C Orthogonal function systems

When complete, orthogonal function systems are suitable for developing functions. A system of polynomials $f_n\left(x\right)$ with $n\in \mathbb {N}$ and $\mathcal {O}\left(f_n\right) = n$ is said to be orthogonal on the interval $\left[a, b\right]$ with respect to the function $w$ if for $n, m\in \mathbb {N}$ with $n\not = m$

\[ \begin{align} \int_{a}^{b}w\left(x\right)f_n\left(x\right)f_m\left(x\right)dx = 0. \end{align} \]

All orthogonal polynomial systems satisfy a Rodrigues formula

\[ \begin{align} f_n\left(x\right) = \frac{1}{e_nw\left(x\right)}\frac{d^n}{dx^n}\left[w\left(x\right)g\left(x\right)^n\right] \end{align} \]

with a polynomial $g\left(x\right)$, which does not depend on $n$.

C.1 Complex exponential functions

The expansion of a function according to complex exponential functions is called Fourier transformation (FT). Let $f\left(x\right)$ be a complex function of a real variable, then one defines the Fourier transform $\newtilde{F}_1\{f\} = \newtilde{f}\left(k\right)$ or the spectrum of $f$ by

\[ \begin{align} \newtilde{F}_1\{f\} = \newtilde{f}\left(k\right) \coloneqq C\int_{-\infty}^{\infty}\exp\left(-ikx\right)f\left(x\right)dx\tag{C.3}\label{eq:def_ft_1d} \end{align} \]

with a real constant $C > 0$. $\newtilde{F}_1$ denotes the operator of the one-dimensional Fourier transform. Eq. (C.3) is, with the exception of a possible factor, the covariance of the function $f$ with the Fourier component $e^{ikx}$. It applies

\[ \begin{align} \newtilde{f}\left(k\right)^\star = C\int_{-\infty}^{\infty}\exp\left(ikx\right)f^\star\left(x\right)dx = \newtilde{f}^\star\left(-k\right). \end{align} \]

If $f$ is real, this becomes

\[ \begin{align} \newtilde{f}\left(k\right)^\star = \newtilde{f}\left(-k\right). \end{align} \]

The inverse Fourier transformation is defined by

\[ \begin{align} \newtilde{F}^{-1}_1\{f\} \coloneqq \newtilde{C}\int_{-\infty}^{\infty}\newtilde{f}\left(k\right)\exp\left(ikx\right)dk\tag{C.6}\label{eq:def_ft_1d_inv} \end{align} \]

with a second constant $\newtilde{C} > 0$. It is a linear combination of the $\newtilde{f}\left(k\right)$ with the corresponding waves. That this is actually the inverse of Eq. (C.3) should now be seen:

\[ \begin{align} \newtilde{C}\int_{-\infty}^{\infty}\newtilde{f}\left(k\right)\exp\left(ikx\right)dk &= \newtilde{C}C\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp\left(-ikx'\right)f\left(x'\right)dx'\exp\left(ikx\right)dk\nonumber\\ &= \newtilde{C}C\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp\left(ik\left(x - x'\right)\right)dkf\left(x'\right)dx'\nonumber\\ & \stackrel{\href{ch-39-derivations-of-some-mathematical-formula.html#eq:delta_distribution_prop_2}{\text{Glg. (A.84)}}}{=} \newtilde{C}C\int_{-\infty}^{\infty}2\pi\delta\left(x - x'\right)'f\left(x'\right)dx' = \newtilde{C}C2\pi f\left(x\right)\\ & \stackrel{\Leftrightarrow\newtilde{C}C = 1/2\pi}{=} & f\left(x\right)\tag{C.8}\label{eq:ft_deriv_0} \end{align} \]

So you have to accept the condition

\[ \begin{align} \newtilde{C} = \frac{1}{2\pi C}\tag{C.9}\label{eq:ft_norm} \end{align} \]

regard. From this you can immediately

\[ \begin{align} \left[\newtilde{F}_1\{\delta\left(x - x_0\right)\}\right]\left(k\right) &= Ce^{-ikx_0},\\ \left[\newtilde{F}_1\{e^{ik'x}\}\right]\left(k\right) &= 2\pi C\delta\left(k' - k\right),\\ \left[\newtilde{F}^{-1}_1\{\delta\left(k - k_0\right)\}\right]\left(x\right) &= \newtilde{C}e^{ik_0x},\\ \left[\newtilde{F}^{-1}_1\{e^{ikx'}\}\right]\left(x\right) &= 2\pi\newtilde{C}\delta\left(x' + x\right) \end{align} \]

derive. The function

\[ \begin{align} \arg\left(\newtilde{f}\left(k\right)\right) \end{align} \]

is called the phase spectrum of $f$, while one

\[ \begin{align} \left|\newtilde{f}\left(k\right)\right| \end{align} \]

referred to as amplitude spectrum. The square of $\left|\newtilde{f}\left(k\right)\right|$ is called power spectrum or power spectrum. The FT is linear because it is defined in terms of an integral and the integration is linear. This applies to functions $f, g$ and constants $a, b$

\[ \begin{align} \newtilde{F}\{af + bg\} = a\newtilde{F}\{f\} + b\newtilde{F}\{g\}. \end{align} \]

Let $N\left(\mu, \sigma\right)$ be the normal distribution with expected value $\mu$ and standard deviation $\sigma.$ Then holds

\[ \begin{align} \newtilde{F}\{N\left(\mu, \sigma\right)\} &= \frac{C}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}\exp\left[-ikx - \frac{\left(x - \mu\right)^2}{2\sigma^2}\right]dx = \frac{C}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}\exp\left[-\frac{1}{2\sigma^2}\left(x^2 + \mu^2 - 2x\left(\mu - ik\sigma^2\right)\right)\right]dx\nonumber\\ &= \frac{C}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}\exp\left[-\frac{1}{2\sigma^2}\left(x - \left(\mu - ik\sigma^2\right)\right)^2 - \frac{1}{2\sigma^2}\left(k^2\sigma^4 + 2\mu i\sigma^2k\right)\right]dx\nonumber\\ &= \exp\left[-\frac{1}{2\sigma^2}\left(k^2\sigma^4 + 2\mu i\sigma^2k\right)\right]\frac{C}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}\exp\left[-\frac{1}{2\sigma^2}x^2\right]dx\nonumber\\ &= C\exp\left(-\frac{k^2}{2/\sigma^2}\right)\exp\left(-\mu ik\right). \end{align} \]

The FT of a normal distribution is therefore an unnormalized normal distribution with a mean value (consider the case $C = 1/\sqrt{2\pi}$) and modified by a complex phase factor

\[ \begin{align} \mu_k = 0 \end{align} \]

and standard deviation

\[ \begin{align} \sigma_k = \frac{1}{\sigma} \end{align} \]

The FT therefore turns a narrow normal distribution into a wide one. In the case of a complex-valued function $f\left(x, y\right)$ of two real variables, we first consider the auxiliary function

\[ \begin{align} f_y\left(x\right) \coloneqq f\left(x, y\right). \end{align} \]

The FT of this is

\[ \begin{align} \newtilde{f}_y\left(k_x\right) = C\int_{-\infty}^{\infty}e^{-ik_xx}f_y\left(x\right)dx. \end{align} \]

If you apply the inverse FT to this, you get $f_y\left(x\right) = f\left(x, y\right)$ again. Now define the Fourier transform $\newtilde{f}\left(k_x, k_y\right)$ of the function $f\left(x, y\right)$ by the Fourier transform in the y direction of $\newtilde{f}_y\left(k_x\right)$, i.e

\[ \begin{align} \newtilde{f}\left(k_x, k_y\right) &\coloneqq C\int_{-\infty}^{\infty}e^{-ik_yy}\newtilde{f}_y\left(k_x\right)dy = C^2\int_{-\infty}^{\infty}e^{-ik_yy}\int_{-\infty}^{\infty}e^{-ik_xx}f\left(x, y\right)dxdy\nonumber\\ &= C^2\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-i\left(k_xx + k_yy\right)}f\left(x, y\right)dxdy. \end{align} \]

Applying the inverse FT twice yields the function $f\left(x, y\right).$ $x$ is not distinguished before $y$ in the derivation, which is why the order in which the FT is applied does not matter. You take notes

\[ \begin{align} \newtilde{F}_2\{f\} &= \newtilde{f}\left(k_x, k_y\right) \coloneqq C_2\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-i\left(k_xx + k_yy\right)}f\left(x, y\right)dxdy,\\ \newtilde{F}^{-1}_2\{f\} &\coloneqq \newtilde{C}_2\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\newtilde{f}\left(k_x, k_y\right)e^{i\left(k_xx + k_yy\right)}dk_xdk_y, \end{align} \]

where

\[ \begin{align} C_2\newtilde{C}_2 = \left(C\newtilde{C}\right)^2 = \frac{1}{\left(2\pi\right)^2} \end{align} \]

must apply. If you continue this inductively, you get the $n-$dimensional FT, where on

\[ \begin{align} C_n\newtilde{C}_n = \left(C\newtilde{C}\right)^n = \frac{1}{\left(2\pi\right)^n} \end{align} \]

must be taken into account.

Eq. (C.8) is

\[ \begin{align} f\left(x\right) &= \newtilde{C}\int_{-\infty}^{\infty}\newtilde{f}\left(k\right)\exp\left(ikx\right)dk, \end{align} \]

If you transfer this to an analogously defined function $g$, you get

\[ \begin{align} g\left(x\right) &= \newtilde{C}\int_{-\infty}^{\infty}\newtilde{g}\left(k'\right)\exp\left(ik'x\right)dk'. \end{align} \]

If you multiply the last two equations together, you get

\[ \begin{align} f\left(x\right)g\left(x\right) &= \newtilde{C}^2\int_{k=-\infty}^{\infty}\int_{k'=-\infty}^{\infty}\newtilde{g}\left(k'\right)\newtilde{f}\left(k\right)\exp\left[i\left(k + k'\right)x\right]dk'dk. \end{align} \]

Integrating this over $x$ gives

\[ \begin{align} \int_{-\infty}^\infty f\left(x\right)g\left(x\right)dx &= \newtilde{C}^2\int_{k = -\infty}^{\infty}\int_{k'=-\infty}^{\infty}\newtilde{g}\left(k'\right)\newtilde{f}\left(k\right)\int_{-\infty}^\infty\exp\left[i\left(k + k'\right)x\right]dxdk'dk\nonumber\\ & \stackrel{\href{ch-39-derivations-of-some-mathematical-formula.html#eq:delta_distribution_prop_2}{\text{Glg. (A.84)}}}{=} \newtilde{C}^2\int_{k = -\infty}^{\infty}\int_{k'=-\infty}^{\infty}\newtilde{g}\left(k'\right)\newtilde{f}\left(k\right)2\pi\delta\left(k + k'\right)dk'dk\nonumber\\ &= 2\pi\newtilde{C}^2\int_{-\infty}^{\infty}\newtilde{g}\left(-k\right)\newtilde{f}\left(k\right)dk \end{align} \]

For $g = f$ you get the so-called Parseval identity

C.1.1 Convolution theorem

Let $f$, $g$ be two continuously differentiable functions $\mathbb{R} \to \mathbb{R}$. The Fourier transforms of these two functions are

\[ \begin{align} \newtilde{f}\left(k\right) = C\int_{-\infty}^{\infty}\exp\left(-ikx\right)f\left(x\right)dx, & {} & \newtilde{g}\left(k\right) = C\int_{-\infty}^{\infty}\exp\left(-ikx\right)g\left(x\right)dx. \end{align} \]

