A viewer suggested floor integral!
Рет қаралды 33,845
Күн бұрынWe look at a nice problem involving an integral, a logarithm, the floor function, and the ceiling function.
Playlist: • Interesting Integrals
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@mooncowtube 3 жыл бұрын
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.
@vardaandua3585 3 жыл бұрын
The last step is wrong...it should be 1/(b-1)
@NagyonCsoki 3 жыл бұрын
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 3 жыл бұрын
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 3 жыл бұрын
@@NagyonCsoki So a/(1-r) = (1/b)/(1-1/b) = (1/b)/((b-1)/b) = 1/(b-1).
@noway2831 3 жыл бұрын
It's always the last step
@candamir26 3 жыл бұрын
Yap, 1/(b-1) should be the final answer.
@backyard282 3 жыл бұрын
Everything so perfect and then that slip at the end :D Great video nevertheless
@jagula 3 жыл бұрын
8:18 that chalk fell xD
@pedromooregaissler6378 3 жыл бұрын
not a true Mathematician if you don't find typos hahahahah
@goodplacetostop2973 3 жыл бұрын
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 3 жыл бұрын
that's a bad place to stop
@goodplacetostart9099 3 жыл бұрын
Lemme tel you 0:01 is the right place to start
@tonyhaddad1394 3 жыл бұрын
What is the last homework you did share it with us ?????? I need it 😥
@goodplacetostop2973 3 жыл бұрын
@@tonyhaddad1394 I’m too lazy right now to dig in KZfaq comments 😂 Just give me one topic you want an homework for and I’ll see what I can do for tomorrow
@tonyhaddad1394 3 жыл бұрын
@@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 😍
@hsjkdsgd 3 жыл бұрын
The little mistake in the end has been deliberately done to see who actually watches the videos carefully.
@gustavowadaslopes2479 3 жыл бұрын
This is the most teacher response ever.
@nagamanikomarla5376 Жыл бұрын
@@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 Жыл бұрын
@@nagamanikomarla5376 pure torture
@Cosine_Wave 3 жыл бұрын
4:02 I like that edit
@ZainAlAazizi 3 жыл бұрын
1/(b-1), and that’s a good place to stop...
@papanujian7758 3 жыл бұрын
Amazing. You just explained this problems solution with a beautiful way. Can't hardly waiting for another videos
@camrouxbg 3 жыл бұрын
That is really slick! And fairly easily accessible to high school students. Thanks!
@themathsgeek8528 3 жыл бұрын
Amazing video Prof!!
@xCorvus7x 3 жыл бұрын
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 3 жыл бұрын
this is called telescoping
@M-F-H 3 жыл бұрын
@@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 3 жыл бұрын
Just watch this impressive Maths channel... it’s very nice like this kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@petermueller162 3 жыл бұрын
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 3 жыл бұрын
Just watch this impressive Maths channel... it’s very nice like this kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@PlutoTheSecond 3 жыл бұрын
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.
@mihaipuiu6231 3 жыл бұрын
The Demonstration of Michael Penn is very simple, very clean, nice, and interesting.
@sephimarzliclican6224 3 жыл бұрын
michael penn is one of the reasons i’m still alive rn
@Div1nePiece 3 жыл бұрын
Isn't it 1/(b-1)? Excellent problem by the way!
@ThePhysicsMathsWizard 3 жыл бұрын
wow, i've never seen anything like this, thanks for making this video
@walidabdelal 3 жыл бұрын
Floogarithms! Excellent brain massager. Cheers Michael!
@raphaeljacobs3518 3 жыл бұрын
So amazing!
@Yiijbyygb 3 жыл бұрын
Excellent problem
@alxjones 2 жыл бұрын
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!
@leickrobinson5186 3 жыл бұрын
Ooo, so close! A swing and a miss!!
@Translacja1 3 жыл бұрын
Brilliant!
@rafael7696 3 жыл бұрын
Very nice problem
@mokkapatisiddharth5793 3 жыл бұрын
I cannot believe I solved a triple floor/ceiling problem all by myself!
@leonardocoroneo6344 3 жыл бұрын
Amazing...as always!
@udic01 3 жыл бұрын
1/(b-1)
@keshavrathore5228 3 жыл бұрын
so underrated channel
@study5133 3 жыл бұрын
Very nice 👍
@sonhoangngan4088 3 жыл бұрын
The result is 1/(b-1), not b/(b-1). This problem is very beautiful. Thank Michael Penn so much!!!
@jacobbills5002 3 жыл бұрын
Just watch this impressive Maths channel... it’s very nice like this kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@wehqwhqwqedqe528 3 жыл бұрын
Beautiful
@BDCOMBO 3 жыл бұрын
Third, awesome geometric series link
@hmafussel94 3 жыл бұрын
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 3 жыл бұрын
Just watch this impressive Maths channel... it’s very nice like this kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@burhan8795 3 жыл бұрын
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
@kevinmartin7760 3 жыл бұрын
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 3 жыл бұрын
Just watch this impressive Maths channel... it’s very nice like this kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@matiasjoaquinbustamantevej3278 3 жыл бұрын
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
@caesar_cipher 3 жыл бұрын
good stuff ! but this will work for any real value of b > 1, b does no need to be integer
@ricardocavalcanti3343 3 жыл бұрын
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.
