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Dominic
Dominic
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Might I ask an interested reader to take the initiative to write the code for the square root function:

Given $f(z,w)=w^2-z=0$, then we have:$\frac{dw}{dt}=-\frac{f_z}{f_w}\frac{dz}{dt}$. Now, letting $z(t)=e^{it}$, numerically solve two IVPs: $z_0=1,w_0=1$ and $z_1=1,w_1=-1$ for t ranging from $(0,2\pi)$ and now check $w(2\pi)$ for both of them. What will you expect to find? How about $w_0\to w_1$ and $w_1\to w_0$. What then does that tell you about the monodromy? And since there are no intervening singular points, this is the same monodromy around infinity. And then using Riemann-Hurwitz, successfully compute, numerically, the genus of the square root function by a very straight-forward means.

I should mention if we find after the first IVP that $w_0\to w_1$ then we need not compute the second IVP right? Also, interested readers may be interested in this blog: Algebraic Functions

Might I ask an interested reader to take the initiative to write the code for the square root function:

Given $f(z,w)=w^2-z=0$, then we have:$\frac{dw}{dt}=-\frac{f_z}{f_w}\frac{dz}{dt}$. Now, letting $z(t)=e^{it}$, numerically solve two IVPs: $z_0=1,w_0=1$ and $z_1=1,w_1=-1$ for t ranging from $(0,2\pi)$ and now check $w(2\pi)$ for both of them. What will you expect to find? How about $w_0\to w_1$ and $w_1\to w_0$. What then does that tell you about the monodromy? And since there are no intervening singular points, this is the same monodromy around infinity. And then using Riemann-Hurwitz, successfully compute, numerically, the genus of the square root function by a very straight-forward means.

I should mention if we find after the first IVP that $w_0\to w_1$ then we need not compute the second IVP right?

Might I ask an interested reader to take the initiative to write the code for the square root function:

Given $f(z,w)=w^2-z=0$, then we have:$\frac{dw}{dt}=-\frac{f_z}{f_w}\frac{dz}{dt}$. Now, letting $z(t)=e^{it}$, numerically solve two IVPs: $z_0=1,w_0=1$ and $z_1=1,w_1=-1$ for t ranging from $(0,2\pi)$ and now check $w(2\pi)$ for both of them. What will you expect to find? How about $w_0\to w_1$ and $w_1\to w_0$. What then does that tell you about the monodromy? And since there are no intervening singular points, this is the same monodromy around infinity. And then using Riemann-Hurwitz, successfully compute, numerically, the genus of the square root function by a very straight-forward means.

I should mention if we find after the first IVP that $w_0\to w_1$ then we need not compute the second IVP right? Also, interested readers may be interested in this blog: Algebraic Functions

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Source Link
Dominic
Dominic
  • 2.9k
  • 1
  • 10
  • 14

Might I ask an interested reader to take the initiative to write the code for the square root function:

Given $f(z,w)=w^2-z=0$, then we have:$\frac{dw}{dt}=-\frac{f_z}{f_w}\frac{dz}{dt}$. Now, letting $z(t)=e^{it}$, numerically solve two IVPs: $z_0=1,w_0=1$ and $z_1=1,w_1=-1$ for t ranging from $(0,2\pi)$ and now check $w(2\pi)$ for both of them. What will you expect to find? How about $w_0\to w_1$ and $w_1\to w_0$. What then does that tell you about the monodromy? And since there are no intervening singular points, this is the same monodromy around infinity. And then using Riemann-Hurwitz, successfully compute, numerically, the genus of the square root function by a very straight-forward means.

I should mention if we find after the first IVP that $w_0\to w_1$ then we need not compute the second IVP right?

Might I ask an interested reader to take the initiative to write the code for the square root function:

Given $f(z,w)=w^2-z=0$, then we have:$\frac{dw}{dt}=-\frac{f_z}{f_w}\frac{dz}{dt}$. Now, letting $z(t)=e^{it}$, numerically solve two IVPs: $z_0=1,w_0=1$ and $z_1=1,w_1=-1$ for t ranging from $(0,2\pi)$ and now check $w(2\pi)$ for both of them. What will you expect to find? How about $w_0\to w_1$ and $w_1\to w_0$. What then does that tell you about the monodromy? And since there are no intervening singular points, this is the same monodromy around infinity. And then using Riemann-Hurwitz, successfully compute, numerically, the genus of the square root function by a very straight-forward means.

Might I ask an interested reader to take the initiative to write the code for the square root function:

Given $f(z,w)=w^2-z=0$, then we have:$\frac{dw}{dt}=-\frac{f_z}{f_w}\frac{dz}{dt}$. Now, letting $z(t)=e^{it}$, numerically solve two IVPs: $z_0=1,w_0=1$ and $z_1=1,w_1=-1$ for t ranging from $(0,2\pi)$ and now check $w(2\pi)$ for both of them. What will you expect to find? How about $w_0\to w_1$ and $w_1\to w_0$. What then does that tell you about the monodromy? And since there are no intervening singular points, this is the same monodromy around infinity. And then using Riemann-Hurwitz, successfully compute, numerically, the genus of the square root function by a very straight-forward means.

I should mention if we find after the first IVP that $w_0\to w_1$ then we need not compute the second IVP right?

Source Link
Dominic
Dominic
  • 2.9k
  • 1
  • 10
  • 14

Might I ask an interested reader to take the initiative to write the code for the square root function:

Given $f(z,w)=w^2-z=0$, then we have:$\frac{dw}{dt}=-\frac{f_z}{f_w}\frac{dz}{dt}$. Now, letting $z(t)=e^{it}$, numerically solve two IVPs: $z_0=1,w_0=1$ and $z_1=1,w_1=-1$ for t ranging from $(0,2\pi)$ and now check $w(2\pi)$ for both of them. What will you expect to find? How about $w_0\to w_1$ and $w_1\to w_0$. What then does that tell you about the monodromy? And since there are no intervening singular points, this is the same monodromy around infinity. And then using Riemann-Hurwitz, successfully compute, numerically, the genus of the square root function by a very straight-forward means.