Hey

[@blackpenredpen]

😮!

[@thefunpolice]

I don't know how to evaluate "shocked face factorial". Is there some Sterling approximation for that?

[@tommyliu7020]

I tried Gamma Function but it didn't work!@@thefunpolice

[@lexinwonderland5741]

Super late (yes, I've been binging the channel for 3 days lol) but I would LOVE a video comparing standard vs exponential vs Lambert vs etc generating functions, and how different generating function TYPES produce different functions with the same sequences. Also, please, more non-elementary functions!! The back-catalogue has content on the polylogarithm and on Ei/li, and I know you posted a video on the Fresnel integrals recently, but still I think your non-elementary function content is some of my favorite! I feel like it expands the mathematical vocabulary we're taught heavily through high school and college in a natural way, and satisfies many curiosities about those topics that education left unsatisfied. Maybe a video comparing various non-elementary functions, or explaining what makes a function non-elementary? In any case, keep up the great work!!

[@sergeipetrov5572]

Thank you so much for the video!

[@jkid1134]

Bprp also calls the power sum closed form expression "best friend", and it's prominent here.

[@evanev7]

The audio balance is really lopsided on this one, unless you have a particular reason to would you consider uploading your videos with mono audio? The audio is also flipped (so when you are on the left, the audio output is louder on the right) which makes it even more offputting!

[@Fleck2011]

yeah, it's impossible for me to listen to this with headphones.

[@Willwantstobeawesome]

I think actual mono often shows up as left only

[@putputlawch6770]

typo at 10:37, should be phi(d)

[@mrphlip]

Similarly at 14:57 should be mu(d)

[@lgooch]

Consequently, at 19:17, it should gamma(d)

[@Horinius]

Yeah, another annoying mistake from Michael. Even though the result is correct, the reasoning is very different. That could lead to students learning the wrong thing.

[@putputlawch6770]

@@Horinius If students didnt know that everyone can make mistakes they better learn that quickly

[@Horinius]

@@putputlawch6770 I never wrote that "students didn't know that everyone can make mistakes". Where? But if nobody points out something is wrong, even the best students cannot know they've been learning things wrong. How could they? Take relativity as an example. Scientists in the past centuries have been wrong about space and time, even though they knew "people could make mistakes" as you said.

[@zathrasyes1287]

You make truly many good videos! Thx for the great content.

[@jkid1134]

I didn't realize this was a Michael Penn video 😂

[@goodplacetostop2973]

7:31 Homework 18:07 Good Place To Stop

[@thefunpolice]

I like these BlackPennRedPenn videos.

[@charleyhoward4594]

not a mathematician, an engineer, but I do get a little out of these videos

[@zuzaaa1998]

If there was a simple way to get back the sequence from Lambert series it would be such a powerful tool. From those simple closed forms that you derived we could get distribution of prime numbers

[@sleepycritical6950]

There is a way though. Just use the Mobius inversion on the sum over divisors of n to get back a_n.

[@zuzaaa1998]

@@sleepycritical6950 Oh I hadn't known about Mobius inversion beforehand, but as you said you need to know divisors of n in order to use it so that wouldn't help in what I was talking about. Nevertheless it is a nice tool, thanks!

[@sleepycritical6950]

@@zuzaaa1998 you could convert it into a straight sum with coefficients equal to 0 if it doesn't divide n and 1 otherwise. From there you could use the Abel summation formula to convert it into some sort of integral over an arithmetic function and then you could use a continuous functions that approximates the arithmetic function to evaluate the integral.

[@Pele-wx5dm]

Let r = # of distinct prime factors. Each divisior is a combination of its prime factors and the amount of them is given by binomial coefficients. Sum (-1)^k (r choose k) from k=0 to r is 0 which can been seen after expanding (1-1)^r. So sum Möbius(n and squarefree) over its divisors is 0. And the extension to non-squarefree will not change the sum since each additional combination containing prime to higher power than one will be zeroed by the Möbius function.

