Solution
Starting with ode (10.4.12) which is \begin {equation} \psi ^{\prime \prime }\relax (y) -y^{2}\psi \relax (y) =-2\epsilon \psi \relax (y) \tag {10.4.12} \end {equation} Where \(\epsilon =\frac {E}{\hbar \omega }\) the energy of the particle. Let the solution be \begin {align} \psi \relax (y) & =u\relax (y) e^{\frac {-y^{2}}{2}}\nonumber \\ & =e^{\frac {-y^{2}}{2}}{\displaystyle \sum \limits _{m=0}^{n}} a_{m}y^{m} \tag {1} \end {align}
Where \begin {equation} u\relax (y) ={\displaystyle \sum \limits _{m=0}^{n}} a_{m}y^{m} \tag {1A} \end {equation} Eq. (1) can be written as\[ \psi \relax (y) =\left \{ \begin {array} [c]{ccc}e^{\frac {-y^{2}}{2}}{\displaystyle \sum \limits _{m=0}^{n}} a_{m}y^{m} & & \lim _{y\rightarrow 0}\\ a_{m}y^{m}e^{\frac {-y^{2}}{2}} & & \lim _{y\rightarrow \infty }\end {array} \right . \] Substituting (1) in 10.4.12 gives\begin {align*} \frac {d^{2}}{dy^{2}}\left (ue^{\frac {-y^{2}}{2}}\right ) -y^{2}ue^{\frac {-y^{2}}{2}} & =-2\epsilon ue^{\frac {-y^{2}}{2}}\\ \frac {d}{dy}\left (u^{\prime }e^{\frac {-y^{2}}{2}}-uye^{\frac {-y^{2}}{2}}\right ) -y^{2}ue^{\frac {-y^{2}}{2}} & =-2\epsilon ue^{\frac {-y^{2}}{2}}\\ \left (u^{\prime \prime }e^{\frac {-y^{2}}{2}}-u^{\prime }ye^{\frac {-y^{2}}{2}}-u^{\prime }ye^{\frac {-y^{2}}{2}}-u\left (e^{\frac {-y^{2}}{2}}-y^{2}e^{\frac {-y^{2}}{2}}\right ) \right ) -y^{2}ue^{\frac {-y^{2}}{2}} & =-2\epsilon ue^{\frac {-y^{2}}{2}}\\ \left (u^{\prime \prime }e^{\frac {-y^{2}}{2}}-u^{\prime }ye^{\frac {-y^{2}}{2}}-u^{\prime }ye^{\frac {-y^{2}}{2}}-ue^{\frac {-y^{2}}{2}}+y^{2}ue^{\frac {-y^{2}}{2}}\right ) -y^{2}ue^{\frac {-y^{2}}{2}} & =-2\epsilon ue^{\frac {-y^{2}}{2}} \end {align*}
Dividing by \(e^{\frac {-y^{2}}{2}}\neq 0\) gives\begin {align*} u^{\prime \prime }-u^{\prime }y-u^{\prime }y-u+y^{2}u-y^{2}u & =-2\epsilon u\\ u^{\prime \prime }-2u^{\prime }y-u & =-2\epsilon u \end {align*}
Which becomes the Hermite ODE as given in 10.4.24\begin {equation} u^{\prime \prime }\relax (y) -2yu^{\prime }\relax (y) +\left ( 2\varepsilon -1\right ) u\relax (y) =0 \tag {10.4.24} \end {equation} From (1A)\begin {align*} u^{\prime } & ={\displaystyle \sum \limits _{m=0}^{n}} ma_{m}y^{m-1}\\ u^{\prime \prime } & ={\displaystyle \sum \limits _{m=0}^{n}} m\left (m-1\right ) a_{m}y^{m-2} \end {align*}
Substituting the above in (10.4.24) gives\begin {align*} {\displaystyle \sum \limits _{m=0}^{n}} m\left (m-1\right ) a_{m}y^{m-2}-2y{\displaystyle \sum \limits _{m=0}^{n}} ma_{m}y^{m-1}+\left (2\varepsilon -1\right ) {\displaystyle \sum \limits _{m=0}^{n}} a_{m}y^{m} & =0\\{\displaystyle \sum \limits _{m=0}^{n}} m\left (m-1\right ) a_{m}y^{m-2}-{\displaystyle \sum \limits _{m=0}^{n}} 2ma_{m}y^{m}+{\displaystyle \sum \limits _{m=0}^{n}} \left (2\varepsilon -1\right ) a_{m}y^{m} & =0\\{\displaystyle \sum \limits _{m=0}^{n}} m\left (m-1\right ) a_{m}y^{m-2}+{\displaystyle \sum \limits _{m=0}^{n}} \left (2\varepsilon -1-2m\right ) a_{m}y^{m} & =0 \end {align*}
The first sum can start from \(m=2\) without affecting the sum, hence the above becomes\[{\displaystyle \sum \limits _{m=2}^{n}} m\left (m-1\right ) a_{m}y^{m-2}+{\displaystyle \sum \limits _{m=0}^{n}} \left (2\varepsilon -1-2m\right ) a_{m}y^{m}=0 \] Let \(m^{\prime }=m-2\) in the first sum, it becomes\[{\displaystyle \sum \limits _{m^{\prime }=0}^{n-2}} \left (m^{\prime }+2\right ) \left (m^{\prime }+1\right ) a_{m^{\prime }+2}y^{m^{\prime }}+{\displaystyle \sum \limits _{m=0}^{n}} \left (2\varepsilon -1-2m\right ) a_{m}y^{m}=0 \] Changing the index in the first sum from \(m^{\prime }\) back to \(m\) gives\[{\displaystyle \sum \limits _{m=0}^{n-2}} \left (m+2\right ) \left (m+1\right ) a_{m+2}y^{m}+{\displaystyle \sum \limits _{m=0}^{n}} \left (2\varepsilon -1-2m\right ) a_{m}y^{m}=0 \] Combining terms gives\begin {equation} {\displaystyle \sum \limits _{m=0}^{n-2}} \left (\left (m+2\right ) \left (m+1\right ) a_{m+2}+\left (2\varepsilon -1-2m\right ) a_{m}\right ) y^{m}+{\displaystyle \sum \limits _{m=n-1}^{n}} \left (2\varepsilon -1-2m\right ) a_{m}y^{m}=0 \tag {1B} \end {equation} Considering the second term above for now. \begin {align*} {\displaystyle \sum \limits _{m=n-1}^{n}} \left (2\varepsilon -1-2m\right ) a_{m}y^{m} & =0\\ \left (2\varepsilon -1-2m\right ) a_{m} & =0\qquad m=n,m=n-1 \end {align*}
Looking at case \(m=n\)\[ \left (2\varepsilon -1-2n\right ) a_{n}=0 \] but \(a_{n}\neq 0\) since that is the highest order of the power series. If \(a_{n}=0\) then the dominant term of the power series is lost. This means \(\left (2\varepsilon -1-2n\right ) =0\) or \begin {equation} \varepsilon =n+\frac {1}{2} \tag {10.4.34} \end {equation} Looking at case \(m=n-1\)\begin {align*} \left (2\varepsilon -1-2\left (n-1\right ) \right ) a_{n-1} & =0\\ \left (2\varepsilon -1-2n+2\right ) a_{n-1} & =0\\ \left (2\varepsilon +1-2n\right ) a_{n-1} & =0 \end {align*}
