How do you take the n'th derivative of Legendre's equation after substituting $\alpha x+\beta=e^t$ in $a_n(\alpha x+\beta)^n y^{(n)}+\dots+a_0y=f(x)$

Asked Sep 04 '17 at 13:17

Active Sep 04 '17 at 13:17

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Background This is from Hobson et al. Mathematical methods.

Taking the $\frac{d^n}{dx^n}$ when changing the independant variable to t with the substitution $\alpha x+\beta=e^t$, I don't see what pattern the dots represent, nor how the final equation was derived. I get for the third derivative: $$\frac{d}{dx}(\frac{d^2y}{dx^2})=\frac{-2\alpha^3}{(\alpha x+\beta)^3}\left (\frac{d^2y}{dt^2}-\frac{dy}{dt}\right )+\frac{\alpha^3}{(\alpha x+\beta)^3}\left (\frac{d^3y}{dt^3}-\frac{d^2y}{dt^2}\right )$$

$$\left (\frac{d^3y}{dx^3}\right )=\frac{\alpha^3}{(\alpha x+\beta)^3}\left (\frac{d^3}{dt^3}-3\frac{d^2y}{dt^2}+2\frac{d}{dt}\right )y$$

What do the dots in the final equation below represent? How were these factored differentials derived?

asked Sep 04 '17 at 13:17

user5389726598465

1,171

1

Induction, chain rule & derivative of a product – Donald Splutterwit Sep 04 '17 at 13:42
@DonaldSplutterwit so what do the dots represent? Possibly that will make things clear – user5389726598465 Sep 04 '17 at 13:57
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\begin{eqnarray} \frac{d}{dt} \left( \frac{d}{dt} -1 \right) \left( \frac{d}{dt} -2 \right)\cdots \left( \frac{d}{dt} -n+2 \right) \left( \frac{d}{dt} -n+1 \right) \end{eqnarray} Like a falling factorial. It is quite easy to prove by induction. – Donald Splutterwit Sep 04 '17 at 14:52
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@DonaldSplutterwit ok, it does make sense now. Thanks :) – user5389726598465 Sep 04 '17 at 15:01

How do you take the n'th derivative of Legendre's equation after substituting $\alpha x+\beta=e^t$ in $a_n(\alpha x+\beta)^n y^{(n)}+\dots+a_0y=f(x)$

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