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Given a Taylor series

$$f(z) = \sum_{k=0}^\infty c_k^{(a)}\frac{(z-a)^k}{k!}$$

of a meromorphic function $f$ in $\mathbb C$ (i.e. analytical except for a set of isolated points) around some value $a\in\mathbb C$, where $c_k^{(a)}=\frac{d^k}{da^k}f(a)$, one can naively apply the translation operator $\exp((b-a)\partial_c)|_{c=a}$ on $f(a+z)$ to obtain the series expansion around another point $b$:

$$\begin{align} f(b+z) &\ =\ e^{(b-a)\partial_c}f(c+z)|_{c=a} \\ &\ =\ \sum_{l=0}^\infty\frac{(b-a)^l\partial_a^l}{l!}\sum_{k=0}^\infty\frac{\partial_a^k f(a)}{k!}z^k \\ &"=" \sum_{k=0}^\infty\underbrace{\left(\sum_{l=0}^\infty\frac{(b-a)^l}{l!}c_{k+l}^{(a)}\right)}_{\stackrel!=c_k^{(b)}}\frac{z^k}{k!} \end{align}$$

The last $"="$ is of course only valid if the sums can actually be swapped - which is precisely my question: Under what conditions is this valid?

I'm fairly sure an absolutely minimum requirement would be the existence of a curve $\Gamma\subset\mathbb C$ connecting $a$ and $b$ such that $f|_\Gamma$ is analytical, since then one can use analytical continuation along $\Gamma$ within the local convergence radius to iteratively "crawl" from $a$ to $b$. However, I'm neither sure whether the formula always remains valid even if there are singularities closer to $a$ or $b$ than these two points are separated, nor whether this criterion alone is sufficient.

edit The bounty will be awarded to an answer at least more soundly proving the jump-case, though a working solution along a continuous curve would be even better. An answer pointing out the flaws in mine will suffice as well... edit2 Looks like the bounty will contribute to global entropy; I'll still award a new bounty of 200 rep once another acceptable answer is posted...

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    $\begingroup$ The existence of such a path isn't sufficient. Consider your archetypical multi-valued functions, $\log$ and powers with non-integer exponents, and take a branch-cut that passes through the circle of convergence of the Taylor series you start with. The translation operator gives you the series you'd get from connecting the two points by a straight line, which may pass through the branch-cut. If you take a path in the domain of the function, that winds (almost, not a closed path) around $0$, you get a different branch and a different Taylor series. $\endgroup$ – Daniel Fischer Sep 24 '13 at 12:57
  • $\begingroup$ @DanielFischer True, but is $\log$ meromorphic (let alone holomorphic before I edited the question)? The branch cut is no set of isolated points but an open one $\endgroup$ – Tobias Kienzler Sep 24 '13 at 12:59
  • $\begingroup$ Before the edit, you had "holomorphic around a point $a \in \mathbb{C}$". That read as if you looked at a very general situation, and a branch of the logarithm/ a power would have qualified. If you're looking at entire meromorphic functions, that's a different situation. I tentatively think it works then, but have to think more about it. $\endgroup$ – Daniel Fischer Sep 24 '13 at 13:04
  • $\begingroup$ @DanielFischer Yes, sorry about that ambiguity, I hope the edit improved that - although that makes me think whether the function being meromorphic on sufficiently large environment of $\Gamma$ would already suffice (though as you mention, crossing a branch cut or other non-isolated singularities would definitely fail) $\endgroup$ – Tobias Kienzler Sep 24 '13 at 13:05
  • $\begingroup$ Related: How to switch to a Laurent series' next “onion” layer? $\endgroup$ – Tobias Kienzler Sep 24 '13 at 13:14
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Since $f$ is meromorphic, its derivatives $f^{(n)}$ have no residues and as long as $b$ is inside the convergence region the fundamental theorem of calculus can be applied to obtain

$$\begin{align} c_n^{(b)} &= f^{(n)}(b) = \underbrace{f^{(n)}(a)}_{=c_n^{(a)}} + \int_a^b f^{(n+1)}(t)\,dt \\ &= c_n^{(a)} + \sum_{k=0}^\infty c_{k+n}^{(a)} \underbrace{\int_a^b \frac{(z-a)^k}{k!}\,dz}_{=\frac{(b-a)^{k+1}}{(k+1)!}} \\ &= \sum_{k=0}^\infty c_{n+k}^{(a)}\frac{(b-a)^k}{k!} (*) \end{align}$$

just as the question suggested.

However, if $b$ lies outside the convergence region, this fails. In that case, one or more intermediate points $c_j$ have to be introduced via which to hop from $a$ to $b$:

Let $c_0=a$ and $c_N=b$, i.e. there will be $N$ jumps connecting $N+1$ $c_j$s. Given the Taylor series $$f_j(z) = \sum_{k=0}^\infty f^{(k)}(c_j)\frac{(z-c_j)^k}{k!}$$ around the point $c_j$, pick a point $c_{j+1}$ in $f_j$'s convergence regoin, i.e. with $|c_j-c_{j+1}|<R_j$ where $R_j^{-1} = \limsup\limits_{k\to\infty}\sqrt[k]{|f^{(k)}(c_j)|}$ is the radius of convergence of $f_j$. Then using aforementioned argument, the coefficients $f^{(k)}(c_{j+1})$ of the series' analytical continuation $f_{j+1}$ around $c_{j+1}$ are $$\begin{align} f^{(k)}(c_{j+1}) &= \sum_{l=0}^\infty f^{(k+l)}(c_j)\frac{(c_{j+1}-c_j)^l}{l!} \\ &= \sum_{l=k}^\infty f^{(l)}(c_j)\frac{(c_{j+1}-c_j)^{l-k}}{(l-k)!} \end{align}$$