But \(\varepsilon =n+\frac {1}{2}\), hence the above becomes\begin {align*} \left (2\left (n+\frac {1}{2}\right ) +1-2n\right ) a_{n-1} & =0\\ \left (2n+1+1-2n\right ) a_{n-1} & =0\\ 2a_{n-1} & =0 \end {align*}
This means \begin {equation} a_{n-1}=0 \tag {2} \end {equation} Now looking at case \(m\leq n-2\) from Eq. (1C) above\begin {align*} {\displaystyle \sum \limits _{m=0}^{n-2}} \left (\left (m+2\right ) \left (m+1\right ) a_{m+2}+\left (2\varepsilon -1-2m\right ) a_{m}\right ) y^{m} & =0\\ \left (m+2\right ) \left (m+1\right ) a_{m+2}+\left (2\varepsilon -1-2m\right ) a_{m} & =0\\ a_{m+2} & =\frac {-\left (2\varepsilon -1-2m\right ) }{\left (m+2\right ) \left (m+1\right ) }a_{m} \end {align*}
But \(\varepsilon =n+\frac {1}{2}\), therefore the above becomes\begin {align} a_{m+2} & =\frac {-\left (2\left (n+\frac {1}{2}\right ) -1-2m\right ) }{\left (m+2\right ) \left (m+1\right ) }a_{m}\nonumber \\ & =\frac {-\left (2n-2m\right ) }{\left (m+2\right ) \left (m+1\right ) }a_{m}\nonumber \\ & =-\frac {2\left (n-m\right ) }{\left (m+2\right ) \left (m+1\right ) }a_{m} \tag {3} \end {align}
If \(n\) is even then \(n-1\) is odd. Then \(a_{n-1}=0\) from (2). But due to the recursive formula (3), this implies \(a_{1}=a_{3}=a_{5}\cdots =0\). Which means all odd terms in the solution polynomial vanish. And if \(n\) is odd, then \(n-1\) is even. Therefore \(a_{n-1}=0\), But due to the recursive formula (3), this implies \(a_{0}=a_{2}=a_{4}\cdots =0\). Which means all even terms in the solution polynomial vanish.
Now Eq. (3) is the recursive relation used to determine all coefficients \(a_{i}\). For \(m=0\), (3) gives\begin {equation} a_{2}=-na_{0} \tag {4} \end {equation} For \(m=1\), (3) gives\begin {equation} a_{3}=\frac {-2\left (n-1\right ) }{3!}a_{1} \tag {5} \end {equation} For \(m=2\), (3) gives \begin {align} a_{4} & =\frac {-2\left (n-2\right ) }{\relax (4) \relax (3) }a_{2}\nonumber \\ & =\frac {-2^{2}\left (n-2\right ) }{4!}a_{2}\nonumber \\ & =\frac {2^{2}\left (n-2\right ) n}{4!}a_{0} \tag {6} \end {align}
For \(m=3,\) (3) gives \begin {align} a_{5} & =\frac {-2\left (n-3\right ) }{\left (3+2\right ) \left ( 3+1\right ) }a_{3}\nonumber \\ & =\frac {-2\left (n-3\right ) }{\relax (5) \relax (4) }\frac {-2^{2}\left (n-1\right ) }{3!}a_{1}\nonumber \\ & =\frac {2^{3}\left (n-3\right ) \left (n-1\right ) }{5!}a_{1} \tag {7} \end {align}
For \(m=4\), (3) gives\begin {align} a_{6} & =\frac {-2\left (n-4\right ) }{\left (4+2\right ) \left ( 4+1\right ) }a_{4}\nonumber \\ & =\frac {-2\left (n-4\right ) }{\relax (6) \relax (5) }\frac {2^{2}\left (n-2\right ) n}{4!}a_{0}\nonumber \\ & =\frac {-2^{3}\left (n-4\right ) \left (n-2\right ) n}{6!}a_{0} \tag {8} \end {align}
And so on. Therefore the solution to the Hermite ODE (2) is \begin {align} u & =\sum _{m=0}^{n}a_{m}y^{m}\nonumber \\ & =a_{0}+a_{1}y+a_{2}y^{2}+a_{3}y^{3}+a_{4}y^{4}+a_{5}y^{5}+a_{6}y^{6}+\cdots \nonumber \\ & =a_{0}+a_{1}y-na_{0}y^{2}-\frac {2\left (n-1\right ) }{3!}a_{1}y^{3}+\frac {2^{2}\left (n-2\right ) n}{4!}a_{0}y^{4}+\frac {2^{3}\left ( n-3\right ) \left (n-1\right ) }{5!}a_{1}y^{5}-\frac {2^{3}\left (n-4\right ) \left (n-2\right ) n}{6!}a_{0}y^{6}+\cdots \tag {9} \end {align}
Which can be written as\begin {align*} u\relax (y) & =a_{0}\left (1-ny^{2}+\frac {2^{2}\left (n-2\right ) n}{4!}y^{4}-\frac {2^{3}\left (n-4\right ) \left (n-2\right ) n}{6!}y^{6}+\cdots \right ) \\ & +a_{1}\left (y-\frac {2\left (n-1\right ) }{3!}y^{3}+\frac {2^{3}\left ( n-3\right ) \left (n-1\right ) }{5!}a_{1}y^{5}+\cdots \right ) \end {align*}
Or\[ u\relax (y) =a_{0}u_{0}+a_{1}u_{1}\] Where \(u_{1},u_{2}\) are two linearly independent solutions for the second order Hermite ODE where\begin {align*} u_{0} & =1-ny^{2}+\frac {2^{2}\left (n-2\right ) n}{4!}y^{4}-\frac {2^{3}\left (n-4\right ) \left (n-2\right ) n}{6!}y^{6}+\cdots \\ u_{1} & =y-\frac {2^{2}\left (n-1\right ) }{3!}y^{3}+\frac {2^{3}\left ( n-3\right ) \left (n-1\right ) }{5!}a_{1}y^{5}+\cdots \end {align*}
For even \(n\) the solution \(u_{0}\relax (y) \) will eventually terminates, and for odd \(n\) the solution \(u_{1}\relax (y) \) eventually terminates. The even Hermite polynomials \(H_{0},H_{2},H_{4},\cdots \) are found from \(u_{0}\relax (y) \) for \(n=0,2,4,\cdots \) and the odd Hermite polynomials \(H_{1},H_{3},H_{5},\cdots \) are found from \(u_{1}\relax (y) \) for \(n=1,3,5,\cdots \). The Hermite polynomials need to also be normalize at the end. The even Hermite polynomials are the following
For \(n=0\)\begin {align*} u_{0}\relax (y) & =a_{0}\left (1-ny^{2}-\frac {2^{2}\left ( n-2\right ) n}{4!}y^{4}-\frac {2^{3}\left (n-4\right ) \left (n-2\right ) n}{6!}y^{6}+\cdots \right ) _{n=0}\\ & =a_{0} \end {align*}
