Mixing
Why does Euler's formula have to be $e^{ix} = cos(x) + isin(x)$
$begingroup$
In part one of this youtube video the uploader goes on to explain the calculus proof for Euler's Formula.
The Formula
$$e^{ix} = cos(x) + isin(x)$$
Differentiate
$$ie^{ix} = f'(x) + i g'(x)$$
Multiply original formula by $i$
$$ie^{ix} = if(x) - g(x)$$
Equate the differentiation and the multiplied version
$$f'(x) + ig'(x) = if(x) - g(x)$$
Equate real and imaginary (and cancel the i)
$$f'(x) = -g(x) qquad g'(x) = f(x)$$
Then he goes on to explain $f(x) = cos(x)$ and $g(x) = sin(x)$. My question is why can't $f(x) = sin(x)$ and $g(x) = -cos(x)$? Can further proof be added to this proof to eliminate $f(x) = sin(x)$ and $g(x) = -cos(x)$?
calculus complex-analysis proof-verification complex-numbers proof-explanation
asked Mar 11 '17 at 15:20
DanDan
1751112
$endgroup$
add a comment |
$begingroup$
In part one of this youtube video the uploader goes on to explain the calculus proof for Euler's Formula.
The Formula
$$e^{ix} = cos(x) + isin(x)$$
Differentiate
$$ie^{ix} = f'(x) + i g'(x)$$
Multiply original formula by $i$
$$ie^{ix} = if(x) - g(x)$$
Equate the differentiation and the multiplied version
$$f'(x) + ig'(x) = if(x) - g(x)$$
Equate real and imaginary (and cancel the i)
$$f'(x) = -g(x) qquad g'(x) = f(x)$$
Then he goes on to explain $f(x) = cos(x)$ and $g(x) = sin(x)$. My question is why can't $f(x) = sin(x)$ and $g(x) = -cos(x)$? Can further proof be added to this proof to eliminate $f(x) = sin(x)$ and $g(x) = -cos(x)$?
calculus complex-analysis proof-verification complex-numbers proof-explanation
asked Mar 11 '17 at 15:20
DanDan
1751112
$endgroup$
$begingroup$
Else the equality would be false at $x=0$, or you wouldn't have a holomorphic function to do complex analysis with, etc. and that all gets messy.
$endgroup$
– Simply Beautiful Art
Mar 11 '17 at 15:22
$begingroup$
Possibly related: How to prove Euler's formula: $e^{it}=cos t +isin t$?
$endgroup$
– Simply Beautiful Art
Mar 11 '17 at 15:23
$begingroup$
@SimplyBeautifulArt It is related but not the same as I already have a proof but want some to explain the end result of it
$endgroup$
– Dan
Mar 11 '17 at 15:27
1
$begingroup$
Yes, of course. Hence I did not have it as possible duplicate :-)
$endgroup$
– Simply Beautiful Art
Mar 11 '17 at 15:30
add a comment |
$begingroup$
In part one of this youtube video the uploader goes on to explain the calculus proof for Euler's Formula.
The Formula
$$e^{ix} = cos(x) + isin(x)$$
Differentiate
$$ie^{ix} = f'(x) + i g'(x)$$
Multiply original formula by $i$
$$ie^{ix} = if(x) - g(x)$$
Equate the differentiation and the multiplied version
$$f'(x) + ig'(x) = if(x) - g(x)$$
Equate real and imaginary (and cancel the i)
$$f'(x) = -g(x) qquad g'(x) = f(x)$$
Then he goes on to explain $f(x) = cos(x)$ and $g(x) = sin(x)$. My question is why can't $f(x) = sin(x)$ and $g(x) = -cos(x)$? Can further proof be added to this proof to eliminate $f(x) = sin(x)$ and $g(x) = -cos(x)$?
calculus complex-analysis proof-verification complex-numbers proof-explanation
asked Mar 11 '17 at 15:20
DanDan
1751112
$endgroup$
In part one of this youtube video the uploader goes on to explain the calculus proof for Euler's Formula.
