Using the complex exponential function to calculate the $sin(30^{circ})$ [closed]
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I need help with the calculation of the following function
$$sin(omega t)=frac{(e^{iωt}-e^{-iωt})}{2j}implies sin(30^{circ})= sinleft(frac pi 6 right) = frac{e^{frac pi 6 i}-e^{-frac pi 6i}}{2i} $$...when I plug in the values for the function in my program code it replays $0.53dots$ But it should do $0.5000000$
Is it because the approximation with two e's and the other numbers with commas divided by $2$ include an such a high error in it(the numbers are not ending) so that i would have to use a Taylor polynomial (it is the aim to use function that can be written as only + and - for schemas. Functions with a sine in it are too abstract and not allowed) for calculating the sine without precalculated values like in a cordic function.
Ps. Please do not write it as Taylor polynomial or cordic.
complex-numbers approximation
edited Jan 13 at 23:22
user376343
3,9834829
asked Jan 13 at 17:18
Wolfgang EspadaWolfgang Espada
41
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closed as unclear what you're asking by Did, Lord Shark the Unknown, Cesareo, metamorphy, José Carlos Santos Jan 14 at 10:11
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I need help with the calculation of the following function
$$sin(omega t)=frac{(e^{iωt}-e^{-iωt})}{2j}implies sin(30^{circ})= sinleft(frac pi 6 right) = frac{e^{frac pi 6 i}-e^{-frac pi 6i}}{2i} $$...when I plug in the values for the function in my program code it replays $0.53dots$ But it should do $0.5000000$
Is it because the approximation with two e's and the other numbers with commas divided by $2$ include an such a high error in it(the numbers are not ending) so that i would have to use a Taylor polynomial (it is the aim to use function that can be written as only + and - for schemas. Functions with a sine in it are too abstract and not allowed) for calculating the sine without precalculated values like in a cordic function.
Ps. Please do not write it as Taylor polynomial or cordic.
complex-numbers approximation
edited Jan 13 at 23:22
user376343
3,9834829
asked Jan 13 at 17:18
Wolfgang EspadaWolfgang Espada
41
$endgroup$
closed as unclear what you're asking by Did, Lord Shark the Unknown, Cesareo, metamorphy, José Carlos Santos Jan 14 at 10:11
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
It may well be that your calculator is doing degrees instead of radians, or the other way around. You must define this with the MODE key on the calculator.
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– DonAntonio
Jan 13 at 17:20
$begingroup$
Welcome to Mathematics Stack Exchange community! The quick tour will help you get the most benefit from your time here. Also, please use MathJax for your equations.
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– dantopa
Jan 13 at 17:31
2
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Note that $30^circ = 30times pi/180 = pi/6$ radians. Therefore, $sin(30^circ) = frac{e^{ipi/6}-e^{-ipi/6}}{2i}$.
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– Math Lover
Jan 13 at 17:47
$begingroup$
I did like this(it was programmed in python and js)....when i put more 0s to i it approximated to 0.5??(from 0.53 to 0.52)...??
$endgroup$
– Wolfgang Espada
Jan 13 at 17:52
$begingroup$
maybe you want to post your script on a programming site instead? this is mainly for mathematics.
$endgroup$
– Wesley Strik
Jan 13 at 18:02
add a comment |
$begingroup$
I need help with the calculation of the following function
$$sin(omega t)=frac{(e^{iωt}-e^{-iωt})}{2j}implies sin(30^{circ})= sinleft(frac pi 6 right) = frac{e^{frac pi 6 i}-e^{-frac pi 6i}}{2i} $$...when I plug in the values for the function in my program code it replays $0.53dots$ But it should do $0.5000000$
Is it because the approximation with two e's and the other numbers with commas divided by $2$ include an such a high error in it(the numbers are not ending) so that i would have to use a Taylor polynomial (it is the aim to use function that can be written as only + and - for schemas. Functions with a sine in it are too abstract and not allowed) for calculating the sine without precalculated values like in a cordic function.
Ps. Please do not write it as Taylor polynomial or cordic.
complex-numbers approximation
edited Jan 13 at 23:22
user376343
3,9834829
asked Jan 13 at 17:18
Wolfgang EspadaWolfgang Espada
41
$endgroup$
I need help with the calculation of the following function
$$sin(omega t)=frac{(e^{iωt}-e^{-iωt})}{2j}implies sin(30^{circ})= sinleft(frac pi 6 right) = frac{e^{frac pi 6 i}-e^{-frac pi 6i}}{2i} $$...when I plug in the values for the function in my program code it replays $0.53dots$ But it should do $0.5000000$
Is it because the approximation with two e's and the other numbers with commas divided by $2$ include an such a high error in it(the numbers are not ending) so that i would have to use a Taylor polynomial (it is the aim to use function that can be written as only + and - for schemas. Functions with a sine in it are too abstract and not allowed) for calculating the sine without precalculated values like in a cordic function.
