Mixing
What is the definition of $e^{ix}$?
$begingroup$
This might seem silly, but in proving Euler Formula taught in Calculus classes, we make the assumption that
$frac{d}{dx}e^{ix} = ie^{ix}$
However $e^{g}$ Pre-Euler’s Formula, only takes in real numbers for g. If we tried to use the Chain Rule where $g = ix$ we have no definition for that.
My question is, how do we know, Pre-Euler’s Formula, that $e^{z}$ exists for complex values of z, and how do we know that $e^{ix}$ is differentiable at $ix$. Basically what is the definition of $e^{ix}$ before knowing Euler's Formula.
calculus complex-analysis
edited Jan 26 at 5:49
Adam Hrankowski
2,094930
asked Jan 26 at 5:03
Black BlastBlack Blast
112
$endgroup$
add a comment |
$begingroup$
This might seem silly, but in proving Euler Formula taught in Calculus classes, we make the assumption that
$frac{d}{dx}e^{ix} = ie^{ix}$
However $e^{g}$ Pre-Euler’s Formula, only takes in real numbers for g. If we tried to use the Chain Rule where $g = ix$ we have no definition for that.
My question is, how do we know, Pre-Euler’s Formula, that $e^{z}$ exists for complex values of z, and how do we know that $e^{ix}$ is differentiable at $ix$. Basically what is the definition of $e^{ix}$ before knowing Euler's Formula.
calculus complex-analysis
edited Jan 26 at 5:49
Adam Hrankowski
2,094930
asked Jan 26 at 5:03
Black BlastBlack Blast
112
$endgroup$
$begingroup$
The function $f$ who differentiate only "takes" real numbers (it "spits" complex numbers out). $fcolon mathbb{R}tomathbb{C}$ is defined as $f(x) = e^{ix}$, so you differentiate it with respect to the real variable $x$.
$endgroup$
– Clement C.
Jan 26 at 5:07
$begingroup$
$frac {mathrm d}{mathrm dx} e^{kx}=ke^{kx}$ where $k$ is a constant.
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 5:07
$begingroup$
@ClementC. The OP is asking for a definition of $e^{z}$ for $zin mathbb{C}$.
$endgroup$
– Mark Viola
Jan 26 at 5:56
$begingroup$
@MarkViola I was reacting to "If we tried to use the Chain Rule [...]" and "differentiable at $ix$" (instead of "at $x$, which is what is used). To be frank, I don't understand the last part: surely, the textbook must have defined $e^{ix}$, otherwise, the whole premise of the question is strange (you don't prove things about quantities not even defined yet).
$endgroup$
– Clement C.
Jan 26 at 5:58
$begingroup$
@ClementC. Yes, I agree; the function must have been defined in the OP's text book.
$endgroup$
– Mark Viola
Jan 26 at 5:59
add a comment |
$begingroup$
This might seem silly, but in proving Euler Formula taught in Calculus classes, we make the assumption that
$frac{d}{dx}e^{ix} = ie^{ix}$
However $e^{g}$ Pre-Euler’s Formula, only takes in real numbers for g. If we tried to use the Chain Rule where $g = ix$ we have no definition for that.
My question is, how do we know, Pre-Euler’s Formula, that $e^{z}$ exists for complex values of z, and how do we know that $e^{ix}$ is differentiable at $ix$. Basically what is the definition of $e^{ix}$ before knowing Euler's Formula.
calculus complex-analysis
edited Jan 26 at 5:49
Adam Hrankowski
2,094930
asked Jan 26 at 5:03
Black BlastBlack Blast
112
$endgroup$
This might seem silly, but in proving Euler Formula taught in Calculus classes, we make the assumption that
$frac{d}{dx}e^{ix} = ie^{ix}$
However $e^{g}$ Pre-Euler’s Formula, only takes in real numbers for g. If we tried to use the Chain Rule where $g = ix$ we have no definition for that.
My question is, how do we know, Pre-Euler’s Formula, that $e^{z}$ exists for complex values of z, and how do we know that $e^{ix}$ is differentiable at $ix$. Basically what is the definition of $e^{ix}$ before knowing Euler's Formula.
calculus complex-analysis
calculus complex-analysis
edited Jan 26 at 5:49
Adam Hrankowski
2,094930
asked Jan 26 at 5:03
Black BlastBlack Blast
112
edited Jan 26 at 5:49
Adam Hrankowski
2,094930
asked Jan 26 at 5:03
Black BlastBlack Blast
112
edited Jan 26 at 5:49
Adam Hrankowski
2,094930
edited Jan 26 at 5:49
Adam Hrankowski
2,094930
edited Jan 26 at 5:49
Adam Hrankowski
2,094930
2,094930
asked Jan 26 at 5:03
Black BlastBlack Blast
112
asked Jan 26 at 5:03
Black BlastBlack Blast
112
asked Jan 26 at 5:03
Black BlastBlack Blast
112
112
$begingroup$
The function $f$ who differentiate only "takes" real numbers (it "spits" complex numbers out). $fcolon mathbb{R}tomathbb{C}$ is defined as $f(x) = e^{ix}$, so you differentiate it with respect to the real variable $x$.
