Mixing
Proof for de Moivre's Formula
$begingroup$
I have a book that has a brief history of the complex numbers and it covers de Moivre's formula:
$(cos(x) + isin(x))^n = cos(nx) + isin(nx)$.
I am very curious as to how this result was originally found, or derived, BEFORE Euler's formula was around. Also, what was the original proof of this?
complex-analysis math-history
edited Feb 6 '15 at 2:59
dustin
6,70892969
asked Feb 6 '15 at 1:40
ChristopherChristopher
393210
$endgroup$
add a comment |
$begingroup$
I have a book that has a brief history of the complex numbers and it covers de Moivre's formula:
$(cos(x) + isin(x))^n = cos(nx) + isin(nx)$.
I am very curious as to how this result was originally found, or derived, BEFORE Euler's formula was around. Also, what was the original proof of this?
complex-analysis math-history
edited Feb 6 '15 at 2:59
dustin
6,70892969
asked Feb 6 '15 at 1:40
ChristopherChristopher
393210
$endgroup$
3
$begingroup$
Probably by induction.
$endgroup$
– dustin
Feb 6 '15 at 1:46
2
$begingroup$
Induction indeed...and basic trigonometric formulas, of course.
$endgroup$
– Timbuc
Feb 6 '15 at 1:52
2
$begingroup$
@Christopher have answer working on putting it together.
$endgroup$
– dustin
Feb 6 '15 at 2:06
$begingroup$
I was puzzled by this result when I saw it because the way I was introduced to multiplication of complex numbers was that one multiplies the absolute values and adds the angles, so this formula seemed like a complete triviality.
$endgroup$
– Michael Hardy
Feb 6 '15 at 2:41
add a comment |
$begingroup$
I have a book that has a brief history of the complex numbers and it covers de Moivre's formula:
$(cos(x) + isin(x))^n = cos(nx) + isin(nx)$.
I am very curious as to how this result was originally found, or derived, BEFORE Euler's formula was around. Also, what was the original proof of this?
complex-analysis math-history
edited Feb 6 '15 at 2:59
dustin
6,70892969
asked Feb 6 '15 at 1:40
ChristopherChristopher
393210
$endgroup$
I have a book that has a brief history of the complex numbers and it covers de Moivre's formula:
$(cos(x) + isin(x))^n = cos(nx) + isin(nx)$.
I am very curious as to how this result was originally found, or derived, BEFORE Euler's formula was around. Also, what was the original proof of this?
complex-analysis math-history
complex-analysis math-history
edited Feb 6 '15 at 2:59
dustin
6,70892969
asked Feb 6 '15 at 1:40
ChristopherChristopher
393210
edited Feb 6 '15 at 2:59
dustin
6,70892969
asked Feb 6 '15 at 1:40
ChristopherChristopher
393210
edited Feb 6 '15 at 2:59
dustin
6,70892969
edited Feb 6 '15 at 2:59
dustin
6,70892969
edited Feb 6 '15 at 2:59
dustin
6,70892969
6,70892969
asked Feb 6 '15 at 1:40
ChristopherChristopher
393210
asked Feb 6 '15 at 1:40
ChristopherChristopher
393210
asked Feb 6 '15 at 1:40
ChristopherChristopher
393210
393210
3
$begingroup$
Probably by induction.
$endgroup$
– dustin
Feb 6 '15 at 1:46
2
$begingroup$
Induction indeed...and basic trigonometric formulas, of course.
$endgroup$
– Timbuc
Feb 6 '15 at 1:52
2
$begingroup$
@Christopher have answer working on putting it together.
$endgroup$
– dustin
Feb 6 '15 at 2:06
$begingroup$
I was puzzled by this result when I saw it because the way I was introduced to multiplication of complex numbers was that one multiplies the absolute values and adds the angles, so this formula seemed like a complete triviality.
$endgroup$
– Michael Hardy
Feb 6 '15 at 2:41
add a comment |
3
$begingroup$
Probably by induction.