The Formula
$$e^{ix} = cos(x) + isin(x)$$
Differentiate
$$ie^{ix} = f'(x) + i g'(x)$$
Multiply original formula by $i$
$$ie^{ix} = if(x) - g(x)$$
Equate the differentiation and the multiplied version
$$f'(x) + ig'(x) = if(x) - g(x)$$
Equate real and imaginary (and cancel the i)
$$f'(x) = -g(x) qquad g'(x) = f(x)$$
Then he goes on to explain $f(x) = cos(x)$ and $g(x) = sin(x)$. My question is why can't $f(x) = sin(x)$ and $g(x) = -cos(x)$? Can further proof be added to this proof to eliminate $f(x) = sin(x)$ and $g(x) = -cos(x)$?
calculus complex-analysis proof-verification complex-numbers proof-explanation
calculus complex-analysis proof-verification complex-numbers proof-explanation
asked Mar 11 '17 at 15:20
DanDan
1751112
asked Mar 11 '17 at 15:20
DanDan
1751112
edited Mar 11 '17 at 15:26
Dan
asked Mar 11 '17 at 15:20
DanDan
1751112
asked Mar 11 '17 at 15:20
DanDan
1751112
asked Mar 11 '17 at 15:20
DanDan
1751112
1751112
$begingroup$
Else the equality would be false at $x=0$, or you wouldn't have a holomorphic function to do complex analysis with, etc. and that all gets messy.
$endgroup$
– Simply Beautiful Art
Mar 11 '17 at 15:22
$begingroup$
Possibly related: How to prove Euler's formula: $e^{it}=cos t +isin t$?
$endgroup$
– Simply Beautiful Art
Mar 11 '17 at 15:23
$begingroup$
@SimplyBeautifulArt It is related but not the same as I already have a proof but want some to explain the end result of it
$endgroup$
– Dan
Mar 11 '17 at 15:27
1
$begingroup$
Yes, of course. Hence I did not have it as possible duplicate :-)
$endgroup$
– Simply Beautiful Art
Mar 11 '17 at 15:30
add a comment |
$begingroup$
Else the equality would be false at $x=0$, or you wouldn't have a holomorphic function to do complex analysis with, etc. and that all gets messy.
$endgroup$
– Simply Beautiful Art
Mar 11 '17 at 15:22
$begingroup$
Possibly related: How to prove Euler's formula: $e^{it}=cos t +isin t$?
$endgroup$
– Simply Beautiful Art
Mar 11 '17 at 15:23
$begingroup$
@SimplyBeautifulArt It is related but not the same as I already have a proof but want some to explain the end result of it
$endgroup$
– Dan
Mar 11 '17 at 15:27
1
$begingroup$
Yes, of course. Hence I did not have it as possible duplicate :-)
$endgroup$
– Simply Beautiful Art
Mar 11 '17 at 15:30
$begingroup$
Else the equality would be false at $x=0$, or you wouldn't have a holomorphic function to do complex analysis with, etc. and that all gets messy.
$endgroup$
– Simply Beautiful Art
Mar 11 '17 at 15:22
$begingroup$
Else the equality would be false at $x=0$, or you wouldn't have a holomorphic function to do complex analysis with, etc. and that all gets messy.
$endgroup$
– Simply Beautiful Art
Mar 11 '17 at 15:22
$begingroup$
Possibly related: How to prove Euler's formula: $e^{it}=cos t +isin t$?
$endgroup$
– Simply Beautiful Art
Mar 11 '17 at 15:23
$begingroup$
Possibly related: How to prove Euler's formula: $e^{it}=cos t +isin t$?
$endgroup$
– Simply Beautiful Art
Mar 11 '17 at 15:23
$begingroup$
@SimplyBeautifulArt It is related but not the same as I already have a proof but want some to explain the end result of it
$endgroup$
– Dan
Mar 11 '17 at 15:27
$begingroup$
@SimplyBeautifulArt It is related but not the same as I already have a proof but want some to explain the end result of it
$endgroup$
– Dan
Mar 11 '17 at 15:27
1
1
$begingroup$
Yes, of course. Hence I did not have it as possible duplicate :-)
$endgroup$
– Simply Beautiful Art
Mar 11 '17 at 15:30
$begingroup$
Yes, of course. Hence I did not have it as possible duplicate :-)
$endgroup$
– Simply Beautiful Art
Mar 11 '17 at 15:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Because we know the initial condition $e^{i0}=1$ holds. As with most differential equations, there's an family of answers that you need to use the initial condition to find the correct one for.
answered Mar 11 '17 at 15:27
Stella BidermanStella Biderman
26.7k63375
$endgroup$
1
$begingroup$
Stella (+1) for the answer. I reject the so-called proof of Euler's formula inasmuch as it uses a result from real analysis (i.e., $frac{de^x}{dx}=e^x$) and presumes that it holds for the yet-to-be-defined object $e^{ix}$.
$endgroup$
– Mark Viola
Mar 11 '17 at 15:36
$begingroup$
@Dr.MV Agreed. My preferred way is to define $e^{ix}$ by the Taylor polynomial and then use complex analysis to prove that the power series preserves the usual properties of the real analysis function $e^x$. Euler's Equation is then simply obtained from comparing power series.