Therefore\[ H_{0}\relax (y) =a_{0}\] To find \(a_{0}\), the normalization \(\int _{-\infty }^{\infty }e^{-y^{2}}H_{n^{\prime }}\relax (y) H_{n}\relax (y) dy=2^{n}n!\sqrt {\pi }\delta _{n,n^{\prime }}\) is used, where \(H_{0}\relax (y) =a_{0}\) in this case. This gives\begin {align*} \int _{-\infty }^{\infty }e^{-y^{2}}H_{0}\relax (y) H_{0}\relax (y) dy & =\sqrt {\pi }\\ \int _{-\infty }^{\infty }e^{-y^{2}}a_{0}^{2}dy & =\sqrt {\pi }\\ a_{0}^{2}\int _{-\infty }^{\infty }e^{-y^{2}}dy & =\sqrt {\pi }\\ a_{0}^{2}\sqrt {\pi } & =\sqrt {\pi }\\ a_{0} & =1 \end {align*}
Hence \(a_{0}=1\) and \[ H_{0}\relax (y) =1 \] For \(n=2\)\begin {align*} u_{0}\relax (y) & =a_{0}\left (1-ny^{2}+\frac {2^{2}\left ( n-2\right ) n}{4!}y^{4}-\frac {2^{3}\left (n-4\right ) \left (n-2\right ) n}{6!}y^{6}+\cdots \right ) _{n=2}\\ & =a_{0}\left (1-2y^{2}\right ) \end {align*}
Therefore\[ H_{2}\relax (y) =a_{0}\left (1-2y^{2}\right ) \] To find \(a_{0}\), There is an easier way to normalize \(H_{n}\relax (x) \) than using the normalization integral equation as was done above. This method will be used for the rest of the problem as it is simpler. It works as follows. \(H_{n}\relax (y) =\left (1-2y^{2}\right ) \) is normalized as follows. The coefficient in front of the largest power in \(y^{n}\) is forced to be \(2^{n}\). In the above, the largest power is \(y^{2}\). Hence \(n=2\). Therefore the coefficient is \(2^{2}=4\). But the coefficient is \(-2\). Therefore the whole expression is multiplied by \(-2\). This means \(a_{0}=-2\). Hence \[ H_{2}\relax (y) =-2\left (1-2y^{2}\right ) \] For \(H_{4}\relax (y) \) (This is not required to find, but found for verification)
For \(n=4\)\begin {align*} u_{0}\relax (y) & =a_{0}\left (1-ny^{2}+\frac {2^{2}\left ( n-2\right ) n}{4!}y^{4}-\frac {2^{3}\left (n-4\right ) \left (n-2\right ) n}{6!}y^{6}+\cdots \right ) _{n=4}\\ & =a_{0}\left (1-4y^{2}+\frac {2^{2}\left (4-2\right ) 4}{4!}y^{4}\right ) \\ & =a_{0}\left (1-4y^{2}+\frac {4}{3}y^{4}\right ) \end {align*}
Therefore\[ H_{4}\relax (y) =a_{0}\left (1-4y^{2}+\frac {4}{3}y^{4}\right ) \] \(H_{4}\relax (y) =a_{0}\left (1-4y^{2}+\frac {4}{3}y^{4}\right ) \) is normalized as follows. The coefficient in front of the largest power in \(y^{n}\) is forced to be \(2^{n}\). In the above, the largest power is \(y^{4}\). Hence \(n=4\). Therefore the coefficient is \(2^{4}=16\). But the coefficient is \(\frac {4}{3}\). Therefore the whole expression is multiplied by \(12\). This means \(a_{0}=12\). Hence
\[ H_{4}\relax (y) =12\left (1-4y^{2}+\frac {4}{3}y^{4}\right ) \] Now the odd Hermite polynomials are found. These are found from \(u_{1}\left ( y\right ) \).
For \(n=1\)\begin {align*} u_{1}\relax (y) & =a_{1}\left (y-\frac {2\left (n-1\right ) }{3!}y^{3}+\frac {2^{3}\left (n-3\right ) \left (n-1\right ) }{5!}a_{1}y^{5}+\cdots \right ) _{n=1}\\ & =a_{1}y \end {align*}
Hence\[ H_{1}\relax (y) =a_{1}y \] \(H_{1}\relax (y) =a_{1}y\) is normalized as follows. The coefficient in front of the largest power in \(y^{n}\) is forced to be \(2^{n}\). In the above, the largest power is \(y^{1}\). Hence \(n=1\). Therefore the coefficient is \(2^{1}=2\). But the coefficient is \(1\). Therefore the whole expression is multiplied by \(2\). This means \(a_{1}=2\). Hence
\[ H_{1}\relax (y) =2y \] For \(n=3\)\begin {align*} u_{2}\relax (y) & =a_{1}\left (y-\frac {2\left (n-1\right ) }{3!}y^{3}+\frac {2^{3}\left (n-3\right ) \left (n-1\right ) }{5!}a_{1}y^{5}+\cdots \right ) _{n=3}\\ & =a_{1}\left (y-\frac {2\left (3-1\right ) }{3!}y^{3}\right ) \end {align*}
Hence\[ H_{3}\relax (y) =a_{1}\left (y-\frac {2}{3}y^{3}\right ) \] \(H_{3}\relax (y) =a_{1}\left (y-\frac {2}{3}y^{3}\right ) \) is normalized as follows. The coefficient in front of the largest power in \(y^{n}\) is forced to be \(2^{n}\). In the above, the largest power is \(y^{3}\). Hence \(n=3\). Therefore the coefficient is \(2^{3}=8\). But the coefficient is \(-\frac {2}{3}\). Therefore the whole expression is multiplied by \(-12\). This means \(a_{1}=-12\). Hence
\[ H_{3}\relax (y) =-12\left (y-\frac {2}{3}y^{3}\right ) \] The following gives the final results\begin {align*} H_{0}\relax (y) & =1\\ H_{1}\relax (y) & =2y\\ H_{2}\relax (y) & =-2\left (1-2y^{2}\right ) \\ H_{3}\relax (y) & =-12\left (y-\frac {2}{3}y^{3}\right ) \\ H_{4}\relax (y) & =12\left (1-4y^{2}+\frac {4}{3}y^{4}\right ) \end {align*}
This part verifies the results obtained in part 1 above for \(m,n\leq 2\) using\begin {equation} \int _{-\infty }^{\infty }e^{-y^{2}}H_{n}\relax (y) H_{m}\relax (y) dy=2^{n}n!\sqrt {\pi }\delta _{n,m} \tag {1} \end {equation} For \(n=0,m=0\)
Eq (1) becomes\begin {align*} \int _{-\infty }^{\infty }e^{-y^{2}}H_{0}\relax (y) H_{0}\relax (y) dy & =\sqrt {\pi }\\ \int _{-\infty }^{\infty }e^{-y^{2}}dy & =\sqrt {\pi } \end {align*}
But \(\int _{-\infty }^{\infty }e^{-y^{2}}dy\) is the Gaussian integral which is \(\sqrt {\pi }\). Hence\[ \sqrt {\pi }=\sqrt {\pi }\] Verified.