The recursion can be solved as $$\begin{align} f^{(l_N)}(b) = f^{(l_N)}(c_N) &= \sum_{l_{N-1}=l_N}^\infty\sum_{l_{N-2}=l_{N-1}}^\infty\cdots\sum_{l_0=l_1}^\infty \underbrace{f^{(l_0)}(c_0)}_{=f^{(l_0)}(a)}\prod_{m=1}^N \frac{(c_{m}-c_{m-1})^{(l_{m-1}-l_{m})}}{(l_{m-1}-l_{m})!} (\dagger) \\ &= \sum_{l_0=l_N}^\infty \sum_{l_1=l_N}^{l_0}\sum_{l_2=l_N}^{l_1}\cdots\sum_{l_{N-1}=l_N}^{l_{N-2}} \underbrace{f^{(l_0)}(c_0)}_{=f^{(l_0)}(a)}\prod_{m=1}^N \frac{(c_{m}-c_{m-1})^{(l_{m-1}-l_{m})}}{(l_{m-1}-l_{m})!} \\ &= \sum_{k_0=0}^\infty\underbrace{f^{(l_N+k_0)}(c_0)}_{=f^{(l_N+k_0)}(a)} \sum_{k_1=0}^{k_0}\sum_{k_2=0}^{k_1}\cdots\sum_{k_{N-1}=0}^{k_{N-2}} \prod_{m=1}^N \frac{(c_{m}-c_{m-1})^{(k_{m-1}-k_{m})}}{(k_{m-1}-k_{m})!} \quad(k_N=0) \end{align}$$


Incomplete part for a continuous curve, almost certainly wrong

Now let $\gamma:[0,1]\to\mathbb C$ be a curve from $a=\gamma(0)$ to $b=\gamma(1)$ and denote $1/N=:dt, j/N=:t_j, c_j:=\gamma(t_j)$ (yes, to be more precise everything should carry an index $N$ as well, but you know where this is heading, right?). Then $$c_j - c_{j-1} = \gamma(t_j) - \gamma(t_j-dt) = \gamma'(t_j)dt - \gamma''(t_j)dt^2/2 + \mathcal O(dt^3).$$

claim 1) The $dt$s in the numerator product can be factored out as $$\prod_{m=1}^N dt^{(l_{m-1}-l_{m})} = dt^{(l_0-l_N)} = dt^{k_0}.$$

claim 2) Since there are $N-1$ sums, this is also the maximum exponent of $dt$ that cannot lead to vanishing terms, so $$k_0 = l_0-l_N \leqslant N-1.$$

claim 3) Since then $N-1\geqslant k_0\geqslant k_1\geqslant...\geqslant k_{N-1} \geqslant 0,$ for a fixed $k_0$ the sums can rewritten into $k_0$ sums that denote which index $1\leqslant j\leqslant N-1$ of the $k_j$ they increase by one - special care has to be taken when multiple of these sums increase the same index.

claim 4) Since there are at most $k_0$ sums and the $dt^{k_0}$ term is unavoidable, all higher order terms of $c_m - c_{m-1}$ will vanish for $N\to\infty, dt\to0$.

claim 5) If two or more sums increase the same index, the effective amount of sums is less than $k_0$ and such terms cannot contribute to the sum due to the $dt^{k_0}$. claim 6) Therefore the only remaining terms when linearizing in $dt$ are those where $k_0$ different indices are increased by exactly one, i.e.

$$f^{(l)}(b) \dot= \sum_{k=0}^{N-1} f^{(l+k)}(a) \underbrace{\sum_{j_1=1}^{N-1}\sum_{j_2=1 \atop j_2\neq j_1}^{N-1}\cdots\sum_{j_{k}}}_{k\,\text{sums}}dt^{k}\prod_{m=1}^{k}\gamma'(c_{j_m})$$

claim 7) The inner sums can be re-expressed as $$\sum_{j_m} = \sum_{j_m=1}^{N-1}\left(1-\sum_{l=1}^{m-1}\delta_{j_l,j_m}\right)$$ i.e. $$f^{(l)}(b) \dot= \sum_{k=0}^{N-1}f^{(l+k)}(a) \prod_{m=1}^k\left(\sum_{j_m=1}^{N-1}\gamma'(c_{j_m})dt\left[1-\sum_{l=1}^{m-1}\delta_{j_m,j_l}\right]\right)$$

claim 8) Now I'm almost convinced the Kronecker deltas don't contribute significantly, but that would result in $$\begin{align} f^{(l)}(b) &\dot= \sum_{k=0}^{N-1}f^{(l+k)}(a) \left(\sum_{j=1}^{N-1}\gamma'(c_j)dt\right)^k \\ &\stackrel{N\to\infty}\longrightarrow \sum_{k=0}^\infty f^{(l+k)}(a)\underbrace{\left(\int\gamma'(t)\,dt\right)^k}_{=(b-a)^k} \end{align}$$

which contradicts $(*)$ and actually has an even smaller convergence radius. Either I must have been too generous in dropping higher order terms or made some other mistake...


It might actually be a better approach to solve the differential equation $$\partial_t f^{(l)}(\gamma(t)) = \gamma'(t) f^{(l+1)}(\gamma(t)),\quad f^{(l)}(\gamma(0)=a) = f^{(l)}(a)$$ but note that this is a system of infinitely many coupled linear ODEs, that are formally solved by $(*)$ but still suffer from the same lack of convergence for all $|\gamma(t)-a|\geqslant R_a$

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