I'd say the decision was right to leave the "incorrect" indexing-reindexing segment in the video. It comes to show that solving a problem is not a simple matter of reciting an answer-as one typically encounters as a student in the academia-but rather a long and iterative process. Sometimes you just make a mess out of the whole thing and have to restart the whole process. Even seasoned veterans happen to make these errors. Kudos to @Michael Penn for not being afraid to show us the "behind the curtain" workings.

@farfa2937

Жыл бұрын

True, tho I'd assume he does the problem first and then shoots the video just copying/explaining the solution.

@SzanyiAtti

Жыл бұрын

I don't think this was a conscious decision, but a simple editing mistake, as it has been already cut out.

@leif1075

Жыл бұрын

I broke up the denominator agor I to m squared plus 3 plus 1 minu@ 2n or us 2n wouldn't many ppl do thst instead kf the n minus 1 or n plus 1 squared thing..it's a matter of luck which is something that suclsabout math..it's not about intelligence sometimes its amateur of picking the lucky way to breal it up there's there's reason to suspect or conclude that breaking it up that way would worlbetter than how I did it..

@leif1075

Жыл бұрын

What the hell rules not holding is he talking about at 6:10..this doesn't make sense. Isn't clear at all..we ar÷ loking for a sum nkt the limit..I get limit will tell you what the values approach but it won't tell you the sum..

@hybmnzz2658

Жыл бұрын

@@leif1075 the formal definition of an infinite product is the limit of partial products. In this case the product up to N.

No clue why this isn’t one of the biggest math channels out there (it sort of is comparatively) but I truly think that Michael Penn and crew are playing and will play a pivotal part in reigniting mathematical interest. Know you’ve got fans rooting for you! Can’t wait for y’all to reach 3.14 MIL subs!!

I had a similar approach. Instead I just took the natural logarithm so that I could work with sums instead of products. Nice problem at the end of the day. Good work Michael!

One of your best videos so far (together with the one about the arithmetic derivative) !

Thank you, Michael! That's just awesome explanation! Got it!

This was super fun for me! Most of your exercises I can follow, but they're way above my pay grade and at no point do I have any illusion that I could solve it on my own. This one was perfectly at the edge of my comfort zone such that I was able to clearly see the next step forward the whole way through, even though I would have been scared of the amount of paper required on my own. Please don't think I'm complaining about other videos though, I love that you're not afraid to expect a lot of the audience and stylistically I think you strike the absolute perfect balance of clarity, thoroughness, and rigor. (and occasional backflip-cut)

love you Michael!

Nice plug to the channel :)

I think there's an editing mistake. You go through the whole reindexing process twice

@blindConjecture

Жыл бұрын

Agreed. And in the first pass (the one that should have been edited out) there's a mistake when you combine the (n-2) and (n+2) products, you should have more terms left over on each end

just clear! Thanks

The skit was very cool! Also plugging your channel once in a while is fine lol.

Love watching and picking up new strategies for solving equations. Reindexing, very cool indeed!

4:41 Dude that's a really crisp "P."

Nice trick factoring *n^3* first in the denominator! If you did not think of that, here's an alternative: *n^6 - 64 = (n^2 - 4) * (n^4 + 4n^2 + 16)* *= (n - 2) * (n + 2) * ( (n^2 + 4)^2 - 4n^2 )* *= " * (n^2 + 4 - 2n) * (n^2 + 4 + 2n)* *= " * ( (n - 1)^2 + 3 ) * ( (n + 1)^2 + 3 )* In the second step we write the last factor as a _difference_ of squares to avoid complex factorization and _a lot_ of extra work!

You'll never reach π*100k, you'll just pass by.

Cool. Good idea to link to another video at the end. That's going to give views I bet.

5:40 We now return you to your regularly-scheduled program.

Fun. I liked the editing. Maybe some math majors found it too. Thanks.

Pretty simple as soon as you get the factorization (although I did look up the factorization)

Gracias.👏🏽

Magic blackboard does it's thing at 6:28!!!

This one was wild

That plug was so adorable!

Just felt like you could appreciate this infinite series, make it into a video maybe ? Sum for n=1 to infinity of 4n/(4n^4+1) The idea is to use Sophie Germain's identity to have a sum of term 4n/(((2n^2)^2+2n^2+1)*((2n^2)^2-2n^2+1)) and then to use partial fraction decomposition and it follows from there somewhat

@leif1075

Жыл бұрын

What if you don't know that darn identity almost ppl dont..I think there is another way to solve right by breaking 4n4 us 1 I to 2n squared us 1 minus 4n and proceeding from there right or something similar probably will work

@alexismiller2349

Жыл бұрын

@@leif1075 Sure I agree, I just thought since this channel is an educational one, maybe he could present the identity. As far as non elementary factorizations go, it's maybe the most famous one ? (except cyclotomic factorization maybe...).

@carstenmeyer7786

Жыл бұрын

If you don't know _Sophie Germain's Identity_ (SG) you can write the denominator as a _difference_ of squares to avoid complex factorization (and _a lot_ of extra work you would otherwise have to do^^). These steps are exactly what you do to prove SG : *f(n) := 4n / (4n^4 + 1) = 4n / ( (2n^2 + 1)^2 - 4n^2 )* *= 4n / ( (2n^2 + 1 - 2n) * (2n^2 + 1 + 2n) )* Via partial fraction decomposition you can rewrite *f(n)* as *f(n) = g(n) - g(n+1) | g(n) := 1 / ( 2n(n - 1) + 1 )* Now the sum you want to calculate has been transformed into a telescoping sum: *\sum_{n = 1}^N f(n) = \sum_{n = 1}^N g(n) - g(n+1) = g(1) - g(N+1) = 1 - g(N+1)* As we take the limit *N -> \infty* the term *g(N+1)* vanishes and the value of the series equals *1* .

