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October 17th, 2017, 01:04 PM   #11
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Quote:
 Originally Posted by zylo If F(z) = zf(z) where f(z) exists, has a derivative everywhere, and maps to entire complex plane, then so does F(z). Again, assume a closed island in w where F(z) undefined. Let F(z) be arbitrarily close to the border. F(z+$\displaystyle \Delta$z) = (z+$\displaystyle \Delta$z)f(z+$\displaystyle \Delta$z)= (z+$\displaystyle \Delta$z}(f(z)+f'(z)$\displaystyle \Delta$z) F(z+$\displaystyle \Delta$z)=F(z)+(f(z)+f'(z)$\displaystyle \Delta$z) where f(z) and f'(z) exist and $\displaystyle \Delta$z is associated with f'(z) so not restricted by the border.
Someone pointed out a while back that in the reals, $f(x) = x$ and $F(x) = x f(x) = x^2$ is a counterexample. But your proof looks like it would apply equally well to the real case, proving a false conclusion. Can you explain this?

Quote:
 Originally Posted by zylo It follows F(z) can cross the border ...
Only because of the lax immigration policies of those commie liberals.

Last edited by Maschke; October 17th, 2017 at 01:08 PM.

October 17th, 2017, 03:24 PM   #12
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Quote:
 Originally Posted by zylo F(z+$\displaystyle \Delta$z) = (z+$\displaystyle \Delta$z)f(z+$\displaystyle \Delta$z)= (z+$\displaystyle \Delta$z}(f(z)+f'(z)$\displaystyle \Delta$z) F(z+$\displaystyle \Delta$z)=F(z)+(f(z)+f'(z)$\displaystyle \Delta$z)
I have a little quibble here, which is that when you replace

$f(z + \Delta z)$ with $f(z) + f'(z) \Delta z$ that is an approximation and not an equality. We may only write

$f(z + \Delta z) \approx f(z) + f'(z) \Delta z$.

You can get them within $\epsilon$ of each other by suitably small choice of $\Delta z$; but there is never equality except in the limit as $\Delta z \to 0$.

To see why this is so let's review the derivation. The derivative of a function of a complex variable is defined by the same formula as that of a function of a real variable:

(*) $\displaystyle f'(z) = \lim_{\Delta z \to 0} \frac{f(z + \Delta z) - f(z)}{\Delta z}$

[Aside: In the complex case the limit is only defined if $\Delta z$ can approach $0$ from any direction in the plane. This turns out to be a profound restriction, making complex-differentiable functions behave much more nicely than real-differentiable ones. For example complex-differentiable functions are infinitely differentiable and always represented by their power series.]

Now we'd like to multiply both sides of (*) by $\Delta z$, but $\Delta z$ appears on both sides of the equation bound by the limit expression. Therefore we must write

$\displaystyle \lim_{\Delta z \to 0} f'(z) \Delta z = \lim_{\Delta z \to 0} (f(z + \Delta z) - f(z)) = \lim_{\Delta z \to 0} f(z + \Delta z) - f(z)$

and then

$\displaystyle \lim_{\Delta z \to 0} f(z + \Delta z) = f(z) + \displaystyle \lim_{\Delta z \to 0} f'(z) \Delta z$

There is only equality in the limit. For $\Delta z \neq 0$ at best we have only approximate equality,

I don't know if this affects the rest of your argument, which is only an outline of a much more rigorous argument that you have not given. Also as noted the same argument proves a falsehood in the real case, which tells us that an error lurks within the vagueness.

2) I have another quibble. You write

Quote:
 Originally Posted by zylo F(z+$\displaystyle \Delta$z) = (z+$\displaystyle \Delta$z)f(z+$\displaystyle \Delta$z)= (z+$\displaystyle \Delta$z}(f(z)+f'(z)$\displaystyle \Delta$z) F(z+$\displaystyle \Delta$z)=F(z)+(f(z)+f'(z)$\displaystyle \Delta$z)
After fixing your '}' typo, the first line is

$F(z+\Delta z) = (z+ \Delta z)f(z+ \Delta z) = (z+ \Delta z)(f(z)+f'(z) \Delta z)$

and using the distributive law the last term is

$(z+ \Delta z)f(z) + (z+ \Delta z)f'(z) \Delta z)$

Then the second line isn't right, it should be

$F(z + \Delta z) = F(z) + (z+ \Delta z) f'(z) \Delta z$

and then changing the $=$ to $\approx$ per the first quibble.

3) I don't follow your border crossing argument but if $F$ misses a point then there's a little region around $F(z)$ for any $z$ that also misses that point. That's because a point is a closed set and its complement is open, so $F(z)$ is always an interior point of the complement of a hole in the plane.

