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Preamble

My question is similar to an earlier MathOverflow question: “Taylor series” is to “Volterra series” as “Padé approximant” is to _________? which I just answered (hopefully my first ever MathOverflow post wasn't too bad!) except I've replaced "Padé approximant" with "Laurent series".

More details

For a function of one real variable the Taylor series can be written as:

\begin{alignat}{2} f(x) &= \sum_{n=0}^\infty c_n(x-x_0)^n &~~~,~~~c_n = \frac{f^{(n)}(x_0)}{n!}\tag{1}~. \end{alignat}

Using the Cauchy integral formula this has been extended to complex variables (for example here):

\begin{alignat}{3} f(z) &= \sum_{n=0}^\infty c_n(z-z_0)^n &~~~,~~~c_n = \frac{f^{(n)}(z_0)}{n!}=\frac{1}{2\pi \textrm{i}}\oint_\gamma \frac{f(z^\prime)\textrm{d}z}{\left(z^\prime - z_0\right)^{n+1}}\tag{2}~. \end{alignat}

Generalizing $f^{(n)}$ to allow $n<0$, we can get the Laurent series:

\begin{alignat}{3} \!\!\!f(z) &= \sum_{n=-\infty}^\infty c_n(z-z_0)^n &~~~,~~~c_n = \frac{f^{(n)}(z_0)}{n!}=\frac{1}{2\pi \textrm{i}} \oint_\gamma \frac{f(z^\prime)\textrm{d}z}{\left(z^\prime - z_0\right)^{n+1}}\tag{3}~. \end{alignat}

This can be generalized to functions with multiple variables (see Theorem 2.7.1 of this PDF for the $z_0=0$ case):

\begin{alignat}{3} {\scriptsize \!\!\! f(z_1,z_2) = \sum_{n_1,n_2=-\infty}^\infty c_{n_1 n_2}(z_1-z_{1,0})^{n_1}(z_2-z_{2,0})^{n_2} ~,~c_{n_1 n_2} = \frac{1}{\left(2\pi \textrm{i}\right)^n} \oint_{\gamma_{1}\times\gamma_2} \frac{f(z_1^\prime,z_2^\prime)\,\textrm{d}z_1\textrm{d}z_2}{\left(z_1^\prime - z_{1,0}\right)^{n_1+1}\left(z_2^\prime - z_{2,0}\right)^{n_2+1}}\tag{4}~.} \end{alignat}

Returning to real-valued variables, if the number of variables is uncountably infinite, for example not just $x_1,x_2$ but all $x_r$, where $r\in \mathbb{R}$, then we can consider expanding a functional $f[x(r)]$ around a specific function $x_0(r)$, which can be written as a Volterra series (similar to a Taylor series but now requiring integration over the real-valued index $r$, an additional time for each successive term in the series):

\begin{align} f[x(r)] &= \sum_{n=0}^\infty \int \int \cdots \int c_n\left(\prod_{i=0}^n \left(x(r)-x_0(r^{(i)})\right) \right) \textrm{d}r^\prime \textrm{d}r^{\prime \prime} \cdots \textrm{d}r^{(n)}\tag{5}\\ c_n &=\frac{f^{(n)}[x_0(r)]}{n!}\tag{6}\\ &= \frac{1}{n!}\frac{\delta f[x_0(r)]}{\delta f[x(r^\prime)]\delta f[x(r^{\prime\prime})]\cdots \delta f[x(r^{(n)})]}. \tag{7}\\ \end{align}

Eq. 5 has been used in practical applications, for example in this 1994 paper in which $r$ represented a 3-dimensional vector so each integral in Eq. 5 was in fact a triple integral.

I can now generalize the Cauchy integral formula to not just be for the derivative of $f(z_0)$ but for the functional derivative of $f[z_0(r)]$ and therefore derive the series expansion of $f[z(r)]$ around the function $z_0(r)$ using Eq. 5 (or Eq. 5 with the sum going from $n=-\infty$ instead of 0) except with Eq. 7 replaced by the generalized Cauchy integral formula for functions. But has this ever been done before? I looked into this a bit in 2009 but didn't find much and wasn't aware of MathOverflow at the time; now searches I've made in 2022 have also not resulted in anything satisfying.

Reason

If you're interested in why I wanted to do this, it's because I was studying generalizations of Born's rule. In Born's rule, the probability as a functional of a complex-valued wavefunction, simply becomes the composite function:

$$ P[\psi(\mathbf{r})] = P\left(\psi(\mathbf{r})\right) = \psi(\mathbf{r})\psi^*(\mathbf{r}) = |\psi(r)|^2.\tag{8} $$

I was doing calculations to supplement a precision test of Born's rule, and under the assumption that the true probability functional in a complete "theory of everything" is some yet-to-be found expression for which Eq. 8 is only the first and most dominant term of a longer series expansion, then I would want to expand my model functionals $P[\psi(\mathbf{r})]$ around the complex function $\psi_0(\mathbf{r}) = \psi(\mathbf{r})\psi^*(\mathbf{r})$, using something like my proposed generalization of Eq. 5.

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  • $\begingroup$ @gmvh sorry I've roll backed your edit too. I request that you do not use the word "obvious" in any communications. $\endgroup$ Commented Feb 9, 2023 at 15:09

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