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HW1, Physics 555A, Spring 2008 (Audit)
spring 2009 Compiled on October 14, 2025 at 5:14pm [public]
Contents
1 Problem
Show that the recurence formula
\begin{equation} C_{q}=-\frac {2\left ( k-q\right ) }{q\left ( q+2l+1\right ) \left ( k+l\right ) }C_{q-1}\tag {1}\end{equation}
can be written as
\begin{equation} C_{q}=\left ( -1\right ) ^{q}\left ( \frac {2}{k+l}\right ) ^{q}\frac {\left ( k-1\right ) !}{\left ( k-q-1\right ) !}\frac {\left ( 2l+1\right ) !}{q!\left ( q+2l+1\right ) !}C_{0}\tag {2}\end{equation}
2 Solution
Proof by induction on \(q\). For \(q=1\), equation (1) becomes
\[ C_{1}=-\frac {2\left ( k-1\right ) }{\left ( 2l+2\right ) \left ( k+l\right ) }C_{0}\]
and equation (2) becomes
\begin{align*} C_{1} & =\left ( -1\right ) \left ( \frac {2}{k+l}\right ) \frac {\left ( k-1\right ) !}{\left ( k-2\right ) !}\frac {\left ( 2l+1\right ) !}{\left ( 2l+2\right ) !}C_{0}\\ & =-\frac {2\left ( k-1\right ) }{\left ( k+l\right ) \left ( 2l+2\right ) !}C_{0}\end{align*}
Hence it is true for \(q=1\). Now assume it is true for \(q=n\), in otherwords, assume that
\begin{equation} C_{n}=-\frac {2\left ( k-n\right ) }{n\left ( n+2l+1\right ) \left ( k+l\right ) }C_{n-1}\tag {3}\end{equation}
implies
\begin{equation} C_{n}=\left ( -1\right ) ^{n}\left ( \frac {2}{k+l}\right ) ^{n}\frac {\left ( k-1\right ) !}{\left ( k-n-1\right ) !}\frac {\left ( 2l+1\right ) !}{n!\left ( n+2l+1\right ) !}C_{0}\tag {4}\end{equation}
Now for the induction step. we need to show that it is true for \(n+1\), i.e. given (4) is true, we need to
show that, by replacing \(n\) by \(n+1\) in the above, that
\begin{equation} C_{n+1}=-\frac {2\left ( k-\left ( n+1\right ) \right ) }{\left ( n+1\right ) \left ( \left ( n+1\right ) +2l+1\right ) \left ( k+l\right ) }C_{n}\tag {5}\end{equation}
implies
\begin{align} C_{n+1} & =\left ( -1\right ) ^{n+1}\left ( \frac {2}{k+l}\right ) ^{n+1}\frac {\left ( k-1\right ) !}{\left ( k-\left ( n+1\right ) -1\right ) !}\frac {\left ( 2l+1\right ) !}{\left ( n+1\right ) !\left ( \left ( n+1\right ) +2l+1\right ) !}C_{0}\nonumber \\ & =\left ( -1\right ) ^{n+1}\left ( \frac {2}{k+l}\right ) ^{n+1}\frac {\left ( k-1\right ) !}{\left ( k-n-2\right ) !}\frac {\left ( 2l+1\right ) !}{\left ( n+1\right ) !\left ( n+2l+2\right ) !}C_{0}\tag {6}\end{align}
We start with (5), and replace the \(C_{n}\) term with what we assumed to be true from (4), hence (5) can
be rewritten as
\[ C_{n+1}=-\frac {2\left ( k-\left ( n+1\right ) \right ) }{\left ( n+1\right ) \left ( \left ( n+1\right ) +2l+1\right ) \left ( k+l\right ) }\overset {C_{n}\text { from (4)}}{\overbrace {\left [ \left ( -1\right ) ^{n}\left ( \frac {2}{k+l}\right ) ^{n}\frac {\left ( k-1\right ) !}{\left ( k-n-1\right ) !}\frac {\left ( 2l+1\right ) !}{n!\left ( n+2l+1\right ) !}C_{0}\right ] }}\]
Simplify the above leads to
\[ C_{n+1}=\left ( -1\right ) ^{n+1}\left ( \frac {2}{k+l}\right ) ^{n+1}\frac {\left ( k-1\right ) !}{\left ( k-n-2\right ) !}\frac {\left ( 2l+1\right ) !}{n!\left ( n+2l+2\right ) !}C_{0}\]
Which is (6). Therefore, the relationship is true for any \(n\). QED