Proving uniform continuity of $f(x) = x sin{frac{1}{x}}$ on $(1,2)$ directly from the definition.
$begingroup$
Before this is marked as a duplicate, I have already looked at the answers here and here.
I believe my question is different because I want to do this proof directly from the definition. The proof given at the first link uses the property of compactness, Lipschitz continuity and derivatives to support the argument. However, my professor's real analysis approach has not been that of metric spaces and, additionally, Lipschitz continuity has not been discussed yet so I'd like to avoid them for now and attempt this proof directly from the definition. I'm interested in a proof similar to the second answer given on the first link but, unfortunately, there are no comments on whether or not it is correct.
The question is to determine whether or not $f(x) = x sin{frac{1}{x}}$ is uniform continuous on $(1,2)$.
I present my own attempt:
Given $epsilon>0$, let $delta= frac{epsilon-1}{2}$ Then $|x-y|<delta implies$ $$left | f(x)-f(y)right | =left | xsin left(frac{1}{x}right)-ysin left(frac{1}{y}right) right|=
left |(x-y) left( sin left(frac{1}{x}right)+sin left(frac{1}{y}right) right) -xsin left(frac{1}{y} right)+ysin left(frac{1}{x} right)right|
leq left |(x-y) left( sin left(frac{1}{x}right)+sin left(frac{1}{y}right) right)right| + left|-xsin left(frac{1}{y} right)+ysin left(frac{1}{x} right)right|$$
$$leq left |x-y right| left |left( sin left(frac{1}{x}right)+sin left(frac{1}{y}right) right)right| + left|-xsin left(frac{1}{y} right)+ysin left(frac{1}{x} right)right|$$
$$leq 2left |x-y right|+1 < 2cdot (frac{epsilon-1}{2})+1=epsilon$$
Any help is appreciated.
real-analysis calculus proof-verification continuity uniform-continuity
edited Dec 23 '18 at 12:01
Namaste
1
asked Dec 23 '18 at 11:49
E.NoleE.Nole
301114
$endgroup$
|
show 1 more comment
$begingroup$
Before this is marked as a duplicate, I have already looked at the answers here and here.
I believe my question is different because I want to do this proof directly from the definition. The proof given at the first link uses the property of compactness, Lipschitz continuity and derivatives to support the argument. However, my professor's real analysis approach has not been that of metric spaces and, additionally, Lipschitz continuity has not been discussed yet so I'd like to avoid them for now and attempt this proof directly from the definition. I'm interested in a proof similar to the second answer given on the first link but, unfortunately, there are no comments on whether or not it is correct.
The question is to determine whether or not $f(x) = x sin{frac{1}{x}}$ is uniform continuous on $(1,2)$.
I present my own attempt:
Given $epsilon>0$, let $delta= frac{epsilon-1}{2}$ Then $|x-y|<delta implies$ $$left | f(x)-f(y)right | =left | xsin left(frac{1}{x}right)-ysin left(frac{1}{y}right) right|=
left |(x-y) left( sin left(frac{1}{x}right)+sin left(frac{1}{y}right) right) -xsin left(frac{1}{y} right)+ysin left(frac{1}{x} right)right|
leq left |(x-y) left( sin left(frac{1}{x}right)+sin left(frac{1}{y}right) right)right| + left|-xsin left(frac{1}{y} right)+ysin left(frac{1}{x} right)right|$$
$$leq left |x-y right| left |left( sin left(frac{1}{x}right)+sin left(frac{1}{y}right) right)right| + left|-xsin left(frac{1}{y} right)+ysin left(frac{1}{x} right)right|$$
$$leq 2left |x-y right|+1 < 2cdot (frac{epsilon-1}{2})+1=epsilon$$
Any help is appreciated.
real-analysis calculus proof-verification continuity uniform-continuity
edited Dec 23 '18 at 12:01
Namaste
1
asked Dec 23 '18 at 11:49
E.NoleE.Nole
301114
$endgroup$
$begingroup$
The answers at the duplicate link are both upvoted, one is accepted by the asker, and the other has 10 upvotes, hence very well received by the community. The fact that a highly upvoted answer has no comments is not sufficient reason to ask the question again.
