Uniform continuity and differentiation.uniform continuity in $(0,infty)$Exploit uniform continuity$f(x) =frac1x^alpha$ continuity and uniform continuityBolzano-Weierstrass and Uniform ContinuityClosed Bounded Intervals and Uniform ContinuityUniform Continuity and ContinuityProving continuity of a function using epsilon and deltaProving continuity in metric spacesContinuity vs. Uniform Continuity in Layman's Terms“Trick” to finding Delta in proving continuity/uniform continuity?
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Uniform continuity and differentiation.
uniform continuity in $(0,infty)$Exploit uniform continuity$f(x) =frac1x^alpha$ continuity and uniform continuityBolzano-Weierstrass and Uniform ContinuityClosed Bounded Intervals and Uniform ContinuityUniform Continuity and ContinuityProving continuity of a function using epsilon and deltaProving continuity in metric spacesContinuity vs. Uniform Continuity in Layman's Terms“Trick” to finding Delta in proving continuity/uniform continuity?
$begingroup$
For (a) I have said
By MVT:
$|f(x)-f(y)|le K|x-y|$
Choose $delta=epsilon/K$
$|f(x)-f(y)|=|(f(x)-f(y))/(x-y)||x-y|=|f'(c)||x-y|le K|x-y| le Kdelta=epsilon$
For (b) I have said
$f'(x)=-4x^3/(1+x^4)^2$ so $|f'(x)| le B$ where be is some B in the real numbers, eg I believe 2 would be a suitable choice for B.
Any help would be appreciated, thank you.
analysis continuity uniform-continuity
edited Mar 29 at 8:05
Glorfindel
3,41381930
asked Mar 12 '14 at 21:55
user127700user127700
287515
$endgroup$
add a comment |
$begingroup$
For (a) I have said
By MVT:
$|f(x)-f(y)|le K|x-y|$
Choose $delta=epsilon/K$
$|f(x)-f(y)|=|(f(x)-f(y))/(x-y)||x-y|=|f'(c)||x-y|le K|x-y| le Kdelta=epsilon$
For (b) I have said
$f'(x)=-4x^3/(1+x^4)^2$ so $|f'(x)| le B$ where be is some B in the real numbers, eg I believe 2 would be a suitable choice for B.
Any help would be appreciated, thank you.
analysis continuity uniform-continuity
edited Mar 29 at 8:05
Glorfindel
3,41381930
asked Mar 12 '14 at 21:55
user127700user127700
287515
$endgroup$
1
$begingroup$
If you add a proof for your belief in b), it will be good.
$endgroup$
– Daniel Fischer
Mar 12 '14 at 21:57
add a comment |
$begingroup$
For (a) I have said
By MVT:
$|f(x)-f(y)|le K|x-y|$
Choose $delta=epsilon/K$
$|f(x)-f(y)|=|(f(x)-f(y))/(x-y)||x-y|=|f'(c)||x-y|le K|x-y| le Kdelta=epsilon$
For (b) I have said
$f'(x)=-4x^3/(1+x^4)^2$ so $|f'(x)| le B$ where be is some B in the real numbers, eg I believe 2 would be a suitable choice for B.
Any help would be appreciated, thank you.
analysis continuity uniform-continuity
edited Mar 29 at 8:05
Glorfindel
3,41381930
asked Mar 12 '14 at 21:55
user127700user127700
287515
$endgroup$
For (a) I have said
By MVT:
$|f(x)-f(y)|le K|x-y|$
Choose $delta=epsilon/K$
$|f(x)-f(y)|=|(f(x)-f(y))/(x-y)||x-y|=|f'(c)||x-y|le K|x-y| le Kdelta=epsilon$
For (b) I have said
$f'(x)=-4x^3/(1+x^4)^2$ so $|f'(x)| le B$ where be is some B in the real numbers, eg I believe 2 would be a suitable choice for B.
