Відео

Last time i was this early

12:07 Isn’t that lower bound still 0?

i wonder if we can do it using generating functions/series

12:47 you have a mistake here : 2^n/10^n = (1/5)^n and not 1/5

BRO CAN I BORROW YOUR BRAIN FOR MY MATH EXAM ?

13:14

How I missed the overkill series!

I got rick rolled in text form by the dastardly Chalk! What a legend!

Use the inverse function derivative rule.

Another cool way to compute ∫ln(sinx)dx is to use Riemann sums, as pi/2n∑ln(sin(k*pi/2n) tends to the value of the integral as n tends to infinity. you then have to compute ∏sin(kpi/2n) which is classic and is equal to n/2^n. You then conclude with the uniqueness of the limit.

The blue region's temperature doesn't change. Why does that not mean that ΔS2 = 0?

You could eazly hust set y=the function and by some simple simplification we would get (sqrt((n+4)÷n)+1)÷2 and bc sqrt is continois we could put the limit inside and the limit of (n+4)÷n is 1 so we wpuld get (1+1)÷2=2÷2=1

This deserves a Rube Goldberg Field's Medal.

I'm slightly confused, the question doesn't seem to limit x and y to a dependency, as far as it's concerned they are freely independent variables. Does the assumption that one is the function applied to the other not limit the potential solutions?

To be rigorous we have to specify when calculating delta T1 that we consider a reversible process bringing the subsystem to the same final temperature as the actual process, and the fact that entropy is a state function. Also the infinitesimal heat change is usually not a differential form so we write lowercase delta Q instead of "dQ". The total entropy change is strictly greater than 0 because the process is not reversible, and since this is a closed system the calculation of Delta S is actually the total entropy created by the process.

You made it worse by assuming that the viewer knows anything about piecewise functions.

I learned some homology theory as a directed study in my undergraduate degree but didn’t get to cohomology and am unfamiliar with it, and I know there is relation to these differential forms and cohomology but do not know what that relation is. All this to say the property that d(df)=0 is reminiscent of a property of defining the boundary maps in homology theory (the boundary of a boundary is the empty set)

Weird enough, one of the authors, Italo Bove, was my professor back in Universidad de la Republica in Montevideo, Uruguay, in the late 1990s.

When you used the formula for geom. series, the geom. series only works if the absolute value of the common ratio is strictly less than 1, though it will not be if x = 0.

bro this is amazing these are genuinely my two only hobbies combined

It’s not a quadratic equation. 🙄

I like the idea of presenting a proof using formal power series, but the example is maybe not the best, and the way the proof is presented seems aimed at scaring the viewer rather then edifying her; it definitely does not give the impression that this is a technique that one could use easily to solve (easy or difficult) problems. It also leaves a lot of question of whether this constitutes a proof at all (why are differentiation formulas derived for real functions valid in the formal poser series world, for instance). I think it would be much easier to understand if you first studied what multiplying with 1-X means for formal power series: for each position (except that of X^0) the coefficient of X^(n-1) gets subtracted from the coefficient of X^n. Doing this once turns you power series into the sum of terms nX^n, and doing it again turns that series into the one with all coefficients 1, except the constant coefficient which remains 0, and a third application turns the series into X (the coefficients of all powers other than X^1 are zero). That shows (without filling a blockboard) that (1-X^3) times the initial series is X, so that the initial series is X/(1-X)^3.

*None* of the examples requires Euler's theorem or Fermat's Little theorem, let alone Chinese Remainder Theorem. Moreover, they are solved easier and faster using basic modular arithmetic. (A). 7^950 mod 100: if you either quickly multiply in your head _or_ read Feynman's "Lucky Numbers" (with a simple method to square numbers around 50), it takes seconds to see 7^4 = 49^2 = (50 - 1)^2 = 50^2 - 2*50*1 + 1 = 2500 - 100 + 1 = 2401. *01* , that is 7^4 ≡ 1 (mod 100) ⇒ 7^950 = 7^2 * 7^948 = 49 * (7^4)^262 ≡ 49 * 1^262 (mod 100) ≡ 49 (mod 100) (B) 3^999,999,999 mod 7: check 3^2, then 3^3 and voila: 3^3 = 27 = 28 - 1 ≡ -1 (mod 7). 3^999,999,999 = (3^3)^333,333,333 ≡ (-1)^333,333,333 (mod 7) ≡ -1 (mod 7) ≡ 6 (mod 7) (C) 3^1000 mod 10: you need to have skipped arithmetic not to know 3^4 = 9*9 = 81 ≡ 1 (mod 10) ⇒ 3^1000 = (3^4)^250 ≡ 1^250 (mod 10) ≡ 1 (mod 10) *P. S. I appreciate your video and thank you for it.*

This is incredible and brilliant.

