Replacing Variables in Limit Defintion
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I am reading an introductory book on complex analysis. They define the neighborhood of infinity as $|z| > 1/epsilon$ for some $epsilon >0$. Then they wish to show the following statement:
$$ lim_{z rightarrow infty}f(z) = w_0 quad text{iff} quad lim_{z rightarrow 0}=fleft(frac{1}{z}right)$$
Their proof goes as follows:
In the forward direction, note by definition we have that $forall epsilon > 0 , exists delta > 0$ st:
$$|z| > 1/delta implies |f(z) - w_0| < epsilon$$
Replacing $z$ by $1/z$ we obtain:
$$0 < |z - 0| > delta implies |fleft(frac{1}{z}right) - w_0| < epsilon$$
Why can they just "replace" $z$ with $1/z$? Can I do this with any function of $z$ like $z^2$?
complex-analysis proof-explanation
asked Dec 25 '18 at 15:16
yoshiyoshi
1,256917
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add a comment |
$begingroup$
I am reading an introductory book on complex analysis. They define the neighborhood of infinity as $|z| > 1/epsilon$ for some $epsilon >0$. Then they wish to show the following statement:
$$ lim_{z rightarrow infty}f(z) = w_0 quad text{iff} quad lim_{z rightarrow 0}=fleft(frac{1}{z}right)$$
Their proof goes as follows:
In the forward direction, note by definition we have that $forall epsilon > 0 , exists delta > 0$ st:
$$|z| > 1/delta implies |f(z) - w_0| < epsilon$$
Replacing $z$ by $1/z$ we obtain:
$$0 < |z - 0| > delta implies |fleft(frac{1}{z}right) - w_0| < epsilon$$
Why can they just "replace" $z$ with $1/z$? Can I do this with any function of $z$ like $z^2$?
complex-analysis proof-explanation
asked Dec 25 '18 at 15:16
yoshiyoshi
1,256917
$endgroup$
add a comment |
$begingroup$
I am reading an introductory book on complex analysis. They define the neighborhood of infinity as $|z| > 1/epsilon$ for some $epsilon >0$. Then they wish to show the following statement:
$$ lim_{z rightarrow infty}f(z) = w_0 quad text{iff} quad lim_{z rightarrow 0}=fleft(frac{1}{z}right)$$
Their proof goes as follows:
In the forward direction, note by definition we have that $forall epsilon > 0 , exists delta > 0$ st:
$$|z| > 1/delta implies |f(z) - w_0| < epsilon$$
Replacing $z$ by $1/z$ we obtain:
$$0 < |z - 0| > delta implies |fleft(frac{1}{z}right) - w_0| < epsilon$$
Why can they just "replace" $z$ with $1/z$? Can I do this with any function of $z$ like $z^2$?
complex-analysis proof-explanation
asked Dec 25 '18 at 15:16
yoshiyoshi
1,256917
$endgroup$
I am reading an introductory book on complex analysis. They define the neighborhood of infinity as $|z| > 1/epsilon$ for some $epsilon >0$. Then they wish to show the following statement:
$$ lim_{z rightarrow infty}f(z) = w_0 quad text{iff} quad lim_{z rightarrow 0}=fleft(frac{1}{z}right)$$
Their proof goes as follows:
In the forward direction, note by definition we have that $forall epsilon > 0 , exists delta > 0$ st:
$$|z| > 1/delta implies |f(z) - w_0| < epsilon$$
Replacing $z$ by $1/z$ we obtain:
$$0 < |z - 0| > delta implies |fleft(frac{1}{z}right) - w_0| < epsilon$$
Why can they just "replace" $z$ with $1/z$? Can I do this with any function of $z$ like $z^2$?
complex-analysis proof-explanation
complex-analysis proof-explanation
asked Dec 25 '18 at 15:16
yoshiyoshi
1,256917
asked Dec 25 '18 at 15:16
yoshiyoshi
1,256917
asked Dec 25 '18 at 15:16
yoshiyoshi
1,256917
asked Dec 25 '18 at 15:16
yoshiyoshi
1,256917
asked Dec 25 '18 at 15:16
yoshiyoshi
1,256917
1,256917
add a comment |
add a comment |
1 Answer
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You can substitute anything you like but you must respect the inequalities.