$endgroup$
– Namaste
Dec 23 '18 at 12:07
1
$begingroup$
@amWhy The answer has 3 upvotes, not 10. The one with 10 is the one that was accepted by the OP which is the one that uses the approach I'm trying to avoid. 3 is definitely not a very highly upvoted answer. It's possible that perhaps 3 other users that used a similar approach upvoted that answer to attract some attention.
$endgroup$
– E.Nole
Dec 23 '18 at 12:12
1
$begingroup$
@amWhy I hope that explains why I had to post a separate question. I really need input on this approach, but now that it has been marked as a duplicate how will I get any assistance on this approach? Again, I'll add that that answer had no comments and only 3 upvotes so surely one is not immediately expected to conclude that it was very well received by the community.
$endgroup$
– E.Nole
Dec 23 '18 at 12:17
1
$begingroup$
@amWhy I get you, but what if I need input on this specific approach? I did, after all, also add a tag of proof verification. And according to the guidelines, if a question has been asked before and already has an answer but those answers do not fully address your question, you can ask a new question.
$endgroup$
– E.Nole
Dec 23 '18 at 12:28
1
$begingroup$
I did not see your proof-verification tag. I will reopen, and perhaps emphasize that you are seeking feedback on your proof.
$endgroup$
– Namaste
Dec 23 '18 at 12:33
|
show 1 more comment
$begingroup$
Before this is marked as a duplicate, I have already looked at the answers here and here.
I believe my question is different because I want to do this proof directly from the definition. The proof given at the first link uses the property of compactness, Lipschitz continuity and derivatives to support the argument. However, my professor's real analysis approach has not been that of metric spaces and, additionally, Lipschitz continuity has not been discussed yet so I'd like to avoid them for now and attempt this proof directly from the definition. I'm interested in a proof similar to the second answer given on the first link but, unfortunately, there are no comments on whether or not it is correct.
The question is to determine whether or not $f(x) = x sin{frac{1}{x}}$ is uniform continuous on $(1,2)$.
I present my own attempt:
Given $epsilon>0$, let $delta= frac{epsilon-1}{2}$ Then $|x-y|<delta implies$ $$left | f(x)-f(y)right | =left | xsin left(frac{1}{x}right)-ysin left(frac{1}{y}right) right|=
left |(x-y) left( sin left(frac{1}{x}right)+sin left(frac{1}{y}right) right) -xsin left(frac{1}{y} right)+ysin left(frac{1}{x} right)right|
leq left |(x-y) left( sin left(frac{1}{x}right)+sin left(frac{1}{y}right) right)right| + left|-xsin left(frac{1}{y} right)+ysin left(frac{1}{x} right)right|$$
$$leq left |x-y right| left |left( sin left(frac{1}{x}right)+sin left(frac{1}{y}right) right)right| + left|-xsin left(frac{1}{y} right)+ysin left(frac{1}{x} right)right|$$
$$leq 2left |x-y right|+1 < 2cdot (frac{epsilon-1}{2})+1=epsilon$$
Any help is appreciated.
real-analysis calculus proof-verification continuity uniform-continuity
edited Dec 23 '18 at 12:01
Namaste
1
asked Dec 23 '18 at 11:49
E.NoleE.Nole
301114
$endgroup$
Before this is marked as a duplicate, I have already looked at the answers here and here.
I believe my question is different because I want to do this proof directly from the definition. The proof given at the first link uses the property of compactness, Lipschitz continuity and derivatives to support the argument. However, my professor's real analysis approach has not been that of metric spaces and, additionally, Lipschitz continuity has not been discussed yet so I'd like to avoid them for now and attempt this proof directly from the definition. I'm interested in a proof similar to the second answer given on the first link but, unfortunately, there are no comments on whether or not it is correct.
The question is to determine whether or not $f(x) = x sin{frac{1}{x}}$ is uniform continuous on $(1,2)$.
