A viewer suggested floor integral!

[@mooncowtube]

A little slip at the very end: when you multiply the numerator and denominator of (1/b)/(1-(1/b)) through by b you wrote b/(b-1), but that should be 1/(b-1) for the final answer.

[@backyard282]

Everything so perfect and then that slip at the end :D Great video nevertheless

[@jagula]

8:18 that chalk fell xD

[@pedromooregaissler6378]

not a true Mathematician if you don't find typos hahahahah

[@vardaandua3585]

The last step is wrong...it should be 1/(b-1)

[@NagyonCsoki]

No, the last step is correct, because the first term of the geometric series is not 1. As it was mentioned, the first term is a=1/b and the common ratio is r=1/b, so the infinite sum is a/(1-r). You may check it here: en.wikipedia.org/wiki/Geometric_series

[@mooncowtube]

Yes, the geometric series sum with a=1/b and r=1/b is done correctly, Levi4234, to give (1/b)/(1-(1/b)), but when Michael multiplies the numerator and denominator of that through by b at the very end he wrote b/(b-1), but that should be 1/(b-1).

[@ricardocavalcanti3343]

@@NagyonCsoki So a/(1-r) = (1/b)/(1-1/b) = (1/b)/((b-1)/b) = 1/(b-1).

[@noway2831]

It's always the last step

[@candamir26]

Yap, 1/(b-1) should be the final answer.

[@goodplacetostop2973]

5:05 Let’s go ahead and do that 11:44 That’s a good place to stop No homework today but don’t forget to subscribe to Good Place To Start, hopfully a future podcast by Michael.

[@Dionisi0]

that's a bad place to stop

[@goodplacetostart9099]

Lemme tel you 0:01 is the right place to start

[@tonyhaddad1394]

What is the last homework you did share it with us ?????? I need it 😥

[@goodplacetostop2973]

@@tonyhaddad1394 I’m too lazy right now to dig in KZbin comments 😂 Just give me one topic you want an homework for and I’ll see what I can do for tomorrow

[@tonyhaddad1394]

@@goodplacetostop2973 dont worry bro i m just kidding about now but when you can give us a homework do it i like math problem a lot 😍

[@Cosine_Wave]

4:02 I like that edit

[@hsjkdsgd]

The little mistake in the end has been deliberately done to see who actually watches the videos carefully.

[@gustavowadaslopes2479]

This is the most teacher response ever.

[@nagamanikomarla5376]

@@gustavowadaslopes2479 honestly, yeah. My math teacher sometimes gives mcq tests where all options are wrong. She expects us to write “none of the given options are correct” as the answer.

[@artemetra3262]

​@@nagamanikomarla5376 pure torture

[@papanujian7758]

Amazing. You just explained this problems solution with a beautiful way. Can't hardly waiting for another videos

[@xCorvus7x]

Instead of summing the rectangles sideways we can also notice that summing up from the right to the left gives 1*(1/b - 1/b^2) + 2*(1/b^2 - 1/b^3) + 3*(1/b^3 - 1/b^4) + 4*(1/b^4 - 1/b^5) + ... + (n-1)*(1/b^(n-1) - 1/b^n) + n*(1/b^n - 1/b^(n+1)) + ... = 1/b - 1/b^2 + 2/b^2 - 2/b^3 + 3/b^3 - 3/b^4 + 4/b^4 - 4/b^5 + ... + (n-1)/b^(n-1) - (n-1)/b^n + n/b^n - n/b^(n+1) + ... . Here every subtracted term is canceled by the next added term yielding the same sum you got by summing along the y-axis.

[@sharpnova2]

this is called telescoping

[@M-F-H]

@@sharpnova2 not exactly, telescoping is (usually) when they completely cancel and you're left only with the first and the last term; here it's n•a(n) - (n-1)•a(n) = a(n) for each pair of neighbouring terms. But yes, roughly the same idea.

[@jacobbills5002]

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[@camrouxbg]

That is really slick! And fairly easily accessible to high school students. Thanks!

[@petermueller162]

Cool trick on rotating the rectangles. I did a u-sub at an earlier step instead of graphing and ended up getting an alternating series that represents the original rectangles. Well done and thanks for the fun problem

[@jacobbills5002]

Just watch this impressive Maths channel... it’s very nice like this kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@PlutoTheSecond]

Interesting that, for your second tool, you had strict inequalities at both bounds, but you could have had a non-strict inequality for the upper bound (i.e. 1/b^(n+1) < x ≤ 1/b^n). That's because f(1/b^n) = floor (log_b(floor(ceil(1/b^n)/(1/b^n)))) = floor(log_b(floor(b^n))) = floor(log_b(b^n)) = floor(n) = n.

from Morocco thank you for your clear full complete explanations you are a genious

[@sephimarzliclican6224]

michael penn is one of the reasons i’m still alive rn

[@mihaipuiu6231]

The Demonstration of Michael Penn is very simple, very clean, nice, and interesting.

