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

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In summary, the Epsilon associated with f(x) cannot be a function of x because it is an ARBITRARY POSITIVE REAL NUMBER and should be different from 0.
  • #1
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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  • #2
No. It shouldn't be function of x.
 
  • #3
learningphysics said:
No. It shouldn't be function of x.

Can it be a function of anything?
 
  • #4
What are you talking about??
Your question is extremely vague.
 
  • #5
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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  • #6
[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].
 
  • #7
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.
 
  • #8
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. independant of x), this is when we say that f is uniformly continuous on [a,b].
 
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1. What are epsilon and delta in an epsilon-delta proof?

Epsilon and delta are variables used in an epsilon-delta proof to define the notion of a limit. Epsilon represents a small positive number, and delta represents a small positive distance. The goal of an epsilon-delta proof is to show that as the distance between the input and the limit point (represented by delta) approaches zero, the output of the function (represented by epsilon) also approaches zero.

2. Why do we use epsilon-delta proofs?

Epsilon-delta proofs are used to formally prove the existence of a limit in a mathematical function. They provide a rigorous and precise way to define the behavior of a function near a certain point, and are an essential tool in the field of calculus.

3. What is the general structure of an epsilon-delta proof?

In general, an epsilon-delta proof consists of three main parts: the setup, the proof, and the conclusion. The setup involves defining the variables epsilon and delta, and stating the limit definition. The proof involves manipulating the inequalities involving epsilon and delta to prove that the limit holds. The conclusion states that the limit has been proven.

4. Are epsilon-delta proofs difficult to understand?

Epsilon-delta proofs can be challenging to understand at first, as they require a solid understanding of calculus and mathematical reasoning. However, with practice and guidance, they can become easier to comprehend. It is important to break down each step and understand the logic behind it in order to fully grasp the concept.

5. Can epsilon-delta proofs be used for all functions?

Yes, epsilon-delta proofs can be used for all functions, as long as the function is defined and continuous at the given point. However, the complexity of the proof may vary depending on the function and the point in question.

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