I present my own attempt:
Given $epsilon>0$, let $delta= frac{epsilon-1}{2}$ Then $|x-y|<delta implies$ $$left | f(x)-f(y)right | =left | xsin left(frac{1}{x}right)-ysin left(frac{1}{y}right) right|=
left |(x-y) left( sin left(frac{1}{x}right)+sin left(frac{1}{y}right) right) -xsin left(frac{1}{y} right)+ysin left(frac{1}{x} right)right|
leq left |(x-y) left( sin left(frac{1}{x}right)+sin left(frac{1}{y}right) right)right| + left|-xsin left(frac{1}{y} right)+ysin left(frac{1}{x} right)right|$$
$$leq left |x-y right| left |left( sin left(frac{1}{x}right)+sin left(frac{1}{y}right) right)right| + left|-xsin left(frac{1}{y} right)+ysin left(frac{1}{x} right)right|$$
$$leq 2left |x-y right|+1 < 2cdot (frac{epsilon-1}{2})+1=epsilon$$
Any help is appreciated.
real-analysis calculus proof-verification continuity uniform-continuity
real-analysis calculus proof-verification continuity uniform-continuity
edited Dec 23 '18 at 12:01
Namaste
1
asked Dec 23 '18 at 11:49
E.NoleE.Nole
301114
edited Dec 23 '18 at 12:01
Namaste
1
asked Dec 23 '18 at 11:49
E.NoleE.Nole
301114
edited Dec 23 '18 at 12:01
Namaste
1
edited Dec 23 '18 at 12:01
Namaste
1
edited Dec 23 '18 at 12:01
Namaste
1
1
asked Dec 23 '18 at 11:49
E.NoleE.Nole
301114
asked Dec 23 '18 at 11:49
E.NoleE.Nole
301114
asked Dec 23 '18 at 11:49
E.NoleE.Nole
301114
301114
$begingroup$
The answers at the duplicate link are both upvoted, one is accepted by the asker, and the other has 10 upvotes, hence very well received by the community. The fact that a highly upvoted answer has no comments is not sufficient reason to ask the question again.
$endgroup$
– Namaste
Dec 23 '18 at 12:07
1
$begingroup$
@amWhy The answer has 3 upvotes, not 10. The one with 10 is the one that was accepted by the OP which is the one that uses the approach I'm trying to avoid. 3 is definitely not a very highly upvoted answer. It's possible that perhaps 3 other users that used a similar approach upvoted that answer to attract some attention.
$endgroup$
– E.Nole
Dec 23 '18 at 12:12
1
$begingroup$
@amWhy I hope that explains why I had to post a separate question. I really need input on this approach, but now that it has been marked as a duplicate how will I get any assistance on this approach? Again, I'll add that that answer had no comments and only 3 upvotes so surely one is not immediately expected to conclude that it was very well received by the community.
$endgroup$
– E.Nole
Dec 23 '18 at 12:17
1
$begingroup$
@amWhy I get you, but what if I need input on this specific approach? I did, after all, also add a tag of proof verification. And according to the guidelines, if a question has been asked before and already has an answer but those answers do not fully address your question, you can ask a new question.
$endgroup$
– E.Nole
Dec 23 '18 at 12:28
1
$begingroup$
I did not see your proof-verification tag. I will reopen, and perhaps emphasize that you are seeking feedback on your proof.
$endgroup$
– Namaste
Dec 23 '18 at 12:33
|
show 1 more comment
$begingroup$
The answers at the duplicate link are both upvoted, one is accepted by the asker, and the other has 10 upvotes, hence very well received by the community. The fact that a highly upvoted answer has no comments is not sufficient reason to ask the question again.
$endgroup$
– Namaste
Dec 23 '18 at 12:07
1
$begingroup$
@amWhy The answer has 3 upvotes, not 10. The one with 10 is the one that was accepted by the OP which is the one that uses the approach I'm trying to avoid. 3 is definitely not a very highly upvoted answer. It's possible that perhaps 3 other users that used a similar approach upvoted that answer to attract some attention.
