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What's the power series of the principal $Ln(z)$?
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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka 7.28(b)
Find a power series for $Ln(z)$ centred at $z_0=1$. (7.28(a) was for $frac1z$ centred at $z_0=1$.)
For $ln(x)$,
$ $ I integrate the power series of a rational function $f$ over some interval $gamma_x$ whose left point is some $x_0 in mathbb R$ and whose right end point is $x$. I could directly the power series for $f(t) = frac 1 t$ over $gamma_x = [1,x], x_0=1$. I could also integrate the power series for $f(t) = frac11+t$ over $[0,x],x_0=0$ and then make a change of variables to get from $ln(1+x)$ to $ln(x)$. These $gamma_x$'s and $x_0$'s ensure that I get $ln(x)$ and not, say, $ln(x)+7$.
For $Ln(z)$,
$ $ the antiderivatives of $frac 1 z$:
- include the principal $Ln$ branch s.t. $Ln(z) = ln|z| + iArg(z)$
- include any other $mathscr Ln$ branch s.t. $mathscr Ln(z) = ln|z| + imathscr Arg(z)$
- include $ln|z|$
- include $int_gamma_zfracdww forall gamma_z subset G$ piecewise smooth paths from $z_0 in G$ to $z$ by Thm 4.15 (refend)
-
(Q1): If $exists gamma_z, z_0$ and $G$ to ensure I have a power series for the principal $Ln$, then what are they?
(Q2): If not, then how else do I get a power series for the principal $Ln$ with the power series of $frac 1 z$? (Or $frac11-z$, $frac11+z$, etc depending on your preference)
(Q3): Is this even possible to do w/o Taylor or Laurent (discussed starting Ch8)?
-
$labelendtagend$
calculus integration complex-analysis convergence power-series
asked Aug 2 at 11:39
BCLC
6,98221973
This question has an open bounty worth +50
reputation from BCLC ending ending at 2018-08-12 13:17:56Z">in 4 days.
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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka 7.28(b)
Find a power series for $Ln(z)$ centred at $z_0=1$. (7.28(a) was for $frac1z$ centred at $z_0=1$.)
For $ln(x)$,
$ $ I integrate the power series of a rational function $f$ over some interval $gamma_x$ whose left point is some $x_0 in mathbb R$ and whose right end point is $x$. I could directly the power series for $f(t) = frac 1 t$ over $gamma_x = [1,x], x_0=1$. I could also integrate the power series for $f(t) = frac11+t$ over $[0,x],x_0=0$ and then make a change of variables to get from $ln(1+x)$ to $ln(x)$. These $gamma_x$'s and $x_0$'s ensure that I get $ln(x)$ and not, say, $ln(x)+7$.
For $Ln(z)$,
$ $ the antiderivatives of $frac 1 z$:
- include the principal $Ln$ branch s.t. $Ln(z) = ln|z| + iArg(z)$
- include any other $mathscr Ln$ branch s.t. $mathscr Ln(z) = ln|z| + imathscr Arg(z)$
- include $ln|z|$
- include $int_gamma_zfracdww forall gamma_z subset G$ piecewise smooth paths from $z_0 in G$ to $z$ by Thm 4.15 (refend)
-
(Q1): If $exists gamma_z, z_0$ and $G$ to ensure I have a power series for the principal $Ln$, then what are they?
(Q2): If not, then how else do I get a power series for the principal $Ln$ with the power series of $frac 1 z$? (Or $frac11-z$, $frac11+z$, etc depending on your preference)
(Q3): Is this even possible to do w/o Taylor or Laurent (discussed starting Ch8)?
-
$labelendtagend$
calculus integration complex-analysis convergence power-series
asked Aug 2 at 11:39
BCLC
6,98221973
This question has an open bounty worth +50
reputation from BCLC ending ending at 2018-08-12 13:17:56Z">in 4 days.
This question has not received enough attention.