[@burhan8795]

at 4:24 x cannot equal to 1. its definitely strictly less than 1 if n is a natural number and b is an interger bigger than 1

[@ThePhysicsMathsWizard]

wow, i've never seen anything like this, thanks for making this video

[@alxjones]

This one was on the MSU problem corner, I remember solving this with my calc 3 professor back in the day. We also did it with the square of the integrand, I remember we got (b+1)/(b-1)^2. Not sure it was right though!

[@Div1nePiece]

Isn't it 1/(b-1)? Excellent problem by the way!

[@walidabdelal]

Floogarithms! Excellent brain massager. Cheers Michael!

[@digxx]

Not sure what I get wrong. Checking numerically for e.g. b=1.1 this integral gives 8.0425 roughly which fits nicely with \sum_{k=1}^\infty 1/\ceil{b^k}. However your result would imply \sum_{k=1}^\infty 1/b^k = 1/(b-1)=10 which is different...

[@sharpnova2]

why are you dropping latex or whatever in a KZbin comment rofl

[@digxx]

@@sharpnova2 Because for everyone familiar it is clear and unambiguous. Anyway, I think at 6:30 the inequality b^n b^n it does not follow \floor(1/x)>=b^n. Oh, I think I missread b is supposed to be an integer; that changes things... But the version above is general.

[@richardlinsley-hood7149]

The best integral is to use the average (mean) instead of delta being either a ceiling or a floor function. f(x + delta), where delta tends to 0 or f(x - delta), where delta tends to 0 are worse than f((x + delta) / 2), where delta tends to 0 This is accurate for straight lines precisely, even with large deltas. It is more accurate than either a floor or a ceiling value for all curved lines for deltas short of delta = 0.

[@jacobbills5002]

Just watch this impressive Maths channel... it’s very nice like this kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@kevinmartin7760]

I think the result is the same without the inner floor function, provided (as stated) that b is an integer. Conversely if you keep the inner floor function the requirement that b be an integer can be removed. When x >= 1 we have (as shown) 1 = 2 then 0 1 so log_b(ceil(x)/x) will only cross an integer boundary when ceil(x)/x does (again because b is an integer, this only happens when ceil(x)/x crosses an integral power of b). If you relax the integral requirement on b, my second observation would not hold because ceil(x)/x could cross a power of b not at an integer boundary, and also b could be between 1 and 2 so my first observation would also not hold. The turning-the-graph-on-its side step is a bit hand-wavy, but I can see how you could first express the integral as a sum of the original (pink?) rectangles, turn that into a sum of sums: sum(all pink rectangles)[f(x)(width of the pink rectangle)] becomes sum(all pink rectangles)[sum(n=1 to f(x))[(width of the pink rectangle)]] then you reverse the summations to get sum(n=1 to infinity)[sum(all the pink rectangles at least as tall as i)[width of the pink rectangle]] finally reducing to sum(n=1 to infinity)[1/b^n]

[@jacobbills5002]

Just watch this impressive Maths channel... it’s very nice like this kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@mokkapatisiddharth5793]

I cannot believe I solved a triple floor/ceiling problem all by myself!

[@matiasjoaquinbustamantevej3278]

Hi great video, but i think the answer is 1/(b-1) because when x=0 the función is indeterminate so you should remplace the 0 with (for example) 1/b^n and use the limit n->infinity. If you do this the answer is 1/(b-1) If i’m wrong i would be very happy to know why.Thanks

[@laszloliptak611]

Nice integral. For the preparatory step I prefer to derive them the other way around to show how you get them to avoid steps that look like as if they come out of the blue. So in the beginning, after noting the the integrand is always an integer, I would find out when its value is n. Then using properties of floor, logarithm, and ceiling one can get the results you started with. Also, I don't see why b needs to be an integer. Wouldn't this hold as long as b>1?

[@jacobbills5002]

Just watch this impressive Maths channel... it’s very nice like this kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@themathsgeek8528]

Amazing video Prof!!

[@sonhoangngan4088]

The result is 1/(b-1), not b/(b-1). This problem is very beautiful. Thank Michael Penn so much!!!

[@jacobbills5002]

Just watch this impressive Maths channel... it’s very nice like this kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@ЛевЯрков-е1ж]

Hello, i have a question. This way integrating functions named Lebesgue integral?

[@CoderboyPB]

Lebesgue integration is simular, but I am not sure, that this is the same.

[@backyard282]

@@herrmarx973 he added up rectangles sideways so it is lebesgue integral

[@joeybf]

That function is Riemann integrable, but Lebesgue integration would give the same result. Since it's fresh in my mind from last semester, there's a nice result about "integrating vertically" for Lebesgue integration. It's in Chapter 2 Exercise 18 of Stein-Shakarchi's Real Analysis textbook.

[@udic01]

1/(b-1)

[@leickrobinson5186]

Ooo, so close! A swing and a miss!!

[@Yiijbyygb]

Excellent problem

[@rafael7696]

Very nice problem

[@michalbotor]

what is the best animation software for maths? blackboard.