Any help would be appreciated, thank you.
analysis continuity uniform-continuity
analysis continuity uniform-continuity
edited Mar 29 at 8:05
Glorfindel
3,41381930
asked Mar 12 '14 at 21:55
user127700user127700
287515
edited Mar 29 at 8:05
Glorfindel
3,41381930
asked Mar 12 '14 at 21:55
user127700user127700
287515
edited Mar 29 at 8:05
Glorfindel
3,41381930
edited Mar 29 at 8:05
Glorfindel
3,41381930
edited Mar 29 at 8:05
Glorfindel
3,41381930
3,41381930
asked Mar 12 '14 at 21:55
user127700user127700
287515
asked Mar 12 '14 at 21:55
user127700user127700
287515
asked Mar 12 '14 at 21:55
user127700user127700
287515
287515
1
$begingroup$
If you add a proof for your belief in b), it will be good.
$endgroup$
– Daniel Fischer
Mar 12 '14 at 21:57
add a comment |
1
$begingroup$
If you add a proof for your belief in b), it will be good.
$endgroup$
– Daniel Fischer
Mar 12 '14 at 21:57
1
1
$begingroup$
If you add a proof for your belief in b), it will be good.
$endgroup$
– Daniel Fischer
Mar 12 '14 at 21:57
$begingroup$
If you add a proof for your belief in b), it will be good.
$endgroup$
– Daniel Fischer
Mar 12 '14 at 21:57
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
(a) Be careful, MVT does not say $$|f(x) - f(y) | leq K|x-y|.$$ It says there exists $c in mathbbR$ (in this case) such that $$|f(x) - f(y) = |f'(c)||x-y|.$$
(b) The maximum of $f'(x)$ occurs at $x = -sqrt[4]3$, so $| f'(x)| leq 3^3/4 $. Use the standard tools in calculus to find this maximum.
answered Mar 13 '14 at 1:52
IAmNoOneIAmNoOne
2,64541221
$endgroup$
add a comment |
$begingroup$
In (a), using MVT we know that given $[a, b] subset mathbbR$, for any $x, y in [a, b]$, $exists c in [a, b]$ such that $|f(x)-f(y)| = |f'(c)| |x-y|$. Since $|f'(c)|<K$, we conclude that $|f(x)-f(y)| leq K |x-y|$ for any $x, y in mathbbR$.
Therefore, $f$ is Lipschitz continuous in $mathbbR$, and therefore uniformly continuous.
answered Mar 13 '14 at 11:30
Imanol Pérez ArribasImanol Pérez Arribas
579314
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
(a) Be careful, MVT does not say $$|f(x) - f(y) | leq K|x-y|.$$ It says there exists $c in mathbbR$ (in this case) such that $$|f(x) - f(y) = |f'(c)||x-y|.$$
(b) The maximum of $f'(x)$ occurs at $x = -sqrt[4]3$, so $| f'(x)| leq 3^3/4 $. Use the standard tools in calculus to find this maximum.
answered Mar 13 '14 at 1:52
IAmNoOneIAmNoOne
2,64541221
$endgroup$
add a comment |
$begingroup$
(a) Be careful, MVT does not say $$|f(x) - f(y) | leq K|x-y|.$$ It says there exists $c in mathbbR$ (in this case) such that $$|f(x) - f(y) = |f'(c)||x-y|.$$
(b) The maximum of $f'(x)$ occurs at $x = -sqrt[4]3$, so $| f'(x)| leq 3^3/4 $. Use the standard tools in calculus to find this maximum.
answered Mar 13 '14 at 1:52
IAmNoOneIAmNoOne
2,64541221
$endgroup$
add a comment |
$begingroup$
(a) Be careful, MVT does not say $$|f(x) - f(y) | leq K|x-y|.$$ It says there exists $c in mathbbR$ (in this case) such that $$|f(x) - f(y) = |f'(c)||x-y|.$$
(b) The maximum of $f'(x)$ occurs at $x = -sqrt[4]3$, so $| f'(x)| leq 3^3/4 $. Use the standard tools in calculus to find this maximum.
