ganeshie8
  • ganeshie8
quick question Are all real differentiable functions when pushed to complex plane are still differentiable ? For example, f(x) = x^2 is differentiable when x is real. Is it still differentiable when x is complex ?
Discrete Math
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Astrophysics
  • Astrophysics
I guess so, maybe this is what you're looking for http://mathworld.wolfram.com/Cauchy-RiemannEquations.html
ganeshie8
  • ganeshie8
\(f(z) = z^2\) \[f'(z_0)=\lim\limits_{\Delta z\to 0}~\dfrac{f(z_0 + \Delta z)-f(z_0)}{\Delta z}\] If i mimic the "real limits" argument, I do get \(f'(z_0) = 2z_0\) was wondering if this is true for all real differentiable functions
Astrophysics
  • Astrophysics
Woah, that's interesting

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Melodious
  • Melodious
hey
ganeshie8
  • ganeshie8
hey doggy knw how to help me
Melodious
  • Melodious
empty can you help me in trignometry
ganeshie8
  • ganeshie8
nope, he is mine for next 30 mnts or so
Empty
  • Empty
Sorry I am not good at trig. Quick answer ganeshie: yeah you can haha.
Melodious
  • Melodious
ok empty sorry
Melodious
  • Melodious
:'(
ganeshie8
  • ganeshie8
for now, im convincing myself saying that if it is differentiable in 1D then it is definitely differentiable in 2D etc etc...
Melodious
  • Melodious
ok empty sorry
ganeshie8
  • ganeshie8
because when we're moving to complex plane from real, we're just literally incrementing the dimension of both domain and range of a function, "i think"
Empty
  • Empty
Yeah, well I mean if we can construct a function out of a power series then we really turn the problem into like understanding: \[f(z)=z^n\] and from there the rest should follow.
ganeshie8
  • ganeshie8
Ahh are you saying power series doesn't bother whether z is real or ccomplex ?
ganeshie8
  • ganeshie8
If that is the case, this must be true \[f'(x) = g(x) \implies f'(z) = g(z)\] where \(x\) is real and \(z\) is complex
Empty
  • Empty
So a quick way to show this: \[f'(z)=\lim_{\Delta z \to 0} \frac{f(z+\Delta z) -f(z)}{\Delta z}\] \[f'(z)=\lim_{\Delta z \to 0} \frac{[z^n+(n-1)z^{n-1}\Delta z +S] -z^n}{\Delta z}\] So S is the sum from the binomial expansion with all the \(\Delta z\) terms with exponents higher than 1. Now we subtract the \(z^n\) terms and divide the \(\Delta z\) to get: \[f'(z)=\lim_{\Delta z \to 0} \left[(n-1)z^{n-1}+\frac{S}{\Delta z} \right]\] Since S has only \(\Delta z\) terms with positive exponents greater than 2, dividing out 1 of them means the limit will take all these guys to zero just leaving us with this other term withohut anything. And then yeah, we can use the power series to extend our functions from real to complex values, I think they call this "analytic contnuation" but I might be wrong in the terminology but that's how they do it. That's why we just can look at the power series to compare for this: \[e^{ix} = \cos x+ i \sin x\] for instance :P
ganeshie8
  • ganeshie8
That looks neat! so proving \(z^n\) is differentiable proves all real functions that can be represented as power series are still differentiable in complex plane xD
Empty
  • Empty
Yeah, exactly, so the weird part is that this seems to be more about differentiation in a direction rather than before, and there's a bunch of weird and cool things that complex differentiation allows us to do that we didn't have before such as if you can differentiate it once, it's infinitely differentiable. I'm probably not the person to describe the rigorous side of this, but I have more or less an intuitive understanding of this. But the main thing is that in the past you would say the left hand and right hand limit have to be equal, in this case we are saying that there are an infinite number of ways to approach the point, and all those give us the same derivative at that value, so in a sense complex differentiability is much more intense. I can describe some intuitive things like the Cauchy Riemann equations (which really are just the same as talking about curl and divergence).