[@noahtaul]

Quicker: for any prime p dividing n, we can pair up the divisors not divisible by p^2 as (d,pd) where d is not divisible by p. But either d is squarefree, in which case d and pd have different parities of prime factors, or p is not squarefree, in which case dp isn’t either. In either case, mu(d)+mu(pd)=0. So all terms cancel out to 0, just from the fact that it has at least one prime factor.

[@Ensivion]

neat, I don't know much number theory, but the key jump seems to be the "reindexing" or as I interpret it as a "reshuffling" of these terms. It took me a while to really sketch out what it was doing. It's amazing how just that change in perspective and intermingling of series creates these three clean results. I think this is probably related to bigger mathematical structures.

[@Alex_Deam]

Different kinds of reindexing/reshuffling like this are used in lots of number theory proofs, so definitely useful to get your head round

[@saboorhalimi]

absolutely nice video by the way does someone know which font is used in his video's thumbnail? I really want to know.

[@Nikolas_Davis]

The Return of Penn *q* (aka, the 'q-bar')

[@dave-bk6vt]

Is that an American thing?

[@franksaved3893]

I suppose the identity Sum(phi(d)) for d|n =n can be derived using the convolution product for arithmetic functions

[@admink8662]

Let k > 1 be a square-free number, and p is a prime factor of k. We can make a partition of the set of all factors of k becomes 2 sets, A and B. A is the set of all factors of k that is divisible by p, and B is the set of all factors of k that is not divisible by p. We can make one-to-one correspondence f:B->A such that f(x) = px for every x in B.

[@SuperYoonHo]

thanks!!!

[@euleri0]

Yo

[@SuperYoonHo]

@@euleri0 hi

[@abrahammekonnen]

4:24 I guess my question here would be why change the divisor from n to d?

[@columbus8myhw]

The Möbius function makes me think of the following exercise. Consider the 100x100 matrix A whose entries A_m,n are 1 if m is a factor of n and 0 otherwise. What is det(A)? (Hint: triangular matrix.) What are the entries of A^(-1)?

[@christianimboden1058]

Can you please get your interns to find the links to all of your previous videos that you reference in your videos (and then put them in the description)? One person hunting for it would be a lot more efficient than dozens of viewers separately looking

[@craig4320]

Answer found in econ 101...

[@chrissch.9254]

This somehow reminds me of modular forms which I have forgotten nearly everything about - have you ever made a video on modular forms? 🙂

[@wavyblade6810]

10:53 does anyone have a link to the video with the proof?

[@dalibormaksimovic6399]

Can we somehow find the sum of k=1 to infinity of zeta(k) * x^k? Zeta is Riemann zeta function.

[@chantlerweatherspoon5273]

It converges when the absolute value of x is less than 1

[@frankwilhoit]

Must did somebody tell you to slow down a bit and make things a little more explicit. I think this is a gain. Please carry on in the new style.

[@aweebthatlovesmath4220]

Hey, does anyone know why is it defined like this?

[@ericthegreat7805]

Shoupdnt it be -q/(1-q)^2?

[@Alex_Deam]

No, because the chain rule means you pick up an extra negative from differentiating 1-q

[@hamzatox]

What is a_n is the Von mongoldt Λ(n)?

[@Alex_Deam]

It works out nicely as log(n)

[@ayoublaghrissi4957]

hello,may i know why do we use this expansion?😊what it is for?

[@IoT_]

Probably you're an engineer or a physicist 😁

[@zuzaaa1998]

@@IoT_ They can be asking about mathematical application, you can come up with many different weirdly looking sums but if they do not show up in any field of maths no one will care

[@IoT_]

@@zuzaaa1998 I know

[@ayoublaghrissi4957]

@@IoT_ ye engineer

[@gouharmaquboolnitp]

Any jee aspirants????