Define the continuously differentiable function $h$ by

\[ \begin{align} h\left(x\right) = f\left(x\right)g\left(x\right). \end{align} \]

The so-called convolution theorem is about how the Fourier transform of $h$ is related to the Fourier transform of $f$ and $g$. For the spectrum of $h$ applies

\[ \begin{align} \newtilde{h}\left(k\right) &= C^2\int_{-\infty}^\infty f\left(x\right)g\left(x\right)\exp\left(-ikx\right)dx.\tag{C.34}\label{eq:faltung_deriv_0} \end{align} \]

Expressing $f$ and $g$ in terms of their spectra, one obtains

\[ \begin{align} f\left(x\right) = \newtilde{C}\int_{-\infty}^{\infty}\exp\left(ikx\right)f\left(k\right)dk, & {} & g\left(x\right) = \newtilde{C}\int_{-\infty}^{\infty}\exp\left(ikx\right)g\left(k\right)dk. \end{align} \]

Putting this into Eq. (C.34), you get

\[ \begin{align} \newtilde{h}\left(k\right) &= C^2\int_{x = -\infty}^\infty\newtilde{C}\int_{k' = -\infty}^{\infty}\exp\left(ik'x\right)f\left(k'\right)dk'\newtilde{C}\int_{k' = -\infty}^{\infty}\exp\left(ik'x\right)g\left(k'\right)dk'\exp\left(-ikx\right)dx\nonumber\\ &= \frac{1}{4\pi^2}\int_{x = -\infty}^\infty\int_{k' = -\infty}^{\infty}\exp\left(ik'x\right)\newtilde{f}\left(k'\right)dk'\int_{k' = -\infty}^{\infty}\exp\left(ik'x\right)\newtilde{g}\left(k'\right)dk'\exp\left(-ikx\right)dx\nonumber\\ &= \frac{1}{4\pi^2}\int_{x = -\infty}^\infty\int_{k' = -\infty}^{\infty}\exp\left(ik'x\right)\newtilde{f}\left(k'\right)dk'\int_{k'' = -\infty}^{\infty}\exp\left(ik''x\right)\newtilde{g}\left(k''\right)dk''\exp\left(-ikx\right)dx\nonumber\\ &= \frac{1}{4\pi^2}\int_{k' = -\infty}^\infty\int_{k'' = -\infty}^{\infty}\newtilde{f}\left(k'\right)\newtilde{g}\left(k''\right)\int_{x = -\infty}^{\infty}\exp\left(ik'x\right)\exp\left(ik''x\right)\exp\left(-ikx\right)dk''dk'dx\nonumber\\ &= \frac{1}{4\pi^2}\int_{k' = -\infty}^\infty\int_{k'' = -\infty}^{\infty}\newtilde{f}\left(k'\right)\newtilde{g}\left(k''\right)\int_{x = -\infty}^{\infty}\exp\left[i\left(k' + k'' - k\right)x\right]dk''dk'dx\nonumber\\ & \stackrel{\href{ch-39-derivations-of-some-mathematical-formula.html#eq:delta_distribution_prop_2}{\text{Glg. (A.84)}}}{=} \frac{1}{4\pi^2}\int_{k' = -\infty}^\infty\int_{k'' = -\infty}^{\infty}\newtilde{f}\left(k'\right)\newtilde{g}\left(k''\right)\int_{x = -\infty}^{\infty}2\pi\delta\left(k' + k'' - k\right)dk''dk'\nonumber\\ & \stackrel{k'' = k - k'}{=} \frac{1}{2\pi}\int_{-\infty}^\infty\newtilde{f}\left(k'\right)\newtilde{g}\left(k - k'\right)dk'. \end{align} \]

The statement

is called the convolution theorem. The special case

\[ \begin{align} g = \frac{f}{2} \end{align} \]

corresponds to the kinetic energy. In this case you get

\[ \begin{align} \newtilde{h}\left(k\right) = \frac{1}{4\pi}\int_{-\infty}^\infty\newtilde{f}\left(k'\right)\newtilde{f}\left(k - k'\right)dk' \not= \pi\newtilde{C}^2\newtilde{f}\left(-k\right)\newtilde{f}\left(k\right). \end{align} \]

Now the question is whether one considers $\newtilde{h}\left(k\right)$ or $\newtilde{f}\left(-k\right)\newtilde{f}\left(k\right)$ as a kinetic energy spectrum, as the Parseval identity Eq. (C.31) suggests. The latter is chosen because $\newtilde{f}\left(-k\right)\newtilde{f}\left(k\right)$ depends exclusively on movements of the wave number $k$, which is not the case with $\newtilde{h}\left(k\right)$.

C.2 Legendre polynomials

Eq. (D.50) is called Legendre'sche differential equation. This is:

\[ \begin{align} \frac{d}{d\mu}\left(1 - \mu^2\right)P'\left(\mu\right) = -\lambda P \end{align} \]

One starts with $P\left(\mu\right)$

\[ \begin{align} P\left(\mu\right) = \sum_{i = 0}^{\infty}U_i\mu^i, \end{align} \]

so that applies

\[ \begin{align} P'\left(\mu\right) &= \sum_{i = 0}^{\infty}iU_i\mu^{i - 1} = \sum_{i = 0}^{\infty}\left(i + 1\right)U_{i + 1}\mu^i,\\ P''\left(\mu\right) &= \sum_{i = 0}^{\infty}i\left(i - 1\right)U_i\mu^{i - 2} = \sum_{i = 0}^{\infty}\left(i + 2\right)\left(i + 1\right)U_{i + 2}\mu^i. \end{align} \]

If you put this in

\[ \begin{align} - 2\mu P'\left(\mu\right) + \left(1 - \mu^2\right)P''\left(\mu\right) = -\lambda P\left(\mu\right) \end{align} \]

one, you get

\[ \begin{align} - 2\sum_{i = 0}^{\infty}iU_i\mu^i + \sum_{i = 0}^{\infty}\left(i + 2\right)\left(i + 1\right)U_{i + 2}\mu^i - \sum_{i = 0}^{\infty}i\left(i - 1\right)U_i\mu^i &= - \lambda\sum_{i = 0}^{\infty}U_i\mu^i\nonumber\\ \Leftrightarrow - 2iU_i+ \left(i + 2\right)\left(i + 1\right)U_{i + 2} - i\left(i - 1\right) &= -\lambda U_i\nonumber\\ \Leftrightarrow U_{i + 2} &= U_i\frac{i\left(i + 1\right) - \lambda}{\left(i + 2\right)\left(i + 1\right)}.\tag{C.45}\label{eq:legendre_iterativ} \end{align} \]

In general, this does not approach zero for $i\to\infty$, so $P$ would diverge. So there must be a $n\in\mathbb{N}$ with $\lambda = n\left(n + 1\right)$, this is a condition on $\lambda$. Additionally you have to set $U_{n + 1} = 0$. $P\left(\mu\right)$ is therefore an $n-$th order polynomial in $\mu$, where $n$ is arbitrary, one writes $P_n\left(\mu\right)$ for this Legendre polynomial. About Eq. (C.45) and a normalization is set $P_n\left(\mu\right)$. Alternatively, one can write the Legendre polynomials as

\[ \begin{align} P_n\left(x\right) = \frac{1}{2^nn!}\frac{d^n}{dx^n}\left(x^2 - 1\right)^n, \tag{C.46}\label{eq:legendre_poly_formel} \end{align} \]

this is now to be shown. Define

\[ \begin{align} \delta_k \coloneqq \begin{cases} 0, k\:\text{ungerade},\\ 1, k\:\text{gerade}. \end{cases} \end{align} \]

Now one can use the binomial theorem Eq. Use (A.17) to

\[ \begin{align} \left(x^2 - 1\right)^n = \sum_{k = 0}^{n}\left(\begin{array}{c} n\\ k \end{array}\right)x^{2k}\left(-1\right)^{n - k} = \sum_{k = 0}^{2n}\delta_k\left(\begin{array}{c} n\\ k/2 \end{array}\right)x^k\left(-1\right)^{n - k/2}\tag{C.48}\label{eq:gen_bin_formula_legendre} \end{align} \]

to note down, so that for the above notation the Legendre polynomials follow Eq. (C.46) with Eq. (A.34)

\[ \begin{align} P_n\left(x\right) &= \frac{1}{2^nn!}\sum_{k = 0}^{n}\frac{\left(k + n\right)!}{k!}\left(\begin{array}{c} n\\ \frac{k + n}{2} \end{array}\right)\left(-1\right)^{n - \frac{k + n}{2}}\delta_{k + n}x^k\nonumber\\ &= \frac{1}{2^nn!}\sum_{k = 0}^{n}\frac{\left(n + k\right)!}{k!}\frac{n!}{\left(\frac{n + k}{2}\right)!\left(n - \frac{n + k}{2}\right)!}\left(-1\right)^{n - \frac{k + n}{2}}\delta_{n + k}x^k\nonumber\\ &= \frac{1}{2^n}\sum_{k = 0}^{n}\frac{\left(n + k\right)!}{k!}\frac{1}{\left(\frac{n + k}{2}\right)!\left(\frac{n - k}{2}\right)!}\left(-1\right)^{\frac{n - k}{2}}\delta_{n + k}x^k. \end{align} \]

For the first two derivatives of the Legendre polynomials follows

\[ \begin{align} P_n'\left(x\right) &= \frac{1}{2^n}\sum_{k = 0}^{n - 1}\left(k + 1\right)\frac{\left(n + k + 1\right)!}{\left(k + 1\right)!}\frac{1}{\left(\frac{n + k + 1}{2}\right)!}\frac{1}{\left(\frac{n - k - 1}{2}\right)!}\left(-1\right)^{\frac{n - k - 1}{2}}\delta_{n + k + 1}x^k\nonumber\\ &= \frac{1}{2^n}\sum_{k = 0}^{n - 1}\frac{\left(n + k + 1\right)!}{k!}\frac{1}{\left(\frac{n + k + 1}{2}\right)!}\frac{1}{\left(\frac{n - k - 1}{2}\right)!}\left(-1\right)^{\frac{n - k - 1}{2}}\delta_{n + k + 1}x^k,\\ P_n''\left(x\right) &= \frac{1}{2^n}\sum_{k = 0}^{n - 2}\left(k + 1\right)\frac{\left(n + k + 2\right)!}{\left(k + 1\right)!}\frac{1}{\left(\frac{n + k + 2}{2}\right)!}\frac{1}{\left(\frac{n - k - 2}{2}\right)!}\left(-1\right)^{\frac{n - k - 2}{2}}\delta_{n + k + 2}x^k\nonumber\\ &= \frac{1}{2^n}\sum_{k = 0}^{n - 2}\frac{\left(n + k + 2\right)!}{k!}\frac{1}{\left(\frac{n + k + 2}{2}\right)!}\frac{1}{\left(\frac{n - k - 2}{2}\right)!}\left(-1\right)^{\frac{n - k - 2}{2}}\delta_{n + k}x^k\nonumber\\ &= \frac{1}{2^n}\sum_{k = 0}^{n - 2}\frac{\left(n + k + 2\right)!}{k!}\frac{1}{\left(\frac{n + k}{2}\right)!}\frac{1}{\left(\frac{n + k}{2} + 1\right)}\frac{\left(\frac{n - k}{2}\right)}{\left(\frac{n - k}{2}\right)!}\left(-1\right)^{\frac{n - k - 2}{2}}\delta_{n + k}x^k\nonumber\\ &= \frac{1}{2^n}\sum_{k = 0}^{n - 2}\frac{1}{\left(\frac{n + k}{2}\right)!}\frac{1}{\left(\frac{n - k}{2}\right)!}\frac{\left(n + k + 1\right)!}{k!}\left(n - k\right)\left(-1\right)^{\frac{n - k - 2}{2}}\delta_{n + k}x^k. \end{align} \]