$endgroup$
– E.Nole
Dec 23 '18 at 12:12
1
$begingroup$
@amWhy I hope that explains why I had to post a separate question. I really need input on this approach, but now that it has been marked as a duplicate how will I get any assistance on this approach? Again, I'll add that that answer had no comments and only 3 upvotes so surely one is not immediately expected to conclude that it was very well received by the community.
$endgroup$
– E.Nole
Dec 23 '18 at 12:17
1
$begingroup$
@amWhy I get you, but what if I need input on this specific approach? I did, after all, also add a tag of proof verification. And according to the guidelines, if a question has been asked before and already has an answer but those answers do not fully address your question, you can ask a new question.
$endgroup$
– E.Nole
Dec 23 '18 at 12:28
1
$begingroup$
I did not see your proof-verification tag. I will reopen, and perhaps emphasize that you are seeking feedback on your proof.
$endgroup$
– Namaste
Dec 23 '18 at 12:33
$begingroup$
The answers at the duplicate link are both upvoted, one is accepted by the asker, and the other has 10 upvotes, hence very well received by the community. The fact that a highly upvoted answer has no comments is not sufficient reason to ask the question again.
$endgroup$
– Namaste
Dec 23 '18 at 12:07
$begingroup$
The answers at the duplicate link are both upvoted, one is accepted by the asker, and the other has 10 upvotes, hence very well received by the community. The fact that a highly upvoted answer has no comments is not sufficient reason to ask the question again.
$endgroup$
– Namaste
Dec 23 '18 at 12:07
1
1
$begingroup$
@amWhy The answer has 3 upvotes, not 10. The one with 10 is the one that was accepted by the OP which is the one that uses the approach I'm trying to avoid. 3 is definitely not a very highly upvoted answer. It's possible that perhaps 3 other users that used a similar approach upvoted that answer to attract some attention.
$endgroup$
– E.Nole
Dec 23 '18 at 12:12
$begingroup$
@amWhy The answer has 3 upvotes, not 10. The one with 10 is the one that was accepted by the OP which is the one that uses the approach I'm trying to avoid. 3 is definitely not a very highly upvoted answer. It's possible that perhaps 3 other users that used a similar approach upvoted that answer to attract some attention.
$endgroup$
– E.Nole
Dec 23 '18 at 12:12
1
1
$begingroup$
@amWhy I hope that explains why I had to post a separate question. I really need input on this approach, but now that it has been marked as a duplicate how will I get any assistance on this approach? Again, I'll add that that answer had no comments and only 3 upvotes so surely one is not immediately expected to conclude that it was very well received by the community.
$endgroup$
– E.Nole
Dec 23 '18 at 12:17
$begingroup$
@amWhy I hope that explains why I had to post a separate question. I really need input on this approach, but now that it has been marked as a duplicate how will I get any assistance on this approach? Again, I'll add that that answer had no comments and only 3 upvotes so surely one is not immediately expected to conclude that it was very well received by the community.
$endgroup$
– E.Nole
Dec 23 '18 at 12:17
1
1
$begingroup$
@amWhy I get you, but what if I need input on this specific approach? I did, after all, also add a tag of proof verification. And according to the guidelines, if a question has been asked before and already has an answer but those answers do not fully address your question, you can ask a new question.
$endgroup$
– E.Nole
Dec 23 '18 at 12:28
$begingroup$
@amWhy I get you, but what if I need input on this specific approach? I did, after all, also add a tag of proof verification. And according to the guidelines, if a question has been asked before and already has an answer but those answers do not fully address your question, you can ask a new question.
$endgroup$
– E.Nole
Dec 23 '18 at 12:28
1
1
$begingroup$
I did not see your proof-verification tag. I will reopen, and perhaps emphasize that you are seeking feedback on your proof.
$endgroup$
– Namaste
Dec 23 '18 at 12:33
$begingroup$
I did not see your proof-verification tag. I will reopen, and perhaps emphasize that you are seeking feedback on your proof.
$endgroup$
– Namaste
Dec 23 '18 at 12:33
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
The problem with your approach is that $delta = frac{epsilon - 1}{2}$ is not a very good choice of $delta$: what if $epsilon-1<0$? Here is a correct solution.
