Toronto Math Forum

APM346-2016F => APM346--Lectures => Chapter 8 => Topic started by: Luyu CEN on November 23, 2016, 04:26:33 PM

Title: Typo
Post by: Luyu CEN on November 23, 2016, 04:26:33 PM
http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter8/S8.1.html#mjx-eqn-eq-8.1.9

If we follow from equation(7) http://www.math.toronto.edu/courses/apm346h1/20169/PDE-textbook/Chapter8/S8.1.html#mjx-eqn-eq-8.1.7
(9) should be
\begin{equation}
\sin^2(\phi)
\Phi'' +\sin(\phi)\cos(\phi)\Phi' = -\bigl(l(l+1)\sin^2(\phi) -m^2\bigr)\Phi,
\end{equation}
instead of
\begin{equation}
\sin^2(\phi)
\Phi'' +2\sin(\phi)\cos(\phi)\Phi' = -\bigl(l(l+1)\sin^2(\phi) -m^2\bigr)\Phi,
\end{equation}
There will be a coefficient 2 if we substitute $\Phi(\phi)=L(\cos(\phi))$ into the equation and use the chain rule to get its derivatives but I think not now.
Title: Re: Typo
Post by: Victor Ivrii on November 24, 2016, 06:00:41 AM
Thanks. There are also more serious omissions. Corrected.