2
The origin is a branch point, so there are no power series centered at the origin. Laurent or Taylor. This is because a converging power series is holomorphic in an annulus, but the existence of a branch cut means that $log$ is not. The usual thing is to get one for $log(1+z)$ by the process you described: integerate the series of $1/(1+z)$. You can, of course, center a series anywhere else. A Tayloir series centered at $z_0$ will then have radius of convergence $R=|z_0|$ when you arrange the branch cut not to disturb that.
– Jyrki Lahtonen
2 days ago
2
It is unclear what you question is. Please explain clearly what you are trying to answer and what you have tried so far.
– Somos
2 days ago
@Somos I put my questions at the end...?
– BCLC
2 days ago
@JyrkiLahtonen Thanks! I checked again, and it was power series around $z_0=1$. For $mathbb R$, I know how to do it. The thing is, how do I know that I'm getting the same answer in $mathbb C$ without Taylor or Laurent series? There's a possibility this could be an error in the textbook, which introduces Taylor and Laurent in the next chapter. But if it's possible to do without Taylor or Laurent, then how do I know integrating and making change of variables will give me a power series for $Ln(z)$ and not any of the other antiderivatives? I mean, what's your $gamma_z$ and $z_0$?
– BCLC
2 days ago
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up vote
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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka 7.28(b)
Find a power series for $Ln(z)$ centred at $z_0=1$. (7.28(a) was for $frac1z$ centred at $z_0=1$.)
For $ln(x)$,
$ $ I integrate the power series of a rational function $f$ over some interval $gamma_x$ whose left point is some $x_0 in mathbb R$ and whose right end point is $x$. I could directly the power series for $f(t) = frac 1 t$ over $gamma_x = [1,x], x_0=1$. I could also integrate the power series for $f(t) = frac11+t$ over $[0,x],x_0=0$ and then make a change of variables to get from $ln(1+x)$ to $ln(x)$. These $gamma_x$'s and $x_0$'s ensure that I get $ln(x)$ and not, say, $ln(x)+7$.
For $Ln(z)$,
$ $ the antiderivatives of $frac 1 z$:
- include the principal $Ln$ branch s.t. $Ln(z) = ln|z| + iArg(z)$
- include any other $mathscr Ln$ branch s.t. $mathscr Ln(z) = ln|z| + imathscr Arg(z)$
- include $ln|z|$
- include $int_gamma_zfracdww forall gamma_z subset G$ piecewise smooth paths from $z_0 in G$ to $z$ by Thm 4.15 (refend)
-
(Q1): If $exists gamma_z, z_0$ and $G$ to ensure I have a power series for the principal $Ln$, then what are they?
(Q2): If not, then how else do I get a power series for the principal $Ln$ with the power series of $frac 1 z$? (Or $frac11-z$, $frac11+z$, etc depending on your preference)
(Q3): Is this even possible to do w/o Taylor or Laurent (discussed starting Ch8)?
-
$labelendtagend$
calculus integration complex-analysis convergence power-series
asked Aug 2 at 11:39
BCLC
6,98221973
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka 7.28(b)
Find a power series for $Ln(z)$ centred at $z_0=1$. (7.28(a) was for $frac1z$ centred at $z_0=1$.)
For $ln(x)$,
$ $ I integrate the power series of a rational function $f$ over some interval $gamma_x$ whose left point is some $x_0 in mathbb R$ and whose right end point is $x$. I could directly the power series for $f(t) = frac 1 t$ over $gamma_x = [1,x], x_0=1$. I could also integrate the power series for $f(t) = frac11+t$ over $[0,x],x_0=0$ and then make a change of variables to get from $ln(1+x)$ to $ln(x)$. These $gamma_x$'s and $x_0$'s ensure that I get $ln(x)$ and not, say, $ln(x)+7$.
