Differentiating polar functions using complex numbers
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I was wondering, given some polar function $r(theta)$ is it possible to convert it into a complex number in exponential form, then differentiate that and then convert it back and have the appropriate derivative of the polar function?
For example take the polar function $r=cos(atheta)$, also known as a rose curve for $ainmathbb{Q}$. Is it possible to 'complexify' this function (not too sure how possible that is) and then take the derivative?
derivatives complex-numbers polar-coordinates
asked Dec 14 '18 at 17:48
Jonathan LowJonathan Low
635
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add a comment |
$begingroup$
I was wondering, given some polar function $r(theta)$ is it possible to convert it into a complex number in exponential form, then differentiate that and then convert it back and have the appropriate derivative of the polar function?
For example take the polar function $r=cos(atheta)$, also known as a rose curve for $ainmathbb{Q}$. Is it possible to 'complexify' this function (not too sure how possible that is) and then take the derivative?
derivatives complex-numbers polar-coordinates
asked Dec 14 '18 at 17:48
Jonathan LowJonathan Low
635
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en.wikipedia.org/wiki/… Euler's identity won't do?
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– mm-crj
Dec 14 '18 at 17:54
add a comment |
$begingroup$
I was wondering, given some polar function $r(theta)$ is it possible to convert it into a complex number in exponential form, then differentiate that and then convert it back and have the appropriate derivative of the polar function?
For example take the polar function $r=cos(atheta)$, also known as a rose curve for $ainmathbb{Q}$. Is it possible to 'complexify' this function (not too sure how possible that is) and then take the derivative?
derivatives complex-numbers polar-coordinates
asked Dec 14 '18 at 17:48
Jonathan LowJonathan Low
635
$endgroup$
I was wondering, given some polar function $r(theta)$ is it possible to convert it into a complex number in exponential form, then differentiate that and then convert it back and have the appropriate derivative of the polar function?
For example take the polar function $r=cos(atheta)$, also known as a rose curve for $ainmathbb{Q}$. Is it possible to 'complexify' this function (not too sure how possible that is) and then take the derivative?
derivatives complex-numbers polar-coordinates
derivatives complex-numbers polar-coordinates
asked Dec 14 '18 at 17:48
Jonathan LowJonathan Low
635
asked Dec 14 '18 at 17:48
Jonathan LowJonathan Low
635
asked Dec 14 '18 at 17:48
Jonathan LowJonathan Low
635
asked Dec 14 '18 at 17:48
Jonathan LowJonathan Low
635
asked Dec 14 '18 at 17:48
Jonathan LowJonathan Low
635
635
$begingroup$
en.wikipedia.org/wiki/… Euler's identity won't do?
$endgroup$
– mm-crj
Dec 14 '18 at 17:54
add a comment |
$begingroup$
en.wikipedia.org/wiki/… Euler's identity won't do?
$endgroup$
– mm-crj
Dec 14 '18 at 17:54
$begingroup$
en.wikipedia.org/wiki/… Euler's identity won't do?
$endgroup$
– mm-crj
Dec 14 '18 at 17:54
$begingroup$
en.wikipedia.org/wiki/… Euler's identity won't do?
$endgroup$
– mm-crj
Dec 14 '18 at 17:54
add a comment |
2 Answers
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Complex numbers are represented in the form $$ z= a e^{itheta} $$ So real part of $ e^{iax} = cos (ax) $ from Euler's identity is applicable here.
answered Dec 14 '18 at 19:03
NarasimhamNarasimham
20.8k52158
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Of course, although I dont think it’d be easier than just taking the derivative in most cases. Using Euler’s identity:
$ cos{(ax)}=frac{e^{iax}+e^{-iax}}{2} $,
$ sin{(ax)}=frac{e^{iax}-e^{-iax}}{2i} $.
answered Dec 14 '18 at 18:25
KirtpoleKirtpole
1012
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add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
Complex numbers are represented in the form $$ z= a e^{itheta} $$ So real part of $ e^{iax} = cos (ax) $ from Euler's identity is applicable here.
answered Dec 14 '18 at 19:03
NarasimhamNarasimham
20.8k52158
$endgroup$
add a comment |
$begingroup$
Complex numbers are represented in the form $$ z= a e^{itheta} $$ So real part of $ e^{iax} = cos (ax) $ from Euler's identity is applicable here.
answered Dec 14 '18 at 19:03
NarasimhamNarasimham
20.8k52158
$endgroup$
add a comment |
$begingroup$
Complex numbers are represented in the form $$ z= a e^{itheta} $$ So real part of $ e^{iax} = cos (ax) $ from Euler's identity is applicable here.
answered Dec 14 '18 at 19:03
NarasimhamNarasimham
20.8k52158
$endgroup$
Complex numbers are represented in the form $$ z= a e^{itheta} $$ So real part of $ e^{iax} = cos (ax) $ from Euler's identity is applicable here.
answered Dec 14 '18 at 19:03
NarasimhamNarasimham
20.8k52158
answered Dec 14 '18 at 19:03
NarasimhamNarasimham
20.8k52158
answered Dec 14 '18 at 19:03
NarasimhamNarasimham
20.8k52158
answered Dec 14 '18 at 19:03
NarasimhamNarasimham
20.8k52158
20.8k52158
add a comment |
add a comment |
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Of course, although I dont think it’d be easier than just taking the derivative in most cases. Using Euler’s identity:
$ cos{(ax)}=frac{e^{iax}+e^{-iax}}{2} $,
$ sin{(ax)}=frac{e^{iax}-e^{-iax}}{2i} $.
answered Dec 14 '18 at 18:25
KirtpoleKirtpole
1012
$endgroup$
add a comment |
$begingroup$
Of course, although I dont think it’d be easier than just taking the derivative in most cases. Using Euler’s identity:
$ cos{(ax)}=frac{e^{iax}+e^{-iax}}{2} $,
$ sin{(ax)}=frac{e^{iax}-e^{-iax}}{2i} $.
answered Dec 14 '18 at 18:25
KirtpoleKirtpole
1012
$endgroup$
add a comment |
$begingroup$
Of course, although I dont think it’d be easier than just taking the derivative in most cases. Using Euler’s identity:
$ cos{(ax)}=frac{e^{iax}+e^{-iax}}{2} $,
$ sin{(ax)}=frac{e^{iax}-e^{-iax}}{2i} $.
answered Dec 14 '18 at 18:25
KirtpoleKirtpole
1012
$endgroup$
Of course, although I dont think it’d be easier than just taking the derivative in most cases. Using Euler’s identity:
$ cos{(ax)}=frac{e^{iax}+e^{-iax}}{2} $,
$ sin{(ax)}=frac{e^{iax}-e^{-iax}}{2i} $.
answered Dec 14 '18 at 18:25
KirtpoleKirtpole
1012
answered Dec 14 '18 at 18:25
KirtpoleKirtpole
1012
answered Dec 14 '18 at 18:25
KirtpoleKirtpole
1012
answered Dec 14 '18 at 18:25
KirtpoleKirtpole
1012
1012
add a comment |
add a comment |
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$begingroup$
en.wikipedia.org/wiki/… Euler's identity won't do?
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– mm-crj
Dec 14 '18 at 17:54