Generating-function representation for scalar products

Abstract

We employ the generating-function representation for an n-dimensional vector in Euclidean or Hilbert space to evaluate scalar products. The generating function is constructed as a power series in a complex variable weighted by the components of a vector. The scalar product is represented by a convolution of the generating functions for the vectors integrated over a closed contour in the complex plane. The analyticity of the generating functions associated with the Laurent theorem reduces the evaluation of the scalar product into counting combinatoric multiplicity factors. As applications, we provide two exemplary computations: the sum of the squares of integers and the normalization of normal modes in a vibrating loaded string. As a byproduct of the latter example, we find a new alternative proof of a famous trigonometric identity that is essential for Fourier analyses.

Appendix A: Proof of Eq. (55)

In this appendix, we prove the identity in Eq. (55) by applying the generating-function approach. We define a sequence $a^{(r)}$ as

$$\begin{aligned} a^{(r)}\equiv & {} \left\{ \left. \sin \frac{rk\pi }{n+1} \right| k=1,~2,~\ldots ,~n. \right\} ,\nonumber \\ r= & {} 1,~2,~\ldots ,~n. \end{aligned}$$

(A1)

The corresponding generating function can be constructed from Eqs. (43) and (54) as

$$\begin{aligned} g\left( z; a^{(r)} \right)= & {} \displaystyle \sum _{k=1}^n z^k\sin \frac{rk\pi }{n+1}= \sin \theta _{r}\frac{z+(-1)^{r-1} z^{n+2}}{(z-z_r^+)(z-z_r^-)}, \nonumber \\ \theta _r= & {} \frac{r\pi }{n+1}, \end{aligned}$$

(A2)

where $z_r^\pm$ are defined in Eq. (44). As is stated in Ref. [6], $g\left( z ,a^{(r)} \right)$ is an entire function: analytic in the entire complex plane. Thus, the function does not have any poles. Then, the left-hand side of Eq. (55) can be expressed in terms of $a^{(r)}$ and $a^{(s)}$ as

$$\begin{aligned}&\sum _{k=1}^n \sin \frac{rk\pi }{n+1}\sin \frac{sk\pi }{n+1}=\sum _{k=1}^n a^{(r)}a^{(s)},\nonumber \\&r,\,s=1,~2,~\ldots ,~n. \end{aligned}$$

(A3)

Substituting $\psi ^\star =a^{(r)}$ and $\chi =a^{(s)}$ into Eq. (18), we can express the series in Eq. (A3) as

$$\begin{aligned}&\sum _{k=1}^n a^{(r)}a^{(s)} = \frac{1}{2\pi i}\oint _C \frac{\mathrm{d}z}{z} g\left( z; a^{(r)} \right) g\left( \frac{1}{z}; a^{(s)}\right) . \end{aligned}$$

(A4)

Substituting the generating function in Eq. (A2) into Eq. (A4) and dividing the expression by $\sin \theta _{r}\sin \theta _{s}$, we find that

$$\begin{aligned} \displaystyle&\!\!\!\frac{\sum _{k=1}^n a^{(r)}a^{(s)}}{\sin \theta _{r}\sin \theta _{s}}\nonumber \\&\,\,=\displaystyle \frac{1}{2\pi i}\oint _C \frac{\mathrm{d}z}{z} \left[ \frac{ z+(-1)^{r-1} z^{n+2} }{(z-z_r^+)(z-z_r^-)} \right] \left[ \frac{ \frac{1}{z}+(-1)^{s-1} \frac{1}{ z^{n+2}} }{\frac{1}{z^2}(z-z_s^+)(z-z_s^-)} \right] \nonumber \\ \,&\,\,=\displaystyle \frac{1}{2\pi i}\oint _C \frac{\mathrm{d}z}{z} \frac{z^2\left[ 1 +(-1)^{r+s} + (-1)^{s-1} z^{-n-1}+ (-1)^{r-1} z^{n+1} \right] }{(z-z^+_r)(z-z^-_r)(z-z^+_s)(z-z^-_s)}.\nonumber \\ \end{aligned}$$

(A5)

According to Eq. (44), $z^\pm _j$ are complex conjugates on the unit circle centered at the complex 0:

$$\begin{aligned} z^\pm _j=e^{\pm i\theta _{j}},\quad \theta _{j}=\frac{j\pi }{n+1},\quad j=r,~s\in \{1,~2,~\ldots ,~n\}.\nonumber \\ \end{aligned}$$

(A6)

Because $g\left( z ,a^{(r)} \right)$ is an entire function, the integrand in Eq. (A5) has a pole only at $z=0$. Thus, the contour integral is invariant under deformation as long as C encloses 0 once counterclockwise. If we choose a contour C entirely inside the unit circle centered at the complex 0, then $|z/z_j^\pm |<1$ at any point z on the contour. Hence, we can make a Taylor-series expansion about $z=0$ as

$$\begin{aligned} \frac{1}{z-z_j^\pm }=-\frac{1}{z_j^\pm } \sum _{k=0}^\infty e^{\mp ik\theta _{j}}z^k,\quad j=r,~s\in \{1,~2,~\ldots ,~n\}.\nonumber \\ \end{aligned}$$