Why does x + y = - x - y imply x + y = 0? Is it because x + y is some number but 0 is the only number that when multiplied by negative 1, it gives it itself?

pi log base e(e) vs e log base e(pi)

My final year undergraduate project involved at one point digging up the Bring-Jerrard reduction to x^5+x+C for some C, and the solution via elliptic functions. Expressing that C directly in terms if the original coefficients of a general monic quintic is complicated.

If you do the substitution x=2(1-cos t) then this becomes integral of the even function ln(2-2 cos t) from 0 to pi, so 1/2 of the same integral from -pi to pi. Now this is a special case of Jensen's Lemma, which says that integral of ln | a - exp(it) | is equal to ln max (1, |a| ), for an complex number a. Just set a=1 to recover the case at hand. (This result is used to define what is called the Mahler measure of a polynomial)

but both integrals can be used to prove it without second law, just by comparing them ( 1/T<=1/e , so the integral of 1/T from e to pi is less than just 1/e from e to pi), but still cool. I like it

You missed the fraction line at the cover picture

Fucking cool

Cool

Fun fact, we dont even know whether π^e is rational or not!

Amazing! Impressive! Wonderful!

After a series of Michael's videos on infinities, I have come to think that equations like *(x−2)² = x²* might have a "legitimate" solution "at infinity" (over the extended real line/complex plane with (a) point(s) at infinity), because both parts of the equation become infinite at *x = ∞,* and it can be solved (quite rigorously) with the definition of what an equation is.

Oh yeah I see it now. I was using 1+1=2 as my starting point. 😑

I have noticed that the infinite element *∞ = 1/0* from _Defense against the "dark art" of mathematics_ (wheel theory) also satisfies the last equation: *x + y = −x − y* *x + y = 0* or *x + y = ∞* but it surely doesn't satisfy the original equation (an extraneous solution).

Seems to work for any sum of first n kth powers. I just did it to find the sum of first n squares.

do not forget to plug an assumption on the second derivative into the conditions of the theorem

I can I can I can solve quintic, I use comptuer 🙂

is the Integral test good place to stop ist it

A nice little problem indeed

would be interesting to prove the same for the sum of first n squares and cubes

Coincidentally, the temperature of the *Cosmic Microwave Background (CMB)* is *2.726 K* - approximately *_e_* kelvins! So deep space can act as a heat reservoir for the experiment described in this video. 🤯

so clear and elegant proof! Thank you

Incompressible solid vs. ideal heat reservoir is clearly frictionless elephants whose masses may be ignored city. I love physics...

This is intense, it is like hitting a thumbtack with a big hammer. I will stick to my easy and cute way of proving it: Let S= 1+2+3+.........+n and also S=n+(n-1)+(n-2)+.........+1 Then adding the two sums gives 2S=(n+1)+(n+1)+..........+(n+1) [n times of (n+1)] 2S=n(n+1) S=n(n+1)/2 Therefore 1+2+3+.........+n=n(n+1)/2

hm. A bit disappointed. I thought the video would be useful -- which it is by no means at all. Usually, polynomials are solved with Durant-Kerner. All the companion and JT stuff is unnecessarily complicated/safeguarded, harder to implement, and slower on real-world hardware. So... with DK the issue is normalization, as is with companion. Hence the issue is always how to acquire an agnostic algorithm to order. Newton/JT first appear to do -- but then not really because of algebraic multiplicity, at which they need to know degree to apply fixture. But your approach hints at no fixture at all. So your presentation is rather the mental masturbation, to quote Lanczos.

Antimatter and black holes were first discovered with a bit of maths, being solutions to differential equations. It's good to see something like this helping to redress the balance.

Entropy!