Astrophysics
  • Astrophysics
That sounds amazing, I think this is also useful, it's not very long either https://en.wikipedia.org/wiki/Holomorphic_function
ganeshie8
  • ganeshie8
I think I'm getting it slowly, both domain and range are 2D, so clearly the left and right limits are not sufficient anymore. for a complex function to be differentiable, the limit of difference quotient must exist and equal in all directions.. this looks more or less like the limit of a function of two varibales |dw:1436609440086:dw|
ganeshie8
  • ganeshie8
as \(z\) approaches \(z_0\) in any direction, \(f'(z)\) must approach the limit of difference quotient. this actually contradicts the previous statement that every real differentiable function is differntiable in complex plane hmm
ganeshie8
  • ganeshie8
@Astrophysics from that wiki link, it seems, for a function \(f(z) = u(x,y) +iv(x,y)\), \(u_x = v_y\) and \(u_y = -v_x\) are sufficent conditions for differentiablity
ganeshie8
  • ganeshie8
*necessary and sufficient
Empty
  • Empty
Hmmm I guess I should be more clear (because I'm not very precise in how good this actually is to myself, I'm not much further than you in some aspects here) that I am saying within the radius of convergence of a power series representation of a function, that's what we can differentiate I think. What those equations are are really just what you get when you differentiate f(z) so it's almost backwards in a sense to think like that. Here I'll show you: \[\frac{\partial f}{\partial x} = \frac{df}{dz} \frac{\partial z}{\partial x}=\frac{df}{dz} \] \[\frac{\partial f}{\partial y} = \frac{df}{dz} \frac{\partial z}{\partial y}=\frac{df}{dz} i\] Now we plug them into each other to get: \[\frac{\partial f}{\partial y} = i\frac{\partial f}{\partial x}\] which just amounts to getting our two equations you've written when you write out f=u+iv.
anonymous
  • anonymous
@ganeshie8 complex differentiability is far more stringent than differentiability in \(\mathbb{R}^2\) with the usual topology, and certainly more so than just partial differentiability in both variables; also @Empty surely you mean \((z+\delta z)^n=z^n+nz^{n-1}\delta z+O((\delta z)^2)\), i.e. \(nz^{n-1}\delta z\) not \((n-1)z^{n-1}\delta z\).
anonymous
  • anonymous
the problem with asking about 'pushing' \(f:\mathbb{R}\to\mathbb{R}\) to \(\tilde f:\mathbb{C}\to\mathbb{C}\) (really 'lifting' -- this is a functor) is that there's not an immediately obvious way to define a unique choice of \(\tilde f\) for \(f\) past that we expect \(\tilde f\) restricted to the canonical embedding of \(\mathbb{R}\) in \(\mathbb{C}\) to be identical to \(f\)
anonymous
  • anonymous
anyways, the reason the requirement for differentiability \(f:\mathbb{C}\to\mathbb{C}\) is stronger than \(f:\mathbb{R}^2\to\mathbb{R}^2\) is that in the complex plane we also constrain the local behavior of \(f\) under small *rotations* in addition to mere translations, which is equivalent to \(f\) preserving angles near a point; this constraint amounts to the Cauchy-Riemann equations in \(f(x+iy)=u(x,y)+iv(x,y)\):$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\\\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$or equivalently in terms of the Wirtinger derivative $$\frac{\partial f}{\partial\bar z}=0$$i.e. \(f\) is functionally independent of the conjugate \(z\) -- this is important as this disallows decoupling the real and imaginary parts of \(z\) in any way which would allow violating the stricter conditions above
anonymous
  • anonymous
now, if we have a function that is real-analytic, then yes, you can 'grow' it off the real line \((-\infty+0i, \infty+0i)\) by expanding the interval of convergence into a disk as there's an elementary result showing \(\sum_{n=0}^\infty a_n z^n\) has the same radius of convergence as a real and complex series and then analytic continuation describes how you can use that to continue 'growing' the function around any singularities you run across
anonymous
  • anonymous
but analytic continuation itself is not going to automatically lift differentiable \(f\in\mathbb{R}^\mathbb{R}\) to \(\tilde f\in\mathbb{C}^\mathbb{C}\) -- for analytic functions you can just use the same power series and consider complex \(z\) but for functions taht aren't analytic it's harder
anonymous
  • anonymous
anyways, if \(f:U\to\mathbb{C}\) where \(U\) is an open set and \(U\cap (-\infty+0i,\infty+0i)\) and \(f\) is differetiable then \(f\) is analytic but that means that \(f\) is analytic on the part of the real line in \(U\), so this type of extending \(f\) only holds for analytic \(f\) and not the general differentiable case
anonymous
  • anonymous
so it follows the answer to your question is no, as was suspected :(
anonymous
  • anonymous
If u think that in terms slope then that would be its slope in an argand plane probably , just a thought though is it correct
anonymous
  • anonymous
So it would be differentiable on a real-co-ordinate system but might be on an argand plane
anonymous
  • anonymous
@Astrophysics ?

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