I don't know if that's relevant, I really can't understand what you're saying. $e^z$ misses $0$ but hits everything arbitrarily close to $0$. For all you know, $F$ does the same thing.

Last edited by Maschke; October 17th, 2017 at 04:20 PM.

October 18th, 2017, 09:06 AM   #13
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Quote:
 Originally Posted by Maschke Someone pointed out a while back that in the reals, $f(x) = x$ and $F(x) = x f(x) = x^2$ is a counterexample. But your proof looks like it would apply equally well to the real case, proving a false conclusion. Can you explain this?
f(z)=z, zf(z)=z$\displaystyle ^{2}$ which maps to entire complex plane, ie, z$\displaystyle ^{2}$ = C has a solution for any C. The polynomial x$\displaystyle ^{2}$+1=0 doesn't have a solution but z$\displaystyle ^{2}$+1=0 does.

Quote:
 Originally Posted by zylo Assume F(z) has a derivative for all z and maps to w plane with a hole in it (area where F(z) doesn't map to). The hole is closed because F'(z) is undefined on it's border. In this case, F'(z) is defined and F'(z)$\displaystyle \Delta$z exists in f(z) for $\displaystyle \Delta$z sufficiently small no matter how close you get to the border. For example, f(z) could map to the inside of a circle for all z and f'(z) exist everywhere inside the circle. The only way out would be to show that at a point near the border F'(z)$\displaystyle \Delta$z is greater than the distance to the border.
F'(z)$\displaystyle \Delta$z can't cross border because it only applies up to the border. f'(z)$\displaystyle \Delta$z can cross border.

October 18th, 2017, 10:53 AM   #14
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Quote:
 Originally Posted by zylo f(z)=z, zf(z)=z$\displaystyle ^{2}$ which maps to entire complex plane, ie, z$\displaystyle ^{2}$ = C has a solution for any C. The polynomial x$\displaystyle ^{2}$+1=0 doesn't have a solution but z$\displaystyle ^{2}$+1=0 does.
Yes but your proof doesn't distinguish between the real and complex cases. You don't seem to be engaging with this point.

Also I pointed out two specific errors in your chain of equalities that you did not address.

October 18th, 2017, 11:51 AM   #15
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Quote:
 Originally Posted by Maschke Yes but your proof doesn't distinguish between the real and complex cases. You don't seem to be engaging with this point.
In other words, there must be a line in your proof that is satisfied in the complex plane but not on the real line.

October 18th, 2017, 06:36 PM   #16
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Quote:
 Originally Posted by v8archie In other words, there must be a line in your proof that is satisfied in the complex plane but not on the real line.
Its much worse than this. His proof is not even wrong. Most of it involves Taylor estimates which don't even apply necessarily for functions which are only assumed differentiable.

This doesn't matter however because he is trying to apply these to prove surjectivity of the map which is absolutely unrelated. After trying to decipher it, the nearest approximation I have to his misunderstanding is that he thinks he can do induction on $\mathbb{C}$ and is trying to begin with the assumption that his function maps onto a punctured neighborhood of a fixed $c \in \mathbb{C}$ and use this to prove it to $c$ as well.

Not only is this assumption complete nonsense, but it has already been pointed out that $f(z) = e^z$ maps onto a punctured neighborhood of 0 yet 0 is not in the image.

This has only caused him to double down on his nonsense and I feel compelled to reply here again only so that unfortunate students who don't know better don't mistake it for math.

 November 10th, 2017, 07:41 AM #17 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100 Theorem: If F'(z) exists for all z, F(z) maps to entire complex plane w. Proof: Assume map of F(z) has a hole in w. Let F(z$\displaystyle _{0}$) be near border of the hole. F(z) defined in a neighborhood of z$\displaystyle _{0}$. F(z$\displaystyle _{0}$+z)$\displaystyle \approx$ F(z$\displaystyle _{0}$)+F'(z$\displaystyle _{0}$)$\displaystyle \Delta$z F(z) crosses the border. Contradiction. F(z) maps to entire complex plane. Which is also Fundamental Theorem of Algebra if F(z) is a complex polynomial. Example: y=cosx has a derivative for all x but does not map to all of y. No solution for y=2. w=cosz has a derivative for all z and maps to all of w. A solution for any w, including 2. REF: Fundamental Theorem of Algebra Proof Post #46 Last edited by zylo; November 10th, 2017 at 07:44 AM. Reason: Add REF
November 10th, 2017, 09:31 AM   #18
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Quote:
 Originally Posted by zylo Theorem: If F'(z) exists for all z, F(z) maps to entire complex plane w.
$e^z$ is a counterexample. How many times has that already been mentioned?

