Complex Analysis 30 | Identity Theorem

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Күн бұрын

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@AptekerS Ай бұрын

Incredibly beautiful theorem. First learned this almost half a century ago, and was amazed by it right then. The amazement is still here.

@brightsideofmaths Ай бұрын

Nice :)

@Ghetto_Bird 9 ай бұрын

What a mind-blowing result this is, just unreal!

@whatitmeans 2 жыл бұрын

This theorem has a really interesting consequence in physics: many will argue that movement never stops, this due classic linear differential eqs. only can stand never-ending solutions in time, even more, no non-piecewise Power Series could stand a finite extinction time due the Identity Theorem, which says that if it have any non-zero measure section equal to zero, then it can only be the trivial zero solution. For having a finite extinction time you need to have a at least one "singular point" in the time variable on the differential equation, so it must be nonlinear, which is interesting when thinking in which mathematical tools are used nowadays for modeling physics phenomena (as example, the classic Schrödinger is linear, so it cannot model any solution with a finite extinction time). For example, in the classical damped pendulum, you need to look for the term "sublinear damping" in order to find solutions that could handle having a finite extinction time. The bad news, is that the possibility of having these finite-time ending points, also breaks uniqueness of solutions, so as mathematical tools are much more complicated to handle.

@synaestheziac 2 жыл бұрын

Very interesting indeed-what sources could I consult, or what should I search for, to learn more about these ideas?

@whatitmeans 2 жыл бұрын

@@synaestheziac Unfortunately I haven't found yet one source talking about it, is like scientists decided to avoid it. Many of what I found is related to singular differential equations, the first related paper I found are "Finite Time Differential Equations" and "Finite Time Controllers" by Vardia T. Haimo, the first one introduces the topic but it have an important typo on one equation, but its resolved and extended on the second one. From my side, I have my recreational research and foundings through some questions in mathstackexchange under the tag [finite-duration]. I have been to busy for updates now but maybe I will add some new questions next month.

@whatitmeans 2 жыл бұрын

@@synaestheziac I would like to make an example of how this topic have been kind of avoided: here on YT you could find the amazing Quantum Mechanic clases by profesor Barton Zwiebach, where on lecture L14.2 Quantization of the energy, min 7:25, he states: "So the requirement that the solution {of the time independent Schrödinger equation for the harmonic oscillator} be normalizable quantizes the energy", here after explaining schrödinger equation is linear, they use a Power Series ansatz which coefficients, in order of having non-exploding solutions in time, leads to the quantization of Energy. Now, since a Power Series solution of a linear differential equation only can have never-ending solutions in time that at best vanishes at infinity, you could find a math mismatch when interpreting the formula E = h v with "h" the planck constant and "v" the frequency of the wavefunction of the particle in question (think for now on, in a photon color frequency): if the photon description is a sine function that last forever, there is no problem since its fourier transform is a delta function with defined frequency, but if the photon is generated and absorved in a finite time extension "T", you will have now that this time-truncated sine function describing the photon is compacted supported in the time variable, so in the frequency domain, its Fourier Transform must be of unlimited bandwidth, so it cannot be mathematically described by just one frequency "v", and this is not due quantum mechanics uncertainty, is more deeper in the math tools used: a power series cannot accurately describe a phenomena with finite time extension, neither a linear ordinary differential equation can... but then, which tool should be used? Thinking in Maxwell's electromagnetic wave equation it is also linear, but in PDEs you could have for a fixed point in space a compacted supported solution in time, but you cannot have a vanishing solution in the spacetime "x" vs "ct" spacetime if there are no singular points in the differential equation, the traveling wave will be always existing in some point in spacetime. So for the causally emitted-absorved photon, you need to have something that is achieving a finite extinction time with unbounded bandwidth, through a nonlinear ODE and/or PDE, whith a non-unique singular solution that cannot be a classic one-piece power series (could be a piecewise defined power series), which is going to be incompatible with energy quantization description of the photon. As an example of a nonlinear ODE that have a finite time solution, think in: x' = -sgn(x) sqrt(|x|), x(0)=1 which stands the answer: x(t)= 1/4 (1-t/2+|1-t/2|)^2 which is singular at t=2, so it have similar non-unique solutions issues like the Norton's Dome example.

@synaestheziac 2 жыл бұрын

@@whatitmeans interesting. I actually watched Zwiebach’s QM lectures a couple years ago!

@mariolemelin27 2 ай бұрын

@whatitmeans Thank you very much for this observation. I will look into it. Very interresting.