For \(n=0,m=1\)
Eq (1) becomes\begin {align*} \int _{-\infty }^{\infty }e^{-y^{2}}H_{0}\relax (y) H_{1}\relax (y) dy & =0\\ \int _{-\infty }^{\infty }e^{-y^{2}}\left (2y\right ) dy & =0\\ 2\int _{-\infty }^{\infty }ye^{-y^{2}}dy & =0 \end {align*}
But \(y\) is odd, and \(e^{-y^{2}}\) is even. Hence the LHS is integral over odd function. Hence it must be zero. Therefore\[ 0=0 \] Verified.
For \(n=0,m=2\)
Eq (1) becomes\begin {align*} \int _{-\infty }^{\infty }e^{-y^{2}}H_{0}\relax (y) H_{2}\relax (y) dy & =0\\ \int _{-\infty }^{\infty }e^{-y^{2}}\left (-2\left (1-2y^{2}\right ) \right ) dy & =0\\ \int _{-\infty }^{\infty }e^{-y^{2}}\left (-2+4y^{2}\right ) dy & =0\\ -2\int _{-\infty }^{\infty }e^{-y^{2}}dy+4\int _{-\infty }^{\infty }y^{2}e^{-y^{2}}dy & =0 \end {align*}
But \(\int _{-\infty }^{\infty }e^{-y^{2}}dy=\sqrt {\pi }\) and \(\int _{-\infty }^{\infty }y^{2}e^{-y^{2}}dy=\frac {\sqrt {\pi }}{2}\), therefore the above becomes\begin {align*} -2\sqrt {\pi }+4\left (\frac {\sqrt {\pi }}{2}\right ) & =0\\ -2\sqrt {\pi }+2\sqrt {\pi } & =0\\ 0 & =0 \end {align*}
Verified.
For \(n=1,m=1\)
Eq (1) becomes\begin {align*} \int _{-\infty }^{\infty }e^{-y^{2}}H_{1}\relax (y) H_{1}\relax (y) dy & =2\sqrt {\pi }\\ \int _{-\infty }^{\infty }e^{-y^{2}}\left (2y\right ) \left (2y\right ) dy & =2\sqrt {\pi }\\ 4\int _{-\infty }^{\infty }y^{2}e^{-y^{2}}dy & =2\sqrt {\pi } \end {align*}
But \(\int _{-\infty }^{\infty }y^{2}e^{-y^{2}}dy=\frac {\sqrt {\pi }}{2}\). The above becomes\begin {align*} 4\frac {\sqrt {\pi }}{2} & =2\sqrt {\pi }\\ 2\sqrt {\pi } & =2\sqrt {\pi } \end {align*}
Verified.
For \(n=1,m=2\)
Eq (1) becomes\begin {align*} \int _{-\infty }^{\infty }e^{-y^{2}}H_{1}\relax (y) H_{2}\relax (y) dy & =0\\ \int _{-\infty }^{\infty }e^{-y^{2}}\left (2y\right ) \left (-2\left ( 1-2y^{2}\right ) \right ) dy & =0\\ \int _{-\infty }^{\infty }e^{-y^{2}}\left (8y^{3}-4y\right ) dy & =0\\ 8\int _{-\infty }^{\infty }y^{3}e^{-y^{2}}dy-4\int _{-\infty }^{\infty }ye^{-y^{2}}dy & =0 \end {align*}
Both integrals in the LHS are zero, since both are odd functions. Therefore\[ 0=0 \] Verified.
For \(n=2,m=2\)
Eq (1) becomes\begin {align*} \int _{-\infty }^{\infty }e^{-y^{2}}H_{2}\relax (y) H_{2}\relax (y) dy & =\relax (4) 2!\sqrt {\pi }\\ \int _{-\infty }^{\infty }e^{-y^{2}}\left (\left (-2\left (1-2y^{2}\right ) \right ) \right ) \left (-2\left (1-2y^{2}\right ) \right ) dy & =8\sqrt {\pi }\\ \int _{-\infty }^{\infty }\left (16y^{4}-16y^{2}+4\right ) e^{-y^{2}}dy & =8\sqrt {\pi }\\ 16\int _{-\infty }^{\infty }y^{4}e^{-y^{2}}dy-16\int _{-\infty }^{\infty }y^{2}e^{-y^{2}}dy+4\int _{-\infty }^{\infty }e^{-y^{2}}dy & =8\sqrt {\pi } \end {align*}
But \(\int _{-\infty }^{\infty }\allowbreak y^{4}e^{-y^{2}}dy=\frac {3}{4}\sqrt {\pi }\) and \(\int _{-\infty }^{\infty }y^{2}e^{-y^{2}}dy=\frac {1}{2}\sqrt {\pi }\ \)and \(\int _{-\infty }^{\infty }e^{-y^{2}}dy=\sqrt {\pi }\). The above becomes\begin {align*} 16\left (\frac {3}{4}\sqrt {\pi }\right ) -16\left (\frac {1}{2}\sqrt {\pi }\right ) +4\sqrt {\pi } & =8\sqrt {\pi }\\ 12\sqrt {\pi }-8\sqrt {\pi }+4\sqrt {\pi } & =8\sqrt {\pi }\\ 8\sqrt {\pi } & =8\sqrt {\pi } \end {align*}
Verified. This completes the solution.