@leif1075

Жыл бұрын

@@carstenmeyer7786 yes that's what I was getting at I was just too lazy and dint care enough to do the rest lol but it's not less smart than using Sophie and since if you don't know Sophie's identity indont think anyone will derive it..wouldn't you just do that? Although donyou need to do partial fraction defompneither..immused to only thinking kf thst in a calculus integrwl.context not just for algebra..aren't you?

@alexismiller2349

Жыл бұрын

@@carstenmeyer7786 Thank you for completing it :D

Very Good Place To Stop: 20:43

I have a feeling that those setting the test question were thinking of Riemann sums instead, converting the logarithm of the product into a (regular, right-endpoint) Riemann sum of the function for the function f(x) = (log(1+x))/x = sum_{j = 1}^\infty (-1)^{j-1} x^{j-1}/j over the interval [0, 1]. The limit of the Riemann sums is an integral that can be computed term by term, leading to sum_j (-1)^{j-1} 1/j^2, which is then finished off as in the video. Seems a little trickier than ordinary for a Math GRE question, as least when I took the Math GRE many years ago.

5:09 bro💀

I like the technique, but the video is riddled with calculation errors.

Most awkward Math-Nerd commercial ever! Love it!

Wow! This seems like one I could do…

10:20 Here we go again, saying 4 while writing 5 and then correct it back to 4. 🤣

@ 9:15 the upper bound on the product should be N+2, not just N. Also, @ 10:22 the lower bound on the product should be 4, not 5.

@omerhalaby2337

Жыл бұрын

He fixed it

@krisbrandenberger544

Жыл бұрын

@@omerhalaby2337 Thanks!

5:40 For a second I thought he said he wanted to reach "π - hundred thousand subscribers", and they are kind of close to that! 😄

@DeletedUser410

Жыл бұрын

He did. You did not mishear

quartic polynomial cancels? 13:11 ; how can u be so sure ?

@PeperazziTube

Жыл бұрын

When N approaches infinity, small additions to N (N+1, N-1) do matter anymore and (N+1)/(N-1) behaves as N/N = 1

@RabbidSloth

Жыл бұрын

You should always check things for yourself on paper if you're unsure. Generally though, this is just what happens when taking limits of rational polynomial expressions. If the degree of the numerator and denominator are equal, they cancel.

@robertveith6383

Жыл бұрын

@@PeperazziTube * *do not matter*

which definition of "infinite product" are we taking here ? Under the definition of lim{k->+inf} product{n=3;n=k} f(n), which is i think the usual definition, like for infinite sums, the reindexing does not make sense. For example if we say f(n)=(n+2)/n , a reindexing would give us (product{n=5;+inf} n)/(product{n=3;+inf} n) = 1/12 . Which is clearly absurd because each term is >1. And in fact if we evaluate the first products : [k=3] 5/3, [k=4]5/3*6/4, [k=5] 5/3*6/4*7/5 = 7/3*6/4, [k=6] 7/3*6/4*8/6 = 7/3*8/4, and in general, 1/12 * (k+1)(k+2). We can see that it does not converge under the usual definition. (maybe for another definition, much like under a certain definition of sum we can say that the sum of natural integers is -1/12, but defining it would certainly be complex) With the function of the video, f(n)=(n³+3n)²/(n^6-64), the first terms of the product from 3 to k are (rounded) : [k=3] 4.38 , [k=4] 13.64 , [k=5] 35.07 , [k=6] 79.01, [k=7] 161.33 ... So it diverges (and as f(n)>1 for n>3 it will not come back down). The definition that seems to be taken in the video is considering the ordered sets of the numerators and the denominators, and "matching them" to get the limit. However it does not seem to me that this is neither a valid definition nor a rigorous one, as we could get any result by associating the different terms (not even commutating) with infinite sets. Am I missing something ? Thank you

2 squared is indeed 4

After this video you really need to change the name of the channel to pinkchalkgreenchalkbluechalkredchalk. ;-)

did you say 500,000 or pi hundred thousand?

@cycklist

Жыл бұрын

It was pi hundred

The answer is -64

The bit where MP splits out the one product series into "many" - can you do that? I worried about cross terms..... Disclaimer: It's Friday and I've had a few drinks :D

@romajimamulo

Жыл бұрын

Yes, if it's a finite product, which it is at this step

@MyOneFiftiethOfADollar

Жыл бұрын

I believe all the algebra he did with product notation is valid due to distributive, commutative and associative axioms. He mentioned it briefly.

I think it wpuld be better for the chanel if u ask "please CONSIDER subscribing" it makes ppl stop to think and its less anoying

Should not feel remiss about promoting your channel occasionally, especially given the attention seeking thumbnails I’ve seen with people claiming Olympiad level problems or genius if you can solve this etc. So many actually rely on KZread income which creates intense competition for page views. “Minor tweaks “ in the algorithm can radically alter one’s page views. Creators really have no idea if they are improving as creators OR if algorithm modifications caused the peaks and valleys

Good place to stop: 13:56

@martinb3000

Жыл бұрын

Good place to restart: 14:13

@SuperYoonHo

Жыл бұрын

yeah right it is 20:43

@SuperYoonHo

Жыл бұрын

2:12 lol

Food for thought: If every subscriber made another account and subscribed with that new account, then we would’ve helped @michaelpenn reach his goal. P.S. Tim is an a alias, I am an avid subscriber of the channel (more than one account 😉), but I wanted to remain anonymous incase my comment offends someone!