Now it is shown that the Legendre polynomials according to Eq. (C.46) actually solve the Legendre differential equation (D.50):

\[ \begin{align} & - 2xP_n'\left(x\right) + \left(1 - x^2\right)P_n''\left(x\right)\hastobe - n\left(n + 1\right)P_n\left(x\right)\nonumber\\ &\Leftrightarrow -2x\frac{1}{2^n}\sum_{k = 0}^{n - 1}\frac{\left(n + k + 1\right)!}{k!}\frac{1}{\left(\frac{n + k + 1}{2}\right)!}\frac{1}{\left(\frac{n - k - 1}{2}\right)!}\left(-1\right)^{\frac{n - k - 1}{2}}\delta_{n + k + 1}x^k\nonumber\\ & + \frac{1}{2^n}\sum_{k = 0}^{n - 2}\frac{1}{\left(\frac{n + k}{2}\right)!}\frac{1}{\left(\frac{n - k}{2}\right)!}\frac{\left(n + k + 1\right)!}{k!}\left(n - k\right)\left(-1\right)^{\frac{n - k - 2}{2}}\delta_{n + k}x^k\nonumber\\ & - \frac{1}{2^n}\sum_{k = 0}^{n - 2}\frac{\left(n + k + 2\right)!}{k!}\frac{1}{\left(\frac{n + k + 2}{2}\right)!}\frac{1}{\left(\frac{n - k - 2}{2}\right)!}\left(-1\right)^{\frac{n - k - 2}{2}}\delta_{n + k}x^{k + 2}\nonumber\\ &= -n\left(n + 1\right)\frac{1}{2^n}\sum_{k = 0}^{n}\frac{\left(n + k\right)!}{k!}\frac{1}{\left(\frac{n + k}{2}\right)!}\frac{1}{\left(\frac{n - k}{2}\right)!}\left(-1\right)^{\frac{n - k}{2}}\delta_{n + k}x^k\nonumber\\ &\Leftrightarrow -2\sum_{k = 0}^{n}k\frac{\left(n + k\right)!}{k!}\frac{1}{\left(\frac{n + k}{2}\right)!}\frac{1}{\left(\frac{n - k}{2}\right)!}\left(-1\right)^{\frac{n - k}{2}}\delta_{n + k}x^k\nonumber\\ & + \sum_{k = 0}^{n - 2}\frac{1}{\left(\frac{n + k}{2}\right)!}\frac{1}{\left(\frac{n - k}{2}\right)!}\frac{\left(n + k + 1\right)!}{k!}\left(n - k\right)\left(-1\right)^{\frac{n - k - 2}{2}}\delta_{n + k}x^k\nonumber\\ & - \sum_{k = 0}^{n}k\left(k - 1\right)\frac{\left(n + k\right)!}{k!}\frac{1}{\left(\frac{n + k}{2}\right)!}\frac{1}{\left(\frac{n - k}{2}\right)!}\left(-1\right)^{\frac{n - k}{2}}\delta_{n + k}x^k\nonumber \end{align} \] \[ \begin{align} &= -n\left(n + 1\right)\sum_{k = 0}^{n}\frac{\left(n + k\right)!}{k!}\frac{1}{\left(\frac{n + k}{2}\right)!}\frac{1}{\left(\frac{n - k}{2}\right)!}\left(-1\right)^{\frac{n - k}{2}}\delta_{n + k}x^k\nonumber\\ &\Leftrightarrow -2\sum_{k = 0}^{n}k\frac{\left(n + k\right)!}{k!}\frac{1}{\left(\frac{n + k}{2}\right)!}\frac{1}{\left(\frac{n - k}{2}\right)!}\left(-1\right)^{\frac{n - k}{2}}\delta_{n + k}x^k\nonumber\\ & - \sum_{k = 0}^{n - 2}\frac{1}{\left(\frac{n + k}{2}\right)!\left(\frac{n - k}{2}\right)!}\frac{\left(n + k\right)!}{k!}\left(n + k + 1\right)\left(n - k\right)\left(-1\right)^{\frac{n - k}{2}}\delta_{n + k}x^k\nonumber\\ & - \sum_{k = 0}^{n}k\left(k - 1\right)\frac{\left(n + k\right)!}{k!}\frac{1}{\left(\frac{n + k}{2}\right)!}\frac{1}{\left(\frac{n - k}{2}\right)!}\left(-1\right)^{\frac{n - k}{2}}\delta_{n + k}x^k\nonumber\\ &= -n\left(n + 1\right)\sum_{k = 0}^{n}\frac{\left(n + k\right)!}{k!}\frac{1}{\left(\frac{n + k}{2}\right)!}\frac{1}{\left(\frac{n - k}{2}\right)!}\left(-1\right)^{\frac{n - k}{2}}\delta_{n + k}x^k \end{align} \]

The equality of the coefficients of the three polynomials is shown. From $k = 0$ to $k = n - 2$ this means

\[ \begin{align} 2k + n^2 - nk + kn - k^2 + n - k + k^2 - k&\hastobe&n^2 + n, \end{align} \]

which is true. For $k = n - 1$ applies

\[ \begin{align} \delta_{n + k} = 0, \end{align} \]

and for $k = n$ holds

\[ \begin{align} -2n - n\left(n - 1\right)\hastobe - n\left(n + 1\right), \end{align} \]

which is also true. Now the normalization of the Legendre polynomials is determined on the set $\left[-1, 1\right]$ that is relevant for this problem.

\[ \begin{align} \int_{-1}^{1}P_n^2\left(x\right)dx &= \frac{1}{2^{2n}n!^2}\int_{-1}^{1}\frac{d^n}{dx^n}\left(x^2 - 1\right)^n\frac{d^n}{dx^n}\left(x^2 - 1\right)^ndx\nonumber\\ &\stackrel{n-\text{fache p.I.}}{=} \frac{1}{2^{2n}n!^2}\left(-1\right)^n\int_{-1}^{1}\left(x^2 - 1\right)^n\frac{d^{2n}}{dx^{2n}}\left(x^2 - 1\right)^ndx\nonumber\\ &= \frac{1}{2^{2n}n!^2}\left(-1\right)^n\left(2n\right)!\int_{-1}^{1}\left(x^2 - 1\right)^ndx \end{align} \]

It still applies

\[ \begin{align} \int_{}^{}\left(x^2 - 1\right)^ndx &= \int\left(x + 1\right)^n\left(x - 1\right)^ndx\stackrel{n\text{ - fache}\:\text{p.I}.}{=}\left(-1\right)^n\int\frac{n!\left(x + 1\right)^{2n}}{\left(2n\right)!}n!dx\nonumber\\ &= \left(-1\right)^n\frac{n!^2}{\left(2n\right)!}\int\left(x + 1\right)^{2n}dx = \left(-1\right)^n\frac{n!^2}{\left(2n\right)!}\left[\frac{\left(x + 1\right)^{2n + 1}}{2n + 1}\right], \end{align} \]

that follows

\[ \begin{align} \int_{-1}^{1}P_n^2\left(x\right)dx = \frac{1}{2^{2n}n!^2}\left(-1\right)^{n}\left(2n\right)!\frac{1}{\left(2n\right)!}\left(-1\right)^nn!^2\frac{2^{2n + 1}}{2n + 1} = \frac{2}{2n + 1}. \end{align} \]

Now the orthogonality of the Legendre polynomials is shown, let $n, m\in \mathbb{N}$ with $n \not = m$, then it holds

\[ \begin{align} \frac{d}{dx}\left[\left(1 - x^2\right)\frac{d}{dx}\right]P_n\left(x\right) = -n\left(n + 1\right)P_n\left(x\right) \end{align} \]

and analogously for $m$. You do the math

\[ \begin{align} \int_{-1}^{1}P_m\left(x\right)\frac{d}{dx}\left[\left(1 - x^2\right)\frac{d}{dx}\right]P_n\left(x\right)dx = -\int_{-1}^{\left(1\right)}P_m'\left(x\right)\left(1 - x^2\right)P_n'\left(x\right)dx \end{align} \]

as well as

\[ \begin{align} \int_{-1}^{1}P_n\left(x\right)\frac{d}{dx}\left[\left(1 - x^2\right)\frac{d}{dx}\right]P_m\left(x\right)dx = -\int_{-1}^{1}P_n'\left(x\right)\left(1 - x^2\right)P_m'\left(x\right)dx, \end{align} \]

also apply

\[ \begin{align} \int_{-1}^{1}P_m\left(x\right)\frac{d}{dx}\left[\left(1 - x^2\right)\frac{d}{dx}\right]P_n\left(x\right)dx &= -n\left(n + 1\right)\int_{-1}^{1}P_m\left(x\right)P_n\left(x\right)dx,\\ \int_{-1}^{\left(1\right)}P_n\left(x\right)\frac{d}{dx}\left[\left(1 - x^2\right)\frac{d}{dx}\right]P_m\left(x\right)\left(x\right)dx &= -m\left(m + 1\right)\int_{-1}^{1}P_n\left(x\right)P_m\left(x\right)dx. \end{align} \]

That's it

\[ \begin{align} -n\left(n + 1\right)\int_{-1}^{1}P_m\left(x\right)P_n\left(x\right)dx = -m\left(m + 1\right)\int_{-1}^{\left(1\right)}P_n\left(x\right)P_m\left(x\right)dx \end{align} \]

and because $n\not = m$

\[ \begin{align} \int_{-1}^{1}P_n\left(x\right)P_m\left(x\right)dx = 0. \end{align} \]

That sums it up

\[ \begin{align} \int_{-1}^{1}P_n\left(x\right)P_m\left(x\right)dx = \frac{2}{2n + 1}\delta_{mn}. \end{align} \]