For $Ln(z)$,
$ $ the antiderivatives of $frac 1 z$:
- include the principal $Ln$ branch s.t. $Ln(z) = ln|z| + iArg(z)$
- include any other $mathscr Ln$ branch s.t. $mathscr Ln(z) = ln|z| + imathscr Arg(z)$
- include $ln|z|$
- include $int_gamma_zfracdww forall gamma_z subset G$ piecewise smooth paths from $z_0 in G$ to $z$ by Thm 4.15 (refend)
-
(Q1): If $exists gamma_z, z_0$ and $G$ to ensure I have a power series for the principal $Ln$, then what are they?
(Q2): If not, then how else do I get a power series for the principal $Ln$ with the power series of $frac 1 z$? (Or $frac11-z$, $frac11+z$, etc depending on your preference)
(Q3): Is this even possible to do w/o Taylor or Laurent (discussed starting Ch8)?
-
$labelendtagend$
calculus integration complex-analysis convergence power-series
asked Aug 2 at 11:39
BCLC
6,98221973
edited 22 hours ago
asked Aug 2 at 11:39
BCLC
6,98221973
asked Aug 2 at 11:39
BCLC
6,98221973
asked Aug 2 at 11:39
BCLC
6,98221973
6,98221973
This question has an open bounty worth +50
reputation from BCLC ending ending at 2018-08-12 13:17:56Z">in 4 days.
This question has not received enough attention.
This question has an open bounty worth +50
reputation from BCLC ending ending at 2018-08-12 13:17:56Z">in 4 days.
This question has not received enough attention.
2
The origin is a branch point, so there are no power series centered at the origin. Laurent or Taylor. This is because a converging power series is holomorphic in an annulus, but the existence of a branch cut means that $log$ is not. The usual thing is to get one for $log(1+z)$ by the process you described: integerate the series of $1/(1+z)$. You can, of course, center a series anywhere else. A Tayloir series centered at $z_0$ will then have radius of convergence $R=|z_0|$ when you arrange the branch cut not to disturb that.
– Jyrki Lahtonen
2 days ago
2
It is unclear what you question is. Please explain clearly what you are trying to answer and what you have tried so far.
– Somos
2 days ago
@Somos I put my questions at the end...?
– BCLC
2 days ago
@JyrkiLahtonen Thanks! I checked again, and it was power series around $z_0=1$. For $mathbb R$, I know how to do it. The thing is, how do I know that I'm getting the same answer in $mathbb C$ without Taylor or Laurent series? There's a possibility this could be an error in the textbook, which introduces Taylor and Laurent in the next chapter. But if it's possible to do without Taylor or Laurent, then how do I know integrating and making change of variables will give me a power series for $Ln(z)$ and not any of the other antiderivatives? I mean, what's your $gamma_z$ and $z_0$?
– BCLC
2 days ago
add a comment |Â
2
The origin is a branch point, so there are no power series centered at the origin. Laurent or Taylor. This is because a converging power series is holomorphic in an annulus, but the existence of a branch cut means that $log$ is not. The usual thing is to get one for $log(1+z)$ by the process you described: integerate the series of $1/(1+z)$. You can, of course, center a series anywhere else. A Tayloir series centered at $z_0$ will then have radius of convergence $R=|z_0|$ when you arrange the branch cut not to disturb that.
– Jyrki Lahtonen
2 days ago
2
It is unclear what you question is. Please explain clearly what you are trying to answer and what you have tried so far.
– Somos
2 days ago
@Somos I put my questions at the end...?
– BCLC
2 days ago
@JyrkiLahtonen Thanks! I checked again, and it was power series around $z_0=1$. For $mathbb R$, I know how to do it. The thing is, how do I know that I'm getting the same answer in $mathbb C$ without Taylor or Laurent series? There's a possibility this could be an error in the textbook, which introduces Taylor and Laurent in the next chapter. But if it's possible to do without Taylor or Laurent, then how do I know integrating and making change of variables will give me a power series for $Ln(z)$ and not any of the other antiderivatives? I mean, what's your $gamma_z$ and $z_0$?