$endgroup$
– Stella Biderman
Mar 11 '17 at 18:07
$begingroup$
Yes, that's a good standard approach.
$endgroup$
– Mark Viola
Mar 11 '17 at 18:12
add a comment |
$begingroup$
Let z=cos(x)+isin(x)
When you 8derive it (in terms of x),
z'=-sin(x)+icos(x)
And since we are in the complex world, -1=i×i
=>z'=i×isin(x)+icos(x)=i(cos(x)+isin(x))=iz
This pretty much makes it a first order differential equation.
After dividing by z and multiplying by dx (z'=dz/dx) both on both sides and integrating both sides, you get:
ln(z)=ix+c
ln(cos(x)+isin(x))=ix+c
If we set x=0, cos(0)=1,sin(0)=0 and ln(1)=0
=> ln(1)=c=>c=0
=> ln(cos(x)+isin(x))=ix
e^ix=cos(x)+isin(x).
Hope this helps.
answered Jan 16 at 5:07
Serouj GhazarianSerouj Ghazarian
11
$endgroup$
$begingroup$
Hi and welcome to the Math.SE. Producing beautiful and beautifully formatted text is one of the points that makes an answer good: if you want to learn how to do this, please have a look at MathJax basic tutorial and quick reference.
$endgroup$
– Daniele Tampieri
Jan 16 at 5:37
add a comment |
Your Answer
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2 Answers
2
active
oldest
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2 Answers
2
active
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votes
$begingroup$
Because we know the initial condition $e^{i0}=1$ holds. As with most differential equations, there's an family of answers that you need to use the initial condition to find the correct one for.
answered Mar 11 '17 at 15:27
Stella BidermanStella Biderman
26.7k63375
$endgroup$
1
$begingroup$
Stella (+1) for the answer. I reject the so-called proof of Euler's formula inasmuch as it uses a result from real analysis (i.e., $frac{de^x}{dx}=e^x$) and presumes that it holds for the yet-to-be-defined object $e^{ix}$.
$endgroup$
– Mark Viola
Mar 11 '17 at 15:36
$begingroup$
@Dr.MV Agreed. My preferred way is to define $e^{ix}$ by the Taylor polynomial and then use complex analysis to prove that the power series preserves the usual properties of the real analysis function $e^x$. Euler's Equation is then simply obtained from comparing power series.
$endgroup$
– Stella Biderman
Mar 11 '17 at 18:07
$begingroup$
Yes, that's a good standard approach.
$endgroup$
– Mark Viola
Mar 11 '17 at 18:12
add a comment |
$begingroup$
Because we know the initial condition $e^{i0}=1$ holds. As with most differential equations, there's an family of answers that you need to use the initial condition to find the correct one for.
answered Mar 11 '17 at 15:27
Stella BidermanStella Biderman
26.7k63375
$endgroup$
1
$begingroup$
Stella (+1) for the answer. I reject the so-called proof of Euler's formula inasmuch as it uses a result from real analysis (i.e., $frac{de^x}{dx}=e^x$) and presumes that it holds for the yet-to-be-defined object $e^{ix}$.
$endgroup$
– Mark Viola
Mar 11 '17 at 15:36
$begingroup$
@Dr.MV Agreed. My preferred way is to define $e^{ix}$ by the Taylor polynomial and then use complex analysis to prove that the power series preserves the usual properties of the real analysis function $e^x$. Euler's Equation is then simply obtained from comparing power series.
$endgroup$
– Stella Biderman
Mar 11 '17 at 18:07
$begingroup$
Yes, that's a good standard approach.
$endgroup$
– Mark Viola
Mar 11 '17 at 18:12
add a comment |
$begingroup$
Because we know the initial condition $e^{i0}=1$ holds. As with most differential equations, there's an family of answers that you need to use the initial condition to find the correct one for.
answered Mar 11 '17 at 15:27
Stella BidermanStella Biderman
26.7k63375
$endgroup$
Because we know the initial condition $e^{i0}=1$ holds. As with most differential equations, there's an family of answers that you need to use the initial condition to find the correct one for.
answered Mar 11 '17 at 15:27
Stella BidermanStella Biderman
26.7k63375
answered Mar 11 '17 at 15:27
Stella BidermanStella Biderman
26.7k63375
answered Mar 11 '17 at 15:27
Stella BidermanStella Biderman
26.7k63375
answered Mar 11 '17 at 15:27
Stella BidermanStella Biderman
26.7k63375
26.7k63375
1
$begingroup$
Stella (+1) for the answer. I reject the so-called proof of Euler's formula inasmuch as it uses a result from real analysis (i.e., $frac{de^x}{dx}=e^x$) and presumes that it holds for the yet-to-be-defined object $e^{ix}$.