Now should

\[ \begin{align} \left(2l + 1\right)xP_l\left(x\right) = \left(l + 1\right)P_{l + 1}\left(x\right) + lP_{l - 1}\left(x\right)\tag{C.67}\label{eq:legendre_poly_prop_1} \end{align} \]

be shown. First, it is clear that on the left and right sides of the equal sign there are polynomials of degree $l + 1$, each with only even or odd powers. You write

\[ \begin{align} P_l\left(x\right) = \frac{1}{2^l}\frac{d^l}{dx^l}\sum_{k = 0}^{l}\frac{1}{k!\left(l - k\right)!}x^{2k}\left(-1\right)^{l - k} = \frac{1}{2^l}\sum_{k = \left(\frac{l}{2}\right)_ + }^{l}\frac{1}{k!\left(l - k\right)!}\frac{\left(2k\right)!}{\left(2k - l\right)!}x^{2k - l}\left(-1\right)^{l - k}, \end{align} \]

where $\left(\frac{l}{2}\right)_ + $ is the smallest natural number $\geq\frac{l}{2}$. Follow from this

\[ \begin{align} \left(2l + 1\right)xP_l\left(x\right) &= \frac{2l + 1}{2^l}\sum_{k = \left(\frac{l}{2}\right)_ + }^{l}\frac{1}{k!\left(l - k\right)!}\frac{\left(2k\right)!}{\left(2k - l\right)!}x^{2k - l + 1}\left(-1\right)^{l - k},\\ \left(l + 1\right)P_{l + 1}\left(x\right) &= -\frac{1}{2^l2}\sum_{k = \left(\frac{l + 1}{2}\right)_ + }^{l + 1}\frac{l + 1}{k!\left(l + 1 - k\right)!}\frac{\left(2k\right)!}{\left(2k - l - 1\right)!}x^{2k - l - 1}\left(-1\right)^{l - k}\\ &= \frac{l + 1}{2^l}\sum_{k = \left(\frac{l - 1}{2}\right)_ + }^{l}\frac{\left(2k + 1\right)}{k!\left(l - k\right)!}\frac{\left(2k\right)!}{\left(2k - l + 1\right)!}x^{2k - l + 1}\left(-1\right)^{l - k}\\ P_{l - 1}\left(x\right) &= -\frac{2l}{2^l}\sum_{k = \left(\frac{l - 1}{2}\right)_ + }^{l - 1}\frac{1}{k!\left(l - 1 - k\right)!}\frac{\left(2k\right)!}{\left(2k - l + 1\right)!}x^{2k - l + 1}\left(-1\right)^{l - k}. \end{align} \]

Two polynomials are equal if and only if all their coefficients are equal.

\[ \begin{align} 4kl - 2l^2 + l + 2k + 1 &= 4kl + l + 2k + 1 - 2l^2\nonumber\\ \Leftrightarrow\left(2l + 1\right)\left(2k - l + 1\right) &= \left(l + 1\right)\left(2k + 1\right) - \left(l - k\right)2l\\ \Leftrightarrow\frac{2l + 1}{l - k}\frac{2k - l + 1}{2k - l + 1} &= \frac{l + 1}{l - k}\frac{2k + 1}{2k - l + 1} - \frac{l - k}{l - k}\frac{2l}{2k - l + 1}\nonumber\\ \Leftrightarrow\frac{2l + 1}{l - k} &= \frac{l + 1}{l - k}\frac{2k + 1}{2k - l + 1} - \frac{2l}{2k - l + 1} \end{align} \]

Now the lower and upper limits of the sums have to be examined. For $k = l$ one obtains

\[ \begin{align} 2l + 1 = \left(l + 1\right)\frac{2l + 1}{2l - l + 1}. \end{align} \]

When it comes to the lower limit, a distinction must be made between cases. If $l$ is even, everything is shown. For $l$ odd one calculates

\[ \begin{align} \left(l + 1\right)\frac{l}{l - \frac{l - 1}{2}} - 2l &= 0. \end{align} \]

For $l = 0$ the statement also applies. This means that Eq. (C.67) shown. From this equation it follows

\[ \begin{align} \cos\left(\theta\right)P_l\left(\cos\left(\theta\right)\right) &= \frac{l + 1}{2l + 1}P_{l + 1}\left(\cos\left(\theta\right)\right) + \frac{l}{2l + 1}P_{l - 1}\left(\cos\left(\theta\right)\right).\tag{C.77}\label{eq:legendre_poly_prop_2} \end{align} \]

Now the solution to Eq. (D.48) for $m\not = 0$ will be discussed. This DGL was:

\[ \begin{align} \frac{d}{d\mu}\left(1 - \mu^2\right)P'\left(\mu\right) = P\left(-\lambda + \frac{m^2}{1 - \mu^2}\right).\tag{C.78}\label{eq:legendre_vorform_2} \end{align} \]

With the approach

\[ \begin{align} P\left(\mu\right) = \left(1 - \mu^2\right)^{m/2}T\left(\mu\right)\tag{C.79}\label{eq:ansatz_legendre_2} \end{align} \]

you get

\[ \begin{align} P'\left(\mu\right) &= -\mu m\left(1 - \mu^2\right)^{m/2 - 1}T\left(\mu\right) + \left(1 - \mu^2\right)^{m/2}T'\left(\mu\right),\\ \left(1 - \mu^2\right)P'\left(\mu\right) &= -\mu m\left(1 - \mu^2\right)^{m/2}T\left(\mu\right) + \left(1 - \mu^2\right)^{m/2 + 1}T'\left(\mu\right). \end{align} \]

Putting this into Eq. (C.78), you get

\[ \begin{align} - m\left(1 - \mu^2\right)^{m/2}T\left(\mu\right) + \mu^2m^2\left(1 - \mu^2\right)^{m/2 - 1}T\left(\mu\right) &- 2\mu\left(m + 1\right)\left(1 - \mu^2\right)^{m/2}T'\left(\mu\right)\nonumber\\ + \left(1 - \mu^2\right)^{m/2 + 1}T''\left(\mu\right) = -\lambda\left(1 - \mu^2\right)^{m/2}T\left(\mu\right) &+ m^2\left(1 - \mu^2\right)^{m/2 - 1}T\left(\mu\right)\nonumber\\ \Leftrightarrow -mT\left(\mu\right) + \mu^2m^2\left(1 - \mu^2\right)^{-1}T\left(\mu\right) &- 2\mu\left(m + 1\right)T'\left(\mu\right) + \left(1 - \mu^2\right)T''\left(\mu\right)\nonumber\\ &= -\lambda T\left(\mu\right) + m^2\left(1 - \mu^2\right)^{-1}T\left(\mu\right)\nonumber\\ \Leftrightarrow\left(1 - \mu^2\right)T''\left(\mu\right) - 2\mu\left(m + 1\right)T'\left(\mu\right) &= \left(-\lambda + m + m^2\frac{1 - \mu^2}{1 - \mu^2}\right)T\left(\mu\right)\nonumber\\ \Leftrightarrow\left(1 - \mu^2\right)T''\left(\mu\right) - 2\mu\left(m + 1\right)T'\left(\mu\right) &= \left(-\lambda + m\left(m + 1\right)\right)T\left(\mu\right).\tag{C.82}\label{eq:legendre_umgeformt} \end{align} \]

It will be a power series approach again

\[ \begin{align} T\left(\mu\right) = \sum_{i = 0}^{\infty}a_i\mu^i \end{align} \]

made. You need the following notations for the first two derivatives:

\[ \begin{align} T'\left(\mu\right) &= \sum_{i = 1}^{\infty}ia_{i}\mu^{i - 1}\\ T''\left(\mu\right) &= \sum_{i = 0}^{\infty}\left(i + 2\right)\left(i + 1\right)a_{i + 2}\mu^i\\ T''\left(\mu\right) &= \sum_{i = 2}^{\infty}i\left(i - 1\right)a_i\mu^{i - 2} \end{align} \]

Putting this into Eq. (C.82), you get a recursion formula again:

\[ \begin{align} \left(i + 2\right)\left(i + 1\right)a_{i + 2} - i\left(i - 1\right)a_i - 2\left(m + 1\right)ia_i &= \left(-\lambda + m\left(m + 1\right)\right)a_i\nonumber\\ \Leftrightarrow a_{i + 2} = a_i\frac{ - \lambda + \left(m + 2i\right)\left(m + 1\right) + i\left(i - 1\right)}{\left(i + 2\right)\left(i + 1\right)} &= a_i\frac{ - \lambda + \left(i + m + 1\right)\left(i + m\right)}{\left(i + 2\right)\left(i + 1\right)} \end{align} \]

Since this doesn't approach zero for high $i$, this has to stop somewhere. So there must be a $I\in \mathbb{N}$ with

\[ \begin{align} \lambda = \left(I + m + 1\right)\left(I + m\right). \end{align} \]

As a product of two integers, $\lambda$ is also an integer, and $\lambda$ is not negative because the two factors cannot have different signs. So there is a $l\in \mathbb{N}$ with

\[ \begin{align} \lambda = l\left(l + 1\right). \end{align} \]

That's it

\[ \begin{align} l = I + m. \end{align} \]

Since $l, I\geq 0$, $m\leq l$. Since $m< - l$ leads to the same DGL as $-m > l$ and there are no solutions for this, there are no solutions for $m< - l$ either. So you have

\[ \begin{align} \lambda = l\left(l + 1\right), & {} & l = 0, 1, 2, 3, \dotsc, & {} & m = 0, \pm 1, \pm 2, \pm 3, \dotsc, \pm l. \end{align} \]

By comparing with Eq. (C.79) you can see that $m + I = l$ is also the degree of $P\left(\mu\right)$. The solutions $T = T_{l, m}$ are therefore determined by two natural numbers $l, m$.

Now we want to derive a closed form for the $T_{l, m}$. The $T_{l, m}$ satisfy the DGL

\[ \begin{align} \frac{d}{d\mu}\left(1 - \mu^2\right)T_{l, m}'\left(\mu\right) &= 2\mu mT_{l, m}'\left(\mu\right) + \left(m\left(m + 1\right) - \lambda\right)T_{l, m}\left(\mu\right).\tag{C.92}\label{eq:dgl_ass_leg_herl} \end{align} \]

For $m = 0$ the solutions are the well-known Legendre polynomials $P_l\left(\mu\right)$ Eq. (C.46). If the solution $T_{l, m}\left(\mu\right)$ is known and $m

\[ \begin{align} T_{l, m + 1}\left(\mu\right) = \frac{d}{d\mu}T_{l, m}\left(\mu\right). \end{align} \]

To show this, one first derives Eq. (C.92) from:

\[ \begin{align} & -2T_{l, m}'\left(\mu\right) - 4\mu T_{l, m}''\left(\mu\right) + \left(1 - \mu^2\right)T_{l, m}'''\left(\mu\right) = 2mT_{l, m}'\left(\mu\right) + 2\mu mT_{l, m}''\left(\mu\right) + \left(m\left(m + 1\right) - \lambda\right)T_{l, m}'\left(\mu\right)\nonumber\\ &\Leftrightarrow -2\mu T_{l, m}''\left(\mu\right) + \left(1 - \mu^2\right)T_{l, m}'''\left(\mu\right)\nonumber\\ &= 2\left(1 + m\right)T_{l, m}'\left(\mu\right) + \left(m\left(m + 1\right) - \lambda\right)T_{l, m}'\left(\mu\right) + 2\mu\left(1 + m\right)T_{l, m}''\left(\mu\right) \end{align} \]

Substituting $T_{l, m + 1} = T_{l, m}'$, it follows

\[ \begin{align} & \frac{d}{d\mu}\left(1 - \mu^2\right)T_{l, m + 1}'\left(\mu\right) = \frac{d}{d\mu}\left(1 - \mu^2\right)\frac{d}{d\mu}T_{l, m}'\left(\mu\right)\nonumber\\ &= -2\mu T_{l, m}''\left(\mu\right) + \left(1 - \mu^2\right)T_{l, m}'''\left(\mu\right) = 2\left(1 + m\right)T_{l, m}'\left(\mu\right) + \left(m\left(m + 1\right) - \lambda\right)T_{l, m}'\left(\mu\right) + 2\mu\left(1 + m\right)T_{l, m}''\left(\mu\right)\nonumber\\ &= 2\mu\left(m + 1\right)T_{l, m + 1}'\left(\mu\right) + \left(\left(m + 1\right)\left(m + 2\right) - \lambda\right)T_{l, m + 1}\left(\mu\right). \end{align} \]

