Happy 233rd 🎂 0:12

Turns out Michael Penn is an alias used by The Mathematician, a member of the long-lost race known as Time Lords

@christianmartin8751

Жыл бұрын

Michael is also the last Highlander, may be one day we get to see his sword...

@ConManAU

Жыл бұрын

It’s been a long time since we had a “which is also my birth year” joke.

@morgengabe1

Жыл бұрын

The ordinators of operations 🎉

@_Heb_

11 ай бұрын

It's his *Penn name*

@RayArias

11 ай бұрын

Are you sure it isn't an alias for Michael Palin the comedian of Monty Python fame?

I love that year of birth running joke

12:05 zero is the new one 😮

So it actually speaks further about the expansion of zeta function near one. It's probably well known that (x-1)ξ(x) ~ 1 as {x->1} , but this limit gives us an extra term in the expansion. Namely lim {x->1+} (ξ(x) - 1/(x-1)) = γ ==> (x-1)ξ(x) = 1 + γ(x-1) + O((x-1)^2) , which may actually be useful in some applications

@sleepycritical6950

Жыл бұрын

Is that supposed to be the xi function or the zeta function?

😮 wait what 1790 is your birth year??

@krabbediem

Жыл бұрын

😂

@kushaldey3003

Жыл бұрын

I think it should be 1970 if anything

@andreyfom-zv3gp

Жыл бұрын

Prof. Michael Penn, 233 years old.

@bhargavsai8014

Жыл бұрын

Fr? I replayed that part again to make sure that my ears are working correctly I think prof Penn was just joking with us lol

vampire Michael makes a return

Mascheroni is Italian. The pronunciation in like Muskeronee

15:53

Now I'm hungry for oily macaroni

@gp-ht7ug

Жыл бұрын

😂😂

@andreyfom-zv3gp

Жыл бұрын

me too bro

guys i'm starting to think michael is not a real person but a transcendent deity. that would explain the consistency of the videos on this channel.

Ah The oiler macaroni constant

The identity that's being proven is really zeta(x) - 1/(x-1) = gamma when x->1. The limit with the sum gave me a bit of a whiplash at first, but you can put it in this pretty, succint way.

Please consider the Italian pronunciation: ma-ske-RO-ni. With a hard "C" like in "rectangle".

@bjornfeuerbacher5514

Жыл бұрын

Not oily macaroni? ;)

@LucaIlarioCarbonini

Жыл бұрын

@@bjornfeuerbacher5514 YUMMYYYYYYYYYYYYYYY!!!!!!!!!!!!!

The usual use of the term "anti-symmetric" has nothing to do with this, but the term "skew-symmetric" comes closer to mainstream usage (if a negative sign results); however, switching n and x actually yields the following: $\lim_{n\to1^+}\Sum_{x=1}^{\infty}\left(\frac1{x^n}-\frac1{n^x} ight)$, which is exactly the same as the Euler-Mascheroni constant.

The (1/x)^n part, looks a bit like geometric series. However, I will keep in mind, when integrating the sums (1/n)^x or (1/x)^n ,that it can be expressed in a fancy way, with eulers constant 😀 (As long the limits stay the same)

Ah yes! The old OILY MACARONI constant!

For a number whose derivation involves the harmonic series and which is found in such frequent proximity to pi and e, I'm mystified that it still hasn't been proven to be transcendental *or* irrational. Given its presence in the graph of the Riemann zeta function, it makes me wonder if the proof of the Riemann hypothesis will involve proving something about the properties of this constant... but I'm just a number nerd who gets irritated by long-unanswered questions. LOL

@13:42 shouldn’t the dx terms be dt?

