Mixing
Show that f(z) = |z| is continuous (epsilon-delta)
$begingroup$
$textbf{Hint:}$ Use $||a|-|b|| leq |a-b|$ in the $epsilon-delta$ analysis.
It's been awhile since I've studied any kind of analysis, I'm not sure if I'm on the right track. I want to show what I've done and see if it makes sense so far.
I need to show that for any $epsilon$, I can find a $delta$ such that $|a-b| < delta implies ||a|-|b|| < epsilon$.
Using the hint, I break the problem into three cases:
Case (i) when $||a|-|b|| = |a-b|$, case (ii) when $||a|-|b|| < |a-b| < epsilon$, case (iii) when $||a|-|b|| < epsilon < |a-b|$.
For case (i), I let $delta = epsilon$. Then $|a-b| < delta implies ||a|-|b|| < epsilon$.
For case (ii), I let $delta = frac{|a-b|+epsilon}{2}$. Then $|a-b| < delta implies ||a|-|b|| < epsilon$ (since $||a|-|b|| < |a-b| < delta < epsilon$) in this case.
I am not sure about the rest. Any hints appreciated.
complex-analysis continuity
asked Jan 24 at 16:29
mXdXmXdX
898
$endgroup$
add a comment |
$begingroup$
$textbf{Hint:}$ Use $||a|-|b|| leq |a-b|$ in the $epsilon-delta$ analysis.
It's been awhile since I've studied any kind of analysis, I'm not sure if I'm on the right track. I want to show what I've done and see if it makes sense so far.
I need to show that for any $epsilon$, I can find a $delta$ such that $|a-b| < delta implies ||a|-|b|| < epsilon$.
Using the hint, I break the problem into three cases:
Case (i) when $||a|-|b|| = |a-b|$, case (ii) when $||a|-|b|| < |a-b| < epsilon$, case (iii) when $||a|-|b|| < epsilon < |a-b|$.
For case (i), I let $delta = epsilon$. Then $|a-b| < delta implies ||a|-|b|| < epsilon$.
For case (ii), I let $delta = frac{|a-b|+epsilon}{2}$. Then $|a-b| < delta implies ||a|-|b|| < epsilon$ (since $||a|-|b|| < |a-b| < delta < epsilon$) in this case.
I am not sure about the rest. Any hints appreciated.
complex-analysis continuity
asked Jan 24 at 16:29
mXdXmXdX
898
$endgroup$
2
$begingroup$
Why cases? $delta=epsilon$ works in general. If $c$ is a point and $|z-c|<delta$ then $||z|-|c||leq |z-c|<delta=epsilon$, using the reverse triangle inequality.
$endgroup$
– LoveTooNap29
Jan 24 at 16:32
$begingroup$
Wait... I don't know why I used cases. So I can just say $delta = epsilon$, so $|a-b| < delta implies ||a|-|b|| < epsilon$, and done?
$endgroup$
– mXdX
Jan 24 at 16:40
1
$begingroup$
yes, using the so-called reverse triangle inequality. Maybe you should also prove this fact in the hint.
$endgroup$
– LoveTooNap29
Jan 24 at 16:42
1
$begingroup$
remember that the idea here is that for any $epsilon$ we need some $delta$. You can chose the $delta$ in this case. You could even do $delta=frac{epsilon}{2}$ or $delta=sqrt{epsilon}$. Any $delta leq epsilon$ works in this example (not in every example).
$endgroup$
– clmundergrad
Jan 24 at 19:51
add a comment |
$begingroup$
$textbf{Hint:}$ Use $||a|-|b|| leq |a-b|$ in the $epsilon-delta$ analysis.
It's been awhile since I've studied any kind of analysis, I'm not sure if I'm on the right track. I want to show what I've done and see if it makes sense so far.
I need to show that for any $epsilon$, I can find a $delta$ such that $|a-b| < delta implies ||a|-|b|| < epsilon$.
Using the hint, I break the problem into three cases:
Case (i) when $||a|-|b|| = |a-b|$, case (ii) when $||a|-|b|| < |a-b| < epsilon$, case (iii) when $||a|-|b|| < epsilon < |a-b|$.
For case (i), I let $delta = epsilon$. Then $|a-b| < delta implies ||a|-|b|| < epsilon$.
For case (ii), I let $delta = frac{|a-b|+epsilon}{2}$. Then $|a-b| < delta implies ||a|-|b|| < epsilon$ (since $||a|-|b|| < |a-b| < delta < epsilon$) in this case.
I am not sure about the rest. Any hints appreciated.
complex-analysis continuity
asked Jan 24 at 16:29
mXdXmXdX
898
$endgroup$
$textbf{Hint:}$ Use $||a|-|b|| leq |a-b|$ in the $epsilon-delta$ analysis.