Ps. Please do not write it as Taylor polynomial or cordic.
complex-numbers approximation
complex-numbers approximation
edited Jan 13 at 23:22
user376343
3,9834829
asked Jan 13 at 17:18
Wolfgang EspadaWolfgang Espada
41
edited Jan 13 at 23:22
user376343
3,9834829
asked Jan 13 at 17:18
Wolfgang EspadaWolfgang Espada
41
edited Jan 13 at 23:22
user376343
3,9834829
edited Jan 13 at 23:22
user376343
3,9834829
edited Jan 13 at 23:22
user376343
3,9834829
3,9834829
asked Jan 13 at 17:18
Wolfgang EspadaWolfgang Espada
41
asked Jan 13 at 17:18
Wolfgang EspadaWolfgang Espada
41
asked Jan 13 at 17:18
Wolfgang EspadaWolfgang Espada
41
41
closed as unclear what you're asking by Did, Lord Shark the Unknown, Cesareo, metamorphy, José Carlos Santos Jan 14 at 10:11
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Did, Lord Shark the Unknown, Cesareo, metamorphy, José Carlos Santos Jan 14 at 10:11
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
It may well be that your calculator is doing degrees instead of radians, or the other way around. You must define this with the MODE key on the calculator.
$endgroup$
– DonAntonio
Jan 13 at 17:20
$begingroup$
Welcome to Mathematics Stack Exchange community! The quick tour will help you get the most benefit from your time here. Also, please use MathJax for your equations.
$endgroup$
– dantopa
Jan 13 at 17:31
2
$begingroup$
Note that $30^circ = 30times pi/180 = pi/6$ radians. Therefore, $sin(30^circ) = frac{e^{ipi/6}-e^{-ipi/6}}{2i}$.
$endgroup$
– Math Lover
Jan 13 at 17:47
$begingroup$
I did like this(it was programmed in python and js)....when i put more 0s to i it approximated to 0.5??(from 0.53 to 0.52)...??
$endgroup$
– Wolfgang Espada
Jan 13 at 17:52
$begingroup$
maybe you want to post your script on a programming site instead? this is mainly for mathematics.
$endgroup$
– Wesley Strik
Jan 13 at 18:02
add a comment |
1
$begingroup$
It may well be that your calculator is doing degrees instead of radians, or the other way around. You must define this with the MODE key on the calculator.
$endgroup$
– DonAntonio
Jan 13 at 17:20
$begingroup$
Welcome to Mathematics Stack Exchange community! The quick tour will help you get the most benefit from your time here. Also, please use MathJax for your equations.
$endgroup$
– dantopa
Jan 13 at 17:31
2
$begingroup$
Note that $30^circ = 30times pi/180 = pi/6$ radians. Therefore, $sin(30^circ) = frac{e^{ipi/6}-e^{-ipi/6}}{2i}$.
$endgroup$
– Math Lover
Jan 13 at 17:47
$begingroup$
I did like this(it was programmed in python and js)....when i put more 0s to i it approximated to 0.5??(from 0.53 to 0.52)...??
$endgroup$
– Wolfgang Espada
Jan 13 at 17:52
$begingroup$
maybe you want to post your script on a programming site instead? this is mainly for mathematics.
$endgroup$
– Wesley Strik
Jan 13 at 18:02
1
1
$begingroup$
It may well be that your calculator is doing degrees instead of radians, or the other way around. You must define this with the MODE key on the calculator.
$endgroup$
– DonAntonio
Jan 13 at 17:20
$begingroup$
It may well be that your calculator is doing degrees instead of radians, or the other way around. You must define this with the MODE key on the calculator.
$endgroup$
– DonAntonio
Jan 13 at 17:20
$begingroup$
Welcome to Mathematics Stack Exchange community! The quick tour will help you get the most benefit from your time here. Also, please use MathJax for your equations.
$endgroup$
– dantopa
Jan 13 at 17:31
$begingroup$
Welcome to Mathematics Stack Exchange community! The quick tour will help you get the most benefit from your time here. Also, please use MathJax for your equations.
$endgroup$
– dantopa
Jan 13 at 17:31
2
2
$begingroup$
Note that $30^circ = 30times pi/180 = pi/6$ radians. Therefore, $sin(30^circ) = frac{e^{ipi/6}-e^{-ipi/6}}{2i}$.
$endgroup$
– Math Lover
Jan 13 at 17:47
$begingroup$
Note that $30^circ = 30times pi/180 = pi/6$ radians. Therefore, $sin(30^circ) = frac{e^{ipi/6}-e^{-ipi/6}}{2i}$.