Solution
The Legendre ODE is given by 10.4.40 as (\(L\) is used instead of \(l\) as it is more clear because \(l\) looks like \(1\), depending on font used.) \begin {equation} \left (1-x^{2}\right ) y^{\prime \prime }-2xy+L\left (L+1\right ) y=0\tag {10.4.40} \end {equation} Let the solution be \[ y=\sum _{n=0}^{\infty }a_{n}x^{n}\] Then\begin {align*} y^{\prime } & =\sum _{n=0}^{\infty }na_{n}x^{n-1}\\ & =\sum _{n=1}^{\infty }na_{n}x^{n-1} \end {align*}
And\begin {align*} y^{\prime \prime } & =\sum _{n=1}^{\infty }n\left (n-1\right ) a_{n}x^{n-2}\\ & =\sum _{n=2}^{\infty }n\left (n-1\right ) a_{n}x^{n-2} \end {align*}
Substituting the above results back in (10.4.40) gives\begin {align} \left (1-x^{2}\right ) \sum _{n=2}^{\infty }n\left (n-1\right ) a_{n}x^{n-2}-2x\sum _{n=1}^{\infty }na_{n}x^{n-1}+L\left (L+1\right ) \sum _{n=0}^{\infty }a_{n}x^{n} & =0\nonumber \\ \sum _{n=2}^{\infty }n\left (n-1\right ) a_{n}x^{n-2}-x^{2}\sum _{n=2}^{\infty }n\left (n-1\right ) a_{n}x^{n-2}-\sum _{n=1}^{\infty }2na_{n}x^{n}+\sum _{n=0}^{\infty }L\left (L+1\right ) a_{n}x^{n} & =0\nonumber \\ \sum _{n=0}^{\infty }\left (n+2\right ) \left (n+1\right ) a_{n+2}x^{n}-\sum _{n=2}^{\infty }n\left (n-1\right ) a_{n}x^{n}-\sum _{n=1}^{\infty }2na_{n}x^{n}+\sum _{n=0}^{\infty }L\left (L+1\right ) a_{n}x^{n} & =0\tag {1} \end {align}
For \(n=0\) only the above gives
\begin {align*} \left (n+2\right ) \left (n+1\right ) a_{n+2}x^{n}+L\left (L+1\right ) a_{n}x^{n} & =0\\ 2a_{2}+L\left (L+1\right ) a_{0} & =0\\ a_{2} & =-\frac {L\left (L+1\right ) }{2}a_{0} \end {align*}
For \(n=1\) only Eq (1) gives
\begin {align*} \left (n+2\right ) \left (n+1\right ) a_{n+2}x^{n}-2na_{n}x^{n}+L\left ( L+1\right ) a_{n}x^{n} & =0\\ \relax (3) \relax (2) a_{3}-2a_{1}+L\left (L+1\right ) a_{1} & =0\\ a_{3} & =\frac {2a_{1}-L\left (L+1\right ) a_{1}}{6}\\ & =\frac {2-L\left (L+1\right ) }{6}a_{1} \end {align*}
And for \(n\geq 2\), Eq(1) gives the recusive relation
\begin {align*} \left (\left (n+2\right ) \left (n+1\right ) a_{n+2}-n\left (n-1\right ) a_{n}-2na_{n}+L\left (L+1\right ) a_{n}\right ) x^{n} & =0\\ \left (n+2\right ) \left (n+1\right ) a_{n+2}-n\left (n-1\right ) a_{n}-2na_{n}+L\left (L+1\right ) a_{n} & =0\\ \left (n+2\right ) \left (n+1\right ) a_{n+2} & =\left (n\left ( n-1\right ) +2n-L\left (L+1\right ) \right ) a_{n} \end {align*}
Hence the two term recursive is\begin {equation} a_{n+2}=\frac {n\left (n-1\right ) +2n-L\left (L+1\right ) }{\left ( n+2\right ) \left (n+1\right ) }a_{n}\tag {1} \end {equation} For \(n=2\)\begin {align*} a_{4} & =\frac {n\left (n-1\right ) +2n-L\left (L+1\right ) }{\left ( n+2\right ) \left (n+1\right ) }a_{2}\\ & =\frac {2\left (2-1\right ) +4-L\left (L+1\right ) }{\relax (4) \relax (3) }a_{2}\\ & =\frac {6-L\left (L+1\right ) }{12}a_{2} \end {align*}
But \(a_{2}=\frac {-L\left (L+1\right ) }{2}a_{0}\) hence the above becomes\[ a_{4}=\frac {6-L\left (L+1\right ) }{12}\left (\frac {-L\left (L+1\right ) }{2}a_{0}\right ) \] For \(n=3\)\begin {align*} a_{5} & =\frac {n\left (n-1\right ) +2n-L\left (L+1\right ) }{\left ( n+2\right ) \left (n+1\right ) }a_{3}\\ & =\frac {3\left (3-1\right ) +6-L\left (L+1\right ) }{\left (3+2\right ) \left (3+1\right ) }a_{3}\\ & =\frac {12-L\left (L+1\right ) }{20}a_{3} \end {align*}
But \(a_{3}=\frac {2-L\left (L+1\right ) }{6}a_{1}\), hence the above becomes\[ a_{5}=\frac {12-L\left (L+1\right ) }{20}\left (\frac {2-L\left (L+1\right ) }{6}a_{1}\right ) \] And so on. The solution becomes
\begin {align*} y & =\sum _{n=0}^{\infty }a_{n}x^{n}\\ & =a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+a_{5}x^{5}+\cdots \\ & =a_{0}+a_{1}x-\frac {L\left (L+1\right ) }{2}a_{0}x^{2}+\frac {2-L\left ( L+1\right ) }{6}a_{1}x^{3}-\left (\frac {6-L\left (L+1\right ) }{12}\right ) \left (\frac {L\left (L+1\right ) }{2}\right ) a_{0}x^{4}+\left ( \frac {12-L\left (L+1\right ) }{20}\right ) \left (\frac {2-L\left ( L+1\right ) }{6}\right ) a_{1}x^{5}+\cdots \\ & =a_{0}\left (1-\frac {L\left (L+1\right ) }{2}x^{2}-\left (\frac {6-L\left (L+1\right ) }{12}\right ) \left (\frac {L\left (L+1\right ) }{2}\right ) x^{4}+\cdots \right ) +a_{1}\left (x+\frac {2-L\left (L+1\right ) }{6}x^{3}+\left (\frac {12-L\left (L+1\right ) }{20}\right ) \left ( \frac {2-L\left (L+1\right ) }{6}\right ) x^{5}+\cdots \right ) \end {align*}
Or\[ y\relax (x) =a_{0}y_{0}\relax (x) +a_{1}y_{1}\relax (x) \] Where\begin {align*} y_{0}\relax (x) & =1-\frac {L\left (L+1\right ) }{2}x^{2}-\left ( \frac {6-L\left (L+1\right ) }{12}\right ) \left (\frac {L\left (L+1\right ) }{2}\right ) x^{4}+\cdots \\ y_{1}\relax (x) & =x+\frac {2-L\left (L+1\right ) }{6}x^{3}+\left ( \frac {12-L\left (L+1\right ) }{20}\right ) \left (\frac {2-L\left ( L+1\right ) }{6}\right ) x^{5}+\cdots \end {align*}
Where \(y_{0},y_{1}\) are two linearly independent solutions. The even Legendre polynomials are obtained from \(y_{0}\relax (x) \) for integer \(L=0,2,4,\cdots \) and the odd Legendre polynomials are obtained from \(y_{1}\relax (x) \) for integer \(L=1,3,5,\cdots \).