(A7)

In the integrand of Eq. (A5), we can discard any contributions that are analytic at $w=0$ because they do not contribute to the residue by applying Cauchy integral formula in Eq. (3). As a result, Eq. (A5) collapses into the sum of the following elementary residues:

$$\begin{aligned}&\!\!\! \frac{\sum _{k=1}^n a^{(r)}a^{(s)}}{\sin \theta _{r}\sin \theta _{s}}\nonumber \\&\,\!\!\!\quad =\displaystyle (-1)^{s-1} \sum _{a=0}^\infty \sum _{b=0}^\infty \sum _{c=0}^\infty \sum _{d=0}^\infty \frac{1}{2\pi i}\oint _C \frac{\mathrm{d}z}{z} z^{1-n+a+b+c+d}e^{i\theta _{r}(a-b)} e^{i\theta _{s}(c-d)}\nonumber \\&\,\!\!\!\quad =\displaystyle (-1)^{s-1} \sum _{a=0}^\infty \sum _{b=0}^\infty \sum _{c=0}^\infty \sum _{d=0}^\infty \delta _{a+b+c+d,n-1}e^{i\theta _{r}(a-b)} e^{i\theta _{s}(c-d)} \nonumber \\&\,\quad = \displaystyle (-1)^{s-1} \sum _{a=0}^{n-1}\quad \sum _{b=0}^{n-1-a}\quad \sum _{c=0}^{n-1-a-b}e^{i\theta _{r}(a-b)} e^{i\theta _{s}(2c+a+b-n+1)} \nonumber \\&\,\quad = \displaystyle \frac{(-1)^{s-1}}{2[\cos \theta _{r}-\cos \theta _{s}]}\left[ \frac{\sin (n+1)\theta _{r}}{\sin \theta _{r}} - \frac{\sin (n+1)\theta _{s}}{\sin \theta _{s}}\right] . \end{aligned}$$

(A8)

If $r\ne s$, then the denominator factors for the two terms $[\cos \theta _{r}-\cos \theta _{s}]\sin \theta _{r}$ and $[\cos \theta _{r}-\cos \theta _{s}]\sin \theta _{s}$ are both non-vanishing. Because the numerator is always vanishing due to the constraints $\sin (n+1)\theta _{r}=\sin (n+1)\theta _{s}=0$, we find that

$$\begin{aligned} \left. \frac{\sum _{k=1}^n a^{(r)}a^{(s)}}{\sin \theta _{r}\sin \theta _{s}} \right| _{r\ne s} =0. \end{aligned}$$

(A9)

If $r= s$, then both the numerator and the denominator vanish simultaneously. Thus, we apply L’Hôpital’s rule by taking the ratio of the first-order derivatives of the numerator and the denominator and applying the identities $\sin (n+1)\theta _{r}=0$ and $\cos (n+1)\theta _{s}=(-1)^s$ in Eq. (36). Equivalently, we can replace $\theta _{s}$ with $\theta _{s} +x$ and take the Taylor-series expansion of both the numerator and the denominator to find that

$$\begin{aligned}&\displaystyle \left. \frac{\sum _{k=1}^n a^{(r)}a^{(s)}}{\sin \theta _{r}\sin \theta _{s}} \right| _{r= s}\nonumber \\&\quad = \displaystyle \frac{(-1)^{s-1} }{2}\lim _{x\rightarrow 0} \frac{ \displaystyle \frac{\sin [(n+1)(\theta _{r}+x)]}{\sin (\theta _{r}+x)} - \frac{\sin (n+1)\theta _{r}}{\sin \theta _{r}} }{ \cos (\theta _{r}+x)-\cos \theta _{r} } \nonumber \\&\quad =\displaystyle \frac{(-1)^{s-1} }{2}\lim _{x\rightarrow 0} \frac{ (n+1)x \sin \theta _{r} \cos [(n+1)\theta _{r}] }{\displaystyle -x\sin ^3\theta _{r} } \nonumber \\&\quad = \displaystyle \frac{ n+1 }{2\sin ^2\theta _{r} }. \end{aligned}$$

(A10)

Combining the results in Eqs. (A9) and (A10), we conclude that

$$\begin{aligned} \sum _{k=1}^n&\sin \frac{rk\pi }{n+1}\sin \frac{sk\pi }{n+1}=\frac{n+1}{2}\,\delta _{rs},\nonumber \\&r,\,s=1,~2,~\ldots ,~n. \end{aligned}$$

(A11)

This completes the proof of Eq. (55).