The solution $T_{l, m}\left(\mu\right)$ is therefore the $mth derivative of the $nth Legendre polynomial

\[ \begin{align} T_{l, m}\left(\mu\right) = \frac{d^m}{d\mu^m}P_l\left(\mu\right). \end{align} \]

The complete solutions $P_{l, m}$ of Eq. (C.78) are called associated Legendre function:

\[ \begin{align} P_{l, m}\left(x\right) = \left(-1\right)^m\left(1 - x^2\right)^{m/2}\frac{d^{m}}{dx^{m}}P_l\left(x\right) = \frac{\left(-\right)^m}{2^ll!}\left(1 - x^2\right)^{m/2}\frac{d^{l + m}}{dx^{l + m}}\left(x^2 - 1\right)^l.\tag{C.97}\label{eq:assoziierte_legendre_funktionen} \end{align} \]

These are no longer polynomials for odd $m$ and reduce to the Legendre polynomials Eq. for $m = 0$. (C.46). It applies

\[ \begin{align} \frac{d}{dx}P_{l, m}\left(x\right) &= -mx\left(-1\right)^m\left(1 - x^2\right)^\frac{m - 2}{2}\frac{d^{m}}{dx^{m}}P_l\left(x\right) + \left(-1\right)^m\left(1 - x^2\right)^{m/2}\frac{d^{m + 1}}{dx^{m + 1}}P_l\left(x\right)\nonumber\\ &= -\frac{mx}{1 - x^2}P_{l, m}\left(x\right) - \frac{1}{\sqrt{1 - x^2}}P_{l, m + 1}\left(x\right).\tag{C.98}\label{eq:legendre_ass_prop_2} \end{align} \]

From Eq. (C.118) follows

\[ \begin{align} \frac{d^n}{dx^n}\left(xf\left(x\right)\right) &= xf^{(n)} + nf^{(n - 1)}\nonumber\\ \Leftrightarrow xf^{(n)} &= \frac{d^n}{dx^n}\left(xf\left(x\right)\right) - nf^{(n - 1)}. \end{align} \]

Furthermore, with Eq. (C.77)

\[ \begin{align} P_l &= \frac{2l + 3}{l + 1}xP_{l + 1} - \frac{l + 2}{l + 1}P_{l + 2}. \end{align} \]

Thus you get

\[ \begin{align} P_{l, m}\left(x\right) &= \left(-1\right)^m\left(1 - x^2\right)^{m/2}\frac{d^{m}}{dx^{m}}P_l\left(x\right) = \frac{\left(-1\right)^m}{l + 1}\left(1 - x^2\right)^{m/2}\frac{d^{m}}{dx^{m}}\left[\left(2l + 3\right)xP_{l + 1}\left(x\right) - \left(l + 2\right)P_{l + 2}\left(x\right)\right], \nonumber\\ &= x\frac{\left(-1\right)^m\left(2l + 3\right)}{l + 1}\left(1 - x^2\right)^{m/2}\frac{d^{m}}{dx^{m}}P_{l + 1}\left(x\right) + m\frac{\left(-1\right)^m\left(2l + 3\right)}{l + 1}\left(1 - x^2\right)^{m/2}\frac{d^{m - 1}}{dx^{m - 1}}P_{l + 1}\left(x\right)\nonumber\\ & - \frac{\left(-1\right)^m\left(l + 2\right)}{l + 1}\left(1 - x^2\right)^{m/2}\frac{d^{m}}{dx^{m}}P_{l + 2}\left(x\right)\nonumber\\ &= \frac{2l + 3}{l + 1}xP_{l + 1, m} - m\frac{2l + 3}{l + 1}\sqrt{1 - x^2}P_{l + 1, m - 1} - \frac{l + 2}{l + 1}P_{l + 2, m}\nonumber\\ \Rightarrow xP_{l + 1, m} &= \frac{l + 1}{2l + 3}P_{l, m} + m\sqrt{1 - x^2}P_{l + 1, m - 1} + \frac{l + 2}{2l + 3}P_{l + 2, m}\nonumber\\ \Rightarrow xP_{l, m} &= \frac{l}{2l + 1}P_{l - 1, m} + m\sqrt{1 - x^2}P_{l, m - 1} + \frac{l + 1}{2l + 1}P_{l + 1, m}.\tag{C.101}\label{eq:legendre_ass_prop_1_vorform} \end{align} \]

It applies

\[ \begin{align} P_{l, m - 1}\left(x\right) &= \left(-\right)^{m - 1}\frac{1}{2^ll!}\left(1 - x^2\right)^{\frac{m - 1}{2}}\frac{d^{l + m - 1}}{dx^{l + m - 1}}\left(x^2 - 1\right)^l. \end{align} \]

Furthermore is

\[ \begin{align} \frac{d^{l + m + 1}}{dx^{l + m + 1}}\left(x^2 - 1\right)^{l + 1} &= \frac{d^{l + m}}{dx^{l + m}}2\left(l + 1\right)x\left(x^2 - 1\right)^{l}\nonumber\\ &= \frac{d^{l + m - 1}}{dx^{l + m - 1}}\left[2\left(l + 1\right)\left(x^2 - 1\right)^{l} + 4x^2l\left(l + 1\right)\left(x^2 - 1\right)^{l - 1}\right]\nonumber\\ &= \frac{d^{l + m - 1}}{dx^{l + m - 1}}\left[2\left(l + 1\right)\left(x^2 - 1\right)^{l} + 4l\left(l + 1\right)\left(x^2 - 1\right)^{l} + 4l\left(l + 1\right)\left(x^2 - 1\right)^{l - 1}\right]\nonumber\\ &= 2\left(1 + 2l\right)\left(l + 1\right)\frac{d^{l + m - 1}}{dx^{l + m - 1}}\left[\left(x^2 - 1\right)^{l} + 4l\left(l + 1\right)\left(x^2 - 1\right)^{l - 1}\right]. \end{align} \]

So it applies

\[ \begin{align} \frac{d^{l + m - 1}}{dx^{l + m - 1}}\left(x^2 - 1\right)^l &= \frac{1}{2\left(1 + 2l\right)\left(l + 1\right)}\frac{d^{l + m + 1}}{dx^{l + m + 1}}\left(x^2 - 1\right)^{l + 1} - \frac{2l}{\left(2l + 1\right)}\frac{d^{l + m - 1}}{dx^{l + m - 1}}\left(x^2 - 1\right)^{l - 1}. \end{align} \]

This follows

\[ \begin{align} & \left(-\right)^{m - 1}\frac{1}{2^ll!}\left(1 - x^2\right)^{\frac{m - 1}{2}}\frac{d^{l + m - 1}}{dx^{l + m - 1}}\left(x^2 - 1\right)^l = \left(-\right)^{m - 1}\frac{1}{2^{l + 1}\left(l + 1\right)!}\left(1 - x^2\right)^{\frac{m - 1}{2}}\frac{1}{\left(1 + 2l\right)}\frac{d^{l + m + 1}}{dx^{l + m + 1}}\left(x^2 - 1\right)^{l + 1}\nonumber\\ & - \frac{1}{2l + 1}\left(-\right)^{m - 1}\frac{1}{2^{l - 1}\left(l - 1\right)!}\left(1 - x^2\right)^{\frac{m - 1}{2}}\frac{d^{l + m - 1}}{dx^{l + m - 1}}\left(x^2 - 1\right)^{l - 1}. \end{align} \]

Thus you get

\[ \begin{align} \sqrt{1 - x^2}P_{l, m - 1} &= -\frac{1}{2l + 1}P_{l + 1, m} + \frac{1}{2l + 1}P_{l - 1, m}. \end{align} \]

It follows

\[ \begin{align} xP_{l, m} &= \frac{l + m}{2l + 1}P_{l - 1, m} + \frac{l + 1 - m}{2l + 1}P_{l + 1, m}.\tag{C.107}\label{eq:legendre_ass_prop_1} \end{align} \]

Herewith and with Eq. (C.101) follows

\[ \begin{align} \frac{l + m}{2l + 1}P_{l - 1, m} + \frac{l + 1 - m}{2l + 1}P_{l + 1, m} &= \frac{l}{2l + 1}P_{l - 1, m} + m\sqrt{1 - x^2}P_{l, m - 1} + \frac{l + 1}{2l + 1}P_{l + 1, m}\nonumber\\ \Rightarrow\sqrt{1 - x^2}P_{l, m - 1} &= \frac{1}{2l + 1}P_{l - 1, m} - \frac{1}{2l + 1}P_{l + 1, m}\nonumber\\ \Rightarrow\sqrt{1 - x^2}P_{l, m} &= \frac{1}{2l + 1}P_{l - 1, m + 1} - \frac{1}{2l + 1}P_{l + 1, m + 1}.\tag{C.108}\label{eq:legendre_ass_prop_3} \end{align} \]

The normalization of the Legendre functions results in:

\[ \begin{align} \int_{-1}^{1}P_{l, m}\left(x\right)^2dx &= \frac{1}{2^{2l}l!^2}\int_{-1}^{\left(1\right)}\left(1 - x^2\right)^m\frac{d^{l + m}}{dx^{l + m}}\left(1 - x^2\right)^l\frac{d^{l + m}}{dx^{l + m}}\left(1 - x^2\right)^ldx\nonumber\\ &\stackrel{m+l-\text{fache p.I}.}{=}\frac{1}{2^{2l}l!^2}\left(-1\right)^{m + l}\left(-1\right)^{m + l}\left(m + l\right)!\frac{\left(2l\right)!}{\left(l - m\right)!}\int_{-1}^{1}\left(1 - x^2\right)^ldx\nonumber\\ &= \frac{1}{2^{2l}l!^2}\frac{\left(m + l\right)!}{\left(l - m\right)!}\left(2l\right)!\int_{-1}^{1}\left(1 + x\right)^l\left(1 - x\right)^ldx\nonumber\\ &\stackrel{l-\text{fache p.I}.}{=}\frac{1}{2^{2l}l!^2}\frac{\left(m + l\right)!}{\left(l - m\right)!}\left(2l\right)!\frac{l!}{\left(2l\right)!}l!\int_{-1}^{1}\left(1 + x\right)^{2l}dx\nonumber\\ &= \frac{1}{2^{2l}}\frac{\left(m + l\right)!}{\left(l - m\right)!}\frac{2^{2l + 1}}{2l + 1} = \frac{2}{2l + 1}\frac{\left(l + m\right)!}{\left(l - m\right)!}. \end{align} \]

Furthermore, for $l\not = l'$ (otherwise one can assume $l

\[ \begin{align} \int_{-1}^{1}P_{l, m}\left(x\right)P_{l', m}\left(x\right)dx &\propto \int_{-1}^{1}\left(1 - x^2\right)^m\frac{d^{l + m}}{dx^{l + m}}\left(1 - x^2\right)^l\frac{d^{l' + m}}{dx^{l' + m}}\left(1 - x^2\right)^{l'}dx. \end{align} \]

