When would I use Cauchy's Integral Formula over Residue
$begingroup$
Just a quick question I've been wondering about. When would I use Cauchy's Integral Formula over the Residue Theorem to solve complex integration problems with poles? To me it seems that Residue theorem is much faster and easier to apply then creating partial fractions for rationals then solving the individual integrals. Any guidance would be greatly appreciated?
complex-analysis residue-calculus complex-integration cauchy-integral-formula
asked Dec 20 '18 at 17:11
SafderSafder
318112
$endgroup$
add a comment |
$begingroup$
Just a quick question I've been wondering about. When would I use Cauchy's Integral Formula over the Residue Theorem to solve complex integration problems with poles? To me it seems that Residue theorem is much faster and easier to apply then creating partial fractions for rationals then solving the individual integrals. Any guidance would be greatly appreciated?
complex-analysis residue-calculus complex-integration cauchy-integral-formula
asked Dec 20 '18 at 17:11
SafderSafder
318112
$endgroup$
$begingroup$
It's a theoretical tool that is useful for proving many things, even though most of the time it might not be the best formula to compute an integral. In some sense it is kind of similar to Cramer's rule for solving linear system: no one uses it for practical purposes.
$endgroup$
– BigbearZzz
Dec 21 '18 at 4:49
add a comment |
$begingroup$
Just a quick question I've been wondering about. When would I use Cauchy's Integral Formula over the Residue Theorem to solve complex integration problems with poles? To me it seems that Residue theorem is much faster and easier to apply then creating partial fractions for rationals then solving the individual integrals. Any guidance would be greatly appreciated?
complex-analysis residue-calculus complex-integration cauchy-integral-formula
asked Dec 20 '18 at 17:11
SafderSafder
318112
$endgroup$
Just a quick question I've been wondering about. When would I use Cauchy's Integral Formula over the Residue Theorem to solve complex integration problems with poles? To me it seems that Residue theorem is much faster and easier to apply then creating partial fractions for rationals then solving the individual integrals. Any guidance would be greatly appreciated?
complex-analysis residue-calculus complex-integration cauchy-integral-formula
complex-analysis residue-calculus complex-integration cauchy-integral-formula
asked Dec 20 '18 at 17:11
SafderSafder
318112
asked Dec 20 '18 at 17:11
SafderSafder
318112
asked Dec 20 '18 at 17:11
SafderSafder
318112
asked Dec 20 '18 at 17:11
SafderSafder
318112
asked Dec 20 '18 at 17:11
SafderSafder
318112
318112
$begingroup$
It's a theoretical tool that is useful for proving many things, even though most of the time it might not be the best formula to compute an integral. In some sense it is kind of similar to Cramer's rule for solving linear system: no one uses it for practical purposes.
$endgroup$
– BigbearZzz
Dec 21 '18 at 4:49
add a comment |
$begingroup$
It's a theoretical tool that is useful for proving many things, even though most of the time it might not be the best formula to compute an integral. In some sense it is kind of similar to Cramer's rule for solving linear system: no one uses it for practical purposes.
$endgroup$
– BigbearZzz
Dec 21 '18 at 4:49
$begingroup$
It's a theoretical tool that is useful for proving many things, even though most of the time it might not be the best formula to compute an integral. In some sense it is kind of similar to Cramer's rule for solving linear system: no one uses it for practical purposes.
$endgroup$
– BigbearZzz
Dec 21 '18 at 4:49
$begingroup$
It's a theoretical tool that is useful for proving many things, even though most of the time it might not be the best formula to compute an integral. In some sense it is kind of similar to Cramer's rule for solving linear system: no one uses it for practical purposes.
$endgroup$
– BigbearZzz
Dec 21 '18 at 4:49
add a comment |
1 Answer
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active
oldest
votes
$begingroup$
It just so happens that the Residue Theorem encompasses the Cauchy's Integral Formula. If you have a analytic function $f$ in the domain D such that $Gamma in D$ and you are asked to evaluate a loop integral of the form $$oint_Gamma frac{f(z)}{(z-z_0)^n}$$ then just use Cauchy's Integral Formula. It turns out its the exact same as the Residue Formula. Note that they both are essentially both looking at the $1/z$ term of the Laurent expansion of a series, and the Residue theorem encompasses everything without doing much preprocessing work.
answered Dec 21 '18 at 4:12
Neeyanth KopparapuNeeyanth Kopparapu
3016
$endgroup$
$begingroup$
Okay that makes sense. It just seemed easier to apply residue on most things.