$endgroup$
– Clement C.
Jan 26 at 5:07
$begingroup$
$frac {mathrm d}{mathrm dx} e^{kx}=ke^{kx}$ where $k$ is a constant.
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 5:07
$begingroup$
@ClementC. The OP is asking for a definition of $e^{z}$ for $zin mathbb{C}$.
$endgroup$
– Mark Viola
Jan 26 at 5:56
$begingroup$
@MarkViola I was reacting to "If we tried to use the Chain Rule [...]" and "differentiable at $ix$" (instead of "at $x$, which is what is used). To be frank, I don't understand the last part: surely, the textbook must have defined $e^{ix}$, otherwise, the whole premise of the question is strange (you don't prove things about quantities not even defined yet).
$endgroup$
– Clement C.
Jan 26 at 5:58
$begingroup$
@ClementC. Yes, I agree; the function must have been defined in the OP's text book.
$endgroup$
– Mark Viola
Jan 26 at 5:59
add a comment |
$begingroup$
The function $f$ who differentiate only "takes" real numbers (it "spits" complex numbers out). $fcolon mathbb{R}tomathbb{C}$ is defined as $f(x) = e^{ix}$, so you differentiate it with respect to the real variable $x$.
$endgroup$
– Clement C.
Jan 26 at 5:07
$begingroup$
$frac {mathrm d}{mathrm dx} e^{kx}=ke^{kx}$ where $k$ is a constant.
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 5:07
$begingroup$
@ClementC. The OP is asking for a definition of $e^{z}$ for $zin mathbb{C}$.
$endgroup$
– Mark Viola
Jan 26 at 5:56
$begingroup$
@MarkViola I was reacting to "If we tried to use the Chain Rule [...]" and "differentiable at $ix$" (instead of "at $x$, which is what is used). To be frank, I don't understand the last part: surely, the textbook must have defined $e^{ix}$, otherwise, the whole premise of the question is strange (you don't prove things about quantities not even defined yet).
$endgroup$
– Clement C.
Jan 26 at 5:58
$begingroup$
@ClementC. Yes, I agree; the function must have been defined in the OP's text book.
$endgroup$
– Mark Viola
Jan 26 at 5:59
$begingroup$
The function $f$ who differentiate only "takes" real numbers (it "spits" complex numbers out). $fcolon mathbb{R}tomathbb{C}$ is defined as $f(x) = e^{ix}$, so you differentiate it with respect to the real variable $x$.
$endgroup$
– Clement C.
Jan 26 at 5:07
$begingroup$
The function $f$ who differentiate only "takes" real numbers (it "spits" complex numbers out). $fcolon mathbb{R}tomathbb{C}$ is defined as $f(x) = e^{ix}$, so you differentiate it with respect to the real variable $x$.
$endgroup$
– Clement C.
Jan 26 at 5:07
$begingroup$
$frac {mathrm d}{mathrm dx} e^{kx}=ke^{kx}$ where $k$ is a constant.
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 5:07
$begingroup$
$frac {mathrm d}{mathrm dx} e^{kx}=ke^{kx}$ where $k$ is a constant.
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 5:07
$begingroup$
@ClementC. The OP is asking for a definition of $e^{z}$ for $zin mathbb{C}$.
$endgroup$
– Mark Viola
Jan 26 at 5:56
$begingroup$
@ClementC. The OP is asking for a definition of $e^{z}$ for $zin mathbb{C}$.
$endgroup$
– Mark Viola
Jan 26 at 5:56
$begingroup$
@MarkViola I was reacting to "If we tried to use the Chain Rule [...]" and "differentiable at $ix$" (instead of "at $x$, which is what is used). To be frank, I don't understand the last part: surely, the textbook must have defined $e^{ix}$, otherwise, the whole premise of the question is strange (you don't prove things about quantities not even defined yet).
$endgroup$
– Clement C.
Jan 26 at 5:58
$begingroup$
@MarkViola I was reacting to "If we tried to use the Chain Rule [...]" and "differentiable at $ix$" (instead of "at $x$, which is what is used). To be frank, I don't understand the last part: surely, the textbook must have defined $e^{ix}$, otherwise, the whole premise of the question is strange (you don't prove things about quantities not even defined yet).
