• Support PF! Buy your school textbooks, materials and every day products Here!

Epsilon Delta and The Triangle Inequality

  • Thread starter DieCommie
  • Start date
  • #26
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
131
Now, the first thing you should do, is to ascertain that what I have done is VALID.
Secondly, if you are unsure about, or don't understand why I have made the particular choices I made, then ask about that.
 
  • #27
156
0
Thank you for sticking around and helping me.

I have gone through the steps, and I see that how it is all valid.

I do have a few particulars I still dont understand, as you predicted.

1)you choose [tex]\delta < \delta_1 = 1 [/tex] arbitrarily.. Is that enable you to solve the delta expression in terms of epsilon?

2)When you did this [tex]\delta^2 + 6x < \delta\delta_1 + 6\delta = 7\delta [/tex] why did you choose to replace only one of the deltas with 1? And why did you choose to replace that particular delta?

3)How do we know this is all true if [tex][|x^2 - 9|<\epsilon[/tex]? Simply because it was stated in the begining?

4)The maxium values [tex]\delta[/tex] can be are [tex] 1[/tex]or[tex] \epsilon/7 [/tex] right? I dont understand why one has epsilon in it and one does not. Why is this and what does it imply?

Thx again for helping me through this. Our teacher just glanced over it and didnt even test us on it, but I still really want to know. It is late and I am tired so I hope this made sense.
 
  • #28
Pyrrhus
Homework Helper
2,178
1
DieCommie, please rest, and read again arildno's reply, it seems you didn't read it as it was presented.
 
  • #29
156
0
What part didnt I read as it was presented?
 
Last edited:
  • #30
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
131
DieCommie said:
Thank you for sticking around and helping me.

I have gone through the steps, and I see that how it is all valid.
I do have a few particulars I still dont understand, as you predicted.
1)you choose [tex]\delta < \delta_1 = 1 [/tex] arbitrarily.. Is that enable you to solve the delta expression in terms of epsilon?
I did it because it would simplify my work further on; it certainly wasn't a NECESSARY step.
2)When you did this [tex]\delta^2 + 6x < \delta\delta_1 + 6\delta = 7\delta [/tex] why did you choose to replace only one of the deltas with 1? And why did you choose to replace that particular delta?
Again, a VALID simplification trick.[/QUOTE]
3)How do we know this is all true if [tex][|x^2 - 9|<\epsilon[/tex]? Simply because it was stated in the begining?
OOOPS!
Here you've misunderstood the issue!
If we CHOOSE [tex]\delta[/tex] to fulfill [tex]\delta<min(1,\frac{\epsilon}{7})[/tex], then [tex]|x-3|<\delta[/tex] IMPLIES that [tex]|x^{2}-9|<\epsilon[/tex]
This is what we're after, not the other way around!!!
The "restrictions" (or conditions) I mentioned are to be understood in the following manner:
"What restrictions/conditions must be placed upon [tex]\delta[/tex] so that [tex]|x-3|<\delta[/tex] IMPLIES [tex]|x^{2}-9|<\epsilon[/tex]?"
4)The maxium values [tex]\delta[/tex] can be are [tex] 1[/tex]or[tex] \epsilon/7 [/tex] right? I dont understand why one has epsilon in it and one does not. Why is this and what does it imply?
Read again, in particular my comment on 3), and if you still have problems, let's tackle those.

Remember that to be able to make valid and smart simplification tricks is a matter of experience (you'll develop a skill in this through practice).
 
Last edited:

Related Threads for: Epsilon Delta and The Triangle Inequality

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
24
Views
5K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
5
Views
13K
Replies
3
Views
978
  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
6
Views
1K
Top