Can a function's continuity be described by a uniform value of delta?

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Discussion Overview

The discussion revolves around the nature of the epsilon-delta definition of continuity in mathematical analysis, specifically whether the epsilon value can be a function of x. Participants explore the implications of defining epsilon and delta in various ways, including their dependence on specific points or other variables.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions if epsilon can be defined as a function of x, suggesting a form like epsilon = delta * (x-2)^2.
  • Several participants assert that epsilon should not be a function of x, emphasizing that it is an arbitrary positive real number.
  • Another participant clarifies that while epsilon is fixed, delta can depend on epsilon in the context of proofs.
  • It is noted that if x refers to a specific point x_0, then delta could be expressed as a function of both epsilon and x_0, allowing for different delta values at different points.
  • A participant mentions that for functions continuous on an interval, delta can be seen as a function of epsilon and x, and if this function is constant over the interval, the function is uniformly continuous.

Areas of Agreement / Disagreement

Participants generally disagree on whether epsilon can be a function of x, with a consensus that it should not be. However, there are varying views on the relationship between epsilon and delta, particularly regarding their dependence on specific points.

Contextual Notes

Some assumptions about the definitions of epsilon and delta are not explicitly stated, and the discussion does not resolve the implications of these definitions on continuity proofs.

pivoxa15
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Can the epsilon associated with f(x) be a function of x?

i.e epsilon = delta * (x-2)^2 valid?
 
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No. It shouldn't be function of x.
 
learningphysics said:
No. It shouldn't be function of x.

Can it be a function of anything?
 
What are you talking about??
Your question is extremely vague.
 
No! it shouldn't be function of anything because it's an ARBITRARY POSITIVE REAL NUMBER ( as such it must be different from 0). When you do an epsilon -- delta proof you must prove that, given an epsilon , which you don't know anything of, some in/equality holds whenever your variable ( x) satisfies some restriction ( i.e. |x-a|< [tex]\delta[/tex])
 
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[itex]\epsilon[/itex] and [itex]\delta[/itex] are constants and, as
Super Leunam said, can't depend on anything. However, in most proofs you are given [itex]\epsilon[/itex] (you can't control its value) and must find a correct [itex]\delta[/itex] for that [itex]\epsilon[/itex]. In that sense, [itex]\delta[/itex] may depend on [itex]\epsilon[/itex].
 
Unless by [tex]x[/tex] the poster means a specific point [tex]x_0[/tex].

So for example if you want to show [tex]f(x)[/tex] is continuous at every single point [tex]x_0[/tex] and you end up with [tex]\delta = x_0\epsilon[/tex], then that is okay.
 
Like Kummer said.

If you want to prove continuity at a point x_0, then your expression for [itex]\delta(\epsilon)[/itex] may depend on x_0 too: [itex]\delta(\epsilon, x_0)[/itex]. And sometimes, like in the proof of the chain rule I think, it becomes necessary to specify that a number delta is the delta associated with epsilon, and x_0 and of the function f, and in that end, we shall write [itex]\delta(\epsilon, x_0, f)[/itex].

Moreover, if a function is continuous on [a,b] say, then it means that given epsilon>0, there is a delta for every x in [a,b] and [itex]\delta(\epsilon, x)[/itex] can be seen as a function sending x in [a,b] to a such proper delta. And if that [itex]\delta(\epsilon, x)[/itex] function can be arranged to be constant over [a,b] (i.e. independent of x), this is when we say that f is uniformly continuous on [a,b].
 
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