In this case we have that $|z| > 1/delta implies |f(z) - w_0| < epsilon$ and a straight replacement gives us $|frac{1}{z}| > 1/delta implies |f(frac{1}{z}) - w_0| < epsilon$.
Taking reciprocals on the left side of the implication, we get $|z|ltdeltaimplies |f(frac{1}{z})-w_0|ltepsilon$
answered Dec 25 '18 at 16:23
John DoumaJohn Douma
5,82021520
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1. Should it be $|f(1/z)-w_0|$ (parenthesis at end is in the wrong place?)? 2. Can you elaborate "you can substitute anything you like so long as it respects the inequality" -- this is the crux of my confusion. Continuity is about "closeness in the range implies closeness in the domain" -- why can I just change variables in the statement?
$endgroup$
– yoshi
Dec 25 '18 at 17:43
1
$begingroup$
Yes. That’s a typo. Variables are just place holders. Why wouldn’t we be able to change variables? When we substitute $z$ with $frac{1}{z}$, we still require the fraction to have the same properties as the original variable.
$endgroup$
– John Douma
Dec 25 '18 at 20:37
add a comment |
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1 Answer
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$begingroup$
You can substitute anything you like but you must respect the inequalities.
In this case we have that $|z| > 1/delta implies |f(z) - w_0| < epsilon$ and a straight replacement gives us $|frac{1}{z}| > 1/delta implies |f(frac{1}{z}) - w_0| < epsilon$.
Taking reciprocals on the left side of the implication, we get $|z|ltdeltaimplies |f(frac{1}{z})-w_0|ltepsilon$
answered Dec 25 '18 at 16:23
John DoumaJohn Douma
5,82021520
$endgroup$
$begingroup$
1. Should it be $|f(1/z)-w_0|$ (parenthesis at end is in the wrong place?)? 2. Can you elaborate "you can substitute anything you like so long as it respects the inequality" -- this is the crux of my confusion. Continuity is about "closeness in the range implies closeness in the domain" -- why can I just change variables in the statement?
$endgroup$
– yoshi
Dec 25 '18 at 17:43
1
$begingroup$
Yes. That’s a typo. Variables are just place holders. Why wouldn’t we be able to change variables? When we substitute $z$ with $frac{1}{z}$, we still require the fraction to have the same properties as the original variable.
$endgroup$
– John Douma
Dec 25 '18 at 20:37
add a comment |
$begingroup$
You can substitute anything you like but you must respect the inequalities.
In this case we have that $|z| > 1/delta implies |f(z) - w_0| < epsilon$ and a straight replacement gives us $|frac{1}{z}| > 1/delta implies |f(frac{1}{z}) - w_0| < epsilon$.
Taking reciprocals on the left side of the implication, we get $|z|ltdeltaimplies |f(frac{1}{z})-w_0|ltepsilon$
answered Dec 25 '18 at 16:23
John DoumaJohn Douma
5,82021520
$endgroup$
$begingroup$
1. Should it be $|f(1/z)-w_0|$ (parenthesis at end is in the wrong place?)? 2. Can you elaborate "you can substitute anything you like so long as it respects the inequality" -- this is the crux of my confusion. Continuity is about "closeness in the range implies closeness in the domain" -- why can I just change variables in the statement?
$endgroup$
– yoshi
Dec 25 '18 at 17:43
1
$begingroup$
Yes. That’s a typo. Variables are just place holders. Why wouldn’t we be able to change variables? When we substitute $z$ with $frac{1}{z}$, we still require the fraction to have the same properties as the original variable.
$endgroup$
– John Douma
Dec 25 '18 at 20:37
add a comment |
$begingroup$
You can substitute anything you like but you must respect the inequalities.