$endgroup$
– Clement C.
Jan 26 at 5:58
$begingroup$
@ClementC. Yes, I agree; the function must have been defined in the OP's text book.
$endgroup$
– Mark Viola
Jan 26 at 5:59
$begingroup$
@ClementC. Yes, I agree; the function must have been defined in the OP's text book.
$endgroup$
– Mark Viola
Jan 26 at 5:59
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A natural way to define $e^z$ for any complex $z$ is by the Taylor expansion $$e^z = sum_{n=0}^{infty}frac{z^n}{n!}$$
When $z$ is purely imaginary then this yields Euler's Formula as
$$e^{ix} = sum_{n=0}^{infty}frac{(ix)^n}{n!}=
sum_{n=0}^{infty}frac{(-1)^nx^{2n}}{(2n)!}+isum_{n=0}^{infty}frac{(-1)^nx^{2n+1}}{(2n+1)!}
=cos(x) + isin(x)$$
answered Jan 26 at 5:13
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
1
$begingroup$
Another way to define $e^z$ for $zin mathbb{C}$ is given by $$e^z=lim_{ntoinfty}left(1+frac znright)^n$$ just as in the limit definition of $e^x$.
$endgroup$
– Mark Viola
Jan 26 at 5:58
add a comment |
$begingroup$
When I was teaching, I always used the series definition, as explained in @ErikParkinson’s answer. But you may also define
$$
e^z=lim_{ntoinfty}left(1+frac znright)^n,.
$$
When you look at this closely, the formula $e^{it}=cos t+isin t$ becomes very reasonable.
answered Jan 26 at 6:08
LubinLubin
45.3k44688
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A natural way to define $e^z$ for any complex $z$ is by the Taylor expansion $$e^z = sum_{n=0}^{infty}frac{z^n}{n!}$$
When $z$ is purely imaginary then this yields Euler's Formula as
$$e^{ix} = sum_{n=0}^{infty}frac{(ix)^n}{n!}=
sum_{n=0}^{infty}frac{(-1)^nx^{2n}}{(2n)!}+isum_{n=0}^{infty}frac{(-1)^nx^{2n+1}}{(2n+1)!}
=cos(x) + isin(x)$$
answered Jan 26 at 5:13
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
1
$begingroup$
Another way to define $e^z$ for $zin mathbb{C}$ is given by $$e^z=lim_{ntoinfty}left(1+frac znright)^n$$ just as in the limit definition of $e^x$.
$endgroup$
– Mark Viola
Jan 26 at 5:58
add a comment |
$begingroup$
A natural way to define $e^z$ for any complex $z$ is by the Taylor expansion $$e^z = sum_{n=0}^{infty}frac{z^n}{n!}$$
When $z$ is purely imaginary then this yields Euler's Formula as
$$e^{ix} = sum_{n=0}^{infty}frac{(ix)^n}{n!}=
sum_{n=0}^{infty}frac{(-1)^nx^{2n}}{(2n)!}+isum_{n=0}^{infty}frac{(-1)^nx^{2n+1}}{(2n+1)!}
=cos(x) + isin(x)$$
answered Jan 26 at 5:13
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
1
$begingroup$
Another way to define $e^z$ for $zin mathbb{C}$ is given by $$e^z=lim_{ntoinfty}left(1+frac znright)^n$$ just as in the limit definition of $e^x$.
$endgroup$
– Mark Viola
Jan 26 at 5:58
add a comment |
$begingroup$
A natural way to define $e^z$ for any complex $z$ is by the Taylor expansion $$e^z = sum_{n=0}^{infty}frac{z^n}{n!}$$
When $z$ is purely imaginary then this yields Euler's Formula as
$$e^{ix} = sum_{n=0}^{infty}frac{(ix)^n}{n!}=
sum_{n=0}^{infty}frac{(-1)^nx^{2n}}{(2n)!}+isum_{n=0}^{infty}frac{(-1)^nx^{2n+1}}{(2n+1)!}
=cos(x) + isin(x)$$
answered Jan 26 at 5:13
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
A natural way to define $e^z$ for any complex $z$ is by the Taylor expansion $$e^z = sum_{n=0}^{infty}frac{z^n}{n!}$$
When $z$ is purely imaginary then this yields Euler's Formula as
$$e^{ix} = sum_{n=0}^{infty}frac{(ix)^n}{n!}=
sum_{n=0}^{infty}frac{(-1)^nx^{2n}}{(2n)!}+isum_{n=0}^{infty}frac{(-1)^nx^{2n+1}}{(2n+1)!}
=cos(x) + isin(x)$$
answered Jan 26 at 5:13
Erik ParkinsonErik Parkinson
1,17519
answered Jan 26 at 5:13
Erik ParkinsonErik Parkinson
1,17519
answered Jan 26 at 5:13
Erik ParkinsonErik Parkinson
1,17519
answered Jan 26 at 5:13
Erik ParkinsonErik Parkinson
1,17519
1,17519
1
$begingroup$
Another way to define $e^z$ for $zin mathbb{C}$ is given by $$e^z=lim_{ntoinfty}left(1+frac znright)^n$$ just as in the limit definition of $e^x$.