We have that $|sin(s) - sin(t)| = left|2overbrace{sinleft(frac{s-t}{2}right)}^{leq frac{s-t}{2}}underbrace{cosleft(frac{s+t}{2}right)}_{leq 1}right| leq |s-t|$ for any real $s,t$.
This implies that $|sin frac{1}{x} - sin frac{1}{y}| leq left|frac{1}{x} - frac{1}{y}right| = left|frac{x-y}{xy}right| leq |x-y|$ since $xy geq 1$, since $x,y in (1,2)$.
Consequently, observe that $$left|xsin frac{1}{x} - y sin frac{1}{y}right| = left|xleft(sin left(frac{1}{x}right) - sinleft(frac{1}{y}right)right) + sinleft(frac{1}{y}right)(x-y)right| leq overbrace{|x|}^{leq 2}underbrace{left|sin left(frac{1}{x}right) - sinleft(frac{1}{y}right)right|}_{leq |x-y|}$$ $$+ underbrace{left|sinleft(frac{1}{y}right)right|}_{leq 1}|x-y| leq 3|x-y|
< epsilon$$ provided we choose $delta = frac{epsilon}{3}$.
answered Dec 23 '18 at 13:25
MathematicsStudent1122MathematicsStudent1122
9,00332668
$endgroup$
$begingroup$
Thanks! Now I see why my delta isn't a good choice. I was thinking, what if, say, I took the absolute value instead? i.e $delta =| frac{epsilon - 1}{2}|$? Then realised there'd be a problem if the numerator were $0$. So my question is that is it a general strategy to not have the choice of $delta$ depend on a value that includes something like $epsilon -a$ where $a>0$?
$endgroup$
– E.Nole
Dec 23 '18 at 14:33
add a comment |
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1 Answer
1
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The problem with your approach is that $delta = frac{epsilon - 1}{2}$ is not a very good choice of $delta$: what if $epsilon-1<0$? Here is a correct solution.
We have that $|sin(s) - sin(t)| = left|2overbrace{sinleft(frac{s-t}{2}right)}^{leq frac{s-t}{2}}underbrace{cosleft(frac{s+t}{2}right)}_{leq 1}right| leq |s-t|$ for any real $s,t$.
This implies that $|sin frac{1}{x} - sin frac{1}{y}| leq left|frac{1}{x} - frac{1}{y}right| = left|frac{x-y}{xy}right| leq |x-y|$ since $xy geq 1$, since $x,y in (1,2)$.
Consequently, observe that $$left|xsin frac{1}{x} - y sin frac{1}{y}right| = left|xleft(sin left(frac{1}{x}right) - sinleft(frac{1}{y}right)right) + sinleft(frac{1}{y}right)(x-y)right| leq overbrace{|x|}^{leq 2}underbrace{left|sin left(frac{1}{x}right) - sinleft(frac{1}{y}right)right|}_{leq |x-y|}$$ $$+ underbrace{left|sinleft(frac{1}{y}right)right|}_{leq 1}|x-y| leq 3|x-y|
< epsilon$$ provided we choose $delta = frac{epsilon}{3}$.
answered Dec 23 '18 at 13:25
MathematicsStudent1122MathematicsStudent1122
9,00332668
$endgroup$
$begingroup$
Thanks! Now I see why my delta isn't a good choice. I was thinking, what if, say, I took the absolute value instead? i.e $delta =| frac{epsilon - 1}{2}|$? Then realised there'd be a problem if the numerator were $0$. So my question is that is it a general strategy to not have the choice of $delta$ depend on a value that includes something like $epsilon -a$ where $a>0$?
$endgroup$
– E.Nole
Dec 23 '18 at 14:33
add a comment |
$begingroup$
The problem with your approach is that $delta = frac{epsilon - 1}{2}$ is not a very good choice of $delta$: what if $epsilon-1<0$? Here is a correct solution.
We have that $|sin(s) - sin(t)| = left|2overbrace{sinleft(frac{s-t}{2}right)}^{leq frac{s-t}{2}}underbrace{cosleft(frac{s+t}{2}right)}_{leq 1}right| leq |s-t|$ for any real $s,t$.