In this case we have that $|z| > 1/delta implies |f(z) - w_0| < epsilon$ and a straight replacement gives us $|frac{1}{z}| > 1/delta implies |f(frac{1}{z}) - w_0| < epsilon$.
Taking reciprocals on the left side of the implication, we get $|z|ltdeltaimplies |f(frac{1}{z})-w_0|ltepsilon$
answered Dec 25 '18 at 16:23
John DoumaJohn Douma
5,82021520
$endgroup$
You can substitute anything you like but you must respect the inequalities.
In this case we have that $|z| > 1/delta implies |f(z) - w_0| < epsilon$ and a straight replacement gives us $|frac{1}{z}| > 1/delta implies |f(frac{1}{z}) - w_0| < epsilon$.
Taking reciprocals on the left side of the implication, we get $|z|ltdeltaimplies |f(frac{1}{z})-w_0|ltepsilon$
answered Dec 25 '18 at 16:23
John DoumaJohn Douma
5,82021520
edited Dec 25 '18 at 20:56
answered Dec 25 '18 at 16:23
John DoumaJohn Douma
5,82021520
answered Dec 25 '18 at 16:23
John DoumaJohn Douma
5,82021520
answered Dec 25 '18 at 16:23
John DoumaJohn Douma
5,82021520
5,82021520
$begingroup$
1. Should it be $|f(1/z)-w_0|$ (parenthesis at end is in the wrong place?)? 2. Can you elaborate "you can substitute anything you like so long as it respects the inequality" -- this is the crux of my confusion. Continuity is about "closeness in the range implies closeness in the domain" -- why can I just change variables in the statement?
$endgroup$
– yoshi
Dec 25 '18 at 17:43
1
$begingroup$
Yes. That’s a typo. Variables are just place holders. Why wouldn’t we be able to change variables? When we substitute $z$ with $frac{1}{z}$, we still require the fraction to have the same properties as the original variable.
$endgroup$
– John Douma
Dec 25 '18 at 20:37
add a comment |
$begingroup$
1. Should it be $|f(1/z)-w_0|$ (parenthesis at end is in the wrong place?)? 2. Can you elaborate "you can substitute anything you like so long as it respects the inequality" -- this is the crux of my confusion. Continuity is about "closeness in the range implies closeness in the domain" -- why can I just change variables in the statement?
$endgroup$
– yoshi
Dec 25 '18 at 17:43
1
$begingroup$
Yes. That’s a typo. Variables are just place holders. Why wouldn’t we be able to change variables? When we substitute $z$ with $frac{1}{z}$, we still require the fraction to have the same properties as the original variable.
$endgroup$
– John Douma
Dec 25 '18 at 20:37
$begingroup$
1. Should it be $|f(1/z)-w_0|$ (parenthesis at end is in the wrong place?)? 2. Can you elaborate "you can substitute anything you like so long as it respects the inequality" -- this is the crux of my confusion. Continuity is about "closeness in the range implies closeness in the domain" -- why can I just change variables in the statement?
$endgroup$
– yoshi
Dec 25 '18 at 17:43
$begingroup$
1. Should it be $|f(1/z)-w_0|$ (parenthesis at end is in the wrong place?)? 2. Can you elaborate "you can substitute anything you like so long as it respects the inequality" -- this is the crux of my confusion. Continuity is about "closeness in the range implies closeness in the domain" -- why can I just change variables in the statement?
$endgroup$
– yoshi
Dec 25 '18 at 17:43
1
1
$begingroup$
Yes. That’s a typo. Variables are just place holders. Why wouldn’t we be able to change variables? When we substitute $z$ with $frac{1}{z}$, we still require the fraction to have the same properties as the original variable.
$endgroup$
– John Douma
Dec 25 '18 at 20:37
$begingroup$
Yes. That’s a typo. Variables are just place holders. Why wouldn’t we be able to change variables? When we substitute $z$ with $frac{1}{z}$, we still require the fraction to have the same properties as the original variable.
$endgroup$
– John Douma
Dec 25 '18 at 20:37
add a comment |
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