– BCLC
2 days ago
2
2
The origin is a branch point, so there are no power series centered at the origin. Laurent or Taylor. This is because a converging power series is holomorphic in an annulus, but the existence of a branch cut means that $log$ is not. The usual thing is to get one for $log(1+z)$ by the process you described: integerate the series of $1/(1+z)$. You can, of course, center a series anywhere else. A Tayloir series centered at $z_0$ will then have radius of convergence $R=|z_0|$ when you arrange the branch cut not to disturb that.
– Jyrki Lahtonen
2 days ago
The origin is a branch point, so there are no power series centered at the origin. Laurent or Taylor. This is because a converging power series is holomorphic in an annulus, but the existence of a branch cut means that $log$ is not. The usual thing is to get one for $log(1+z)$ by the process you described: integerate the series of $1/(1+z)$. You can, of course, center a series anywhere else. A Tayloir series centered at $z_0$ will then have radius of convergence $R=|z_0|$ when you arrange the branch cut not to disturb that.
– Jyrki Lahtonen
2 days ago
2
2
It is unclear what you question is. Please explain clearly what you are trying to answer and what you have tried so far.
– Somos
2 days ago
It is unclear what you question is. Please explain clearly what you are trying to answer and what you have tried so far.
– Somos
2 days ago
@Somos I put my questions at the end...?
– BCLC
2 days ago
@Somos I put my questions at the end...?
– BCLC
2 days ago
@JyrkiLahtonen Thanks! I checked again, and it was power series around $z_0=1$. For $mathbb R$, I know how to do it. The thing is, how do I know that I'm getting the same answer in $mathbb C$ without Taylor or Laurent series? There's a possibility this could be an error in the textbook, which introduces Taylor and Laurent in the next chapter. But if it's possible to do without Taylor or Laurent, then how do I know integrating and making change of variables will give me a power series for $Ln(z)$ and not any of the other antiderivatives? I mean, what's your $gamma_z$ and $z_0$?
– BCLC
2 days ago
@JyrkiLahtonen Thanks! I checked again, and it was power series around $z_0=1$. For $mathbb R$, I know how to do it. The thing is, how do I know that I'm getting the same answer in $mathbb C$ without Taylor or Laurent series? There's a possibility this could be an error in the textbook, which introduces Taylor and Laurent in the next chapter. But if it's possible to do without Taylor or Laurent, then how do I know integrating and making change of variables will give me a power series for $Ln(z)$ and not any of the other antiderivatives? I mean, what's your $gamma_z$ and $z_0$?
– BCLC
2 days ago
add a comment |Â
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2
The origin is a branch point, so there are no power series centered at the origin. Laurent or Taylor. This is because a converging power series is holomorphic in an annulus, but the existence of a branch cut means that $log$ is not. The usual thing is to get one for $log(1+z)$ by the process you described: integerate the series of $1/(1+z)$. You can, of course, center a series anywhere else. A Tayloir series centered at $z_0$ will then have radius of convergence $R=|z_0|$ when you arrange the branch cut not to disturb that.
– Jyrki Lahtonen
2 days ago
2
It is unclear what you question is. Please explain clearly what you are trying to answer and what you have tried so far.
– Somos
2 days ago
@Somos I put my questions at the end...?
– BCLC
2 days ago
@JyrkiLahtonen Thanks! I checked again, and it was power series around $z_0=1$. For $mathbb R$, I know how to do it. The thing is, how do I know that I'm getting the same answer in $mathbb C$ without Taylor or Laurent series? There's a possibility this could be an error in the textbook, which introduces Taylor and Laurent in the next chapter. But if it's possible to do without Taylor or Laurent, then how do I know integrating and making change of variables will give me a power series for $Ln(z)$ and not any of the other antiderivatives? I mean, what's your $gamma_z$ and $z_0$?
– BCLC
2 days ago