$endgroup$
– dustin
Feb 6 '15 at 1:46
2
$begingroup$
Induction indeed...and basic trigonometric formulas, of course.
$endgroup$
– Timbuc
Feb 6 '15 at 1:52
2
$begingroup$
@Christopher have answer working on putting it together.
$endgroup$
– dustin
Feb 6 '15 at 2:06
$begingroup$
I was puzzled by this result when I saw it because the way I was introduced to multiplication of complex numbers was that one multiplies the absolute values and adds the angles, so this formula seemed like a complete triviality.
$endgroup$
– Michael Hardy
Feb 6 '15 at 2:41
3
3
$begingroup$
Probably by induction.
$endgroup$
– dustin
Feb 6 '15 at 1:46
$begingroup$
Probably by induction.
$endgroup$
– dustin
Feb 6 '15 at 1:46
2
2
$begingroup$
Induction indeed...and basic trigonometric formulas, of course.
$endgroup$
– Timbuc
Feb 6 '15 at 1:52
$begingroup$
Induction indeed...and basic trigonometric formulas, of course.
$endgroup$
– Timbuc
Feb 6 '15 at 1:52
2
2
$begingroup$
@Christopher have answer working on putting it together.
$endgroup$
– dustin
Feb 6 '15 at 2:06
$begingroup$
@Christopher have answer working on putting it together.
$endgroup$
– dustin
Feb 6 '15 at 2:06
$begingroup$
I was puzzled by this result when I saw it because the way I was introduced to multiplication of complex numbers was that one multiplies the absolute values and adds the angles, so this formula seemed like a complete triviality.
$endgroup$
– Michael Hardy
Feb 6 '15 at 2:41
$begingroup$
I was puzzled by this result when I saw it because the way I was introduced to multiplication of complex numbers was that one multiplies the absolute values and adds the angles, so this formula seemed like a complete triviality.
$endgroup$
– Michael Hardy
Feb 6 '15 at 2:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Euler was the first to write down de Moirve's formula which I found in The Elementary Mathematical Works of Leonhard Euler starting on page 29. Euler found
$$
(cos(x)pm isin(x))(cos(y)pm isin(y))(cos(z)pm isin(z)) = cos(x pm y pm z)pm isin(xpm ypm z)
$$
from here he deduce what we call de Moirve's theorem
$$
(cos(x)pm isin(x))^n = cos(nx)pm isin(nx)
$$
Then it follows that
begin{align}
cos(nx) &= 1/2[(cos(x)+ isin(x))^n+(cos(x) - isin(x))^n]tag{1}\
sin(nx) &= 1/2[(cos(x)+ isin(x))^n-(cos(x) - isin(x))^n]tag{2}
end{align}
Euler than expanded equations $(1)$ and $(2)$ by the binomial theorem. He then let $x$ be a small angle approximation and $n$ be infinitely large so that $xn$ was finite. Let $xn=v$. Then $sin(x) = x = v/n$ and we get
begin{align}
cos(v) &= 1 - frac{v^2}{2!} + cdots\
sin(v) &= v - frac{v^3}{3!} + cdots
end{align}
Using equations $(1)$ and $(2)$ and letting $j = n$ be the infinitely large number so that $jx=v$ so that $nx = v = v/j$. Now we have $sin(x) = v/j$ and $cos(x)=1$. With these substitutions, we get
begin{align}
cos(v) &=frac{(1+iv/j)^j+(1-iv/j)^j}{2}\
sin(v) &=frac{(1+iv/j)^j-(1-iv/j)^j}{2i}
end{align}
In the linked document, the author who compiled the document (Euler) previously has shown that $(1+z/j)^j = e^z$. Letting $z=pm iv$, we get
begin{align}
e^{iv} &= cos(v) + isin(v)\
e^{-iv} &= cos(v) - isin(v)
end{align}
answered Feb 6 '15 at 2:25
dustindustin
6,70892969
$endgroup$
1
$begingroup$
Great answer and great link! Thank you so much.