$endgroup$
– Mark Viola
Mar 11 '17 at 15:36
$begingroup$
@Dr.MV Agreed. My preferred way is to define $e^{ix}$ by the Taylor polynomial and then use complex analysis to prove that the power series preserves the usual properties of the real analysis function $e^x$. Euler's Equation is then simply obtained from comparing power series.
$endgroup$
– Stella Biderman
Mar 11 '17 at 18:07
$begingroup$
Yes, that's a good standard approach.
$endgroup$
– Mark Viola
Mar 11 '17 at 18:12
add a comment |
1
$begingroup$
Stella (+1) for the answer. I reject the so-called proof of Euler's formula inasmuch as it uses a result from real analysis (i.e., $frac{de^x}{dx}=e^x$) and presumes that it holds for the yet-to-be-defined object $e^{ix}$.
$endgroup$
– Mark Viola
Mar 11 '17 at 15:36
$begingroup$
@Dr.MV Agreed. My preferred way is to define $e^{ix}$ by the Taylor polynomial and then use complex analysis to prove that the power series preserves the usual properties of the real analysis function $e^x$. Euler's Equation is then simply obtained from comparing power series.
$endgroup$
– Stella Biderman
Mar 11 '17 at 18:07
$begingroup$
Yes, that's a good standard approach.
$endgroup$
– Mark Viola
Mar 11 '17 at 18:12
1
1
$begingroup$
Stella (+1) for the answer. I reject the so-called proof of Euler's formula inasmuch as it uses a result from real analysis (i.e., $frac{de^x}{dx}=e^x$) and presumes that it holds for the yet-to-be-defined object $e^{ix}$.
$endgroup$
– Mark Viola
Mar 11 '17 at 15:36
$begingroup$
Stella (+1) for the answer. I reject the so-called proof of Euler's formula inasmuch as it uses a result from real analysis (i.e., $frac{de^x}{dx}=e^x$) and presumes that it holds for the yet-to-be-defined object $e^{ix}$.
$endgroup$
– Mark Viola
Mar 11 '17 at 15:36
$begingroup$
@Dr.MV Agreed. My preferred way is to define $e^{ix}$ by the Taylor polynomial and then use complex analysis to prove that the power series preserves the usual properties of the real analysis function $e^x$. Euler's Equation is then simply obtained from comparing power series.
$endgroup$
– Stella Biderman
Mar 11 '17 at 18:07
$begingroup$
@Dr.MV Agreed. My preferred way is to define $e^{ix}$ by the Taylor polynomial and then use complex analysis to prove that the power series preserves the usual properties of the real analysis function $e^x$. Euler's Equation is then simply obtained from comparing power series.
$endgroup$
– Stella Biderman
Mar 11 '17 at 18:07
$begingroup$
Yes, that's a good standard approach.
$endgroup$
– Mark Viola
Mar 11 '17 at 18:12
$begingroup$
Yes, that's a good standard approach.
$endgroup$
– Mark Viola
Mar 11 '17 at 18:12
add a comment |
$begingroup$
Let z=cos(x)+isin(x)
When you 8derive it (in terms of x),
z'=-sin(x)+icos(x)
And since we are in the complex world, -1=i×i
=>z'=i×isin(x)+icos(x)=i(cos(x)+isin(x))=iz
This pretty much makes it a first order differential equation.
After dividing by z and multiplying by dx (z'=dz/dx) both on both sides and integrating both sides, you get:
ln(z)=ix+c
ln(cos(x)+isin(x))=ix+c
If we set x=0, cos(0)=1,sin(0)=0 and ln(1)=0
=> ln(1)=c=>c=0
=> ln(cos(x)+isin(x))=ix
e^ix=cos(x)+isin(x).
Hope this helps.
answered Jan 16 at 5:07
Serouj GhazarianSerouj Ghazarian
11
$endgroup$
$begingroup$
Hi and welcome to the Math.SE. Producing beautiful and beautifully formatted text is one of the points that makes an answer good: if you want to learn how to do this, please have a look at MathJax basic tutorial and quick reference.
$endgroup$
– Daniele Tampieri
Jan 16 at 5:37
add a comment |
$begingroup$
Let z=cos(x)+isin(x)
When you 8derive it (in terms of x),
z'=-sin(x)+icos(x)
And since we are in the complex world, -1=i×i
=>z'=i×isin(x)+icos(x)=i(cos(x)+isin(x))=iz
This pretty much makes it a first order differential equation.