$endgroup$
– Mark Viola
Jan 26 at 5:58
add a comment |
1
$begingroup$
Another way to define $e^z$ for $zin mathbb{C}$ is given by $$e^z=lim_{ntoinfty}left(1+frac znright)^n$$ just as in the limit definition of $e^x$.
$endgroup$
– Mark Viola
Jan 26 at 5:58
1
1
$begingroup$
Another way to define $e^z$ for $zin mathbb{C}$ is given by $$e^z=lim_{ntoinfty}left(1+frac znright)^n$$ just as in the limit definition of $e^x$.
$endgroup$
– Mark Viola
Jan 26 at 5:58
$begingroup$
Another way to define $e^z$ for $zin mathbb{C}$ is given by $$e^z=lim_{ntoinfty}left(1+frac znright)^n$$ just as in the limit definition of $e^x$.
$endgroup$
– Mark Viola
Jan 26 at 5:58
add a comment |
$begingroup$
When I was teaching, I always used the series definition, as explained in @ErikParkinson’s answer. But you may also define
$$
e^z=lim_{ntoinfty}left(1+frac znright)^n,.
$$
When you look at this closely, the formula $e^{it}=cos t+isin t$ becomes very reasonable.
answered Jan 26 at 6:08
LubinLubin
45.3k44688
$endgroup$
add a comment |
$begingroup$
When I was teaching, I always used the series definition, as explained in @ErikParkinson’s answer. But you may also define
$$
e^z=lim_{ntoinfty}left(1+frac znright)^n,.
$$
When you look at this closely, the formula $e^{it}=cos t+isin t$ becomes very reasonable.
answered Jan 26 at 6:08
LubinLubin
45.3k44688
$endgroup$
add a comment |
$begingroup$
When I was teaching, I always used the series definition, as explained in @ErikParkinson’s answer. But you may also define
$$
e^z=lim_{ntoinfty}left(1+frac znright)^n,.
$$
When you look at this closely, the formula $e^{it}=cos t+isin t$ becomes very reasonable.
answered Jan 26 at 6:08
LubinLubin
45.3k44688
$endgroup$
When I was teaching, I always used the series definition, as explained in @ErikParkinson’s answer. But you may also define
$$
e^z=lim_{ntoinfty}left(1+frac znright)^n,.
$$
When you look at this closely, the formula $e^{it}=cos t+isin t$ becomes very reasonable.
answered Jan 26 at 6:08
LubinLubin
45.3k44688
answered Jan 26 at 6:08
LubinLubin
45.3k44688
answered Jan 26 at 6:08
LubinLubin
45.3k44688
answered Jan 26 at 6:08
LubinLubin
45.3k44688
45.3k44688
add a comment |
add a comment |
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$begingroup$
The function $f$ who differentiate only "takes" real numbers (it "spits" complex numbers out). $fcolon mathbb{R}tomathbb{C}$ is defined as $f(x) = e^{ix}$, so you differentiate it with respect to the real variable $x$.
$endgroup$
– Clement C.
Jan 26 at 5:07
$begingroup$
$frac {mathrm d}{mathrm dx} e^{kx}=ke^{kx}$ where $k$ is a constant.
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 5:07
$begingroup$
@ClementC. The OP is asking for a definition of $e^{z}$ for $zin mathbb{C}$.
$endgroup$
– Mark Viola
Jan 26 at 5:56
$begingroup$
@MarkViola I was reacting to "If we tried to use the Chain Rule [...]" and "differentiable at $ix$" (instead of "at $x$, which is what is used). To be frank, I don't understand the last part: surely, the textbook must have defined $e^{ix}$, otherwise, the whole premise of the question is strange (you don't prove things about quantities not even defined yet).
$endgroup$
– Clement C.
Jan 26 at 5:58
$begingroup$
@ClementC. Yes, I agree; the function must have been defined in the OP's text book.
$endgroup$
– Mark Viola
Jan 26 at 5:59