This implies that $|sin frac{1}{x} - sin frac{1}{y}| leq left|frac{1}{x} - frac{1}{y}right| = left|frac{x-y}{xy}right| leq |x-y|$ since $xy geq 1$, since $x,y in (1,2)$.
Consequently, observe that $$left|xsin frac{1}{x} - y sin frac{1}{y}right| = left|xleft(sin left(frac{1}{x}right) - sinleft(frac{1}{y}right)right) + sinleft(frac{1}{y}right)(x-y)right| leq overbrace{|x|}^{leq 2}underbrace{left|sin left(frac{1}{x}right) - sinleft(frac{1}{y}right)right|}_{leq |x-y|}$$ $$+ underbrace{left|sinleft(frac{1}{y}right)right|}_{leq 1}|x-y| leq 3|x-y|
< epsilon$$ provided we choose $delta = frac{epsilon}{3}$.
answered Dec 23 '18 at 13:25
MathematicsStudent1122MathematicsStudent1122
9,00332668
$endgroup$
$begingroup$
Thanks! Now I see why my delta isn't a good choice. I was thinking, what if, say, I took the absolute value instead? i.e $delta =| frac{epsilon - 1}{2}|$? Then realised there'd be a problem if the numerator were $0$. So my question is that is it a general strategy to not have the choice of $delta$ depend on a value that includes something like $epsilon -a$ where $a>0$?
$endgroup$
– E.Nole
Dec 23 '18 at 14:33
add a comment |
$begingroup$
The problem with your approach is that $delta = frac{epsilon - 1}{2}$ is not a very good choice of $delta$: what if $epsilon-1<0$? Here is a correct solution.
We have that $|sin(s) - sin(t)| = left|2overbrace{sinleft(frac{s-t}{2}right)}^{leq frac{s-t}{2}}underbrace{cosleft(frac{s+t}{2}right)}_{leq 1}right| leq |s-t|$ for any real $s,t$.
This implies that $|sin frac{1}{x} - sin frac{1}{y}| leq left|frac{1}{x} - frac{1}{y}right| = left|frac{x-y}{xy}right| leq |x-y|$ since $xy geq 1$, since $x,y in (1,2)$.
Consequently, observe that $$left|xsin frac{1}{x} - y sin frac{1}{y}right| = left|xleft(sin left(frac{1}{x}right) - sinleft(frac{1}{y}right)right) + sinleft(frac{1}{y}right)(x-y)right| leq overbrace{|x|}^{leq 2}underbrace{left|sin left(frac{1}{x}right) - sinleft(frac{1}{y}right)right|}_{leq |x-y|}$$ $$+ underbrace{left|sinleft(frac{1}{y}right)right|}_{leq 1}|x-y| leq 3|x-y|
< epsilon$$ provided we choose $delta = frac{epsilon}{3}$.
answered Dec 23 '18 at 13:25
MathematicsStudent1122MathematicsStudent1122
9,00332668
$endgroup$
The problem with your approach is that $delta = frac{epsilon - 1}{2}$ is not a very good choice of $delta$: what if $epsilon-1<0$? Here is a correct solution.
We have that $|sin(s) - sin(t)| = left|2overbrace{sinleft(frac{s-t}{2}right)}^{leq frac{s-t}{2}}underbrace{cosleft(frac{s+t}{2}right)}_{leq 1}right| leq |s-t|$ for any real $s,t$.
This implies that $|sin frac{1}{x} - sin frac{1}{y}| leq left|frac{1}{x} - frac{1}{y}right| = left|frac{x-y}{xy}right| leq |x-y|$ since $xy geq 1$, since $x,y in (1,2)$.