For \(L=0\)\[ y\relax (x) =a_{0}\relax (1) \] Since all higher terms vanish. Choosing \(a_{0}=1\) then \[ P_{0}\relax (x) =1 \] For \(L=2\)\begin {align*} y\relax (x) & =a_{0}\left (1-\frac {L\left (L+1\right ) }{2}x^{2}-\left (\frac {6-L\left (L+1\right ) }{12}\right ) \left ( \frac {L\left (L+1\right ) }{2}\right ) x^{4}+\cdots \right ) \\ & =a_{0}\left (1-\frac {2\left (2+1\right ) }{2}x^{2}-\left (\frac {6-2\left (2+1\right ) }{12}\right ) \left (\frac {2\left (2+1\right ) }{2}\right ) x^{4}+\cdots \right ) \\ & =a_{0}\left (1-3x^{2}\right ) \end {align*}
Since all higher terms vanish. Choosing \(a_{0}=-\frac {1}{2}\) then\[ P_{2}\relax (x) =\frac {1}{2}\left (3x^{2}-1\right ) \] For\(\ L=1\)
Since \(L\) is odd, then \(y_{1}\relax (x) \) is used now.\begin {align*} y\relax (x) & =a_{1}\left (x+\frac {2-L\left (L+1\right ) }{6}x^{3}+\left (\frac {12-L\left (L+1\right ) }{20}\right ) \left ( \frac {2-L\left (L+1\right ) }{6}\right ) x^{5}+\cdots \right ) \\ & =a_{1}\left (x+\frac {2-\left (1+1\right ) }{6}x^{3}+\left ( \frac {12-\left (1+1\right ) }{20}\right ) \left (\frac {2-\left (1+1\right ) }{6}\right ) x^{5}+\cdots \right ) \\ & =a_{1}x \end {align*}
Since all higher terms vanish. Choosing \(a_{1}=1\) then\[ P_{1}\relax (x) =x \] For \(L=3\)\begin {align*} y\relax (x) & =a_{1}\left (x+\frac {2-L\left (L+1\right ) }{6}x^{3}+\left (\frac {12-L\left (L+1\right ) }{20}\right ) \left ( \frac {2-L\left (L+1\right ) }{6}\right ) x^{5}+\cdots \right ) \\ & =a_{1}\left (x+\frac {2-3\left (3+1\right ) }{6}x^{3}+\left ( \frac {12-3\left (3+1\right ) }{20}\right ) \left (\frac {2-3\left ( 3+1\right ) }{6}\right ) x^{5}+\cdots \right ) \\ & =a_{1}\left (x-\frac {5}{3}x^{3}\right ) \end {align*}
Since all higher terms vanish. Choosing \(a_{1}=-\frac {3}{2}\) then\begin {align*} P_{3}\relax (x) & =-\frac {3}{2}\left (x-\frac {5}{3}x^{3}\right ) \\ & =\frac {1}{2}\left (5x^{3}-3x\right ) \end {align*}
Summary\begin {align*} P_{0}\relax (x) & =1\\ P_{1}\relax (x) & =x\\ P_{2}\relax (x) & =\frac {1}{2}\left (3x^{2}-1\right ) \\ P_{3}\relax (x) & =\frac {1}{2}\left (5x^{3}-3x\right ) \end {align*}
To show any two are orthogonal over \(-1\leq x\leq 1\). Selecting \(P_{0}\left ( x\right ) \) and \(P_{1}\relax (x) \), then \begin {align*} \int _{-1}^{1}P_{0}\relax (x) P_{1}\relax (x) dx & =\int _{-1}^{1}xdx\\ & =\frac {1}{2}\left [ x^{2}\right ] _{-1}^{1}\\ & =\frac {1}{2}\left (1-1\right ) \\ & =0 \end {align*}
Hence \(P_{0}\relax (x) \) and \(P_{1}\relax (x) \) are orthogonal to each others. Verified.
Solution
Let\[ \left \{ |x\rangle \right \} =\left \{ 1,x,x^{2},x^{3},\cdots \right \} \] Where \(|x_{1}\rangle =1,|x_{2}\rangle =x,|x_{3}\rangle =x^{2}\) and so on. Let \begin {align*} P_{0} & =|x_{1}\rangle \\ & =1 \end {align*}
Normalizing gives\[ P_{0}=\frac {P_{0}}{\left \Vert P_{0}\right \Vert }=\frac {1}{\sqrt {\int _{-1}^{1}dx}}=\sqrt {\frac {1}{2}}\] And\begin {align*} P_{1} & =|x_{2}\rangle -P_{0}\langle P_{0}|x_{2}\rangle \\ & =x-\sqrt {\frac {1}{2}}\langle \sqrt {\frac {1}{2}}|x_{2}\rangle \\ & =x-\frac {1}{2}\int _{-1}^{1}xdx\\ & =x-0\\ & =x \end {align*}
Normalizing gives\[ P_{1}=\frac {P_{1}}{\left \Vert P_{1}\right \Vert }=\frac {x}{\sqrt {\int _{-1}^{1}x^{2}dx}}=\frac {x}{\sqrt {\frac {2}{3}}}=\sqrt {\frac {3}{2}}x \] And\begin {align*} P_{2} & =|x_{3}\rangle -\left (P_{0}\langle P_{0}|x_{3}\rangle +P_{1}\langle P_{1}|x_{3}\rangle \right ) \\ & =x^{2}-\left (\sqrt {\frac {1}{2}}\langle \sqrt {\frac {1}{2}}|x_{3}\rangle +\sqrt {\frac {3}{2}}x\langle \sqrt {\frac {3}{2}}x|x_{3}\rangle \right ) \\ & =x^{2}-\left (\frac {1}{2}\int _{-1}^{1}x^{2}dx+\frac {3}{2}x\int _{-1}^{1}xx^{2}dx\right ) \\ & =x^{2}-\left (\frac {1}{2}\left [ \frac {x^{3}}{3}\right ] _{-1}^{1}+\frac {3}{2}x\int _{-1}^{1}x^{3}dx\right ) \\ & =x^{2}-\left (\frac {1}{2}\frac {1}{3}\left [ 1-\left (-1\right ) ^{3}\right ] +0\right ) \\ & =x^{2}-\left (\frac {1}{2}\frac {1}{3}\relax (2) \right ) \\ & =x^{2}-\frac {1}{3} \end {align*}
Normalizing\begin {align*} P_{2} & =\frac {P_{2}}{\left \Vert P_{2}\right \Vert }\\ & =\frac {x^{2}-\frac {1}{3}}{\sqrt {\int _{-1}^{1}\left (x^{2}-\frac {1}{3}\right ) \left (x^{2}-\frac {1}{3}\right ) dx}}\\ & =\frac {x^{2}-\frac {1}{3}}{\sqrt {\frac {8}{45}}}\\ & =\sqrt {\frac {45}{8}}\left (x^{2}-\frac {1}{3}\right ) \\ & =\sqrt {\frac {5}{8}}3\left (x^{2}-\frac {1}{3}\right ) \\ & =\sqrt {\frac {5}{8}}\left (3x^{2}-1\right ) \end {align*}
And\begin {align*} P_{3} & =|x_{4}\rangle -\left (P_{0}\langle P_{0}|x_{4}\rangle +P_{1}\langle P_{1}|x_{4}\rangle +P_{2}\langle P_{2}|x_{4}\rangle \right ) \\ & =x^{3}-\left (\sqrt {\frac {1}{2}}\langle \sqrt {\frac {1}{2}}|x_{4}\rangle +\sqrt {\frac {3}{2}}x\langle \sqrt {\frac {3}{2}}x|x_{4}\rangle +\sqrt {\frac {5}{8}}\left (3x^{2}-1\right ) \langle \sqrt {\frac {5}{8}}\left ( 3x^{2}-1\right ) |x_{4}\rangle \right ) \\ & =x^{3}-\left (\frac {1}{2}\int _{-1}^{1}x^{3}dx+\frac {3}{2}x\int _{-1}^{1}xx^{3}dx+\frac {5}{8}\left (3x^{2}-1\right ) \int _{-1}^{1}\left ( 3x^{2}-1\right ) x^{3}dx\right ) \\ & =x^{3}-\left (\frac {1}{2}\int _{-1}^{1}x^{3}dx+\frac {3}{2}x\int _{-1}^{1}x^{4}dx+\frac {5}{8}\left (3x^{2}-1\right ) \int _{-1}^{1}\left ( 3x^{5}-x^{3}\right ) dx\right ) \\ & =x^{3}-\left (\frac {1}{2}\left [ \frac {x^{4}}{4}\right ] _{-1}^{1}+\frac {3}{2}x\left [ \frac {x^{5}}{5}\right ] _{-1}^{1}+\frac {5}{8}\left ( 3x^{2}-1\right ) \relax (0) \right ) \\ & =x^{3}-\left (\frac {1}{8}\left [ x^{4}\right ] _{-1}^{1}+\frac {3}{10}x\left [ x^{5}\right ] _{-1}^{1}\right ) \\ & =x^{3}-\left (\frac {1}{8}\left [ 1-\left (-1\right ) ^{4}\right ] +\frac {3}{10}x\left [ 1-\left (-1\right ) ^{5}\right ] \right ) \\ & =x^{3}-\left (\frac {1}{8}\left [ 0\right ] +\frac {3}{10}x\left [ 2\right ] \right ) \\ & =x^{3}-\frac {3}{5}x \end {align*}
Normalizing\begin {align*} P_{3} & =\frac {P_{3}}{\left \Vert P_{3}\right \Vert }\\ & =\frac {x^{3}-\frac {3}{5}x}{\sqrt {\int _{-1}^{1}\left (x^{3}-\frac {3}{5}x\right ) \left (x^{3}-\frac {3}{5}x\right ) dx}}\\ & =\frac {x^{3}-\frac {3}{5}x}{\sqrt {\frac {8}{175}}}\\ & =\sqrt {\frac {175}{8}}\left (x^{3}-\frac {3}{5}x\right ) \\ & =\sqrt {\frac {\left (25\right ) \relax (7) }{8}}\left (x^{3}-\frac {3}{5}x\right ) \\ & =\sqrt {\frac {7}{8}}\left (5x^{3}-3x\right ) \end {align*}
These are the first 4 Legenrdre polynomials. The scaling is different from the last problem due to difference in method used to normalize them. The following table shows the final result and difference in scaling.