This is obtained by $m + l-$fold partial integration, in which $\left(1 - x^2\right)^m\frac{d^{l + m}}{dx^{l + m}}$ is derived and $\frac{d^{l' + m}}{dx^{l' + m}}\left(1 - x^2\right)^{l'}$ is integrated

\[ \begin{align} \int_{-1}^{1}P_{l, m}\left(x\right)P_{l', m}\left(x\right)dx &\propto \int_{-1}^{1}\frac{d^{l' - l}}{dx^{l' - l}}\left(1 - x^2\right)^{l'}dx = 0.\tag{C.111}\label{eq:ortho_ass_lgen} \end{align} \]

The term without an integral is omitted for each integration step, since it is evaluated at the edges $\pm 1$ and terms $\propto\left(1 - x^2\right)$ always appear there as factors. So it applies

\[ \begin{align} \int_{-1}^{1}P_{l, m}\left(x\right)P_{l', m}\left(x\right)dx = \frac{2}{2l + 1}\frac{\left(l + m\right)!}{\left(l - m\right)!}\delta_{l, l'}. \end{align} \]

C.3 Hermite polynomials

The Hermite polynomials are defined by their Rodrigues formula

\[ \begin{align} H_n\left(x\right) \coloneqq \left(-\right)^ne^{x^2}\frac{d^n}{dx^n}e^{-x^2}.\tag{C.113}\label{eq:hermite_poly_geschlossen} \end{align} \]

If you put in Eq. (4.69) $\newtilde{E} - 1 = 2n$, you get the Hermite's differential equation

\[ \begin{align} P_n'' - 2xP' + 2nP = 0.\tag{C.114}\label{eq:hermite_dgl} \end{align} \]

Now we want to show that the Hermite polynomials according to Eq. (C.113) solve this DGL and thus also the recursion formula Eq. (4.77). Yes, they apply

\[ \begin{align} \frac{dH_n}{dx}\left(x\right) &= \left(-\right)^ne^{x^2}\left[2x\frac{d^n}{dx^n}e^{-x^2} + \frac{d^{n + 1}}{dx^{n + 1}}e^{-x^2}\right],\\ \frac{d^2H_n}{dx^2}\left(x\right) &= \left(-\right)^ne^{x^2}\left[4x^2\frac{d^n}{dx^n}e^{-x^2} + 4x\frac{d^{n + 1}}{dx^{n + 1}}e^{-x^2} + \frac{d^{n + 2}}{dx^{n + 2}}e^{-x^2} + 2\frac{d^n}{dx^{n}}e^{-x^2}\right]. \end{align} \]

Plugging all of this into the left side of Hermite's DGL, you get

\[ \begin{align} \left(\frac{d^n}{dx^n}e^{-x^2}\right)\left[2 + 2n\right] + \left(\frac{d^{n + 1}}{dx^{n + 1}}e^{-x^2}\right)\left[-2x + 4x\right] + \left(\frac{d^{n + 2}}{dx^{n + 2}}e^{-x^2}\right) &= 0\nonumber\\ \Leftrightarrow \left[2\left(1 + n\right)\frac{d^{n}}{dx^n} + 2x\frac{d^{n + 1}}{dx^{n + 1}} + \frac{d^{n + 2}}{dx^{n + 2}}\right]\exp\left(-x^2\right) &= 0. \end{align} \]

This is true for $n = 0$. By differentiating the equation in terms of $x$ one obtains the same statement for $n + 1$. The Hermite polynomials thus solve the Hermite differential equation, you can set $P_n\left(x\right) = H_n\left(x\right)$. Furthermore, it is clear that this applies

\[ \begin{align} \frac{d^n}{dx^n}\left(f\left(x\right)g\left(x\right)\right) &= \frac{d^{n - 1}}{dx^{n - 1}}\left[f'g + fg'\right] = \frac{d^{n - 2}}{dx^{n - 2}}\left[f''g + 2f'g' + fg''\right]\nonumber\\ &= \sum_{k = 0}^{n}\left(\begin{array}{c} n\\ k \end{array}\right)f^{(k)}g^{(n - k)} \tag{C.118}\label{eq:produktregel_n} \end{align} \]

for $n\in\mathbb{N}$ and $f, g:\mathbb{R}\to\mathbb{R}$, $n-$-fold differentiable. This can be shown formally via complete induction. For $n = 0$ the statement is correct. The statement applies to an $n\in\mathbb{N}$. Then applies

\[ \begin{align} \frac{d^{n + 1}}{dx^{n + 1}}\left(fg\right) &= \frac{d}{dx}\left(\frac{d^n}{dx^n}\left(fg\right)\right) = \frac{d}{dx}\sum_{k = 0}^{n}\left(\begin{array}{c} n\\ k \end{array}\right)f^{(k)}g^{(n - k)}\nonumber\\ &= \sum_{k = 0}^{n}\left(\begin{array}{c} n\\ k \end{array}\right)\left[f^{(k + 1)}g^{(n - k)} + f^{(k)}g^{(n + 1 - k)}\right], \end{align} \]

the rest follows analogous to the proof of the binomial theorem Eq. (A.17). This can be used to

\[ \begin{align} H_n'\left(x\right) &= 2x\left(-\right)^ne^{x^2}\frac{d^n}{dx^n}e^{-x^2} + \left(-\right)^ne^{x^2}\frac{d^{n + 1}}{dx^{n + 1}}e^{-x^2}\nonumber\\ &= 2x\left(-\right)^ne^{x^2}\frac{d^n}{dx^n}e^{-x^2} + \left(-\right)^ne^{x^2}\frac{d^n}{dx^n}\left(-2xe^{-x^2}\right) \end{align} \]

to simplify. In the second summand, put in terms of Eq. (C.118) $f = -2x$ and $g = e^{-x^2}$, then only the terms with $k = 0$ and $k = 1$ play a role in the sum, so

\[ \begin{align} \frac{d^n}{dx^n}\left(-2xe^{-x^2}\right) = -2x\frac{d^n}{dx^n}e^{-x^2} - 2n\frac{d^{n - 1}}{dx^{n - 1}}e^{-x^2}. \end{align} \]

It therefore applies

\[ \begin{align} H_n'\left(x\right) = 2nH_{n - 1}\left(x\right).\tag{C.122}\label{eq:herm_pol_prop_1} \end{align} \]

It is immediately clear that this applies

\[ \begin{align} H_n' &= 2xH_n - H_{n + 1}\Leftrightarrow nH_{n - 1} + \frac{H_{n + 1}}{2} = xH_n.\tag{C.123}\label{eq:herm_pol_prop_2} \end{align} \]

The coefficient $C_n$ before the highest power is $C_n = 2^n$, as can be easily seen. The following applies to the normalization of Hermite polynomials

\[ \begin{align} \int_{ - \infty}^{\infty}H_n^2\left(x\right)e^{-x^2}dx &= \sqrt{\pi}2^nn!.\tag{C.124}\label{eq:hermite_polynome_normierung} \end{align} \]

To show this, first make it clear that:

\[ \begin{align} \int_{-\infty}^{\infty}H_n\left(x\right)H_m\left(x\right)e^{-x^2}dx = 0, \end{align} \]

for $n\not = m$, since in this case $H_n\left(x\right)\exp\left(-\frac{x^2}{2}\right)$ and $H_m\left(x\right)\exp\left(-\frac{x^2}{2}\right)$ are two eigenfunctions of a Hermitian operator with different eigenvalues, and therefore orthogonal.

Now one can show the statement about complete induction. For $n = 0$, $H_n = 1$, so

\[ \begin{align} \int_{ - \infty}^{\infty}H_0^2\left(x\right)e^{-x^2}dx &= \sqrt{\pi}. \end{align} \]

For $n = 1$, $H_1\left(x\right) = -2x$, so holds

\[ \begin{align} \int_{ - \infty}^{\infty}H_1^2e^{-x^2}dx &= 4\int_{ - \infty}^{\infty}x^2e^{-x^2}dx = 2\int_{ - \infty}^{\infty}e^{-x^2}dx = 2\sqrt{\pi}. \end{align} \]

So for $n = 0$ and $n = 1$ the statement is true. Now the statement already applies to an $n$ and also to $n - 1$. Then the equations (C.122) and (C.123) also follow

\[ \begin{align} & \int_{ - \infty}^{\infty}H_{n + 1}^2e^{-x^2}dx = 4\int_{ - \infty}^{\infty}\left[x^2H_n^2 + n^2H_{n - 1}^2 - 2xnH_{n - 1}H_n\right]e^{-x^2}dx\nonumber\\ &= 4n^2\sqrt{\pi}2^{n - 1}\left(n - 1\right)! + 4\int_{ - \infty}^{\infty}x^2H_n^2e^{-x^2}dx - 4n\int_{ - \infty}^{\infty}\left[H_{n - 1}'H_{n} + H_{n - 1}H_n'\right]e^{-x^2}dx\nonumber\\ &= 2n\sqrt{\pi}2^nn! + 2\int_{ - \infty}^{\infty}\left[2xH_n'H_n + H_n^2\right]e^{-x^2}dx - 8n^2\sqrt{\pi}2^{n - 1}\left(n - 1\right)!\nonumber\\ &= \left(2n - 4n\right)\sqrt{\pi}2^nn! + 2\int_{ - \infty}^{\infty}H_n'^2e^{-x^2}dx + 2\sqrt{\pi}2^nn!\nonumber\\ &= \sqrt{\pi}2^nn!\left(2n\right) + 2\sqrt{\pi}2^nn! = \sqrt{\pi}2^{n + 1}\left(n + 1\right)!. \end{align} \]

So the statement also applies to $n + 1$ and therefore to all $n\in \mathbb{N}$.

C.4 Laguerre polynomials

Laguerre's differential equation is:

\[ \begin{align} xP'' + \left(k + 1 - x\right)P' + nP = 0\tag{C.129}\label{eq:laguerre_dgl} \end{align} \]

for $k\in \mathbb{N}$. A power series approach is made for $P\left(x\right)$,

\[ \begin{align} P\left(x\right) = \sum_{i = 0}^{\infty}a_ix^i. \end{align} \]

Follow with that

\[ \begin{align} P'\left(x\right) &= \sum_{i = 0}^{\infty}ia_ix^{i - 1} = \sum_{i = 0}^{\infty}\left(i + 1\right)a_{i + 1}x^i,\\ P''\left(x\right) &= \sum_{i = 0}^{\infty}i\left(i + 1\right)a_{i + 1}x^{i - 1}. \end{align} \]

If you put this in, you get

\[ \begin{align} i\left(i + 1\right)a_{i + 1} + \left(k + 1\right)\left(i + 1\right)a_{i + 1} - ia_i + na_i &= 0 \Leftrightarrow a_{i + 1} = a_i\frac{i - n}{\left(i + 1\right)\left(i + k + 1\right)}.\tag{C.133}\label{eq:rek_laguerre} \end{align} \]