It's been awhile since I've studied any kind of analysis, I'm not sure if I'm on the right track. I want to show what I've done and see if it makes sense so far.
I need to show that for any $epsilon$, I can find a $delta$ such that $|a-b| < delta implies ||a|-|b|| < epsilon$.
Using the hint, I break the problem into three cases:
Case (i) when $||a|-|b|| = |a-b|$, case (ii) when $||a|-|b|| < |a-b| < epsilon$, case (iii) when $||a|-|b|| < epsilon < |a-b|$.
For case (i), I let $delta = epsilon$. Then $|a-b| < delta implies ||a|-|b|| < epsilon$.
For case (ii), I let $delta = frac{|a-b|+epsilon}{2}$. Then $|a-b| < delta implies ||a|-|b|| < epsilon$ (since $||a|-|b|| < |a-b| < delta < epsilon$) in this case.
I am not sure about the rest. Any hints appreciated.
complex-analysis continuity
complex-analysis continuity
asked Jan 24 at 16:29
mXdXmXdX
898
asked Jan 24 at 16:29
mXdXmXdX
898
asked Jan 24 at 16:29
mXdXmXdX
898
asked Jan 24 at 16:29
mXdXmXdX
898
asked Jan 24 at 16:29
mXdXmXdX
898
898
2
$begingroup$
Why cases? $delta=epsilon$ works in general. If $c$ is a point and $|z-c|<delta$ then $||z|-|c||leq |z-c|<delta=epsilon$, using the reverse triangle inequality.
$endgroup$
– LoveTooNap29
Jan 24 at 16:32
$begingroup$
Wait... I don't know why I used cases. So I can just say $delta = epsilon$, so $|a-b| < delta implies ||a|-|b|| < epsilon$, and done?
$endgroup$
– mXdX
Jan 24 at 16:40
1
$begingroup$
yes, using the so-called reverse triangle inequality. Maybe you should also prove this fact in the hint.
$endgroup$
– LoveTooNap29
Jan 24 at 16:42
1
$begingroup$
remember that the idea here is that for any $epsilon$ we need some $delta$. You can chose the $delta$ in this case. You could even do $delta=frac{epsilon}{2}$ or $delta=sqrt{epsilon}$. Any $delta leq epsilon$ works in this example (not in every example).
$endgroup$
– clmundergrad
Jan 24 at 19:51
add a comment |
2
$begingroup$
Why cases? $delta=epsilon$ works in general. If $c$ is a point and $|z-c|<delta$ then $||z|-|c||leq |z-c|<delta=epsilon$, using the reverse triangle inequality.
$endgroup$
– LoveTooNap29
Jan 24 at 16:32
$begingroup$
Wait... I don't know why I used cases. So I can just say $delta = epsilon$, so $|a-b| < delta implies ||a|-|b|| < epsilon$, and done?
$endgroup$
– mXdX
Jan 24 at 16:40
1
$begingroup$
yes, using the so-called reverse triangle inequality. Maybe you should also prove this fact in the hint.
$endgroup$
– LoveTooNap29
Jan 24 at 16:42
1
$begingroup$
remember that the idea here is that for any $epsilon$ we need some $delta$. You can chose the $delta$ in this case. You could even do $delta=frac{epsilon}{2}$ or $delta=sqrt{epsilon}$. Any $delta leq epsilon$ works in this example (not in every example).
$endgroup$
– clmundergrad
Jan 24 at 19:51
2
2
$begingroup$
Why cases? $delta=epsilon$ works in general. If $c$ is a point and $|z-c|<delta$ then $||z|-|c||leq |z-c|<delta=epsilon$, using the reverse triangle inequality.
$endgroup$
– LoveTooNap29
Jan 24 at 16:32
$begingroup$
Why cases? $delta=epsilon$ works in general. If $c$ is a point and $|z-c|<delta$ then $||z|-|c||leq |z-c|<delta=epsilon$, using the reverse triangle inequality.
$endgroup$
– LoveTooNap29
Jan 24 at 16:32
$begingroup$
Wait... I don't know why I used cases. So I can just say $delta = epsilon$, so $|a-b| < delta implies ||a|-|b|| < epsilon$, and done?
$endgroup$
– mXdX
Jan 24 at 16:40
$begingroup$
Wait... I don't know why I used cases. So I can just say $delta = epsilon$, so $|a-b| < delta implies ||a|-|b|| < epsilon$, and done?
$endgroup$
– mXdX
Jan 24 at 16:40
1
1
$begingroup$
yes, using the so-called reverse triangle inequality. Maybe you should also prove this fact in the hint.