answered Mar 13 '14 at 1:52
IAmNoOneIAmNoOne
2,64541221
$endgroup$
(a) Be careful, MVT does not say $$|f(x) - f(y) | leq K|x-y|.$$ It says there exists $c in mathbbR$ (in this case) such that $$|f(x) - f(y) = |f'(c)||x-y|.$$
(b) The maximum of $f'(x)$ occurs at $x = -sqrt[4]3$, so $| f'(x)| leq 3^3/4 $. Use the standard tools in calculus to find this maximum.
answered Mar 13 '14 at 1:52
IAmNoOneIAmNoOne
2,64541221
answered Mar 13 '14 at 1:52
IAmNoOneIAmNoOne
2,64541221
answered Mar 13 '14 at 1:52
IAmNoOneIAmNoOne
2,64541221
answered Mar 13 '14 at 1:52
IAmNoOneIAmNoOne
2,64541221
2,64541221
add a comment |
add a comment |
$begingroup$
In (a), using MVT we know that given $[a, b] subset mathbbR$, for any $x, y in [a, b]$, $exists c in [a, b]$ such that $|f(x)-f(y)| = |f'(c)| |x-y|$. Since $|f'(c)|<K$, we conclude that $|f(x)-f(y)| leq K |x-y|$ for any $x, y in mathbbR$.
Therefore, $f$ is Lipschitz continuous in $mathbbR$, and therefore uniformly continuous.
answered Mar 13 '14 at 11:30
Imanol Pérez ArribasImanol Pérez Arribas
579314
$endgroup$
add a comment |
$begingroup$
In (a), using MVT we know that given $[a, b] subset mathbbR$, for any $x, y in [a, b]$, $exists c in [a, b]$ such that $|f(x)-f(y)| = |f'(c)| |x-y|$. Since $|f'(c)|<K$, we conclude that $|f(x)-f(y)| leq K |x-y|$ for any $x, y in mathbbR$.
Therefore, $f$ is Lipschitz continuous in $mathbbR$, and therefore uniformly continuous.
answered Mar 13 '14 at 11:30
Imanol Pérez ArribasImanol Pérez Arribas
579314
$endgroup$
add a comment |
$begingroup$
In (a), using MVT we know that given $[a, b] subset mathbbR$, for any $x, y in [a, b]$, $exists c in [a, b]$ such that $|f(x)-f(y)| = |f'(c)| |x-y|$. Since $|f'(c)|<K$, we conclude that $|f(x)-f(y)| leq K |x-y|$ for any $x, y in mathbbR$.
Therefore, $f$ is Lipschitz continuous in $mathbbR$, and therefore uniformly continuous.
answered Mar 13 '14 at 11:30
Imanol Pérez ArribasImanol Pérez Arribas
579314
$endgroup$
In (a), using MVT we know that given $[a, b] subset mathbbR$, for any $x, y in [a, b]$, $exists c in [a, b]$ such that $|f(x)-f(y)| = |f'(c)| |x-y|$. Since $|f'(c)|<K$, we conclude that $|f(x)-f(y)| leq K |x-y|$ for any $x, y in mathbbR$.
Therefore, $f$ is Lipschitz continuous in $mathbbR$, and therefore uniformly continuous.
answered Mar 13 '14 at 11:30
Imanol Pérez ArribasImanol Pérez Arribas
579314
answered Mar 13 '14 at 11:30
Imanol Pérez ArribasImanol Pérez Arribas
579314
answered Mar 13 '14 at 11:30
Imanol Pérez ArribasImanol Pérez Arribas
579314
answered Mar 13 '14 at 11:30
Imanol Pérez ArribasImanol Pérez Arribas
579314
579314
add a comment |
add a comment |
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$begingroup$
If you add a proof for your belief in b), it will be good.
$endgroup$
– Daniel Fischer
Mar 12 '14 at 21:57