@StratosFair 2 жыл бұрын

A beautiful theorem with a short and sweet proof. Just what I like :)

@realcirno1750 Жыл бұрын

my favourite complex analysis theorem

@brightsideofmaths Жыл бұрын

Mine too! Thanks for the support :)

@ferry7185 10 ай бұрын

can you explain a bit more on 10:40, why you conclude that h(z) will be non-zero if a_m is not zero, how about if a_m * (z-c)^m and the rest of term the convergence of series will be 0? so we have h(z) = 0

@brightsideofmaths 10 ай бұрын

Good question. For this implication, one has to put some stuff together. First h cannot be constant and it consist of powers (z-c)^n. However if (z-c) is very small, then the largest contribution comes from the smallest power.

@eamon_concannon 4 ай бұрын

@@brightsideofmathsI see that first term dominates each of the remaining terms for sufficiently small z-c but it seems that we require that magnitude of first power be larger (or unequal to) than magnitude of sum of remaining terms in power series in order for series not to converge to zero. I don't know how to prove this.

@eamon_concannon 4 ай бұрын

@@brightsideofmaths Another approach (for m≠ 0) is to write h(z) = (z-c)^m g(z) where g(z) is a power series with g(c) = a_m ≠ 0. Since g(z) is a continuous function, there must be a neighbourhood of c that does not contain any zeros of g(z). Therefore the zeros of h(z) are c and points outside the aforementioned neighbourhood of c. This means that c is an isolated zero of h(z) i.e. c is not an accumulation point of M. If m =0 (so a_0 ≠ 0), then c is not a zero of h(z) and since h(z) is continuous there is a neighbourhood of c that does contain any zeros of h(z) so again, c is not an accumulation point of M.

@brightsideofmaths 4 ай бұрын

@@eamon_concannon We know that the the whole power series is convergent, so we can make (z-c) as small as we want. And we make it smaller and smaller until the first power dominates the rest.

@brightsideofmaths 4 ай бұрын

@@eamon_concannon I like your second approach!

@synaestheziac 2 жыл бұрын

Is analytic continuation coming up next?

@somethingstuff4975 4 ай бұрын

Hi, thanks for the video, just a quick question about 9:11. is this the mth derivative or nth derivative?

@brightsideofmaths 4 ай бұрын

Good point! This should be the m-th derivative :)

@Sarah-mp9lb 8 ай бұрын

Extremely helpful, thanks! I downloaded the script via steady 👍

@brightsideofmaths 8 ай бұрын

Glad it helped! And thanks for the support!

@AlpakaAntifa 5 ай бұрын

@@brightsideofmathsYou have a complete script for complex analysis? :o I assume it is only available for steady supporters?

@brightsideofmaths 5 ай бұрын

@@AlpakaAntifa Yes, it's the bonus for supporters. Are you interested?

@AlpakaAntifa 5 ай бұрын

@@brightsideofmaths Do you have a table of contents for it somewhere on your website? :) Generally: Yes, I really prefer your methods of explaining topics over those of my professors plus you really helped me pass past mathematical modules. :)

@brightsideofmaths 5 ай бұрын

@@AlpakaAntifa I am happy about every support on Steady because then I am able to continue this work here :)

@RSLT 2 жыл бұрын

Great, that is fantastic simple proof. Thank you very much for explaining it very well. Also, I need your help. How can I say this in exact math terms: Assume zeta (s)-zeta(1-s)=0. Therefore because zeta (s)-zeta(1-s) =0 the Sum 1/n^s - Sum 1/n^(1-s) or sum( 1/n^s) -1/n^(1-s) converge to zero in the critical strip. I want to use the identity theorem or Uniqueness of the analytic continuation. The challenge is sum 1/n^s is a divergence power series in the critical strip. Please advise.

@struyep Жыл бұрын

For 3 => 2 why can't we taylor expand at c to get that the function is zero throughout the domain it's holomorphic on?

@brightsideofmaths Жыл бұрын

Taylor expansion is just local. So you need an argument why it holds for the whole domain.

@struyep Жыл бұрын

Thanks, that makes sense. My essential motivation is that checking closed vs openness is quite "clever" and I don't understand the reason/intuition behind coming up with the proof idea. This is in contrast to what is being proven which seems quite trivial to me, if all the derivatives of a function exist and are zero then the function must be constant, because the derivatives are all also zero in the neighbourhood. I think this might work: by taylor expansion, the function and all its derivatives are zero in any disc in D containing c (because in the proof of analyticity, the same series works for the entire disc). Now we can taylor expand from the boundary of this disc to prove more points are zero. More formally, we can consider the path between c and and any other point z, and show a sequence of intersecting discs overlapping this path. By openness, the infinum of the maximum disc size around any point in the path is still finite, so we will reach z within finitely many discs, and by induction we can show that f is zero on all discs and so is zero on the final point z.

@brightsideofmaths Жыл бұрын

@@struyep I like the idea and it captures the correct properties. However, you have to show that finite disc suffice. I guess, you can make this into a correct proof.