After dividing by z and multiplying by dx (z'=dz/dx) both on both sides and integrating both sides, you get:
ln(z)=ix+c
ln(cos(x)+isin(x))=ix+c
If we set x=0, cos(0)=1,sin(0)=0 and ln(1)=0
=> ln(1)=c=>c=0
=> ln(cos(x)+isin(x))=ix
e^ix=cos(x)+isin(x).
Hope this helps.
answered Jan 16 at 5:07
Serouj GhazarianSerouj Ghazarian
11
$endgroup$
$begingroup$
Hi and welcome to the Math.SE. Producing beautiful and beautifully formatted text is one of the points that makes an answer good: if you want to learn how to do this, please have a look at MathJax basic tutorial and quick reference.
$endgroup$
– Daniele Tampieri
Jan 16 at 5:37
add a comment |
$begingroup$
Let z=cos(x)+isin(x)
When you 8derive it (in terms of x),
z'=-sin(x)+icos(x)
And since we are in the complex world, -1=i×i
=>z'=i×isin(x)+icos(x)=i(cos(x)+isin(x))=iz
This pretty much makes it a first order differential equation.
After dividing by z and multiplying by dx (z'=dz/dx) both on both sides and integrating both sides, you get:
ln(z)=ix+c
ln(cos(x)+isin(x))=ix+c
If we set x=0, cos(0)=1,sin(0)=0 and ln(1)=0
=> ln(1)=c=>c=0
=> ln(cos(x)+isin(x))=ix
e^ix=cos(x)+isin(x).
Hope this helps.
answered Jan 16 at 5:07
Serouj GhazarianSerouj Ghazarian
11
$endgroup$
Let z=cos(x)+isin(x)
When you 8derive it (in terms of x),
z'=-sin(x)+icos(x)
And since we are in the complex world, -1=i×i
=>z'=i×isin(x)+icos(x)=i(cos(x)+isin(x))=iz
This pretty much makes it a first order differential equation.
After dividing by z and multiplying by dx (z'=dz/dx) both on both sides and integrating both sides, you get:
ln(z)=ix+c
ln(cos(x)+isin(x))=ix+c
If we set x=0, cos(0)=1,sin(0)=0 and ln(1)=0
=> ln(1)=c=>c=0
=> ln(cos(x)+isin(x))=ix
e^ix=cos(x)+isin(x).
Hope this helps.
answered Jan 16 at 5:07
Serouj GhazarianSerouj Ghazarian
11
edited Jan 16 at 5:20
answered Jan 16 at 5:07
Serouj GhazarianSerouj Ghazarian
11
answered Jan 16 at 5:07
Serouj GhazarianSerouj Ghazarian
11
answered Jan 16 at 5:07
Serouj GhazarianSerouj Ghazarian
11
11
$begingroup$
Hi and welcome to the Math.SE. Producing beautiful and beautifully formatted text is one of the points that makes an answer good: if you want to learn how to do this, please have a look at MathJax basic tutorial and quick reference.
$endgroup$
– Daniele Tampieri
Jan 16 at 5:37
add a comment |
$begingroup$
Hi and welcome to the Math.SE. Producing beautiful and beautifully formatted text is one of the points that makes an answer good: if you want to learn how to do this, please have a look at MathJax basic tutorial and quick reference.
$endgroup$
– Daniele Tampieri
Jan 16 at 5:37
$begingroup$
Hi and welcome to the Math.SE. Producing beautiful and beautifully formatted text is one of the points that makes an answer good: if you want to learn how to do this, please have a look at MathJax basic tutorial and quick reference.
$endgroup$
– Daniele Tampieri
Jan 16 at 5:37
$begingroup$
Hi and welcome to the Math.SE. Producing beautiful and beautifully formatted text is one of the points that makes an answer good: if you want to learn how to do this, please have a look at MathJax basic tutorial and quick reference.
$endgroup$
– Daniele Tampieri
Jan 16 at 5:37
add a comment |
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$begingroup$
Else the equality would be false at $x=0$, or you wouldn't have a holomorphic function to do complex analysis with, etc. and that all gets messy.
$endgroup$
– Simply Beautiful Art
Mar 11 '17 at 15:22
$begingroup$
Possibly related: How to prove Euler's formula: $e^{it}=cos t +isin t$?
$endgroup$
– Simply Beautiful Art
Mar 11 '17 at 15:23
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@SimplyBeautifulArt It is related but not the same as I already have a proof but want some to explain the end result of it
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– Dan
Mar 11 '17 at 15:27
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Yes, of course. Hence I did not have it as possible duplicate :-)
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– Simply Beautiful Art
Mar 11 '17 at 15:30