Sequences of complex numbers -- Complex Analysis 6

[@sillymel]

(20:29) Although the proof of convergentCauchy itself doesn’t change for complex-valued sequences, the proof of one of the theorems used to _prove_ that result, the Bolzano-Weierstrass Theorem (every bounded sequence has a convergent subsequence) _does_ change for complex-valued sequences. The exact way it changes depends on how exactly you proved the original theorem, so instead, I’ll present a general method to prove the Belzano-Weierstrass Theorem for complex-valued sequences using the equivalent theorem for real valued sequences. Let {c_n} be a bounded complex-valued sequence. Then for some B∈ℝ, |c_n|

[@cosmicnomad8575]

Excellent video on Complex Analysis and actually helped me understand some Real Analysis concepts even better too

[@DoktorApe]

Are you going to cover analytic continuation at some point? Its been a long time since I saw it worked out, but to me it was satisfying to know complex "versions" of a real function are unique (at least if you insist on enough continuity).

[@jokarmaths7771]

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[@schweinmachtbree1013]

12:57 lower limit should be k=0

[@schweinmachtbree1013]

and there should be absolute values on the LHS when taking the square root at 14:32 (which we want anyway so this is helpful; it means that we don't need to care about the fact that n^(1/n) is bigger than 1)

[@jokarmaths7771]

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[@emmanuelweiss8672]

Thank you very much for this series on complex analysis!

[@jokarmaths7771]

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[@aweebthatlovesmath4220]

This series is amazing keep it up!

[@jokarmaths7771]

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[@Grecks75]

23:06 |Re(z)| and |Im(z)| are less than *or equal* to |z|, not strictly less than. One of them will always be equal when z is on the real or imaginary axis. Edit: Watching on, I saw that you corrected it, so nevermind my comment.

[@samuelmarger9031]

Love this complex analysis series!

[@jokarmaths7771]

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[@iabervon]

It'd be interesting to see how this lecture would go assuming the students have had metric spaces and (R^2, d_2) in particular. A bunch of things are obvious, but then products and quotients are probably not operations you've considered on the isomorphic space.

[@schweinmachtbree1013]

Good observation. All of the bookwork about sequences in this video holds for metric spaces in general (but not the examples, e.g. z^n -> 0 for |z| < 1 since general metric spaces don't have a multiplication). In particular, convergent Cauchy being a property of metric spaces called completeness, so *R* and *C* with their absolute values as their metrics are complete metric spaces. However the "algebra of limits" in the warm-up exercises doesn't make sense in general metric spaces, because not all metric spaces have algebraic operations - the context in which the algebra of limits makes sense is normed algebraic structures, e.g. normed vector spaces (over a normed field), normed rings, and normed algebras (over a normed ring). the proofs of the sum/product/scalar multiplication rules in these settings are the same (mutatis mutandis) as for real or complex sequences (they usually involving taking two lots of epsilon/2).

[@jokarmaths7771]

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[@pharaohgarmar5611]

Love the illustrations with Mathematica. Unfortunately i couldn’t reproduce the Riemann plots because I think they need a more recent version of Mathematica than I have.

[@edwardjcoad]

This is my problem with all these limit definitions; I'm left unsatisfied as it only seems to tell me what I already know...like Jack said in 30 Rock, its like the Huffington Post!! Can you please demonstrate a limit definition where the process breaks and therefore shows there isn't a limit. I don't need to know that the lim of x-> 2 of x^2 = 4. What I want to see what happens with a crazy broken function.

[@strikeemblem2886]

Here's a standard example where the limit does not exist: lim sin(1/x), where we take x->0

[@schweinmachtbree1013]

and an examples of a limit that does exist but not "trivially": lim[x sin(1/x)] as x->0. As a consequence, the function defined by f(x) = x sin(1/x) for x=/=0 and f(0)=0 has a limit at 0 but its derivative does not (i.e. it is not differentiable at 0), and the function defined by g(x) = x^2 sin(1/x) for x=/=0 and g(0) = 0 has a limit at 0 and a derivative at 0, but doesn't have a second derivative at 0. In general, for a positive integer k the function defined by h(x) = x^k sin(1/x) for x=/=0 and h(0) = 0 is (k-1)-times differentiable, but not k-times differentiable (at 0). If this interests you, you should graph these functions and their secant lines at 0 in desmos or geogebra.

[@iabervon]

If you'd like a complex sequence example, e^(in) is bounded but doesn't converge. You can build a similar sequence on the harmonic series if you want one that's doesn't converge but isn't obviously not Cauchy (pairs of adjacent terms get closer together without bound, but there's always a term later on too far away from any term).

[@jokarmaths7771]

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[@edwardjcoad]

@@strikeemblem2886 so if I follow the standard process with the opening line of e > 0; |sin(1/x) - L| < e (epsilon) and |x-0| < d (delta) -> then what? Is it because if e -> 0 then the sin(1/x) - L > e for values of L and therefore L must change to satisfy the opening statement therefore L doesn't exist? For examples where functions are described piecewise its clear based on the function there isn't a limit - but what would the scratch work look like.

[@ecourt93]

Could someone briefly explain how to show that the sequence at the end of the video (with binomial coefficients of complex numbers) is bounded precisely when the real part of alpha is -1 or greater? Not sure where to start.

[@liyi-hua2111]

i think it’s ratio test. turning alpha to a+bi and C(alpha,n) to (r_0*r_1*…*r_(n-1)*e^(theta_0+…+theta_(n-1))/n! where r_k = sqrt((a-k+1)^2+b^2)) wish there exists K s.t. for all k > K, r_k/k =< 1 so we can take product from term 1 to K to be our bound M

[@mourad3078]

Warm up problems sollotions ?

[@mathismind]

Nice!

[@jokarmaths7771]

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[@Mathskylive]

Bài tập giới hạn cửa ngọ của giải tích.

[@bennobrueck9610]

Does anybody know, how to proof the last warm up problem?

[@MacHooolahan]

Bit stumped visualizing that last proof. Can someone please indulge an amateur? Looking at this in the complex plane, If the real part converges does it matter that the imaginary part keeps spirally round *without* converging? Doesn't it still land on some well defined point as you go to infinity?

[@adityansingla5656]

If you are talking about the convergence of im and real part, I personally visualised ot as an xy plane with y axis as imaginary part Now for a sequence of points on the plane to converge, both x and y coordinates should converge (as if either doesn't converge, we get a blowing up to infinity or some sort of weord oscillation type thing going on) And as both x and y converge, then the point should also converge (the converse) And x and y are real and im parts Therefore real and im parts must converge and vice versa

[@MacHooolahan]

@@adityansingla5656 You're right - thanks! I was (mistakenly) picturing this as the argument to an exponential, which would of course be a whole different story....

[@noelani976]

First to "like" and to "comment". A good feat!

[@jokarmaths7771]

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[@artificialresearching4437]

I might be wrong, but this could work: kzbin.info/www/bejne/hqe4o5WmqL2Ngrc P.S. A piece of advice: make video 1.5 times faster, I speak very slowly)

[@jokarmaths7771]

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