former equation. Lastly, equate coefficients in the result to determine in succession $A,B,C,\&\text{c}.$

Ex. - To expand $\sin m\theta$ and $\cos m\theta$ in ascending powers of $\sin\theta$ or $\cos\theta$. Put $x=\sin\theta$, and $y=\sin m\theta=\sin(m\sin^{-1}x)$. Therefore$$y'=\cos(m\sin^{-1}x)\frac m{\sqrt{1-x^2}}\tag1$$$$y''=-\sin(m\sin^{-1}x)\frac {m^2}{\sqrt{1-x^2}}+\cos(m\sin^{-1}x)\frac {mx}{\left(1-x^2\right)^{3/2}}\tag2$$Therefore, eliminating $\cos(m\sin^{-1}x)$,$$(1-x^2)y''-xy'+m^2y=0\tag3$$Let$$y=A+A_1x+A_2x^2+A_3x^3+\ldots+A_{n-1}x^{n-1}+A_nx^n\tag4$$Differentiate twice, and put the values of $y$, $y'$, and $y''$ in equation $(3)$$$0=m^2\left(A+A_1x+\ldots+A_nx^n\right)-x\left(A_1+2A_2x+3A_3x^2+\ldots+nA_nx^{n-1}\right)+(1-x^2)\left(2A_2+6A_3x+\ldots+n(n-1)A_nx^{n-2}+n(n+1)A_{n+1}x^{n-1}+(n+1)(n+2)A_{n+2}x^n\right)$$Equate the collected coefficients of $x^n$ to zero, we get the relation$$A_{n+2}=\frac {n^2-m^2}{(n+1)(n+2)}A_n\tag5$$ etc.

Questions:

  1. Why do you stop at $x^{n-1}$ for $y'$? Why do you not add $(n+1)A_{n+1}x^{n}$?
  2. Where does the relation for $(5)$ come from? If the book got 1. right, then I'm not sure how $(5)$ is derived.

I've spent some time going over the book and looking at each step closely. But I still can't tell whether the last term of $y'$ is a mistake, or is intentional.

If it is a mistake, it'll affect question 2 too. I'm wondering if you guys can help me a bit.

Crescendo
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  • I think that $(2)$ is not correct. It should be $$y''=\frac{m x \cos \left(m \sin ^{-1}(x)\right)}{\left(1-x^2\right)^{3/2}}-\frac{m^2 \sin \left(m \sin ^{-1}(x)\right)}{1-x^2}$$ – Claude Leibovici Jul 25 '17 at 08:13

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I hope that the process would be clearer using summations instead.

Considering the differential equation $$(1-x^2)y''-xy'+m^2y=0$$ and using $$y=\sum_{i=0}^k a_ix^i\qquad y'=\sum_{i=0}^k ia_ix^{i-1}\qquad y''=\sum_{i=0}^k i(i-1)a_ix^{i-2}$$ we then have (expanding $(1-x^2)y''$ as $y''-x^2y''$) $$\sum_{i=0}^k i(i-1)a_ix^{i-2}-\sum_{i=0}^k i(i-1)a_ix^{i}-\sum_{i=0}^k ia_ix^{i}+\sum_{i=0}^k m^2a_ix^i=0 $$ that be compacted (notice that the last three summations are involving $x^i$) as $$\sum_{i=0}^k i(i-1)a_ix^{i-2}-\sum_{i=0}^k(i^2-m^2)a_ix^i=0$$ Now, for the same degree $n$, we then have $$(n+1)(n+2)a_{n+2}-(n^2-m^2)a_n=0\implies a_{n+2}=\frac{n^2-m^2}{(n+1)(n+2)}$$ and since it is a second order differential equation $a_0$ and $a_1$ are just undefined for the time being.

Claude Leibovici
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