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Tex problems

  1. Oct 11, 2007 #1
    First of all: I love this forum! I'm trying to write a blog to help me study for my mid-term.

    I'm having trouble posting tex to the blog... I'm going to try it here and see if it works... It keeps showing equations that I used in older posts...

    Okay here is the pasted text from my blog post:

    The function f(x) = u(x,y) + iv(x,y) is differentiable at a point z= x +iy of a region in the complex plane if and only if the partial derivatives [tex]U_{x}[/tex],[tex]U_{y}[/tex],[tex]V_{x}[/tex],[tex]V_{y}[/tex] are continuos and satisfy the Cauchy-Riemann conditions....

    ----

    okay it worked here... and now it is working at my blog... but I don't know why this keeps happening?
     
  2. jcsd
  3. Oct 11, 2007 #2
    The function f(x) = u(x,y) + iv(x,y) is differentiable at a point z= x +iy of a region in the complex plane if and only if the partial derivatives [tex]U_{x}[/tex],[tex]U_{y}[/tex],[tex]V_{x}[/tex],[tex]V_{y}[/tex] are continuos and satisfy the Cauchy-Riemann conditions.

    Just like in calculus the derivative of f(z) is given by the following limit:

    [tex]f'(z)=\mathop{\lim}\limits_{\Delta z \to 0}\frac{f(z+\Delta z)-f(z)}{\Delta z}[/tex]

    We write f(z) = u(x,y) + iv(x,y)

    And then compute the limit along the real and then the imaginary axis. To do this first we set [tex]\Delta z = \Delta x[/tex], then we set [tex]\Delta z = i \Delta y[/tex].

    With [tex]\Delta z = \Delta x[/tex] we have:

    [tex]f'(z)=\mathop{\lim}\limits_{\Delta x \to 0}\frac{u(x+\Delta x, y)-u(x,y)}{\Delta x} + i\frac{v(x+\Delta x, y)-v(x,y)}{\Delta x}[/tex]

    [tex]u_{x}(x,y) + iv_{x}(x,y)[/tex]

    and with [tex]\Delta z = i \Delta y[/tex] we have:

    [tex]f'(z)=\mathop{\lim}\limits_{\Delta y \to 0}\frac{u(x,y + \Delta y)-u(x,y)}{i \Delta y} + i\frac{v(x, y+ \Delta y)-v(x,y)}{i \Delta y}[/tex]

    This is the one to watch. Remember that [tex]\frac{1}{i}=-i[/tex]

    [tex]-iu_{y}(x,y) + v_{y}(x,y)[/tex]

    Now set the real and imaginary parts of the two limits equal to each other since they both define f'(z). From this we get the Cauchy-Riemann conditions.

    This shows that C-R is necessary, but now we much show that it is also sufficient: that is we must show that if the partials meet the C-R condition then f(z) is differentiable. Once we show this we will have proved the theorem.

    If [tex]U_{x}[/tex],[tex]U_{y}[/tex],[tex]V_{x}[/tex],[tex]V_{y}[/tex] are continuous at the point (x, y) then:

    [tex]\Delta u = u_{x} \Delta x + u_{y} \Delta y + \epsilon_{1}| \Delta z|[/tex]
    [tex]\Delta v = v_{x} \Delta x + v_{y} \Delta y + \epsilon_{2}| \Delta z|[/tex]

    Where [tex]| \Delta z|=sqrt{\Delta x^{2}+\Delta y^{2}}[/tex]
    [tex]\mathop{\lim}\limits_{\Delta z \to 0}\epsilon_{1} =\mathop{\lim}\limits_{\Delta z \to 0}\epsilon_{2}=0[/tex]

    and

    [tex]\Delta u = u(x+ \Delta x, y+ \Delta y)-u(x,y)[/tex]
    [tex]\Delta v = v(x+ \Delta x, y+ \Delta y)-v(x,y)[/tex]

    Calling [tex]\Delta f = \Delta u + i \Delta v[/tex], we have

    [tex]\frac{\Delta f}{\Delta z}=\frac{\Delta u}{\Delta z}+i\frac{\Delta v}{\Delta z}[/tex]
    [tex]=u_{x}\frac{\Delta x}{\Delta z}+u_{y}\frac{\Delta y}{\Delta z}[/tex]
    [tex] + iv_{x}\frac{\Delta x}{\Delta z}+iv_{y}\frac{\Delta y}{\Delta z}+ (\epsilon_{1} +i\epsilon_{2})\frac{|\Delta z|}{\Delta z}[/tex]
     
    Last edited: Oct 11, 2007
  4. Oct 11, 2007 #3

    Moonbear

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    If it's not displaying correctly, try clearing your cache. Sometimes that causes problems on the main pages if you edit things, and I'm not sure if that happens in the blogs too (I didn't even remember that we had LaTex enabled in the blogs).
     
  5. Oct 11, 2007 #4
    Please don't turn it off I'm having such a good time!
     
  6. Oct 11, 2007 #5
    I tried clearing the cache and that didn't work... the od thing is everything looks find here, but not on the blog post...
     
  7. Oct 11, 2007 #6

    robphy

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    Last edited: Oct 11, 2007
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