@laszloliptak611 3 жыл бұрын
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 3 жыл бұрын
Just watch this impressive Maths channel... it’s very nice like this kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@ikocheratcr 3 жыл бұрын
I think Michael makes this last step "mistakes" to see if we are paying attention ...
@noelani976 3 жыл бұрын
Really? I don't think so.
@peterklenner2563 3 жыл бұрын
I wonder how much of this solution actually started at the end with a plot of the integrand?
@nablahnjr.6728 3 жыл бұрын
such a good integral
@jacobbills5002 3 жыл бұрын
Just watch this impressive Maths channel... it’s very nice like this kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@nintukumardas2484 3 жыл бұрын
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 2 жыл бұрын
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.
@bilalahmed-fe5xt 3 жыл бұрын
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
@richardlinsley-hood7149 3 жыл бұрын
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 3 жыл бұрын
Just watch this impressive Maths channel... it’s very nice like this kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@digxx 3 жыл бұрын
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 3 жыл бұрын
why are you dropping latex or whatever in a KZfaq comment rofl
@digxx 3 жыл бұрын
@@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.
@taladon101 3 жыл бұрын
Is the infinity symbol in the thumbnail a side ways 8?
@michalbotor 3 жыл бұрын
what is the best animation software for maths? blackboard.
@AlephThree 3 жыл бұрын
Wow, that is a tough problem!
@michalbotor 3 жыл бұрын
wow. i absolutely love it! 😍 also. lebesgue > riemann
@atreidesson Жыл бұрын
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
@user-fw9ej9gj1h 3 жыл бұрын
Hello, i have a question. This way integrating functions named Lebesgue integral?
@CoderboyPB 3 жыл бұрын
Lebesgue integration is simular, but I am not sure, that this is the same.
@backyard282 3 жыл бұрын
@@herrmarx973 he added up rectangles sideways so it is lebesgue integral
@hybmnzz2658 3 жыл бұрын
No this is not Lebesgue integral. Integrating along the y-axis does not make it Lebesgue.
@joeybeauvais-feisthauer3137 3 жыл бұрын
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.
@AlchemistOfNirnroot Жыл бұрын
But what is b?
@wafizariar8555 3 жыл бұрын
Isn't it 1/(b-1)?
@Thidos 3 жыл бұрын
And what about x
@shonakduggal3855 3 жыл бұрын
Please do some multivariable calculus
@sharpnova2 3 жыл бұрын
are you kidding? he has done a fuckton of multivariate calculus. up to and including differential forms chains etc etc
@jacobbills5002 3 жыл бұрын
Just watch this impressive Maths channel... it’s very nice like this kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@berzerksharma 3 жыл бұрын
B^-1 / 1 -b^-1 = 1 / b-1 ?
@djvalentedochp 3 жыл бұрын
nice
@randomcubing7106 3 жыл бұрын
FLOOR GANG OWW
@levonnigogoosian7547 3 жыл бұрын
The question does not specify that x>= 1. If x is less than 1 then the solution does not apply.
@sharpnova2 3 жыл бұрын
isn't it 1/(b-1)?
@TheHuesSciTech 3 жыл бұрын
Yes, see other comments.
@abderrahmanyousfi5565 3 жыл бұрын
👍🏻👍🏻
@wesleydeng71 3 жыл бұрын
So, today's homework is to correct the last step.😃
@pineco74 3 жыл бұрын
squigle notation, make it happen
@seroujghazarian6343 3 жыл бұрын
It's 1/(b-1), not b/(b-1)
@TheKluVerKamp 3 жыл бұрын
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!
@TheKluVerKamp 3 жыл бұрын
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 3 жыл бұрын
Under certain conditions, continuity is a sufficient, but not necessary, condition for integrability of a function. (See any book on Calculus.)
@stephenbeck7222 3 жыл бұрын
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 3 жыл бұрын
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 3 жыл бұрын
@@zachbills8112 That's not true for the Riemann integral.
@Reza_Audio 3 жыл бұрын
Michael, this time you were wrong about the good place to stop :(
@youssefallani6407 3 жыл бұрын
2nd comment good Michael penn
@Bennici 3 жыл бұрын
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 3 жыл бұрын
Who doesn’t love an infinite integral of a function with three different floor/ceiling operators?
@sporksto4372 3 жыл бұрын
Why is your forehead so wide?
@darthmath1071 3 жыл бұрын
because math. (stupid questions need stupid answers)
@jbtechcon7434 3 жыл бұрын
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.