| \(P_{n}\) | Problem 10.4.5 result | Problem 10.4.4 result |
| \(P_{0}\relax (x) \) | \(\sqrt {\frac {1}{2}}\) | \(1\) |
| \(P_{1}\relax (x) \) | \(\sqrt {\frac {3}{2}}x\) | \(x\) |
| \(P_{2}\relax (x) \) | \(\sqrt {\frac {5}{8}}\left (3x^{2}-1\right ) \) | \(\frac {1}{2}\left (3x^{2}-1\right ) \) |
| \(P_{3}\relax (x) \) | \(\sqrt {\frac {7}{8}}\left (5x^{3}-3x\right ) \) | \(\frac {1}{2}\left (5x^{3}-3x\right ) \) |
Solution
Since the ODE is singular at \(x=0\) then Frobenius series is used. Let \begin {align*} y & =x^{s}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =\sum _{n=0}^{\infty }c_{n}x^{n+s}\qquad c_{0}\neq 0 \end {align*}
Hence\begin {align*} y^{\prime } & =\sum _{n=0}^{\infty }\left (n+s\right ) c_{n}x^{n+s-1}\\ y^{\prime \prime } & =\sum _{n=0}^{\infty }\left (n+s\right ) \left ( n+s-1\right ) c_{n}x^{n+s-2} \end {align*}
Substituting this in the ODE (10.4.65) gives\begin {align*} x\sum _{n=0}^{\infty }\left (n+s\right ) \left (n+s-1\right ) c_{n}x^{n+s-2}+\left (1-x\right ) \sum _{n=0}^{\infty }\left (n+s\right ) c_{n}x^{n+s-1}+m\sum _{n=0}^{\infty }c_{n}x^{n+s} & =0\\ \sum _{n=0}^{\infty }\left (n+s\right ) \left (n+s-1\right ) c_{n}x^{n+s-1}+\sum _{n=0}^{\infty }\left (n+s\right ) c_{n}x^{n+s-1}-\sum _{n=0}^{\infty }\left (n+s\right ) c_{n}x^{n+s}+m\sum _{n=0}^{\infty }c_{n}x^{n+s} & =0\\ \sum _{n=0}^{\infty }\left (\left (n+s\right ) \left (n+s-1\right ) +\left ( n+s\right ) \right ) c_{n}x^{n+s-1}+\sum _{n=0}^{\infty }\left (m-\left ( n+s\right ) \right ) c_{n}x^{n+s} & =0 \end {align*}
To make all power on \(x\) the same, the second sum is rewritten by shifting the index. This gives\[ \sum _{n=0}^{\infty }\left (\left (n+s\right ) \left (n+s-1\right ) +\left ( n+s\right ) \right ) c_{n}x^{n+s-1}+\sum _{n=1}^{\infty }\left (m-\left ( n-1+s\right ) \right ) c_{n-1}x^{n+s-1}=0 \] For \(n=0\)\begin {align*} \left (\left (n+s\right ) \left (n+s-1\right ) +\left (n+s\right ) \right ) c_{n}x^{n+s-1} & =0\\ \left (\left (n+s\right ) \left (n+s-1\right ) +\left (n+s\right ) \right ) c_{0} & =0 \end {align*}
But by definition \(c_{0}\neq 0\). Therefore the indicial equation is \[ \left (n+s\right ) \left (n+s-1\right ) +\left (n+s\right ) =0 \] But \(n=0\). This becomes\begin {align*} s\left (s-1\right ) +s & =0\\ s^{2}-s+s & =0\\ s^{2} & =0 \end {align*}
Hence root is \(s=0\) (repeated root). Since there is a repeated root, then this is degenerate case. First solution \(y_{1}\relax (x) \) is the assumed form but with \(s=0\). This means\begin {align*} y_{1}\relax (x) & =x^{0}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =\sum _{n=0}^{\infty }c_{n}x^{n} \end {align*}
And the second solution is \begin {align*} y_{2}\relax (x) & =y_{1}\ln x+x^{s}\sum _{n=0}^{\infty }b_{n}x^{n}\\ & =y_{1}\ln x+\sum _{n=0}^{\infty }b_{n}x^{n} \end {align*}
But this solution \(y_{2}\relax (x) \) is not bounded at \(x=0\) due to \(\ln x\) blowing up at origin. The regular solution is only \(y_{1}\relax (x) \). So \(y_{1}\relax (x) \) will be used from now on and not \(y_{2}\relax (x) \). Therefore\begin {align*} y_{1}^{\prime }\relax (x) & =\sum _{n=0}^{\infty }nc_{n}x^{n-1}\\ y_{1}^{\prime \prime }\relax (x) & =\sum _{n=0}^{\infty }n\left ( n-1\right ) c_{n}x^{n-2} \end {align*}
Substituting the above in ODE (10.4.65) gives\begin {align*} x\sum _{n=0}^{\infty }n\left (n-1\right ) c_{n}x^{n-2}+\left (1-x\right ) \sum _{n=0}^{\infty }nc_{n}x^{n-1}+m\sum _{n=0}^{\infty }c_{n}x^{n} & =0\\ \sum _{n=0}^{\infty }n\left (n-1\right ) c_{n}x^{n-1}+\sum _{n=0}^{\infty }nc_{n}x^{n-1}-\sum _{n=0}^{\infty }nc_{n}x^{n}+\sum _{n=0}^{\infty }mc_{n}x^{n} & =0\\ \sum _{n=0}^{\infty }\left (n\left (n-1\right ) +n\right ) c_{n}x^{n-1}+\sum _{n=0}^{\infty }\left (m-n\right ) c_{n}x^{n} & =0 \end {align*}