Last edited by Maschke; November 10th, 2017 at 09:33 AM.

November 10th, 2017, 05:22 PM   #19
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Quote:
 Originally Posted by zylo Theorem: If F'(z) exists for all z, F(z) maps to entire complex plane w.
$F(z) = e^z$ is an obvious counterexample. But let's walk through this step by step.

Quote:
 Originally Posted by zylo Proof: Assume map of F(z) has a hole in w.
With $F(z) = e^z$, we have that $F$ is defined and complex-differentiable on the entire plane, and fails to hit $0$. Is that what you mean by a "hole?" You haven't really said, but I'll assume this fits your definition. If not, say what a hole is.

Quote:
 Originally Posted by zylo Let F(z$\displaystyle _{0}$) be near border of the hole.
Ok. Let's take some number really really close to $0$ in the $w$-plane, say $e^{-10^6}$. This is a number on the positive real axis, very close to $0$. In this case $z_0 = -10^6$, or negative one million. A number way out there to the left on the real axis. In fact the only way to get $e^z$ close to zero is for $z$ to approach $-\infty$.

I hope you can see that numbers very far to the left on the negative real axis get mapped to a small interval to the right of $0$ on the positive real axis.

To visualize how neighborhoods work, it's helpful to think in terms of rectangles. A rectangle about a point gets mapped to a wedge shape truncated by concentric circles. It never crosses any border, never hits $0$. It's helpful to play around with this visualization app.

https://www.geogebra.org/m/Ac43DNNP

Here's the image of a rectangle around the origin.

As you move the rectangle to the left, it gets smaller. As you move it up or down, it rotates around a circle. If you moved the rectangle leftward so that it was centered on $-10^6$, the wedge would be very small and its left end would be very close to $0$. But $0$ would still be outside the wedge. Because $e^z$ never hits $0$.

Quote:
 Originally Posted by zylo F(z) defined in a neighborhood of z$\displaystyle _{0}$.
Yes that's fine. You can take any neighborhood of $-10^6$ and map it to an interval to the right of $0$ on the real axis. But you can see that no "border crossing," whatever the heck that means, is occurring. Rectangular neighborhoods of points way out to the left get mapped to small wedges to the right of $0$.

[Circular neighborhoods are very tricky, I'm still trying to visualize those!]

Quote:
 Originally Posted by zylo F(z$\displaystyle _{0}$+z)$\displaystyle \approx$ F(z$\displaystyle _{0}$)+F'(z$\displaystyle _{0}$)$\displaystyle \Delta$z
Ok fine. What does that prove? Let's plug in $z_0 = -10^6$. Then we have

$F(z - 10^6) \approx e^{-10^6} + e^{-10^6} \Delta z$.

Which tells us what, exactly?

Quote:
 Originally Posted by zylo F(z) crosses the border.
What does that mean? What border has been crossed? There's nothing in this example or in the visualization of the image of a neighborhood of $z_0$ that fits a picture of "crossing a border." On the contrary, the wedges get close to zero and rotate around zero but no wedge ever contains zero.

Quote:
 Originally Posted by zylo Contradiction. F(z) maps to entire complex plane.
What is the contradiction? Since we already know that $F$ misses $0$, your conclusion is false.

Can you shed some light on your argument? It really makes no sense and has a very obvious counterexample.

Last edited by Maschke; November 10th, 2017 at 05:33 PM.

 November 13th, 2017, 07:13 AM #20 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100 Let w=F(z) have a derivative for all of z. Assume there is a point w$\displaystyle _{p}$ which F(z) doesn't map to. I can define F'{z) at any point w$\displaystyle _{0}$ in a neighborhood of w$\displaystyle _{p}$, and w$\displaystyle _{0}$ has a neighborhood which doesn't include w$\displaystyle _{p}$. Visually, draw a circle around w$\displaystyle _{p}$. Draw w$\displaystyle _{0}$ inside the circle. There is a circle around w$\displaystyle _{0}$ which doesn't include w$\displaystyle _{p}$. ie, the initial assumption doesn't lead to a contradiction, and F'(z) exists for all z doesn't imply F maps z to all of w. Particularly,from definition of derivative: $\displaystyle |\Delta z||F'-\epsilon|< |\Delta F| < |\Delta z||F'+\epsilon|$ if $\displaystyle |\Delta z| < \delta$ ie, if $\displaystyle |\Delta F|$ sufficiently small for F to be defined. Same argument applies for zf(z) where f(z) has a derivative for all z, and f(z) maps to entire complex plane. Because you don't know that (z+$\displaystyle \Delta$z)f(z+$\displaystyle \Delta$z) maps to w$\displaystyle _{p}$.

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