@krisbrandenberger544

Жыл бұрын

Yes

@Happy_Abe

Жыл бұрын

@@krisbrandenberger544 thanks

Really nice symmetry!

i've got a question guys, maybe someone can help me. I do understand that lim(ln(n)-ln(n+1) = 0 But, why does this entail that you can just replace ln(n) by ln(n+1) in the definition of the Euler-Maschoni constant? Both the sum 1 + ... +1/n as well as ln(n) are divergent, right? The nice thing is that it becomes convergent, when you subtract both sequences from each other? So we are not talking about of the sum of two convergent sequences. That would be simple.

@AlcyonEldara

Жыл бұрын

Add and subtract ln(N+1) in the definition. Then split in two: 1) The def with ln(N+1) instead of ln(N) 2) ln(N+1)-ln(N) Since the LHS converges and 2) also converges, 1) converges and we can split the limit.

@nevoitzhak2092

Жыл бұрын

Because within the limit n+1 is equivalent to n so it doesn't matter.

@UdssRAP

Жыл бұрын

But it is the sum of two convergent sequences c_n=1+...+1/n-ln(n) = 1+...+1/n-ln(n+1) + ln(n+1)-ln(n) = a_n + b_n with a_n = 1+...+1/n - ln(n+1), b_n = ln(n+1) - ln(n) Since (b_n)_n is convergent, you get that (a_n)_n is convergent if and only if (c_n)_n is convergent and there limit is equal since b_n goes to 0.

@kilianklaiber6367

Жыл бұрын

@@UdssRAP O.K. Great, so assuming you know that the sequence as a whole converges then you can prove the equality in this manner. Thanks a lot!

Good pedagogical fun to tell students, Ln (x) is BASICALLY the sum of the reciprocals of the integers, and so it equals the sum of the harmonic series. They both diverge, but very slowly. What? They aren't EXACTLY the same? Well, what exactly is the difference between them?

Nice video

One of my favourite numbers. I'd bet any money that it's transcendental, but there doesn't seem to be any progress in proving that.

12:16 ---> true, given that 0 = 1

How is it valid to replace ln(n) with ln(n+1) in the limit? Yes they approach each-other in the limit, but what then allows us to replace it when combining it with a sum that doesn’t converge such as the harmonic sum here?

@UdssRAP

Жыл бұрын

c_n=1+...+1/n-ln(n) = 1+...+1/n-ln(n+1) + ln(n+1)-ln(n) = a_n + b_n with a_n = 1+...+1/n - ln(n+1), b_n = ln(n+1) - ln(n) Since (b_n)_n is convergent, you get that (a_n)_n is convergent if and only if (c_n)_n is convergent and there limit is equal since b_n goes to 0.

@TheEternalVortex42

Жыл бұрын

If lim (a_n - b_n) = 0 then lim (a_n + c_n) = lim (b_n + c_n) This is because lim(a_n + c_n) = lim(a_n + b_n - b_n + c_n) = lim (b_n + c_n) + lim(a_n - b_n) = lim(b_n + c_n) You can split up the limits in that step since each of the corresponding limits converges.

@Happy_Abe

Жыл бұрын

@@TheEternalVortex42 this makes sense if we can assume certain limits converge such as an+cn and bn+cn

14:07 I don't think you can do that? The antiderivate of n^(-x) is -n^(-x)/log(n), and with x in the exponent you can't just get that it's constant over a length of 1? In fact by integrating since it's a decreasing function you're actually making the value smaller so even if you replace '=' with '

@TheEternalVortex42

Жыл бұрын

There was a typo, it's supposed to be an integral with respect to t, not x.

I feel like if we just used commutativity/associativity, we could do away with limits. At that point continuity is just a question of commutator application.

I don't get why at 13:56 Michael pushes that term inside the integral without much reasoning behind it. I'm probably not getting something easy I guess...

@thomashoffmann8857

Жыл бұрын

It sounded like he assumed that it's a constant to x 🤔

@italyball2166

Жыл бұрын

@@thomashoffmann8857 but it contains x so it doesn't really work as a constant

@r.maelstrom4810

Жыл бұрын

@@italyball2166 Because he mistakenly replaced dt for dx. So 1/n^x is a constant inside the integral with respect to t.