Consequently, observe that $$left|xsin frac{1}{x} - y sin frac{1}{y}right| = left|xleft(sin left(frac{1}{x}right) - sinleft(frac{1}{y}right)right) + sinleft(frac{1}{y}right)(x-y)right| leq overbrace{|x|}^{leq 2}underbrace{left|sin left(frac{1}{x}right) - sinleft(frac{1}{y}right)right|}_{leq |x-y|}$$ $$+ underbrace{left|sinleft(frac{1}{y}right)right|}_{leq 1}|x-y| leq 3|x-y|
< epsilon$$ provided we choose $delta = frac{epsilon}{3}$.
answered Dec 23 '18 at 13:25
MathematicsStudent1122MathematicsStudent1122
9,00332668
answered Dec 23 '18 at 13:25
MathematicsStudent1122MathematicsStudent1122
9,00332668
answered Dec 23 '18 at 13:25
MathematicsStudent1122MathematicsStudent1122
9,00332668
answered Dec 23 '18 at 13:25
MathematicsStudent1122MathematicsStudent1122
9,00332668
9,00332668
$begingroup$
Thanks! Now I see why my delta isn't a good choice. I was thinking, what if, say, I took the absolute value instead? i.e $delta =| frac{epsilon - 1}{2}|$? Then realised there'd be a problem if the numerator were $0$. So my question is that is it a general strategy to not have the choice of $delta$ depend on a value that includes something like $epsilon -a$ where $a>0$?
$endgroup$
– E.Nole
Dec 23 '18 at 14:33
add a comment |
$begingroup$
Thanks! Now I see why my delta isn't a good choice. I was thinking, what if, say, I took the absolute value instead? i.e $delta =| frac{epsilon - 1}{2}|$? Then realised there'd be a problem if the numerator were $0$. So my question is that is it a general strategy to not have the choice of $delta$ depend on a value that includes something like $epsilon -a$ where $a>0$?
$endgroup$
– E.Nole
Dec 23 '18 at 14:33
$begingroup$
Thanks! Now I see why my delta isn't a good choice. I was thinking, what if, say, I took the absolute value instead? i.e $delta =| frac{epsilon - 1}{2}|$? Then realised there'd be a problem if the numerator were $0$. So my question is that is it a general strategy to not have the choice of $delta$ depend on a value that includes something like $epsilon -a$ where $a>0$?
$endgroup$
– E.Nole
Dec 23 '18 at 14:33
$begingroup$
Thanks! Now I see why my delta isn't a good choice. I was thinking, what if, say, I took the absolute value instead? i.e $delta =| frac{epsilon - 1}{2}|$? Then realised there'd be a problem if the numerator were $0$. So my question is that is it a general strategy to not have the choice of $delta$ depend on a value that includes something like $epsilon -a$ where $a>0$?
$endgroup$
– E.Nole
Dec 23 '18 at 14:33
add a comment |
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$begingroup$
The answers at the duplicate link are both upvoted, one is accepted by the asker, and the other has 10 upvotes, hence very well received by the community. The fact that a highly upvoted answer has no comments is not sufficient reason to ask the question again.
$endgroup$
– Namaste
Dec 23 '18 at 12:07
1
$begingroup$
@amWhy The answer has 3 upvotes, not 10. The one with 10 is the one that was accepted by the OP which is the one that uses the approach I'm trying to avoid. 3 is definitely not a very highly upvoted answer. It's possible that perhaps 3 other users that used a similar approach upvoted that answer to attract some attention.
$endgroup$
– E.Nole
Dec 23 '18 at 12:12
1
$begingroup$
@amWhy I hope that explains why I had to post a separate question. I really need input on this approach, but now that it has been marked as a duplicate how will I get any assistance on this approach? Again, I'll add that that answer had no comments and only 3 upvotes so surely one is not immediately expected to conclude that it was very well received by the community.
$endgroup$
– E.Nole
Dec 23 '18 at 12:17
1
$begingroup$
@amWhy I get you, but what if I need input on this specific approach? I did, after all, also add a tag of proof verification. And according to the guidelines, if a question has been asked before and already has an answer but those answers do not fully address your question, you can ask a new question.
$endgroup$
– E.Nole
Dec 23 '18 at 12:28
1
$begingroup$
I did not see your proof-verification tag. I will reopen, and perhaps emphasize that you are seeking feedback on your proof.
$endgroup$
– Namaste
Dec 23 '18 at 12:33