$endgroup$
– Safder
Dec 22 '18 at 0:01
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It just so happens that the Residue Theorem encompasses the Cauchy's Integral Formula. If you have a analytic function $f$ in the domain D such that $Gamma in D$ and you are asked to evaluate a loop integral of the form $$oint_Gamma frac{f(z)}{(z-z_0)^n}$$ then just use Cauchy's Integral Formula. It turns out its the exact same as the Residue Formula. Note that they both are essentially both looking at the $1/z$ term of the Laurent expansion of a series, and the Residue theorem encompasses everything without doing much preprocessing work.
answered Dec 21 '18 at 4:12
Neeyanth KopparapuNeeyanth Kopparapu
3016
$endgroup$
$begingroup$
Okay that makes sense. It just seemed easier to apply residue on most things.
$endgroup$
– Safder
Dec 22 '18 at 0:01
add a comment |
$begingroup$
It just so happens that the Residue Theorem encompasses the Cauchy's Integral Formula. If you have a analytic function $f$ in the domain D such that $Gamma in D$ and you are asked to evaluate a loop integral of the form $$oint_Gamma frac{f(z)}{(z-z_0)^n}$$ then just use Cauchy's Integral Formula. It turns out its the exact same as the Residue Formula. Note that they both are essentially both looking at the $1/z$ term of the Laurent expansion of a series, and the Residue theorem encompasses everything without doing much preprocessing work.
answered Dec 21 '18 at 4:12
Neeyanth KopparapuNeeyanth Kopparapu
3016
$endgroup$
$begingroup$
Okay that makes sense. It just seemed easier to apply residue on most things.
$endgroup$
– Safder
Dec 22 '18 at 0:01
add a comment |
$begingroup$
It just so happens that the Residue Theorem encompasses the Cauchy's Integral Formula. If you have a analytic function $f$ in the domain D such that $Gamma in D$ and you are asked to evaluate a loop integral of the form $$oint_Gamma frac{f(z)}{(z-z_0)^n}$$ then just use Cauchy's Integral Formula. It turns out its the exact same as the Residue Formula. Note that they both are essentially both looking at the $1/z$ term of the Laurent expansion of a series, and the Residue theorem encompasses everything without doing much preprocessing work.
answered Dec 21 '18 at 4:12
Neeyanth KopparapuNeeyanth Kopparapu
3016
$endgroup$
It just so happens that the Residue Theorem encompasses the Cauchy's Integral Formula. If you have a analytic function $f$ in the domain D such that $Gamma in D$ and you are asked to evaluate a loop integral of the form $$oint_Gamma frac{f(z)}{(z-z_0)^n}$$ then just use Cauchy's Integral Formula. It turns out its the exact same as the Residue Formula. Note that they both are essentially both looking at the $1/z$ term of the Laurent expansion of a series, and the Residue theorem encompasses everything without doing much preprocessing work.
answered Dec 21 '18 at 4:12
Neeyanth KopparapuNeeyanth Kopparapu
3016
answered Dec 21 '18 at 4:12
Neeyanth KopparapuNeeyanth Kopparapu
3016
answered Dec 21 '18 at 4:12
Neeyanth KopparapuNeeyanth Kopparapu
3016
answered Dec 21 '18 at 4:12
Neeyanth KopparapuNeeyanth Kopparapu
3016
3016
$begingroup$
Okay that makes sense. It just seemed easier to apply residue on most things.
$endgroup$
– Safder
Dec 22 '18 at 0:01
add a comment |
$begingroup$
Okay that makes sense. It just seemed easier to apply residue on most things.
$endgroup$
– Safder
Dec 22 '18 at 0:01
$begingroup$
Okay that makes sense. It just seemed easier to apply residue on most things.
$endgroup$
– Safder
Dec 22 '18 at 0:01
$begingroup$
Okay that makes sense. It just seemed easier to apply residue on most things.
$endgroup$
– Safder
Dec 22 '18 at 0:01
add a comment |
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$begingroup$
It's a theoretical tool that is useful for proving many things, even though most of the time it might not be the best formula to compute an integral. In some sense it is kind of similar to Cramer's rule for solving linear system: no one uses it for practical purposes.
$endgroup$
– BigbearZzz
Dec 21 '18 at 4:49