To make powers on \(x\) the same, the index of the first sum is shifted to give\[ \sum _{n=-1}^{\infty }\left (\left (n+1\right ) n+\left (n+1\right ) \right ) c_{n+1}x^{n}+\sum _{n=0}^{\infty }\left (m-n\right ) c_{n}x^{n}=0 \] But when \(n=-1\) the first sum is zero. So the first sum index can start \(n=0\) which gives\[ \sum _{n=0}^{\infty }\left (\left (n+1\right ) n+\left (n+1\right ) \right ) c_{n+1}x^{n}+\sum _{n=0}^{\infty }\left (m-n\right ) c_{n}x^{n}=0 \] Now the sums are combined to give\[ \sum _{n=0}^{\infty }\left [ \left (\left (n+1\right ) n+\left (n+1\right ) \right ) c_{n+1}+\left (m-n\right ) c_{n}\right ] x^{n}=0 \] Hence recursive relation is\begin {align*} \left (\left (n+1\right ) n+\left (n+1\right ) \right ) c_{n+1}+\left ( m-n\right ) c_{n} & =0\\ c_{n+1} & =\frac {n-m}{\left (\left (n+1\right ) n+\left (n+1\right ) \right ) }c_{n}\\ & =\frac {n-m}{n^{2}+2n+1}c_{n} \end {align*}
For \(n=0\)\[ c_{1}=-mc_{0}\] For \(n=1\)\begin {align*} c_{2} & =\frac {1-m}{1+2+1}c_{1}\\ & =\frac {1-m}{4}c_{1}\\ & =\frac {1-m}{4}\left (-mc_{0}\right ) \\ & =\frac {m^{2}-m}{4}c_{0} \end {align*}
For \(n=2\)\begin {align*} c_{3} & =\frac {2-m}{2^{2}+4+1}c_{2}\\ & =\frac {2-m}{9}c_{2}\\ & =\frac {2-m}{9}\left (\frac {m^{2}-m}{4}c_{0}\right ) \\ & =\frac {\left (2-m\right ) \left (m^{2}-m\right ) }{36}c_{0}\\ & =\frac {-m^{3}+3m^{2}-2m}{36}c_{0} \end {align*}
For \(n=3\)\begin {align*} c_{4} & =\frac {n-m}{n^{2}+2n+1}c_{3}\\ & =\frac {3-m}{9+6+1}c_{3}\\ & =\frac {3-m}{16}\left (\frac {-m^{3}+3m^{2}-2m}{36}c_{0}\right ) \\ & =\frac {\left (3-m\right ) \left (-m^{3}+3m^{2}-2m\right ) }{\left ( 16\right ) \left (36\right ) }c_{0}\\ & =\frac {m^{4}-6m^{3}+11m^{2}-6m}{576}c_{0} \end {align*}
And so on. The solution becomes\begin {align*} y_{1}\relax (x) & =c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+c_{4}x^{4}+\cdots \\ & =c_{0}-mc_{0}x+\frac {m^{2}-m}{4}c_{0}x^{2}+\frac {-m^{3}+3m^{2}-2m}{36}c_{0}x^{3}+\frac {m^{4}-6m^{3}+11m^{2}-6m}{576}c_{0}x^{4}+\cdots \\ & =c_{0}\left (1-mx+\frac {m^{2}-m}{4}x^{2}+\frac {-m^{3}+3m^{2}-2m}{36}x^{3}+\frac {m^{4}-6m^{3}+11m^{2}-6m}{576}x^{4}+\cdots \right ) \end {align*}
Setting \(c_{0}=1\), the solution is\[ y_{1}\relax (x) =1-mx+\frac {m^{2}-m}{4}x^{2}+\frac {-m^{3}+3m^{2}-2m}{36}x^{3}+\frac {m^{4}-6m^{3}+11m^{2}-6m}{576}x^{4}+\cdots \] For integer \(m\) these are polynomials given by
For \(m=0\)\[ L_{0}\relax (x) =1 \] Since rest of terms are zero.
For \(m=1\)\[ L_{1}\relax (x) =1-x \] Since rest of terms are zero.
For \(m=2\)\[ L_{2}\relax (x) =1-2x+\frac {1}{2}x^{2}\] Since rest of terms are zero.
For \(m=3\)\begin {align*} L_{3}\relax (x) & =1-3x+\frac {3^{2}-3}{4}x^{2}+\frac {-3^{3}+3\left ( 3^{2}\right ) -6}{36}x^{3}\\ & =1-3x+\frac {3}{2}x^{2}-\frac {1}{6}x^{3} \end {align*}
Since rest of terms are zero. Hence
\begin {align*} L_{0}\relax (x) & =1\\ L_{1}\relax (x) & =1-x\\ L_{2}\relax (x) & =1-2x+\frac {1}{2}x^{2}\\ L_{3}\relax (x) & =1-3x+\frac {3}{2}x^{2}-\frac {1}{6}x^{3} \end {align*}
Or\begin {align*} L_{0}\relax (x) & =1\\ L_{1}\relax (x) & =1-x\\ L_{2}\relax (x) & =\frac {1}{2}\left (2-4x+x^{2}\right ) \\ L_{3}\relax (x) & =\frac {1}{6}\left (6-18x+9x^{2}-x^{3}\right ) \end {align*}
The following shows that \(L_{1}\relax (x) ,L_{2}\relax (x) \) are orthogonal on \(0\leq x\leq \infty \) with weight \(e^{-x}\)\begin {align*} \int _{0}^{\infty }L_{1}\relax (x) L_{2}\relax (x) e^{-x}dx & =\int _{0}^{\infty }\left (1-x\right ) \left (\frac {1}{2}\left ( 2-4x+x^{2}\right ) \right ) e^{-x}dx\\ & =\int _{0}^{\infty }\left (-\frac {1}{2}x^{3}+\frac {5}{2}x^{2}-3x+1\right ) e^{-x}dx\\ & =-\frac {1}{2}\int _{0}^{\infty }x^{3}e^{-x}dx+\frac {5}{2}\int _{0}^{\infty }x^{2}e^{-x}dx-3\int _{0}^{\infty }xe^{-x}dx+\int _{0}^{\infty }e^{-x}dx \end {align*}
To evaluate these integrals the following relation will be used \[ \int _{0}^{\infty }x^{n}e^{-x}=n! \] Therefore \begin {align*} \int _{0}^{\infty }x^{3}e^{-x}dx & =3!=6\\ \int _{0}^{\infty }x^{2}e^{-x}dx & =2!=2\\ \int _{0}^{\infty }xe^{-x}dx & =1!=1 \end {align*}
And \[ \int _{0}^{\infty }e^{-x}dx=-\left [ e^{-x}\right ] _{0}^{\infty }=-\left ( 0-1\right ) =1 \] Using these results gives\begin {align*} \int _{0}^{\infty }L_{1}\relax (x) L_{2}\relax (x) e^{-x}dx & =-\frac {1}{2}\relax (6) +\frac {5}{2}\relax (2) -3\left ( 1\right ) +1\\ & =0 \end {align*}
This shows that \(L_{1}\relax (x) ,L_{2}\relax (x) \) are orthogonal on \(0\leq x\leq \infty \) with weight \(e^{-x}\). This complete the solution.