$endgroup$
– Christopher
Feb 6 '15 at 2:39
$begingroup$
@Christopher you probably want to go through the link since I left out the some details as well as the binomial expansion of $(1)$ and $(2)$.
$endgroup$
– dustin
Feb 6 '15 at 2:39
1
$begingroup$
@dustin Very nice approach!
$endgroup$
– Math-fun
Feb 16 '15 at 20:49
$begingroup$
@Mehdi unfortunately, Euler gets all the credit here. I have no idea why this was named after de Moirve though. It mentions he was working on some equations but I don't see the relation or why he gets the credit.
$endgroup$
– dustin
Feb 16 '15 at 20:50
$begingroup$
@dustin well he has been great (and is). By the way I just found an interesting youtube video on Euler: youtube.com/watch?v=h-DV26x6n_Q
$endgroup$
– Math-fun
Feb 16 '15 at 20:54
|
show 1 more comment
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1 Answer
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$begingroup$
Euler was the first to write down de Moirve's formula which I found in The Elementary Mathematical Works of Leonhard Euler starting on page 29. Euler found
$$
(cos(x)pm isin(x))(cos(y)pm isin(y))(cos(z)pm isin(z)) = cos(x pm y pm z)pm isin(xpm ypm z)
$$
from here he deduce what we call de Moirve's theorem
$$
(cos(x)pm isin(x))^n = cos(nx)pm isin(nx)
$$
Then it follows that
begin{align}
cos(nx) &= 1/2[(cos(x)+ isin(x))^n+(cos(x) - isin(x))^n]tag{1}\
sin(nx) &= 1/2[(cos(x)+ isin(x))^n-(cos(x) - isin(x))^n]tag{2}
end{align}
Euler than expanded equations $(1)$ and $(2)$ by the binomial theorem. He then let $x$ be a small angle approximation and $n$ be infinitely large so that $xn$ was finite. Let $xn=v$. Then $sin(x) = x = v/n$ and we get
begin{align}
cos(v) &= 1 - frac{v^2}{2!} + cdots\
sin(v) &= v - frac{v^3}{3!} + cdots
end{align}
Using equations $(1)$ and $(2)$ and letting $j = n$ be the infinitely large number so that $jx=v$ so that $nx = v = v/j$. Now we have $sin(x) = v/j$ and $cos(x)=1$. With these substitutions, we get
begin{align}
cos(v) &=frac{(1+iv/j)^j+(1-iv/j)^j}{2}\
sin(v) &=frac{(1+iv/j)^j-(1-iv/j)^j}{2i}
end{align}
In the linked document, the author who compiled the document (Euler) previously has shown that $(1+z/j)^j = e^z$. Letting $z=pm iv$, we get
begin{align}
e^{iv} &= cos(v) + isin(v)\
e^{-iv} &= cos(v) - isin(v)
end{align}
answered Feb 6 '15 at 2:25
dustindustin
6,70892969
$endgroup$
1
$begingroup$
Great answer and great link! Thank you so much.
$endgroup$
– Christopher
Feb 6 '15 at 2:39
$begingroup$
@Christopher you probably want to go through the link since I left out the some details as well as the binomial expansion of $(1)$ and $(2)$.
$endgroup$
– dustin
Feb 6 '15 at 2:39
1
$begingroup$
@dustin Very nice approach!
$endgroup$
– Math-fun
Feb 16 '15 at 20:49
$begingroup$
@Mehdi unfortunately, Euler gets all the credit here. I have no idea why this was named after de Moirve though. It mentions he was working on some equations but I don't see the relation or why he gets the credit.