The factor $\frac{i - n}{\left(i + 1\right)\left(i + k + 1\right)}$ approaches zero for high $i$ like $1/i$, the recursion must therefore terminate. So $n\in \mathbb{N}$ must be. $n$ is the degree of the polynomial $P$. The polynomials

\[ \begin{align} L_{n, k}\left(x\right) \coloneqq \sum_{i = 0}^{n}\left(\begin{array}{c} n + k\\ n - i \end{array}\right)\frac{\left(-\right)^{(i)}}{i!}x^{(i)}.\tag{C.134}\label{eq:laguerre_polynome_explizit} \end{align} \]

solve Eq. (C.129) also, to show this, check their recurrence relation. First, they are polynomials of degree $n$. The coefficient before the highest power of $x$ is $\frac{\left(-\right)^n}{n!}$, this implies normalization. It applies

\[ \begin{align} a_i &= \frac{\left(n + k\right)!}{\left(n - i\right)!\left(k + i\right)!i!}\left(-\right)^{(i)}\nonumber\\ \Leftrightarrow \frac{a_{i + 1}}{a_i} &= -\frac{\left(n - i\right)!\left(k + i\right)!i!}{\left(n - i - 1\right)!\left(k + i + 1\right)!\left(i + 1\right)!} = \frac{i - n}{\left(k + 1 + i\right)\left(i + 1\right)}. \end{align} \]

Therefore, one can use Eq. View (C.134) as the definition of the Laguerre polynomials. You can do all of this

\[ \begin{align} n \to n_r,& {} & k \to 2l + 1 \end{align} \]

rewrite to

\[ \begin{align} L_{n_r, 2l + 1}^{}\left(x\right) \coloneqq \sum_{i = 0}^{n_r}\left(\begin{array}{c} n_r + 2l + 1\\ n_r - i \end{array}\right)\frac{\left(-\right)^{(i)}}{i!}x^{(i)} \end{align} \]

and

\[ \begin{align} xL_{n_r, 2l + 1}'' + \left(2l + 2 - x\right)L_{n_r, 2l + 1}' + n_rL_{n_r, 2l + 1} = 0. \end{align} \]

This is the formulation used in Section 4.8. The Rodrigues formula of Laguerre polynomials is

\[ \begin{align} P_{n, k}\left(x\right) = \frac{e^x}{n!x^k}\frac{d^n}{dx^n}\left[e^{-x}x^{n + k}\right]. \end{align} \]

You can immediately see that these are polynomials of degree $n$. The coefficient in front of $x^n$ is $\frac{\left(-\right)^n}{n!}$. For $l\in \mathbb{N}$ with $l\leq n$ holds

\[ \begin{align} \frac{d^l}{dx^l}x^{n + k} = \frac{\left(n + k\right)!}{\left(n + k - l\right)!}x^{n + k - l} \end{align} \]

It follows

\[ \begin{align} \frac{d^n}{dx^n}\left[e^{-x}x^{n + k}\right] &= \sum_{i = 0}^{n}\left(-\right)^{i}e^{-x}\left(\begin{array}{c} n\\ i \end{array}\right)\frac{\left(n + k\right)!}{\left(k + i\right)!}x^{k + i} = \sum_{i = 0}^{n}\left(-\right)^{i}e^{-x}\frac{n!}{i!\left(n - i\right)!}\frac{\left(n + k\right)!}{\left(k + i\right)!}x^{k + i}. \end{align} \]

This gives you

\[ \begin{align} P_{n, k}\left(x\right) &= \sum_{i = 0}^{n}\left(-\right)^{i}\frac{1}{i!\left(n - i\right)!}\frac{\left(n + k\right)!}{\left(k + i\right)!}x^{i} = \sum_{i = 0}^{n}\left(\begin{array}{c} n + k\\ n - i \end{array}\right)\frac{\left(-\right)^{(i)}}{i!}x^{(i)}. \end{align} \]

The Rodrigues formula therefore corresponds to the explicit representation Eq. (C.134). The Laguerre polynomials satisfy the integral property

\[ \begin{align} \int_{0}^{\infty}x^{k + 1}e^{-x}L_{n, k}\left(x\right)^2dx &= \frac{\left(n + k\right)!}{n!}\left(2n + k + 1\right)\tag{C.143}\label{eq:laguerre_integral}. \end{align} \]

This is what you count on

\[ \begin{align} & \int_{0}^{\infty}x^{k + 1}e^{-x}L_{n, k}\left(x\right)^2dx = \int_{0}^{\infty}x^{k + 1}e^{-x}L_{n, k}\left(x\right)\frac{e^x}{n!x^k}\frac{d^n}{dx^n}\left[e^{-x}x^{n + k}\right]dx\nonumber\\ &= \frac{1}{n!}\int_{0}^{\infty}xL_{n, k}\left(x\right)\frac{d^n}{dx^n}\left[e^{-x}x^{n + k}\right]dx = \frac{1}{n!}\left(-1\right)^n\int_{0}^{\infty}e^{-x}x^{n + k}\frac{d^n}{dx^n}xL_{n, k}\left(x\right)dx. \end{align} \]

According to the explicit representation Eq. (C.134) applies

\[ \begin{align} \frac{d^n}{dx^n}xL_{n, k}\left(x\right) &= \left(-\right)^n\left(n + 1\right)x + n!\frac{\left(-\right)^{n - 1}}{\left(n - 1\right)!}\frac{\left(n + k\right)!}{\left(n + k - 1\right)!}\nonumber\\ &= \left(-\right)^n\left(n + 1\right)x + n\left(-\right)^{n - 1}\left(n + k\right). \end{align} \]

With Eq. (A.96) follows

\[ \begin{align} \int_0^\infty\dotsc dx &= \frac{\left(-\right)^n}{n!}\left[\left(-\right)^n\left(n + 1\right)\left(n + k + 1\right)! - n\left(-\right)^n\left(n + k\right)\left(n + k\right)!\right]\nonumber\\ &= \frac{\left(n + k\right)!}{n!}\left[\left(n + 1\right)\left(n + k + 1\right) - n\left(n + k\right)\right] = \frac{\left(n + k\right)!}{n!}\left(2n + k + 1\right). \end{align} \]

Furthermore, $k\geq2$ applies

\[ \begin{align} \int_{0}^{\infty}x^{k - 2}\left[L_{n, k}\left(x\right)\right]^2\exp\left(-x\right)dx = \frac{\left(n + k\right)!}{n!}\frac{2n + k + 1}{\left(k - 1\right)k\left(k + 1\right)}\tag{C.147}\label{eq:laguerre_int_prop_2}. \end{align} \]

This is taken into account

\[ \begin{align} & \int_{0}^{\infty}x^{k - 2}e^{-x}L_{n, k}\left(x\right)^2dx = \int_{0}^{\infty}x^{k - 2}e^{-x}L_{n, k}\left(x\right)\frac{e^x}{n!x^k}\frac{d^n}{dx^n}\left[e^{-x}x^{n + k}\right]dx\nonumber\\ &= \frac{1}{n!}\int_{0}^{\infty}\frac{1}{x^2}L_{n, k}\left(x\right)\frac{d^n}{dx^n}\left[e^{-x}x^{n + k}\right]dx = \frac{1}{n!}\left(-1\right)^n\int_{0}^{\infty}e^{-x}x^{n + k}\frac{d^n}{dx^n}\left[\frac{1}{x^2}L_{n, k}\left(x\right)\right]dx. \end{align} \]

$L_{n, k}$ ist ein Polynom vom Grad $n$, also ist $L_{n, k}/x^2$ vom Grad $n - 2$. Schreibe

\[ \begin{align} L_{n, k} = a + bx + \mathcal{O}\left(x^2\right), \end{align} \]

applies accordingly

\[ \begin{align} \frac{d}{dx^n}\frac{L_{n, k}}{x^2} &= \frac{d}{dx^n}\left(\frac{a}{x^2} + \frac{b}{x}\right) = \left(-\right)^n\left(a\left(n + 1\right)!x^{-2 - n} + n!bx^{-n - 1}\right). \end{align} \]

According to the explicit representation Eq. (C.134) apply

\[ \begin{align} a = \frac{\left(n + k\right)!}{n!k!}, & {} & b = -\frac{\left(n + k!\right)}{\left(n - 1\right)!\left(k + 1\right)!}. \end{align} \]

Thus follows

\[ \begin{align} \frac{d}{dx^n}\frac{L_{n, k}}{x^2} &= \left(-\right)^n\left(\frac{\left(n + k\right)!}{n!k!}\left(n + 1\right)!x^{-2 - n} - \frac{\left(n + k\right)!}{\left(n - 1\right)!\left(k + 1\right)!}n!x^{-n - 1}\right)\nonumber\\ &= \left(-\right)^n\left(n + k\right)!\left[\frac{\left(n + 1\right)}{k!}x^{-2 - n} - \frac{n}{\left(k + 1\right)!}x^{-n - 1}\right]. \end{align} \]

This gives you

\[ \begin{align} e^{-x}x^{n + k}\frac{d}{dx^n}\frac{L_{n, k}}{x^2} &= e^{-x}\left(-\right)^n\left(n + k\right)!\left[\frac{\left(n + 1\right)}{k!}x^{k - 2} - \frac{n}{\left(k + 1\right)!}x^{k - 1}\right]. \end{align} \]

With Eq. (A.96) follows

\[ \begin{align} \int_0^\infty\dotsc dx &= \frac{\left(n + k\right)!}{n!}\left[\frac{\left(n + 1\right)}{k\left(k - 1\right)} - \frac{n}{k\left(k + 1\right)}\right]\nonumber\\ &= \frac{\left(n + k\right)!}{n!}\left[\frac{\left(n + 1\right)\left(k + 1\right)}{k\left(k - 1\right)\left(k + 1\right)} - \frac{n\left(k - 1\right)}{k\left(k + 1\right)\left(k - 1\right)}\right]\nonumber\\ &= \frac{\left(n + k\right)!}{n!}\frac{2n + k + 1}{k\left(k + 1\right)\left(k - 1\right)}. \end{align} \]

C.5 Spherical surface functions

The dependencies of the solution Eq. (D.42) of $\cos\left(\theta\right)$ and $\phi$ are combined to $P_{l, m}\left(\cos\left(\theta\right)\right)\exp\left(im\phi\right)$. The spherical harmonics are defined by

\[ \begin{align} Y_{l, m}\left(\theta, \phi\right) = \sqrt{\frac{2l + 1}{4\pi}\frac{\left(l - m\right)!}{\left(l + m\right)!}}P_{l, m}\left(\cos\left(\theta\right)\right)\exp\left(im\phi\right).\tag{C.155}\label{eq:def_spherical_harmonics} \end{align} \]

with the associated Legendre functions $P_{l, m}\left(x\right)$, Eq. (C.97). The preliminary factors result from the normalization

\[ \begin{align} \int_{\theta = 0}^\pi\int_{\phi = 0}^{2\pi}\left|Y_{l, m}\left(\theta, \phi\right)\right|^2\sin\left(\theta\right) d\phi d\theta\hastobe1, \end{align} \]

which is now verified:

\[ \begin{align} & \frac{2l + 1}{4\pi}\frac{\left(l - m\right)!}{\left(l + m\right)!}\int_{\theta = 0}^\pi\int_{\phi = 0}^{2\pi}\left(P_{l, m}\left(\cos\left(\theta\right)\right)\right)^2\sin\left(\theta\right) d\phi d\theta\nonumber\\ &= \frac{2l + 1}{2}\frac{\left(l - m\right)!}{\left(l + m\right)!}\int_{-1}^{1}P_{l, m}\left(z\right)^2dz = 1 \end{align} \]

