# Meditation on Taylor Series

[March 30, 2018]

(Notes. Definitely not interesting unless, at minimum, you really really liked calculus.)

## 1

We can often write a differentiable function $$f(x)$$ as a Taylor series around a point $$x$$, approximating it in terms of its derivatives at that point:

$f(x+a) = \sum_{0}^{\infty} \frac{a^{n} f^{(n)}(x) }{n!}$

And, under certain conditions, this series will converge exactly to the values of the function at nearby points.

(It may be that there is a certain radius of convergence around $$x$$ in which this approximation is valid. For the remainder of this page, assume we’re dealing with $$f$$ and $$x$$ such that $$f$$ has a series which is convergent around $$x$$ and we’re staying close enough for that to be valid.)

(You may be more used to seeing this as $$f(x) = \sum \frac{(x - x_0)^{n} f^{(n)}(x_0)}{n!}$$. They’re equivalent, of course, but for our purposes it will be cleaner to write out it as an approximation for a displacement $$a$$ from a point $$x$$, rather than having to write the displacement as $$(x - x_0)$$.)

This can be written in a cleaner notation if we let ourselves treat the derivative operator $$\p_{x}$$ as a variable (sometimes we will omit the subscript $$x$$ to keep things uncluttered) and then treat the whole summation as an operator acting on f:

$f(x+a) = [\sum \frac{(a \p_{x})^{n}}{n!}]f(x)$

And it’s cleaner still if we recognize the summation as the Taylor series for $$e^{x}$$ (neglecting, perhaps, to define this rigorously):

$f(x+a) = e^{a\p_{x}} f(x)$

(In physics we look at this and say that $$\p_{x}$$ is the ‘generator of translations’, in the sense of generators of Lie Groups, and that $$e^{a \p_{x}}$$ is the translation operator.)

This form is especially nice because it lets us translate by one variable at a time when working with multivariate functions:

$f(x+a, y) = e^{a\p_{x}} f(x,y)$

Or translate by complex variables, using $$\p_{z} = \frac{1}{2}(\p_{x} - i \p_{y})$$:1

$f(z + a) = e^{a \p_{z}} f(z)$

Or calculate higher-dimensional Taylor series, using $$\nabla = \p_{x} + \p_{y}$$:

$f(\b{x} + \b{v}) = e^{\mathbf{v} \cdot \nabla} f(\mathbf{x}) = \Big[ \sum \frac{(v_{x} \p_{x} + v_{y} \p_{y})^{n}}{n!} \Big] f(x,y)$

Or write out multiple translations in a row:2

$f(x + a + b) = e^{b\p_{x}} e^{a\p_{x}} f(x) = e^{(b + a)\p_{x}} f(x)$

Or implement a time-translation operator for wave functions in (non-relativistic) quantum mechanics to compute how systems evolve in time while preserving total probability by construction3:

$\psi(x, t) = e^{t \p_{t}} \psi(x, 0) = e^{- \frac{i t}{\hbar} H} \psi(x, 0)$

So it’s just all really great, when it works and the series converge and everything commutes the way you expect, and you can take integrals and derivatives term-by-term and everything’s somehow peachy.

(In physics we tend to, instead of carefully proving things converge, just do the calculations and see if they match what they should be afterwards, and then wave our hands and conclude that it works, because it’s easier that way and because (I suspect) getting a coherent calculus of operators is an analytical nightmare, and definitely not in immediate reach of the curious undergraduate.)

## 2

Assume $$F(x) = \int f(x) dx$$ exists, and consider antidifferentiation as a left inverse of the differentation operator:

$\p^{-1} f(x) = \int_{0}^{x} f(x') dx' = F(x) - F(0)$

(Why left? because $$(\p \circ \p^{-1}) f(x) = f$$, but $$(\p^{-1} \circ \p) f = f + c$$ is only equal up to a constant.)

What can we do with $$\p^{-1}$$? Well, we can produce the $$\frac{a^{n}}{n!}$$ term in our Taylor expansion:

\begin{aligned} \p^{-1} (1) &= x \\ \p^{-2} (1) &= \frac{x^{2}}{2} \\ \p^{-n} (1) &= \frac{x^{n}}{n!} \\ \p_{a}^{-n} (1) = \p^{-n} (1) \Big|_{x = a} &= \frac{a^{n}}{n!} \\ \end{aligned}

(That’s a derivative with respect to $$a$$ instead of $$x$$. They’re just symbols, after all.)

