Some thoughts on Taylor series.

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.

(Sometimes this is \(f(x) = \sum \frac{(x - x_0)^{n} f^{(n)}(x_0)}{n!}\). 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 \((x - x_0)\). Also, 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.

This can be written in a cleaner notation if we let ourselves treat the derivative operator \(\p_{x}\) as a variable, and then treat the whole summation as an operator acting on \(f\):

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

(Seomtimes we’ll omit the subscript and just write \(\p\) to keep things uncluttered.)

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

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

This shows that \(e^{\p_x}\) implements the translation operator acting on functions, and \(\p_x\) acts like a “generator of translations” in the sense of generators of Lie Groups.

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 \(\p_{z} = \frac{1}{2}(\p_{x} - i \p_{y})\).

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

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

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:

\[f(x + a + b) = e^{b\p_{x}} e^{a\p_{x}} f(x) \? e^{(b + a)\p_{x}} f(x)\]

Whether \(e^{a\p_{x}}e^{b\p_{y}} = e^{b\p_{y}}e^{a\p_{x}}\) is true depends on whether the (underlying coordinate system has any curvature. If it does then the LHS \(f(x + a + b)\) doesn’t make sense as you would need to specify the path of each translation, rather than just a displacement from \(x\).)

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2

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

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

(It’s a left inverse 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 \(a^n/ n!\) 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 makes sense, because in some sense it is one. It basically means: project \(f(x)\) onto the basis spanned by each polynomial order \(\frac{x^{n}}{n!}\), and then write it in terms of those basis vectors 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 \(f(x+a)\). It is similar to writing \(\b{v} = v_x \b{x} + v_y \b{y} + \ldots\).

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3. Misc

in differential geometry one things about about \(f(x)\) as a sort of abstract object \(f\) that doesn’t know about any coordinate systems yet. Before you have coordinates, it’s just a map between two spaces, for instance \(f: A \mapsto B\). To get a coordinate representation there needs to be another mapping \(\hat{x}: \bb{R}^n \mapsto A\) which locates points on \(A\) according to some Euclidean coordinate system (or a subset of it, or whatever). We can write this as

\[f(x) = f \circ \hat{x} : \bb{R}^n \ra B\]

Although it’s nice to treat \(\hat{x}\) as an operator acting on \(f\): the “evaluate at \(x\) operator”:

\[\hat{x} f \equiv f \circ \hat{x} = f(x)\]

Translating by a vector \(\vec{a}\) in \(\bb{R^n}\) is then written as the operation of addition when it takes place in \(\bb{R}^n\), although I guess we can be weird and write it as function that is applied like this:

\[\begin{aligned} (+ \vec{a})(\hat{x}) &= \hat{x} \circ (+ \vec{a}) \equiv \hat{x} + \vec{a} \\ (\hat{x} + \vec{a}) f &= f \circ [\hat{x} \circ (+ \vec{a})] = f(x + \vec{a}) \end{aligned}\]

We can also write it as an operator that acts on \(f\):

\[\hat{x} (T_{\vec{a}} f) = (\hat{x} + \vec{a}) f = f(x + \vec{a})\]

All of this is pretty messy in math notation; it makes more sense by just writing out the whole chain:

\[x \ra x + \vec{a} \ra f(x+\vec{a})\]

And then you get a differently named object depending on where you put the parentheses.

Anyway, I find it interesting that the Taylor series operators from earlier give ways of writing these various things (for certain classes of functions, etc):

\[\begin{aligned} (\hat{x} + \vec{a}) f &= \hat{x} (T_{\vec{a}} f) = \Big[ \sum (\p^{-1}_{a} \p_{x})^{n} \Big] [ \hat{x} f] \\ \end{aligned}\]

I dunno. Guess that’s a bit interesting.

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For posterity, the other way of doing this is:

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 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 corresponds to just integrating the derivative of \(f\):

\[f(x + a) = f(x) + \int_{0}^{a} f'(x + y) \d y\]

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

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

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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.