Furthermore, the spherical surface functions are orthogonal. For $m\not = m'$ holds:

\[ \begin{align} & \int_{\theta = 0}^{\pi}\int_{\phi = 0}^{2\pi}Y_{l, m}\left(\theta, \phi\right)^\star Y_{l', m'}\left(\theta, \phi\right)\sin\left(\theta\right) d\phi d\theta\nonumber\\ &\propto \int_{\theta = 0}^{\pi}\int_{\phi = 0}^{2\pi}\exp\left(i\left(m - m'\right)\phi\right)d\phi P_{l, m}^{\star}\left(\cos\left(\theta\right)\right)P_{l', m'}\left(\cos\left(\theta\right)\right)\sin\left(\theta\right) d\theta = 0. \end{align} \]

For $l\not = l'$ one obtains

\[ \begin{align} & \int_{\theta = 0}^{\pi}\int_{\phi = 0}^{2\pi}Y_{l, m}\left(\theta, \phi\right)^\star Y_{l', m}\left(\theta, \phi\right)\sin\left(\theta\right) d\phi d\theta = 2\pi\int_{-1}^{1}P_{l, m}\left(z\right)P_{l', m}\left(z\right)dz\stackrel{\href{#eq:ortho_ass_lgen}{\text{Glg. (C.111)}}}{=}0. \end{align} \]

In summary you get

\[ \begin{align} \int_{\theta = 0}^{\pi}\int_{\phi = 0}^{2\pi}Y_{l, m}\left(\theta, \phi\right)^\star Y_{l', m'}\left(\theta, \phi\right)\sin\left(\theta\right) d\phi d\theta = \delta_{l, l'}\delta_{m, m'}. \end{align} \]

Still applies

\[ \begin{align} \Delta_{\theta, \phi} Y_{l, m}\left(\theta, \phi\right) &= -l\left(l + 1\right)Y_{l, m}\left(\theta, \phi\right) \end{align} \]

with

\[ \begin{align} \Delta_{\theta, \phi} = \Delta_{\mu, \phi} = \frac{\partial}{\partial\mu}\left(1 - \mu^2\right)\frac{\partial}{\partial\mu} + \frac{1}{1 - \mu^2}\frac{\partial^2}{\partial\phi^2}, \end{align} \]

as the angle component of the Laplace operator and $\mu = \cos\left(\theta\right)$. It applies

\[ \begin{align} \frac{\partial^2}{\partial\phi^2}Y_{l, m} &= -m^2Y_{l, m}\left(\theta, \phi\right), \end{align} \]

thus is here

\[ \begin{align} \Delta_{\mu, \phi} = \frac{\partial}{\partial\mu}\left(1 - \mu^2\right)\frac{\partial}{\partial\mu} - \frac{m^2}{1 - \mu^2}. \end{align} \]

For the associated Legendre polynomials $P_{l, m}\left(\mu\right)$ the DGL (C.78) applies

\[ \begin{align} \frac{d}{d\mu}\left(1 - \mu^2\right)P_{l, m}'\left(\mu\right) - \frac{m^2}{1 - \mu^2}P_{l, m}\left(\mu\right) = -l\left(l + 1\right)P_{l, m}\left(\mu\right). \end{align} \]

Thus follows

From Eq. (C.98) follows

Still applies

This follows

\[ \begin{align} & \exp\left(\pm i\phi\right)\left(i\cot\left(\theta\right)\frac{\partial}{\partial\phi}\pm\frac{\partial}{\partial\theta}\right)Y_{l, m}\left(\theta, \phi\right) = -m\cot\left(\theta\right)\exp\left(\pm i\phi\right)Y_{l, m}\left(\theta, \phi\right)\nonumber\\ &\pm \exp\left(\pm i\phi\right)m\cot\left(\theta\right) Y_{l, m}\left(\cos\left(\theta\right)\right)\pm\sqrt{l^2 - m^2 + l - m}\exp\left(\pm i\phi\right)Y_{l, m + 1}\left(\theta, \phi\right)\exp\left(-i\phi\right)\nonumber \end{align} \]

So it applies

\[ \begin{align} \newhat{L}_ + Y_{l, m}\left(\theta, \phi\right) &= \sqrt{l^2 - m^2 + l - m}Y_{l, m + 1}\left(\theta, \phi\right). \end{align} \]

This means that for the eigenstates $|n, l, m\rangle$ in the hydrogen atom

\[ \begin{align} \newhat{L}_ + |n, l, m\rangle = \sqrt{l^2 - m^2 + l - m}|n, l, m\rangle. \end{align} \]

From Eq. (C.107) follows

\[ \begin{align} \cos\left(\theta\right)Y_{l, m} &= \sqrt{\frac{2l + 1}{4\pi}\frac{\left(l - m\right)!}{\left(l + m\right)!}}\cos\left(\theta\right)P_{l, m}\left(\cos\left(\theta\right)\right)\exp\left(im\phi\right)\nonumber\\ &= \sqrt{\frac{2l + 1}{4\pi}\frac{\left(l - m\right)!}{\left(l + m\right)!}}\left[\frac{l + m}{2l + 1}P_{l - 1, m} + \frac{l - m + 1}{2l + 1}P_{l + 1, m}\right]\exp\left(im\phi\right)\nonumber\\ &= \sqrt{\frac{2l + 1}{4\pi}\frac{\left(l - m\right)!}{\left(l + m\right)!}}\frac{l + m}{2l + 1}P_{l - 1, m}\exp\left(im\phi\right) + \sqrt{\frac{2l + 1}{4\pi}\frac{\left(l - m\right)!}{\left(l + m\right)!}}\frac{l - m + 1}{2l + 1}P_{l + 1, m}\exp\left(im\phi\right)\nonumber\\ &= \frac{l + m}{2l + 1}\sqrt{\frac{2l + 1}{2l - 1}\frac{l - m}{l + m}}\sqrt{\frac{2l - 1}{4\pi}\frac{\left(l - 1 - m\right)!}{\left(l - 1 + m\right)!}}P_{l - 1, m}\exp\left(im\phi\right)\nonumber\\ & + \frac{l - m + 1}{2l + 1}\sqrt{\frac{2l + 1}{2l + 3}\frac{l + 1 + m}{l + 1 - m}}\sqrt{\frac{2l + 3}{4\pi}\frac{\left(l + 1 - m\right)!}{\left(l + 1 + m\right)!}}P_{l + 1, m}\exp\left(im\phi\right)\nonumber\\ &= \frac{l + m}{2l + 1}\sqrt{\frac{2l + 1}{2l - 1}\frac{l - m}{l + m}}Y_{l - 1, m} + \frac{l - m + 1}{2l + 1}\sqrt{\frac{2l + 1}{2l + 3}\frac{l + 1 + m}{l + 1 - m}}Y_{l + 1, m}.\nonumber \end{align} \]

Therefore applies

From Eq. (C.108) follows

\[ \begin{align} \sin\left(\theta\right)Y_{l, m} &= \sqrt{\frac{2l + 1}{4\pi}\frac{\left(l - m\right)!}{\left(l + m\right)!}}\sin\left(\theta\right)P_{l, m}\left(\cos\left(\theta\right)\right)\exp\left(im\phi\right)\nonumber\\ &= \frac{1}{2l + 1}\sqrt{\frac{2l + 1}{4\pi}\frac{\left(l - m\right)!}{\left(l + m\right)!}}\left[P_{l - 1, m + 1}\left(\cos\left(\theta\right)\right) - P_{l + 1, m + 1}\left(\cos\left(\theta\right)\right)\right]\exp\left(im\phi\right)\nonumber\\ &= \frac{1}{2l + 1}\sqrt{\frac{2l + 1}{4\pi}\frac{\left(l - m\right)!}{\left(l + m\right)!}}P_{l - 1, m + 1}\left(\cos\left(\theta\right)\right)\exp\left(im\phi\right)\nonumber\\ & -\frac{1}{2l + 1}\sqrt{\frac{2l + 1}{4\pi}\frac{\left(l - m\right)!}{\left(l + m\right)!}}P_{l + 1, m + 1}\left(\cos\left(\theta\right)\right)\exp\left(im\phi\right)\nonumber\\ &= \frac{1}{2l + 1}\sqrt{\frac{2l + 1}{2l - 1}\left(l - m - 1\right)\left(l - m\right)}\sqrt{\frac{2l - 1}{4\pi}\frac{\left(l - m - 2\right)!}{\left(l + m\right)!}}P_{l - 1, m + 1}\left(\cos\left(\theta\right)\right)\exp\left(im\phi\right)\nonumber\\ & -\frac{1}{2l + 1}\sqrt{\frac{2l + 1}{2l + 3}\left(l + m + 1\right)\left(l + m + 2\right)}\sqrt{\frac{2l + 3}{4\pi}\frac{\left(l - m\right)!}{\left(l + m + 2\right)!}}P_{l + 1, m + 1}\left(\cos\left(\theta\right)\right)\exp\left(im\phi\right).\nonumber \end{align} \]

Therefore applies

C.5.1 Products

C.5.2 Transformation to geographical coordinates

Now the derived equations should be transformed to geographical coordinates $\left(\phi, \lambda\right)$. Previously, $\theta = \pi/2 - \phi$ was used instead of latitude $\phi$. So they apply

\[ \begin{align} \cos\left(\theta\right) &= \cos\left(\frac{\pi}{2} - \phi\right) = \sin\left(\phi\right),\\ \sin\left(\theta\right) &= \sin\left(\frac{\pi}{2} - \phi\right) = \cos\left(\phi\right),\\ \cot\left(\theta\right) &= \tan\left(\phi\right),\\ \frac{\partial}{\partial\phi} &= \frac{\partial\theta}{\partial\phi}\frac{\partial}{\partial\theta} = -\frac{\partial}{\partial\theta},\\ \phi_\text{sph. coord.} &= \lambda. \end{align} \]

From Eq. (C.167) follows

\[ \begin{align} \frac{\partial}{\partial\phi}Y_{l, m}\left(\theta, \phi\right) &= -m\tan\left(\phi\right)Y_{l, m} - \sqrt{l^2 - m^2 + l - m}Y_{l, m + 1}\exp\left(-i\lambda\right).\tag{C.178}\label{eq:spherical_harmonic_deriv_theta_geo} \end{align} \]

From Eq. (C.168) follows

\[ \begin{align} \frac{\partial}{\partial\lambda}Y_{l, m} &= imY_{l, m}.\tag{C.179}\label{eq:spherical_harmonic_deriv_phi_geo} \end{align} \]

From Eq. (C.171) will

\[ \begin{align} \sin\left(\phi\right)Y_{l, m} &= \sqrt{\frac{l^2 - m^2}{4l^2 - 1}}Y_{l - 1, m} + \sqrt{\frac{\left(l + 1\right)^2 - m^2}{4\left(l + 1\right)^2 - 1}}Y_{l + 1, m}.\tag{C.180}\label{eq:spherical_harm_prop_2_geo} \end{align} \]

From Eq. (C.172) will

\[ \begin{align} \cos\left(\phi\right)Y_{l, m} &= \sqrt{\frac{1}{4l^2 - 1}\left(l - m - 1\right)\left(l - m\right)}e^{-i\lambda}Y_{l - 1, m + 1}\nonumber\\ &- \sqrt{\frac{1}{4l^2 + 8l + 3}\left(l + m + 1\right)\left(l + m + 2\right)}e^{-i\lambda}Y_{l + 1, m + 1}.\tag{C.181}\label{eq:spherical_harm_prop_3_geo} \end{align} \]