$endgroup$
– dustin
Feb 16 '15 at 20:50
$begingroup$
@dustin well he has been great (and is). By the way I just found an interesting youtube video on Euler: youtube.com/watch?v=h-DV26x6n_Q
$endgroup$
– Math-fun
Feb 16 '15 at 20:54
|
show 1 more comment
$begingroup$
Euler was the first to write down de Moirve's formula which I found in The Elementary Mathematical Works of Leonhard Euler starting on page 29. Euler found
$$
(cos(x)pm isin(x))(cos(y)pm isin(y))(cos(z)pm isin(z)) = cos(x pm y pm z)pm isin(xpm ypm z)
$$
from here he deduce what we call de Moirve's theorem
$$
(cos(x)pm isin(x))^n = cos(nx)pm isin(nx)
$$
Then it follows that
begin{align}
cos(nx) &= 1/2[(cos(x)+ isin(x))^n+(cos(x) - isin(x))^n]tag{1}\
sin(nx) &= 1/2[(cos(x)+ isin(x))^n-(cos(x) - isin(x))^n]tag{2}
end{align}
Euler than expanded equations $(1)$ and $(2)$ by the binomial theorem. He then let $x$ be a small angle approximation and $n$ be infinitely large so that $xn$ was finite. Let $xn=v$. Then $sin(x) = x = v/n$ and we get
begin{align}
cos(v) &= 1 - frac{v^2}{2!} + cdots\
sin(v) &= v - frac{v^3}{3!} + cdots
end{align}
Using equations $(1)$ and $(2)$ and letting $j = n$ be the infinitely large number so that $jx=v$ so that $nx = v = v/j$. Now we have $sin(x) = v/j$ and $cos(x)=1$. With these substitutions, we get
begin{align}
cos(v) &=frac{(1+iv/j)^j+(1-iv/j)^j}{2}\
sin(v) &=frac{(1+iv/j)^j-(1-iv/j)^j}{2i}
end{align}
In the linked document, the author who compiled the document (Euler) previously has shown that $(1+z/j)^j = e^z$. Letting $z=pm iv$, we get
begin{align}
e^{iv} &= cos(v) + isin(v)\
e^{-iv} &= cos(v) - isin(v)
end{align}
answered Feb 6 '15 at 2:25
dustindustin
6,70892969
$endgroup$
1
$begingroup$
Great answer and great link! Thank you so much.
$endgroup$
– Christopher
Feb 6 '15 at 2:39
$begingroup$
@Christopher you probably want to go through the link since I left out the some details as well as the binomial expansion of $(1)$ and $(2)$.
$endgroup$
– dustin
Feb 6 '15 at 2:39
1
$begingroup$
@dustin Very nice approach!
$endgroup$
– Math-fun
Feb 16 '15 at 20:49
$begingroup$
@Mehdi unfortunately, Euler gets all the credit here. I have no idea why this was named after de Moirve though. It mentions he was working on some equations but I don't see the relation or why he gets the credit.
$endgroup$
– dustin
Feb 16 '15 at 20:50
$begingroup$
@dustin well he has been great (and is). By the way I just found an interesting youtube video on Euler: youtube.com/watch?v=h-DV26x6n_Q
$endgroup$
– Math-fun
Feb 16 '15 at 20:54
|
show 1 more comment
$begingroup$
Euler was the first to write down de Moirve's formula which I found in The Elementary Mathematical Works of Leonhard Euler starting on page 29. Euler found
$$
(cos(x)pm isin(x))(cos(y)pm isin(y))(cos(z)pm isin(z)) = cos(x pm y pm z)pm isin(xpm ypm z)
$$
from here he deduce what we call de Moirve's theorem
$$
(cos(x)pm isin(x))^n = cos(nx)pm isin(nx)
$$
Then it follows that
begin{align}
cos(nx) &= 1/2[(cos(x)+ isin(x))^n+(cos(x) - isin(x))^n]tag{1}\
sin(nx) &= 1/2[(cos(x)+ isin(x))^n-(cos(x) - isin(x))^n]tag{2}
end{align}