@italyball2166

Жыл бұрын

@@r.maelstrom4810 Ah, this makes so much sense, thank you 😅

@ 14:08 Should be dt, not dx.

So which level is this maths ? PhD or Masters?

It's not obvious to me at the end (@13:30) why the difference of the two terms is always positive, though I guess it doesn't matter, as it seems we'd just need it to be nonzero to apply the M-test.

@minamagdy4126

Жыл бұрын

Ignoring the ^x as a monotonically increasing operator, for n

@xizar0rg

Жыл бұрын

@@minamagdy4126 thx

Happy 233rd birthday. You look hardly a day over 150!

In which cases is this γ constant used? Thanks

@bjornfeuerbacher5514

Жыл бұрын

Do you mean "used in mathematics"? In physics? In engineering? Or what? In mathematics this constant appears in a lot of places, especially in connection with the Gamma function and the Zeta function. In physics, it appears mostly when these functions are important, for exampe, in the statistical mechanics of quantum gases. In engineering? Probably never. ;)

@christianmartin8751

Жыл бұрын

@@bjornfeuerbacher5514 Or may be in pasta engineering ?

Hi, Hi, 12:06 : x->1+ and not x->0+

Г ' (1) = -γ

How can the difference of these series be finite when the two series diverge at vastly different rates? One becomes the harmonic series and the other becomes 1+1+1... Surely the latter would overpower the former? Computing out a million terms agrees with this.

@j.d.kurtzman7333

Жыл бұрын

You’ve now discovered why you can’t willy nilly change the order of limits and sums. Note that x>1 (strict) so it isn’t the harmonic series and a bunch of 1+1+1… it’s two series that are of same order, then you take limit

@MathematicFanatic

Жыл бұрын

@@j.d.kurtzman7333 How can one show this result computationally? I am picking x=1.00001 and then evaluating those series to N=10^6 terms each and then subtracting. The result is a very large negative. I don't see how the order of the sum or limit is the problem, it doesn't seem to matter whether I shrink x to zero first or grow N first, the result is the same.

@j.d.kurtzman7333

Жыл бұрын

@@MathematicFanatic honestly don’t know, you might need more terms. Harmonic series diverges slowly, so sum[1/n^1.00001] probably converges quite slowly and the later terms may matter. Eventually 1/n^1.000001 > 1/1.000001^n

@vkessel

Жыл бұрын

That's what confuses me too. It seems like the result should be negative infinity. Perhaps more terms are needed but when you graph it, it seems like there is definitely an asymptote going to -inf from the right at 1. Edit: Analyzed the results. The first of the anti-symmetric terms doesn't become greater than the second until nearly a million terms, and then it has to make up the distance but does so veeeery slowly since it's logarithmic growth. Too slowly to compute effectively. Despite some trivial errors others have pointed out, the steps in this proof are correct and I'm confident that it would indeed reach the constant as it approaches infinity.

@TheEternalVortex42

Жыл бұрын

You have to do the sum first before taking the limit, not the other way around.

Euler-Mascarpone constant

@bjornfeuerbacher5514

Жыл бұрын

Oily macaroni, you mean. ;)

I pressed the like button the 666th time. 😈

When I first read the title of this video I was confused because I misread ‘anti-symmetric’ as ‘anti-Semitic’, then I realised that it’s obviously anti-symmetric.

It looks like you were cheating at 12:10. When you take the difference between 2 diverging sums you can construct any number you want.

@TheEternalVortex42

Жыл бұрын

x>1 so both sums are convergent

@grafrotz5286

11 ай бұрын

@@TheEternalVortex42 but the limits are divergent if ou treat them separately. As a physicist i have no problem interchange limit and sum, but the mathematicians always tell me you have to proof very carefully if it is allowed. The convergence is only in the sense of principal value.

Mascheroni - wrong pronunciation - UNEDUCATED!