Holomorphic?
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From: I.N. Galidakis on 20 Jul 2010 13:32 David C. Ullrich wrote: > On Mon, 19 Jul 2010 19:57:08 +0300, "I.N. Galidakis" > <morpheus(a)olympus.mons> wrote: > >> David C. Ullrich wrote: >>> On Mon, 19 Jul 2010 08:11:36 +0300, "I.N. Galidakis" >>> <morpheus(a)olympus.mons> wrote: >>> >>>> I am checking the holomorphicity for the complex function >>>> f(r,phi)=x(r,phi)+y(r,phi)*i on the disk D={z:|z|<r+eps}. x and y are given >>>> as: >>>> >>>> x(r,phi)=r*sin(n*phi)*cos(m*phi) >>>> y(r,phi)=r*sin(n*phi)*sin(m*phi), >>>> >>>> m,n\in N. >>>> >>>> Checking the Polar form of the Cauchy-Riemann equations for these, I find: >>>> >>>> dx/dr=sin(n*phi)*cos(m*phi) >>>> >>>> while, >>>> >>>> 1/r*dy/dphi=cos(n*phi)*n*sin(m*phi)+sin(n*phi)*cos(m*phi)*m >>>> >>>> and these don't look equal. Does this mean that f(r,phi) is not holomorphic >>>> on the disk D? >>>> >>>> Am I missing something? >>> >>> Yes. You're using the "polar form of the C-R equations". That's >>> the form they take in standard polar coordinates. But your >>> r and phi are _not_ standard polar coordinates! >>> >>> If x = r cos(phi) and y = r sin(phi) then you should find that >>> x + iy _does_ satisfy the "polar form of the C-R equations". >> >> If I understand what you are saying, if I set: >> >> theta=m*phi >> R=r*sin(n/m*theta) >> >> then, under this transformation, >> >> x=R*cos(theta) >> y=R*sin(theta) >> >> I will find that the C-R equations (for R and theta) are satisfied? Right? > > No. If R=r*sin(n/m*theta) then R and theta are not independent. Let me try once more: set: x=r*cos(A) then, A=arccos(sin(n*phi)*cos(m*phi)) and then r*sin(A)=sin(arccos(sin(n*phi)*cos(m*phi))) =r*sqrt(1-sin(n*phi)^2*cos(m*phi)^2) =r*sqrt(1-(x/r)^2) =sqrt(r^2-x^2) =y and then, dx/dr=sin(n*phi)*cos(m*phi) and, 1/r*dy/dA=1/r*(dy/dphi)*(dphi/dA) =sin(n*phi)*cos(m*phi) and they agree. Hopefully that's what you meant? One question if I may: Can you enlighten us on how did you see that this curve was holomorphic to begin, with without doing the calculations? >> Thanks, -- I.
From: David C. Ullrich on 21 Jul 2010 09:28 On Tue, 20 Jul 2010 20:32:54 +0300, "I.N. Galidakis" <morpheus(a)olympus.mons> wrote: >David C. Ullrich wrote: >> On Mon, 19 Jul 2010 19:57:08 +0300, "I.N. Galidakis" >> <morpheus(a)olympus.mons> wrote: >> >>> David C. Ullrich wrote: >>>> On Mon, 19 Jul 2010 08:11:36 +0300, "I.N. Galidakis" >>>> <morpheus(a)olympus.mons> wrote: >>>> >>>>> I am checking the holomorphicity for the complex function >>>>> f(r,phi)=x(r,phi)+y(r,phi)*i on the disk D={z:|z|<r+eps}. x and y are given >>>>> as: >>>>> >>>>> x(r,phi)=r*sin(n*phi)*cos(m*phi) >>>>> y(r,phi)=r*sin(n*phi)*sin(m*phi), >>>>> >>>>> m,n\in N. >>>>> >>>>> Checking the Polar form of the Cauchy-Riemann equations for these, I find: >>>>> >>>>> dx/dr=sin(n*phi)*cos(m*phi) >>>>> >>>>> while, >>>>> >>>>> 1/r*dy/dphi=cos(n*phi)*n*sin(m*phi)+sin(n*phi)*cos(m*phi)*m >>>>> >>>>> and these don't look equal. Does this mean that f(r,phi) is not holomorphic >>>>> on the disk D? >>>>> >>>>> Am I missing something? >>>> >>>> Yes. You're using the "polar form of the C-R equations". That's >>>> the form they take in standard polar coordinates. But your >>>> r and phi are _not_ standard polar coordinates! >>>> >>>> If x = r cos(phi) and y = r sin(phi) then you should find that >>>> x + iy _does_ satisfy the "polar form of the C-R equations". >>> >>> If I understand what you are saying, if I set: >>> >>> theta=m*phi >>> R=r*sin(n/m*theta) >>> >>> then, under this transformation, >>> >>> x=R*cos(theta) >>> y=R*sin(theta) >>> >>> I will find that the C-R equations (for R and theta) are satisfied? Right? >> >> No. If R=r*sin(n/m*theta) then R and theta are not independent. > >Let me try once more: > >set: > >x=r*cos(A) > >then, > >A=arccos(sin(n*phi)*cos(m*phi)) I'm already totally lost. Why does x=r*cos(A) imply A=arccos(sin(n*phi)*cos(m*phi)) ? (It can't, because you haven't said anything about what phi is.) >and then > >r*sin(A)=sin(arccos(sin(n*phi)*cos(m*phi))) >=r*sqrt(1-sin(n*phi)^2*cos(m*phi)^2) >=r*sqrt(1-(x/r)^2) >=sqrt(r^2-x^2) >=y > >and then, > >dx/dr=sin(n*phi)*cos(m*phi) > >and, > >1/r*dy/dA=1/r*(dy/dphi)*(dphi/dA) >=sin(n*phi)*cos(m*phi) > >and they agree. > >Hopefully that's what you meant? What I meant was exactly what I said, no more and no less. If x = r cos(phi) and y = r sin(phi) then you should find that x + iy _does_ satisfy the "polar form of the C-R equations". >One question if I may: Can you enlighten us on how did you see that this curve >was holomorphic to begin, with without doing the calculations? x + iy is what's typiically known as "z", and dz/dz exists. >>> Thanks,
From: I.N. Galidakis on 22 Jul 2010 21:57 David C. Ullrich wrote: [snip for brevity] > What I meant was exactly what I said, no more and no less. > > If x = r cos(phi) and y = r sin(phi) then you should find that > x + iy _does_ satisfy the "polar form of the C-R equations". Yes, that's what you said, except that in this case I don't see how it applies, since x and y are _not_ given by these equations. x and y are given by the equations I give in my first post (also below). The curve in "standard" Polar coordinates is descibed by: r=cos(k*phi), with k=m/n\in Q and I cannot see what your "if x = r cos(phi)..." has to do with the above. >> One question if I may: Can you enlighten us on how did you see that this >> curve was holomorphic to begin, with without doing the calculations? > > x + iy is what's typiically known as "z", and dz/dz exists. I assume you mean x+yi, with: x = r*cos(n*phi)*sin(m*phi) and y = r*cos(n*phi)*cos(m*phi) Can you then show us what's "dz/dz"? Thanks, -- I.
From: I.N. Galidakis on 22 Jul 2010 22:05 I.N. Galidakis wrote: > David C. Ullrich wrote: [snip] >> x + iy is what's typiically known as "z", and dz/dz exists. > > I assume you mean x+yi, with: > > x = r*cos(n*phi)*sin(m*phi) > > and > > y = r*cos(n*phi)*cos(m*phi) > > Can you then show us what's "dz/dz"? Make the above: x = r*sin(n*phi)*cos(m*phi), and y = r*sin(n*phi)*cos(m*phi) I inadvertently changed sines and cosines, but in the end it doesn't matter. The changed x and y give a similar curve. > Thanks, -- I.
From: David C. Ullrich on 23 Jul 2010 04:30 On Fri, 23 Jul 2010 04:57:23 +0300, "I.N. Galidakis" <morpheus(a)olympus.mons> wrote: >David C. Ullrich wrote: >[snip for brevity] > >> What I meant was exactly what I said, no more and no less. >> >> If x = r cos(phi) and y = r sin(phi) then you should find that >> x + iy _does_ satisfy the "polar form of the C-R equations". > >Yes, that's what you said, except that in this case I don't see how it applies, >since x and y are _not_ given by these equations. x and y are given by the >equations I give in my first post (also below). The curve in "standard" Polar >coordinates is descibed by: > >r=cos(k*phi), with k=m/n\in Q > >and I cannot see what your "if x = r cos(phi)..." has to do with the above. You were applying the C-R equations "in polar form" in a situation where they're simply not applicable, since your coordinates are not polar coordinates. I was trying to clarify in which situation those equations _would_ imply the function was holomorphic. >>> One question if I may: Can you enlighten us on how did you see that this >>> curve was holomorphic to begin, with without doing the calculations? >> >> x + iy is what's typiically known as "z", and dz/dz exists. > >I assume you mean x+yi, with: > >x = r*cos(n*phi)*sin(m*phi) When I say "if x = r cos(phi)"if x = r cos(phi)" why would you assume that I meant x = r*cos(n*phi)*sin(m*phi)? Especially after I clarified that what I meant was what I wrote, not what you assume I meant? As of about the second line of your last post I said I didn't follow what you meant since you hadn't specified the relation between something and something else. Instead of explaining you "snipped for brevity". Good luck. >and > >y = r*cos(n*phi)*cos(m*phi) > >Can you then show us what's "dz/dz"? > >Thanks,
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