$endgroup$
– Math Lover
Jan 13 at 17:47
$begingroup$
I did like this(it was programmed in python and js)....when i put more 0s to i it approximated to 0.5??(from 0.53 to 0.52)...??
$endgroup$
– Wolfgang Espada
Jan 13 at 17:52
$begingroup$
I did like this(it was programmed in python and js)....when i put more 0s to i it approximated to 0.5??(from 0.53 to 0.52)...??
$endgroup$
– Wolfgang Espada
Jan 13 at 17:52
$begingroup$
maybe you want to post your script on a programming site instead? this is mainly for mathematics.
$endgroup$
– Wesley Strik
Jan 13 at 18:02
$begingroup$
maybe you want to post your script on a programming site instead? this is mainly for mathematics.
$endgroup$
– Wesley Strik
Jan 13 at 18:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$sin(30^circ) = sin frac{pi}{6}$. So in your formula, use $omega t = frac{pi}{6}$.
begin{align}
e^{jpi/6} &approx 0.8660254040+j ;0.5000000000
\
e^{-jpi/6} &approx 0.8660254040-j ;0.5000000000
\
frac{e^{jpi/6}-e^{-jpi/6}}{2j} &approx 0.5000000000
end{align}
answered Jan 13 at 17:50
GEdgarGEdgar
63.6k269176
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$sin(30^circ) = sin frac{pi}{6}$. So in your formula, use $omega t = frac{pi}{6}$.
begin{align}
e^{jpi/6} &approx 0.8660254040+j ;0.5000000000
\
e^{-jpi/6} &approx 0.8660254040-j ;0.5000000000
\
frac{e^{jpi/6}-e^{-jpi/6}}{2j} &approx 0.5000000000
end{align}
answered Jan 13 at 17:50
GEdgarGEdgar
63.6k269176
$endgroup$
add a comment |
$begingroup$
$sin(30^circ) = sin frac{pi}{6}$. So in your formula, use $omega t = frac{pi}{6}$.
begin{align}
e^{jpi/6} &approx 0.8660254040+j ;0.5000000000
\
e^{-jpi/6} &approx 0.8660254040-j ;0.5000000000
\
frac{e^{jpi/6}-e^{-jpi/6}}{2j} &approx 0.5000000000
end{align}
answered Jan 13 at 17:50
GEdgarGEdgar
63.6k269176
$endgroup$
add a comment |
$begingroup$
$sin(30^circ) = sin frac{pi}{6}$. So in your formula, use $omega t = frac{pi}{6}$.
begin{align}
e^{jpi/6} &approx 0.8660254040+j ;0.5000000000
\
e^{-jpi/6} &approx 0.8660254040-j ;0.5000000000
\
frac{e^{jpi/6}-e^{-jpi/6}}{2j} &approx 0.5000000000
end{align}
answered Jan 13 at 17:50
GEdgarGEdgar
63.6k269176
$endgroup$
$sin(30^circ) = sin frac{pi}{6}$. So in your formula, use $omega t = frac{pi}{6}$.
begin{align}
e^{jpi/6} &approx 0.8660254040+j ;0.5000000000
\
e^{-jpi/6} &approx 0.8660254040-j ;0.5000000000
\
frac{e^{jpi/6}-e^{-jpi/6}}{2j} &approx 0.5000000000
end{align}
answered Jan 13 at 17:50
GEdgarGEdgar
63.6k269176
answered Jan 13 at 17:50
GEdgarGEdgar
63.6k269176
answered Jan 13 at 17:50
GEdgarGEdgar
63.6k269176
answered Jan 13 at 17:50
GEdgarGEdgar
63.6k269176
63.6k269176
add a comment |
add a comment |
1
$begingroup$
It may well be that your calculator is doing degrees instead of radians, or the other way around. You must define this with the MODE key on the calculator.
$endgroup$
– DonAntonio
Jan 13 at 17:20
$begingroup$
Welcome to Mathematics Stack Exchange community! The quick tour will help you get the most benefit from your time here. Also, please use MathJax for your equations.
$endgroup$
– dantopa
Jan 13 at 17:31
2
$begingroup$
Note that $30^circ = 30times pi/180 = pi/6$ radians. Therefore, $sin(30^circ) = frac{e^{ipi/6}-e^{-ipi/6}}{2i}$.
$endgroup$
– Math Lover
Jan 13 at 17:47
$begingroup$
I did like this(it was programmed in python and js)....when i put more 0s to i it approximated to 0.5??(from 0.53 to 0.52)...??
$endgroup$
– Wolfgang Espada
Jan 13 at 17:52
$begingroup$
maybe you want to post your script on a programming site instead? this is mainly for mathematics.
$endgroup$
– Wesley Strik
Jan 13 at 18:02