Euler than expanded equations $(1)$ and $(2)$ by the binomial theorem. He then let $x$ be a small angle approximation and $n$ be infinitely large so that $xn$ was finite. Let $xn=v$. Then $sin(x) = x = v/n$ and we get
begin{align}
cos(v) &= 1 - frac{v^2}{2!} + cdots\
sin(v) &= v - frac{v^3}{3!} + cdots
end{align}
Using equations $(1)$ and $(2)$ and letting $j = n$ be the infinitely large number so that $jx=v$ so that $nx = v = v/j$. Now we have $sin(x) = v/j$ and $cos(x)=1$. With these substitutions, we get
begin{align}
cos(v) &=frac{(1+iv/j)^j+(1-iv/j)^j}{2}\
sin(v) &=frac{(1+iv/j)^j-(1-iv/j)^j}{2i}
end{align}
In the linked document, the author who compiled the document (Euler) previously has shown that $(1+z/j)^j = e^z$. Letting $z=pm iv$, we get
begin{align}
e^{iv} &= cos(v) + isin(v)\
e^{-iv} &= cos(v) - isin(v)
end{align}
answered Feb 6 '15 at 2:25
dustindustin
6,70892969
$endgroup$
Euler was the first to write down de Moirve's formula which I found in The Elementary Mathematical Works of Leonhard Euler starting on page 29. Euler found
$$
(cos(x)pm isin(x))(cos(y)pm isin(y))(cos(z)pm isin(z)) = cos(x pm y pm z)pm isin(xpm ypm z)
$$
from here he deduce what we call de Moirve's theorem
$$
(cos(x)pm isin(x))^n = cos(nx)pm isin(nx)
$$
Then it follows that
begin{align}
cos(nx) &= 1/2[(cos(x)+ isin(x))^n+(cos(x) - isin(x))^n]tag{1}\
sin(nx) &= 1/2[(cos(x)+ isin(x))^n-(cos(x) - isin(x))^n]tag{2}
end{align}
Euler than expanded equations $(1)$ and $(2)$ by the binomial theorem. He then let $x$ be a small angle approximation and $n$ be infinitely large so that $xn$ was finite. Let $xn=v$. Then $sin(x) = x = v/n$ and we get
begin{align}
cos(v) &= 1 - frac{v^2}{2!} + cdots\
sin(v) &= v - frac{v^3}{3!} + cdots
end{align}
Using equations $(1)$ and $(2)$ and letting $j = n$ be the infinitely large number so that $jx=v$ so that $nx = v = v/j$. Now we have $sin(x) = v/j$ and $cos(x)=1$. With these substitutions, we get
begin{align}
cos(v) &=frac{(1+iv/j)^j+(1-iv/j)^j}{2}\
sin(v) &=frac{(1+iv/j)^j-(1-iv/j)^j}{2i}
end{align}
In the linked document, the author who compiled the document (Euler) previously has shown that $(1+z/j)^j = e^z$. Letting $z=pm iv$, we get
begin{align}
e^{iv} &= cos(v) + isin(v)\
e^{-iv} &= cos(v) - isin(v)
end{align}
answered Feb 6 '15 at 2:25
dustindustin
6,70892969
edited Feb 6 '15 at 4:01
answered Feb 6 '15 at 2:25
dustindustin
6,70892969
answered Feb 6 '15 at 2:25
dustindustin
6,70892969
answered Feb 6 '15 at 2:25
dustindustin
6,70892969
6,70892969
1
$begingroup$
Great answer and great link! Thank you so much.
$endgroup$
– Christopher
Feb 6 '15 at 2:39
$begingroup$
@Christopher you probably want to go through the link since I left out the some details as well as the binomial expansion of $(1)$ and $(2)$.
$endgroup$
– dustin
Feb 6 '15 at 2:39
1
$begingroup$
@dustin Very nice approach!
$endgroup$
– Math-fun
Feb 16 '15 at 20:49
$begingroup$
@Mehdi unfortunately, Euler gets all the credit here. I have no idea why this was named after de Moirve though. It mentions he was working on some equations but I don't see the relation or why he gets the credit.
