Talk:Dirichlet conditions

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Edit Request - Dirichlet's Theorem for 1-Dimensional Fourier Series

The section: Dirichlet's Theorem for 1-Dimensional Fourier Series, the following sentence: "The analogous statement holds..." appears right in the middle of the explanation of what the Theorem is. I am left wondering which "statement" it is referring to?? Apparently for a periodic function of any period, (even an infinite period?? doesn't this contradict the condition of boundedness??) The more I read this section the worse it appears. "For all x..." - are we supposed to know that "x" is a Real number? The Fourier Series is generally a FINITE integral, which contradicts (seemingly) the -∞ to +∞ integration in this section. "We state..." is about as pompous as you can get, but a trivial point. How does a function "oscillate" at a point?? BY DEFINITION, a function must be single valued at any point (in its domain). Why does the definition assume the given period? It either IS necessary or is NOT necessary. (Apparently it is not). Can someone either delete or move the offending sentence and possibly rewrite the entire section in a more clear fashion? e.g. "Dirichlet's theorem: If (a function [f:x ∈ R → C (?)] satisfies Dirichlet conditions, then for all x, the Fourier series, given by [insert formula here] is convergent. Where etc. etc." (talk) 05:17, 7 July 2014 (UTC)


"These three conditions are satisfied if f is a function of bounded variation over a period."

This seems to be a false statement. Here are some examples.

(1) f=0 has bounded variation, yet it does not satisfy the three conditions. Indeed, it has an infinite number of maxima and minima in each finite interval. So the thrid condition is violated.

(2) FIx a decreasing sequence of positive numbers x_i -->0. Let g be a function with g(0)=0, g(x_i) = (-1/2)^i for i=0,1,2,..., and monotone between successive x_i. Then g is BV, in fact even continuous, so the Fourier series actually converges uniformly, but g has an infinite number of maxima and minima, so again the third condition is violated.

(3) A BV function can easily have a countably infinite number of jumps, so the second condition is easy to violate. (talk) 03:32, 30 March 2016 (UTC)

The above mentioned statement regarding BV is valid only for the 2nd Condition "finite number of extrema" out of the 3 conditions. I think, this condition is not violated for the 2nd condition as you have stated in (3) because a BV function can not have infinite number of maxima or minima unless it is a constant function.
So, I suggest, we modify the condition 2
2. f must have a finite number of extrema in any given bounded interval, i.e. there must be a finite number of maxima and minima in the interval.
2. f must be of bounded variation in any given bounded interval, i.e. there must be a finite number of maxima and minima in the interval.
Please refer to this book p.198 [1] available at [1]
Pardhu Madipalli talk 20:49, 23 December 2017 (UTC)
Updated the article with respect to the above mentioned points.
Pardhu Madipalli talk 11:11, 25 December 2017 (UTC)

Bounded Variation = Finite Extrema?

I believe the use of "i.e." in the second condition is incorrect. There exist BV functions with infinitely many extrema on a closed interval. For example, consider the function on any closed interval of positive length containing zero (after filling a removable singularity at zero). This function is BV with infinitely many critical points in the interval. While this example is not periodic, it may be easily modified to serve as a counterexample to the implied statement in condition (2). I feel that the use of "e.g." would be more appropriate here. Samuel Li (talk) 20:45, 30 March 2018 (UTC)

  1. ^ Alan V, Oppenheim (1997). Signals & Systems. Prentice Hall. p. 198. 
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