Anonymous at Sun, 21 Apr 2024 09:42:16 UTC No. 16138606

You can easily create a continious function around x0, for which epsilon remains the same despite lowering delta. So your assumption goes into the trash

Anonymous at Sun, 21 Apr 2024 10:01:53 UTC No. 16138615

>>16138606
Ok, but it never gets larger right?

Anonymous at Sun, 21 Apr 2024 10:31:53 UTC No. 16138638

epislon_2 <= epislon_1 implies delta_2 <= delta_1 if we assume that delta is always the maximum possible delta. That would be your intuition. However, the definition doesn't force you to choose the maximum possible delta. It just says there exists a delta. So you may choose delta arbitrarily to violate this intuition.

Anonymous at Sun, 21 Apr 2024 10:33:56 UTC No. 16138641

>>16138638
Does delta_2 <= delta_1 imply epislon_2 <= epislon_1 also? If we choose the largest delta?

Anonymous at Sun, 21 Apr 2024 10:47:48 UTC No. 16138659

>>16138641
For all practical intuitive purposes you may think of it like this. However, I'd like to emphasize that this direction goes against the definition. You choose delta depending on epsilon, not epsilon depending on delta.

Anonymous at Sun, 21 Apr 2024 11:00:19 UTC No. 16138679

for every delta there's a set of all viable epsilons for which the implication holds. you're asking whether delta_2 < delta_1 implies that the infimum of the set of all viable epsilon_2 is smaller than the infimum of the set of all viable epsilon_1. this is true, but the inequality is non-strict (counter example is the constant function}.

Anonymous at Sun, 21 Apr 2024 11:01:38 UTC No. 16138681

>>16138679
Why is it true that for every δ there exists an ε? I thought it's for every ε there exists a δ? How do we know the other way around is true?

Anonymous at Sun, 21 Apr 2024 11:05:44 UTC No. 16138688

>>16138679
>delta_2 < delta_1 implies that the infimum of the set of all viable epsilon_2 is smaller than the infimum of the set of all viable epsilon_1. this is true,
How do you show that?

Anonymous at Sun, 21 Apr 2024 12:06:58 UTC No. 16138741

>>16138688
if delta1 works for epsilon then delta2 works for epsilon

Anonymous at Sun, 21 Apr 2024 12:08:48 UTC No. 16138743

>>16138741
Yes, but how does that imply that inf(ε_2) < inf(ε_1)?

Anonymous at Sun, 21 Apr 2024 12:22:33 UTC No. 16138758

>>16138587
counter-example: [math]f(x) = c [/math]

Anonymous at Sun, 21 Apr 2024 12:37:49 UTC No. 16138786

>>16138758
Ok, what about less than or equal to?

Anonymous at Sun, 21 Apr 2024 12:58:20 UTC No. 16138808

>>16138786
Then it's true since if [math]0 < \delta_1 < \delta_2[/math] then if
[math] A_i := \{f(x) \quad | \quad 0 < |x - a| < \delta_i \}, \quad i = 1, 2[/math]
We have
[math]A _1 \subset A_2[/math]
Not even continuity is needed since we have inclusion of the range subsets over which we're taking supremums to obtain our epsilons

Anonymous at Sun, 21 Apr 2024 13:16:49 UTC No. 16138835

>>16138808
How does that coincide with ε_1<=ε_2?

Anonymous at Sun, 21 Apr 2024 13:29:36 UTC No. 16138850

>>16138835
I messed up, the definitions should have been
[math]
A_i := \{|f(x) - f(a)| \, : \, 0 < |x - a| < \delta_i \}
[/math]
then [math]\epsilon_i := \sup A_i[/math].

This is essentially what you're asking right, we get a smaller band on the y axis when we push our x's closer on the x-axis? In that case we see that since [math]A_1 \subseteq A_2[/math] we have to have [math]\epsilon_1 \leq \epsilon_2[/math]

Anonymous at Sun, 21 Apr 2024 13:55:07 UTC No. 16138883

>>16138850
Is [math]ε_i[/math] the [math]\inf[/math] of the possible [math]ε[/math] such that for [math]m \subset A_i[/math] we have [math]m \leq ε[/math]?

Anonymous at Sun, 21 Apr 2024 14:07:39 UTC No. 16138894

homework thread

Anonymous at Sun, 21 Apr 2024 14:10:11 UTC No. 16138895

>>16138883
the supremum not the infimum, we're looking at the greatest distance on the y-axis that our limitation on the x axis can give us, but yes.
that would be the meaning of pushing delta closer to a to obtain a smaller or equal epsilon on the y axis.

Anonymous at Sun, 21 Apr 2024 14:18:25 UTC No. 16138900

>>16138895
What i meant is, there exists several ε > |f(x) - f(a)| for |x-a|<δ, but the set A_i doesn't contain all of them. For example, there exists ε > ε_i such that when |x-a|<δ_i we have ε > ε_i > |f(x) - f(a)|. Actually, should it be |f(x)-f(a)|<=ε_i as ε_i is the supremum?

Anonymous at Sun, 21 Apr 2024 14:53:47 UTC No. 16138929

>>16138606
Trivially true for the constant function sure, but can you provide a non-trivial example?

Anonymous at Sun, 21 Apr 2024 15:30:16 UTC No. 16138979

>>16138929
f(x)=x*e^(-x)*(sgn(x)+1)
x0 = 0

Anonymous at Sun, 21 Apr 2024 17:45:28 UTC No. 16139129

>>16138979
>sgn(x)
not real math