$endgroup$
– dustin
Feb 16 '15 at 20:50
$begingroup$
@dustin well he has been great (and is). By the way I just found an interesting youtube video on Euler: youtube.com/watch?v=h-DV26x6n_Q
$endgroup$
– Math-fun
Feb 16 '15 at 20:54
|
show 1 more comment
1
$begingroup$
Great answer and great link! Thank you so much.
$endgroup$
– Christopher
Feb 6 '15 at 2:39
$begingroup$
@Christopher you probably want to go through the link since I left out the some details as well as the binomial expansion of $(1)$ and $(2)$.
$endgroup$
– dustin
Feb 6 '15 at 2:39
1
$begingroup$
@dustin Very nice approach!
$endgroup$
– Math-fun
Feb 16 '15 at 20:49
$begingroup$
@Mehdi unfortunately, Euler gets all the credit here. I have no idea why this was named after de Moirve though. It mentions he was working on some equations but I don't see the relation or why he gets the credit.
$endgroup$
– dustin
Feb 16 '15 at 20:50
$begingroup$
@dustin well he has been great (and is). By the way I just found an interesting youtube video on Euler: youtube.com/watch?v=h-DV26x6n_Q
$endgroup$
– Math-fun
Feb 16 '15 at 20:54
1
1
$begingroup$
Great answer and great link! Thank you so much.
$endgroup$
– Christopher
Feb 6 '15 at 2:39
$begingroup$
Great answer and great link! Thank you so much.
$endgroup$
– Christopher
Feb 6 '15 at 2:39
$begingroup$
@Christopher you probably want to go through the link since I left out the some details as well as the binomial expansion of $(1)$ and $(2)$.
$endgroup$
– dustin
Feb 6 '15 at 2:39
$begingroup$
@Christopher you probably want to go through the link since I left out the some details as well as the binomial expansion of $(1)$ and $(2)$.
$endgroup$
– dustin
Feb 6 '15 at 2:39
1
1
$begingroup$
@dustin Very nice approach!
$endgroup$
– Math-fun
Feb 16 '15 at 20:49
$begingroup$
@dustin Very nice approach!
$endgroup$
– Math-fun
Feb 16 '15 at 20:49
$begingroup$
@Mehdi unfortunately, Euler gets all the credit here. I have no idea why this was named after de Moirve though. It mentions he was working on some equations but I don't see the relation or why he gets the credit.
$endgroup$
– dustin
Feb 16 '15 at 20:50
$begingroup$
@Mehdi unfortunately, Euler gets all the credit here. I have no idea why this was named after de Moirve though. It mentions he was working on some equations but I don't see the relation or why he gets the credit.
$endgroup$
– dustin
Feb 16 '15 at 20:50
$begingroup$
@dustin well he has been great (and is). By the way I just found an interesting youtube video on Euler: youtube.com/watch?v=h-DV26x6n_Q
$endgroup$
– Math-fun
Feb 16 '15 at 20:54
$begingroup$
@dustin well he has been great (and is). By the way I just found an interesting youtube video on Euler: youtube.com/watch?v=h-DV26x6n_Q
$endgroup$
– Math-fun
Feb 16 '15 at 20:54
|
show 1 more comment
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3
$begingroup$
Probably by induction.
$endgroup$
– dustin
Feb 6 '15 at 1:46
2
$begingroup$
Induction indeed...and basic trigonometric formulas, of course.
$endgroup$
– Timbuc
Feb 6 '15 at 1:52
2
$begingroup$
@Christopher have answer working on putting it together.
$endgroup$
– dustin
Feb 6 '15 at 2:06
$begingroup$
I was puzzled by this result when I saw it because the way I was introduced to multiplication of complex numbers was that one multiplies the absolute values and adds the angles, so this formula seemed like a complete triviality.
$endgroup$
– Michael Hardy
Feb 6 '15 at 2:41