And therefore:

$e^{x} = \Big[ \sum_{n = 0}^{\infty} \p^{-n} \Big] 1 = 1 + x + \frac{x^{2}}{2!} + \ldots$ $e^{a\p_{x}} = \sum \p_{a}^{-n} (1) \p_{x}^{n} = 1 + a \p + \frac{a^{2}}{2!} \p^{2} + \ldots$ $f(x+a) = e^{a \p_{x}} f(x) = \Big[ \sum \p_{a}^{-n}(1) \p_{x}^{n} \Big] f(x)$

Really, since $$\p_{a}$$ refers to a different variable, it will just treat $$\p_{x} f(x)$$ as a constant – so we can just write:

$f(x+a)= \Big[ \sum (\p^{-1}_{a} \p_{x})^{n} \Big] f(x)$

Which has a nice symmetry to it. It reminds me of a change of basis, which, in some sense, it is.

It basically means: project $$f(x)$$ onto its behavior at each polynomial order $$\frac{x^{n}}{n!}$$, and then write it literally in terms of those polynomial orders using $$\frac{a^{n}}{n!}$$. If (1) $$f$$ is truly entirely constructible entirely from polynomials, and (2) the resulting sum converges, this should be equivalent to $$f(x+a)$$.

## 3 Misc

If we consider $$f$$ as an abstract function object which only takes on a value $$f(x)$$ when composed with a point $$\hat{x} \circ f = f(x)$$, then we can write a suggestive (but probably not too meaningful) equations like:

$(\hat{x} + \vec{a}) \circ f = \hat{x} \circ \Big[ \sum (\p^{-1}_{a} \p_{x})^{n} \Big] \circ f$ $(\hat{x} + \vec{a}) = \hat{x} \circ \sum (\p^{-1}_{a} \p_{x})^{n}$ $\hat{x} \circ (+a) = \hat{x} \circ \sum (\p^{-1}_{a} \p_{x})^{n}$

This says, approximately, that $$T_{\vec{a}}$$, translation by $$\vec{a}$$, is associative, and can be implemented in either x-space or ‘operators on functions’-space.

$(\hat{x} \circ T_{\vec{a}}) \circ f = \hat{x} \circ (T_{\vec{a}} \circ f)$

Of course it’s associative even if you can’t write $$f$$ as a Taylor series, but, this gives a sort of ‘implementation’ for it, when it is.

There are other ways to conceptually ‘implement’ $$T_{\vec{a}} f$$:

Since the derivative operator gives the value of $$f(x + \e) \approx f(x) + \e f'(x)$$, at a point slightly displaced from $$x$$, we can presumably in principle do this infinitely many times to move a finite distance $$a$$:

$f(x + a) = \lim_{\D x \ra 0} f(x) + f'(x)\D x + f'(x + \D x)\D x + \ldots$

Which of course corresponds to just integrating the derivative of $$f$$:

$f(x + a) = f(x) + \int_{0}^{a} f'(x + y) dy$

Alternatively we may write this as applying an infinitesimal translation operator $$T_{\e}$$ infinitely many times, which just leads back to the exponential expression:

$f(x + a) = \lim_{\e \ra 0} T^{\frac{a}{\e}}_{\e} f(x) = \lim_{\e \ra 0} (1 + (T_{\e} - 1))^{\frac{a}{\e}} f(x)$ $= \lim_{\e \ra 0} (1 + \D_{\e})^{\frac{a}{\e}} f(x) = e^{a \p_{x}} f(x)$

## 4

I don’t know, maybe this will be useful to someone, someday. I needed to write it down to keep various thoughts bundled together for later.

1. The $$\frac{1}{2}$$ and the negative sign are required so that $$\p_{z} \cdot dz =$$ $$\p_{z} \cdot (dx + i dy) = 1$$.

2. Whether $$e^{a\p_{x}}e^{b\p_{y}} = e^{b\p_{y}}e^{a\p_{x}}$$ gets into whether the underlying manifold has any curvature.

3. Given Schrödinger’s equation $$H\psi = i \hbar \p_{t} \psi$$, expand $$e^{t \p_{t}} \psi$$ term-by-term, substitute, and unexpand.