Eulerology? Euleristics? Euleritics?

In today's finding of Euler's class, we will learn ...

or Cauchy, Riemann or Gauss

@@petterituovinem8412 i don’t think that will “cauch” on.

Euleritics or Eulerology?

"Flammable!"

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There's no good place to Stop 😢

It is true, but it requires the dominated convergence theorem, it's not trivial

@@giovanni1946 il faut appliquer le théorème de convergence dominée à la fonction fn(x)=0 si x>n et fn(x)=(1-x/n)^n si x

@@Raphael-wg7zi Exact

There are references below to the DCT being "required" to understand what is going on. In fact this is a legitimate question that arose in classical analysis long before Lebesgue. In his book " A Course of Pure Mathematics" Hardy deals with relative rates of convergence (see problem 16 pages 167-8) which are at the heart of Tauberian theory. Littlewood's 3 principles ( Every (measurable) set is nearly a finite sum of intervals; every function (of class Lp) is nearly continuous; every convergent sequence of functions is nearly uniformly convergent) ensure that most of the time you can get away with the "obvious" (assuming of course that you have checked that the hypotheses are satisfied).

f(x)=(1+x/n)^n, n tends to infinity is a formal definition of e^x, proven by taking derivative.

Yes. That is correct.

He's right bro

Totally! ^.^

Based replier🙏

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3) plug in n=99

1. Other constants include the Mertens constant, closely related to the Euler Mascheroni constant. 2. Both of these numbers often appear in number theory, specifically when dealing with prime numbers. 3. There are like many other ways of approximating gamma by manipulating the original series or limit but yes.

It appears everytime in Quantum Field Theory. When you have to compute Feynman Diagrams with 1 or more loops you need to approximate the Gamma function. Since the derivative of Gamma(z) at z=1 is -γ you see it often

I pronunce it with a Spanish "sh" also know as German "Sch"

@@friedrichhayek4862 it's a common mistake, but that's italian

@@friedrichhayek4862 Complete nonsense! It´s neither German nor Spanish, but Italian!

@@friedrichhayek4862 Your pronunciation is wrong. It is a Italian name. Pizzamid Head is right with his commentary.

Just separate the "s" from "ch": Mas-cheroni ("Mas-keroni") (English native speakers have their own way of pronunciation of foreign words / names: mostly false *lol*)

it's increasing as you add the terms, not each term separately

which is about the point I fell off the oxcart, and from the edge of the road watched it recede into the distance

Correct, think about it. If n=2, you'd only have 1 integral, from 1-2. If n=3, you'd only have 2: 1-2 and 2-3. For arbitrary n, n-1 integrals

Exactly my thought when watching this for the first time: seen from a standpoint of formal logic, it seems he's renaming a bound variable (the one in the limit for e^-x, which SHOULD be different from n) to a name that's (locally) free inside the integral - which COULD be an invalid rename!

Another viewer with more mathematical knowledge in measure theory than me commented that it is correct, but it requires the dominated convergence theorem and is thus non-trivial. If you wanna look for that comment. I looked up the theorem on wikipedia and although the theorem seems logical enough, I'm not sure how it explains that taking the limit this way is correct.

Just like pi, e, zeta(3) and other constants, it often appears as a part of answers for weird/non-elementary integrals, sums and other problems.

The random wikipedia equations which mathematicians cook up for no reason

I hope other people will chime in with uses! I like Mathologer's video on the harmonic series, which also deals with the Euler-Mascheroni constant. Essentially, if you want to compute the nth partial sum of the harmonic series, this is quite tricky to do since there is no known nice formula for the nth partial sum. However, you can approximate it using the natural log function (which we can compute to arbitrary precision!). The error between this natural log approximation and the partial sums gets closer and closer to the Euler-Mascheroni constant. So if you want to approximate 1+1/2+1/3+...+1/n for some large positive integer n, you can compute ln(n+1)+γ, and this will be a very good approximation. But this is just one application, and it's not too hard to believe based on the limit Michael proved in this video. I would also love to hear what other people say!

One application of this constant, is the Laplace transform of natural log. A method that is commonly used for converting Calculus into Algebra, as a strategy for solving differential equations. I've tried to find an example problem where it would be practical to solve, that starts with natural log, and uses its Laplace transform to solve, but I can't seem to come up with one. If anyone can suggest one that works, please let me know.

rubbish

The sum 1/n is more than the integral 1/x dx from 1 to n+1 (draw a box of height 1,1/2,...,1/n on graph y=1/x), and integral is ln(n+1) which is more than ln(n), for all n. Therefore the constant is strictly more than 0. The proof that it's less than 1 is in video 5:06 (it's also strict inequality)

@@warmpianist What you just said are arguments to show that a_n is strictly between 0 and 1. The limit then could still be equal to 0 or 1.

@@eaglesquishy a_n is more than 0 for all n, and it is strictly increasing (also proven), therefore the limit is more than 0. And from the video the limit (not a_n) is proven to be less than 1 in the video

​@@warmpianist a_n is actually decreasing (not proven or disproven in the video). Also, if you look closely, the limit was shown to be less than or equal to 1.

Segue (Italian for 'follows') - a Segway is an electric vehicle! 😉

Oh I think I did at least some extra work

I think u mean 1/k not 1/(k+1), pls be careful about the +- sign of the decreasing function

You mean, "oily macaroni". ;)

Props to Papa Flammy for coining the alias "oily macaroni constant"

A sign of missing education.,

Michael gets it slightly wrong: the C is hard, so it starts like 'mask', not 'mash'.

@@davidgould9431 thx

One application of this constant, is the Laplace transform of natural log. A method that is commonly used for converting Calculus into Algebra, as a strategy for solving differential equations.

@@carultch quite interesting, Math is about building tools to simplify analysis. Its always nice to know where do these tools are applied. Not just about learning about random symbolic facts. So thx for the application!

@@CppExpedition Indeed. Laplace Transforms are an awesome tool. I've tried to come up with an example where you could use the Laplace transform of natural log to solve a DiffEQ, but I haven't had any success thus far. Every example I try, seems to stump Wolfram Alpha. It's much easier to use the method of Laplace transforms when the DiffEq uses trig, exponentials, algebraic functions, and impulse/step/ramp functions, since they all have algebraic Laplace transforms that are practical to untangle.

@@carultch don't worry, either way i would calculate any fourier transform through numeric FFT.

no, both are true

@@Pablo360able ...kinda like asking equivalent geometric question about translating parallel lines (or the 2 lines endpoint extensions to infinity ...does a pair of parallel lines remain parallel(=) ...or.....notParallel(~=)....@ infinity

@@abdonecbishop it's like exactly none of that. what you are saying is mathematical word salad.

@@Pablo360able you sound really confident...to bad you are wrong .. kzfaq.info/get/bejne/erWen8akzs6pk4E.html

@@abdonecbishop gonna need to timestamp the relevant part because I don't see what singularities of pairs has to do with you thinking that an expression involving rational numbers can't have an irrational value

It is! I think he's made a video about it before. You can also verify it by seeing how each side of the equation arises as the sum of the entries in a multiplication table.