$endgroup$
– LoveTooNap29
Jan 24 at 16:42
$begingroup$
yes, using the so-called reverse triangle inequality. Maybe you should also prove this fact in the hint.
$endgroup$
– LoveTooNap29
Jan 24 at 16:42
1
1
$begingroup$
remember that the idea here is that for any $epsilon$ we need some $delta$. You can chose the $delta$ in this case. You could even do $delta=frac{epsilon}{2}$ or $delta=sqrt{epsilon}$. Any $delta leq epsilon$ works in this example (not in every example).
$endgroup$
– clmundergrad
Jan 24 at 19:51
$begingroup$
remember that the idea here is that for any $epsilon$ we need some $delta$. You can chose the $delta$ in this case. You could even do $delta=frac{epsilon}{2}$ or $delta=sqrt{epsilon}$. Any $delta leq epsilon$ works in this example (not in every example).
$endgroup$
– clmundergrad
Jan 24 at 19:51
add a comment |
1 Answer
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$begingroup$
No need for cases. Let $epsilon >0$ and $delta =epsilon$. Then if $cin mathbb{C}$ and $|z-c|<delta$, by the reverse triangle inequality,
$$||z|-|c||leq |z-c|<delta=epsilon,$$
hence the function $f(z)=|z|$ is continuous at $c$. The link contains a proof of the fact used here, by the way.
answered Jan 24 at 18:19
LoveTooNap29LoveTooNap29
1,1331614
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add a comment |
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$begingroup$
No need for cases. Let $epsilon >0$ and $delta =epsilon$. Then if $cin mathbb{C}$ and $|z-c|<delta$, by the reverse triangle inequality,
$$||z|-|c||leq |z-c|<delta=epsilon,$$
hence the function $f(z)=|z|$ is continuous at $c$. The link contains a proof of the fact used here, by the way.
answered Jan 24 at 18:19
LoveTooNap29LoveTooNap29
1,1331614
$endgroup$
add a comment |
$begingroup$
No need for cases. Let $epsilon >0$ and $delta =epsilon$. Then if $cin mathbb{C}$ and $|z-c|<delta$, by the reverse triangle inequality,
$$||z|-|c||leq |z-c|<delta=epsilon,$$
hence the function $f(z)=|z|$ is continuous at $c$. The link contains a proof of the fact used here, by the way.
answered Jan 24 at 18:19
LoveTooNap29LoveTooNap29
1,1331614
$endgroup$
add a comment |
$begingroup$
No need for cases. Let $epsilon >0$ and $delta =epsilon$. Then if $cin mathbb{C}$ and $|z-c|<delta$, by the reverse triangle inequality,
$$||z|-|c||leq |z-c|<delta=epsilon,$$
hence the function $f(z)=|z|$ is continuous at $c$. The link contains a proof of the fact used here, by the way.
answered Jan 24 at 18:19
LoveTooNap29LoveTooNap29
1,1331614
$endgroup$
No need for cases. Let $epsilon >0$ and $delta =epsilon$. Then if $cin mathbb{C}$ and $|z-c|<delta$, by the reverse triangle inequality,
$$||z|-|c||leq |z-c|<delta=epsilon,$$
hence the function $f(z)=|z|$ is continuous at $c$. The link contains a proof of the fact used here, by the way.
answered Jan 24 at 18:19
LoveTooNap29LoveTooNap29
1,1331614
answered Jan 24 at 18:19
LoveTooNap29LoveTooNap29
1,1331614
answered Jan 24 at 18:19
LoveTooNap29LoveTooNap29
1,1331614
answered Jan 24 at 18:19
LoveTooNap29LoveTooNap29
1,1331614
1,1331614
add a comment |
add a comment |
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$begingroup$
Why cases? $delta=epsilon$ works in general. If $c$ is a point and $|z-c|<delta$ then $||z|-|c||leq |z-c|<delta=epsilon$, using the reverse triangle inequality.
$endgroup$
– LoveTooNap29
Jan 24 at 16:32
$begingroup$
Wait... I don't know why I used cases. So I can just say $delta = epsilon$, so $|a-b| < delta implies ||a|-|b|| < epsilon$, and done?
$endgroup$
– mXdX
Jan 24 at 16:40
1
$begingroup$
yes, using the so-called reverse triangle inequality. Maybe you should also prove this fact in the hint.
$endgroup$
– LoveTooNap29
Jan 24 at 16:42
1
$begingroup$
remember that the idea here is that for any $epsilon$ we need some $delta$. You can chose the $delta$ in this case. You could even do $delta=frac{epsilon}{2}$ or $delta=sqrt{epsilon}$. Any $delta leq epsilon$ works in this example (not in every example).
$endgroup$
– clmundergrad
Jan 24 at 19:51