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Post by mondo on Jul 20, 2023 7:06:42 GMT
I did the exercise at the end of page 518, the claim is "If $f$ is analytic function then $\overline{\nabla}f = 2f'$" but what is $f'$? Is it a derivative in respect to $x$ or $y$? Base on the calculation I did I think author really means $2\frac{d}{dx}f$ or $2\frac{d}{dy}f$. Do you agree Vasco?
There is one more problem I have with this, if you do $\frac{d}{dy}f$ you should get $\partial_y{u} + i(\partial_y{v})$ however I get $\partial_y{v} -i(\partial_y{u})$ - real and imaginary parts are swapped and sign is inverted. Here are my calculations:
$\overline{\nabla}f = (\partial_x - i\partial_y)(u+iv) = \partial_x{u} + i\partial_x{v} - i\partial_y{u} + \partial_y{v} = [\partial_x{u} + \partial_y{v}] + i[\partial_x{v} - \partial_y{u}]$ Now if Cauchy-Riemann equations were to hold we have equal terms in first square bracket and equal and opposite in sign terms in the second braces. CR equations are $\frac{\partial_u}{\partial_x} = \frac{\partial_v}{\partial_y}$ and $\frac{\partial_u}{\partial_y} = -\frac{\partial_v}{\partial_x}$ So if we first substitute for $\partial_x$ in my equation we get: $2\partial_x{u} + i2\partial_x{v} = 2\frac{d}{dx}f$
However when I substitute for $\partial_y$ I get $2\partial_y{v} + i2(-2\partial_y{u}) \ne \frac{d}{dy}f$
So where is the catch?
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Post by Admin on Jul 20, 2023 7:23:18 GMT
Mondo
$f$ is an analytic function of $z$ so $f'$ means the derivative of $f$ with respect to $z$. This is the notation throughout the whole of the book!
So no I don't agree.
Vasco
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Post by mondo on Jul 20, 2023 8:09:27 GMT
But where is $z$ in $f$? $f$ is said to be a vector with two components $u$ and $v$. Neither partials is calculated in regards to $z$ either.
Beside that, why $\partial_y{f}$ doesn't match the one I calculated using $\overline{\nabla}f$ and substituting CR equations?
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Post by Admin on Jul 20, 2023 8:35:08 GMT
Mondo
An example would be $f=z^2=(x+iy)^2=x^2-y^2+2xyi=u+iv$
So $u=x^2-y^2$ and $v=2xy$
Vasco
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Post by mondo on Jul 20, 2023 8:53:23 GMT
Yes I know this, it was used so many times before. So knowing that both $u$ and $v$ depend on $z$, $f' = \frac{d}{dz}u + i(\frac{d}{dz}v)$ but this is not connected to our partials in regards to $x$ and $y$ so how can we say that they are equal to $2f'$?
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Post by Admin on Jul 20, 2023 9:02:01 GMT
Mondo
I will produce a short document which shows this and post a link to it later today.
Vasco
PS Where are you situated in the world? I am in the UK.
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Post by Admin on Jul 20, 2023 9:05:36 GMT
mondo
i will post a link to it within the next 30 minutes.
Vasco
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Post by Admin on Jul 20, 2023 10:04:13 GMT
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Post by Admin on Jul 20, 2023 13:47:32 GMT
Mondo
I just corrected a typo in my document on line 14. I had missed out the $f$.
Don't forget that $f'=f'(z)=df/dz$ only exists if $f=f(z)$ is analytic. See section VIII of chapter 4 on pages 207-210.
Vasco
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Post by mondo on Jul 20, 2023 16:59:53 GMT
Vasco, thank you. However I must confess I saw this before, you posted a larger notes document in the other thread here. While it found it helpful in other regards it doesn't satisfy my question in this particular example. Here is the problem - You say, "If we use $\overline{\nabla}$ and the CR equations and follow the same procedure again we find that $\overline{\nabla}f = 2f'$" - well that's what I try to prove here, your document just restates the exercise, not solving it.
I my first post I made an attempt to show that $\overline{\nabla}f = 2f'$. While it works for $\frac{d}{dx}f$ it doesn't for $\frac{d}{dy}f$
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Post by Admin on Jul 20, 2023 17:55:43 GMT
Mondo
$\overline{\nabla}=\partial_x-i\partial_y$
So
$\overline{\nabla}f=(\partial_x-i\partial_y)(u+iv)$
$\overline{\nabla}f=\partial_xu+i\partial_xv-i\partial_yu+\partial_yv$
$\overline{\nabla}f=\partial_x(u+iv)+\partial_y(v-iu)=f'+f'=2f'$
Vasco
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Post by mondo on Jul 20, 2023 18:22:21 GMT
Mondo $\overline{\nabla}=\partial_x-i\partial_y$ So $\overline{\nabla}f=(\partial_x-i\partial_y)(u+iv)$ $\overline{\nabla}f=\partial_xu+i\partial_xv-i\partial_yu+\partial_yv$ $\overline{\nabla}f=\partial_x(u+iv)+\partial_y(v-iu)=f'+f'=2f'$ Vasco It looks like you got the same result as I did, the only difference is you concluded $\partial_y(v-iu) = f'$ why is it true? What is $\partial_y(u+iv)$ then? Is it also $f'$? |
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Post by Admin on Jul 20, 2023 18:29:01 GMT
Mondo
They are both equal to $f'$. See page 210.
Vasco
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Post by Admin on Jul 20, 2023 19:04:21 GMT
Mondo
$f'$ by definition must be the same in all directions through a given point $z$, and so the derivative along the direction of the $x$-axis must equal the derivative along the $y$-axis.
Vasco
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Post by mondo on Jul 20, 2023 19:37:23 GMT
Mondo $f'$ by definition must be the same in all directions through a given point $z$, and so the derivative along the direction of the $x$-axis must equal the derivative along the $y$-axis. Vasco I am rereading pages 207-210 now. As to what you said above - for $\partial_x$ you, and the book, seem to calculate it for $u + iv$ while for $\partial_y$ you do it for $v - iu$ which is clearly not the same, I try to understand why. |
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