Error function (defined on the whole complex plane) is entire

Click For Summary

Homework Help Overview

The discussion revolves around the error function \(\text{erf}(z) = \int_{0}^{z} e^{-t^{2}} dt\) and its classification as an entire function. Participants are seeking a proof or justification for this property, particularly focusing on the function's analyticity across the complex plane.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Some participants suggest using the Taylor series expansion of the integrand and integrating term by term to demonstrate convergence everywhere. Others express uncertainty about the Taylor series expansion process and inquire about relevant references or theorems, such as Cauchy's Theorem. Additionally, there is mention of differentiating the error function to show that its derivative is analytic, which would imply that the function itself is entire.

Discussion Status

The discussion is active, with various approaches being explored, including series expansion and differentiation. Participants are questioning assumptions and seeking clarification on methods, but no consensus has been reached regarding the proof of the error function's entire nature.

Contextual Notes

Participants have noted the challenge of applying the Fundamental Theorem of Calculus in this context, as well as the need to consider whether the function has an analytic continuation across the entire complex plane.

julypraise
Messages
104
Reaction score
0

Homework Statement


The wiki page says that error function \mbox{erf}(z) = \int_{0}^{z} e^{-t^{2}} dt is entire. But I cannot find anywhere its proof. Could you give me some stcratch proof of this?


Homework Equations





The Attempt at a Solution


I've tried to use Fundamental Theorem of Calculus but as it is the line integral I couldn't use it.
 
Physics news on Phys.org
You can just expand the integrand as a Taylor series, integrate by terms (since everything is finite) and then convince yourself that the resulting series converges everywhere.
 
clamtrox said:
You can just expand the integrand as a Taylor series, integrate by terms (since everything is finite) and then convince yourself that the resulting series converges everywhere.

Ah.. frankly I'm not sure of how to expand by Taylor series... Is it by using Cauchy's Theorem? Could you give me any reference textbook where I can look up the theorem for this Taylor series expansion?
 
julypraise said:
Ah.. frankly I'm not sure of how to expand by Taylor series... Is it by using Cauchy's Theorem? Could you give me any reference textbook where I can look up the theorem for this Taylor series expansion?

If you pick any analysis textbook at random, the odds of it containing a chapter on Taylor series is very very high. The extension to complex variables is straightforward.

You can also proceed by doing the standard proof that the function is holomorphic, by using the Cauchy-Riemann equations, but then you also have to consider whether the function has an analytic continuation to the entire complex plane. For example logarithm is holomorphic in its domain, but is not an entire function. That's why the Taylor series route is more straightforward: you can show that the series converges everywhere, which automatically shows you that the function is entire.
 
julypraise said:

Homework Statement


The wiki page says that error function \mbox{erf}(z) = \int_{0}^{z} e^{-t^{2}} dt is entire. But I cannot find anywhere its proof. Could you give me some stcratch proof of this?


Homework Equations





The Attempt at a Solution


I've tried to use Fundamental Theorem of Calculus but as it is the line integral I couldn't use it.

I don't see what's wrong with just differentiating it, showing the derivative is analytic throughout the complex plane, then concluding it's entire. That is, since

\frac{d}{dz} \text{erf}(z)=e^{-z^2}

and e^{-z^2} is analytic, thus the error function is entire.
 

Similar threads

Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
32
Views
4K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
2
Views
2K