# For the mathematically inclined: help identify a power series

July 18, 2023 1:21 PM Subscribe

While playing around with some math stuff related to finding (complex) roots of polynomial functions, this funky little power series-like thing popped out and I'm not sure I know what it is.

This is about as close as I can get to proper notation:

For some function f:

Σₙ₌₀ ((xi)ⁿ f⁽ⁿ⁾(a)) / n!

And in case the unicode doesn't show up properly for anybody that's:

The sum from 'n' equals zero to infinity of one over 'n' factorial, times (xi) to the nth power, times the nth derivative of 'f' at 'a'. Where 'x' is the variable, 'a' and 'x' are real numbers and 'i' is the imaginary unit.

It

I'm comfortable with the math but I'm no mathematician so there's all sorts of stuff out there that I'm just not familiar with. So I'm wondering if the above series looks familiar to anyone, if it's an actual named thing that I can look up, or if it's just kind of a neat thing that I found of no real significance. That would be totally fine. I still think it's neat! And it's similarity to Taylor series may just be because I'm messing around with polynomials?

In case it helps I get the series by taking a general polynomial function of degree 'm' and feeding it the complex number (a + bi), basically turning it into a function of two variables 'a' and 'b'. Then fixing 'a' at some value and allowing 'b' (changed to 'x' in the series above) to vary. Is this a thing? Or not so much. Or maybe I just messed up the math. Also entirely possible.

This is about as close as I can get to proper notation:

For some function f:

Σₙ₌₀ ((xi)ⁿ f⁽ⁿ⁾(a)) / n!

And in case the unicode doesn't show up properly for anybody that's:

The sum from 'n' equals zero to infinity of one over 'n' factorial, times (xi) to the nth power, times the nth derivative of 'f' at 'a'. Where 'x' is the variable, 'a' and 'x' are real numbers and 'i' is the imaginary unit.

It

*looks*kind of like a Taylor series of 'f' around 'a' but if it was it would need an (x - a)^n term in there, right? And the 'i' wouldn't be there.

I'm comfortable with the math but I'm no mathematician so there's all sorts of stuff out there that I'm just not familiar with. So I'm wondering if the above series looks familiar to anyone, if it's an actual named thing that I can look up, or if it's just kind of a neat thing that I found of no real significance. That would be totally fine. I still think it's neat! And it's similarity to Taylor series may just be because I'm messing around with polynomials?

In case it helps I get the series by taking a general polynomial function of degree 'm' and feeding it the complex number (a + bi), basically turning it into a function of two variables 'a' and 'b'. Then fixing 'a' at some value and allowing 'b' (changed to 'x' in the series above) to vary. Is this a thing? Or not so much. Or maybe I just messed up the math. Also entirely possible.

Response by poster:

Nice! I'm glad I wasn't imagining that. It's been a while since I've taken calculus so I wasn't even entirely sure I was remembering Taylor series correctly. I had to look it up to be sure before posting.

Yeah, I wrote it as a finite sum originally. But then I didn't want to have to care what 'm' was so figured I'd just leave it out since it's the same thing either way. Anyway, thanks! Greatly appreciated!

posted by Mister_Sleight_of_Hand at 2:44 PM on July 18, 2023

*It is the Taylor series of f, but analytically continued into the complex plane!*Nice! I'm glad I wasn't imagining that. It's been a while since I've taken calculus so I wasn't even entirely sure I was remembering Taylor series correctly. I had to look it up to be sure before posting.

*Technically...the sum terminates at n=m*Yeah, I wrote it as a finite sum originally. But then I didn't want to have to care what 'm' was so figured I'd just leave it out since it's the same thing either way. Anyway, thanks! Greatly appreciated!

posted by Mister_Sleight_of_Hand at 2:44 PM on July 18, 2023

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isthe Taylor series of f, but analytically continued into the complex plane! Consider an analytic function f(z), where z is a complex number. You're essentially taking the Taylor expansion of f about the point z_0, where z_0 = a (a real number), but then looking at its values only along the "line" z = a + ix (with a fixed).Expressed in a slightly more familiar way, your series is:

f(z) = Sum_{n=0}^{infinity} (1/n!) (d^n f / d z^n)|_{z=z_0} (z - z_0)^n

(Technically, if f(z) is a polynomial of finite integer degree m, the sum terminates at n=m; derivatives beyond that all yield zero.)

posted by heatherlogan at 1:52 PM on July 18, 2023 [5 favorites]