[@caesar_cipher]

good stuff ! but this will work for any real value of b > 1, b does no need to be integer

[@ricardocavalcanti3343]

Not so fast! At a certain point (c. 5:45) Michael used the identity floor(b^n) = b^n, which is valid only if b is integer.

[@ikocheratcr]

I think Michael makes this last step "mistakes" to see if we are paying attention ...

[@noelani976]

Really? I don't think so.

[@raphaeljacobs3518]

So amazing!

[@atreidesson]

that would be some kind of interesting if the floors weren't there and b=e, and I guess the lower integral limit should be 1 then

[@taladon101]

Is the infinity symbol in the thumbnail a side ways 8?

[@nintukumardas2484]

Sir if you solve this problem I will be very grateful to you Statement:Find all (a,n) where a and n are integers and such that a^n|n^a - 1

[@RexxSchneider]

For positive integers: (1, n) and (a, 1) trivially. You'll find that (2, 3) seems to be the only non-trivial solution. Negative values of n produce lots more solutions, but those are not very interesting.

[@keshavrathore5228]

so underrated channel

[@levonnigogoosian7547]

The question does not specify that x>= 1. If x is less than 1 then the solution does not apply.

[@AlchemistOfNirnroot]

But what is b?

[@KarlMarX_93]

Brilliant!

[@hmafussel94]

Hmm very nice problem and great solution. When i tried i had also the idea that for x => 1 its 0, but for the swcond part i intervalled it between 1/n+1 and 1/n and couldnt figure it out after that and was not patient enough to try more xD

[@jacobbills5002]

Just watch this impressive Maths channel... it’s very nice like this kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@bilalahmed-fe5xt]

Sir can u solve this problem? Question Find n belongs to complex s.t for all x belong to real number this equation holds x^n (2x+n/x)=1

[@Thidos]

And what about x

[@berzerksharma]

B^-1 / 1 -b^-1 = 1 / b-1 ?

[@peterklenner2563]

I wonder how much of this solution actually started at the end with a plot of the integrand?

[@michalbotor]

wow. i absolutely love it! 😍 also. lebesgue > riemann

[@nablahnjr.6728]

such a good integral

[@jacobbills5002]

Just watch this impressive Maths channel... it’s very nice like this kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@The_lenny]

Amazing...as always!

[@shonakduggal3855]

Please do some multivariable calculus

[@sharpnova2]

are you kidding? he has done a fuckton of multivariate calculus. up to and including differential forms chains etc etc

[@jacobbills5002]

Just watch this impressive Maths channel... it’s very nice like this kzbin.info/door/ZDkxpcvd-T1uR65Feuj5Yg

[@sharpnova2]

isn't it 1/(b-1)?

[@TheHuesSciTech]

Yes, see other comments.

[@BDCOMBO]

Third, awesome geometric series link

[@wehqwhqwqedqe528]

Beautiful

[@randomcubing7106]

FLOOR GANG OWW

[@wesleydeng71]

So, today's homework is to correct the last step.😃

[@study5133]

Very nice 👍

[@AmenAmenzo]

This is amazing! I have a question though, shouldn't we carefully talk about the continuity of the function f on the interval [0,1) ? because if it is not continuous we aren't allowed to calculate the integral!

[@AmenAmenzo]

and the jumps in values that f is doing from 1/b to 1/b² doesn't make it seem continuous out of the box

[@ricardocavalcanti3343]

Under certain conditions, continuity is a sufficient, but not necessary, condition for integrability of a function. (See any book on Calculus.)

[@stephenbeck7222]

Functions don’t have to be continuous to be integral. Michael never even directly calculated the value of the function at any of the endpoints. Because single points don’t contribute to the value of the integral.

[@zachbills8112]

There are clearly only countably many points of discontinuity, which makes the set of points of discontinuity measure zero, so they can't effect the integral.

[@M-F-H]

@@zachbills8112 That's not true for the Riemann integral.

[@seroujghazarian6343]

It's 1/(b-1), not b/(b-1)

[@djvalentedochp]

nice

[@pineco74]

squigle notation, make it happen

[@abderrahmanyousfi5565]

👍🏻👍🏻

[@AlephThree]

Wow, that is a tough problem!

[@Reza_Audio]

Michael, this time you were wrong about the good place to stop :(

[@Bennici]

Just 24 hours ago I responded to someone jokingly remarking that you overdo these floor/inequality problems. I defended you, and now this. I am disappointed ;)

[@stephenbeck7222]

Who doesn’t love an infinite integral of a function with three different floor/ceiling operators?

[@youssefallani6407]

2nd comment good Michael penn

[@jbtechcon7434]

Should have ignored this viewer's suggestion. This is the first problem I've seen on this channel that I figured out at a glance and was a bit dismayed that it got made into a video.

[@sporksto4372]

Why is your forehead so wide?

[@darthmath1071]

because math. (stupid questions need stupid answers)

[@wafizariar8555]

Isn't it 1/(b-1)?