1. The problem statement, all variables and given/known data
For [itex]y=sin(m \arcsin x) [/itex], prove that
$$(1-x^2)y_{n+2} - (2n+1)xy_{n+1}+(m^2-n^2)y_n=0 $$
2. Relevant equations
3. The attempt at a solution
My first approach would be to find a general expression for y_n. Starting with
$$y_1=m\dfrac{\cos(m \arcsin x)}{\sqrt{1-x^2}} \\
y_2 = m\dfrac{\cos(m \arcsin x)}{\sqrt{1-x^2}(1-x^2)} - m^2 \dfrac{\sin(m \arcsin x)}{1-x^2} $$
The expression seems to get uglier as n increases and it's too difficult to calculate y_3. Also it's not easy to generalize an expression for y_n just by looking at the results obtained.
For [itex]y=sin(m \arcsin x) [/itex], prove that
$$(1-x^2)y_{n+2} - (2n+1)xy_{n+1}+(m^2-n^2)y_n=0 $$
2. Relevant equations
3. The attempt at a solution
My first approach would be to find a general expression for y_n. Starting with
$$y_1=m\dfrac{\cos(m \arcsin x)}{\sqrt{1-x^2}} \\
y_2 = m\dfrac{\cos(m \arcsin x)}{\sqrt{1-x^2}(1-x^2)} - m^2 \dfrac{\sin(m \arcsin x)}{1-x^2} $$
The expression seems to get uglier as n increases and it's too difficult to calculate y_3. Also it's not easy to generalize an expression for y_n just by looking at the results obtained.
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