We present an interesting integral involving the floor function. Playlist: • Interesting Integrals www.michael-penn.net www.researchgate.net/profile/... www.randolphcollege.edu/mathem...
Пікірлер: 169
@TheOneSevenNine4 жыл бұрын
I've never seen anyone write a summation sign with such ferocity
@blackpenredpen4 жыл бұрын
You opened my eyes to new and crazy things!! And that camera angle switch at 6:55 was fresh!
@MichaelPennMath4 жыл бұрын
Thanks, I found another tripod at the thrift store a few months ago and finally put it to good use!
@paillote4 жыл бұрын
This channel should be called "theFloorFuncionGuy"
@FireStormOOO_3 жыл бұрын
Indeed. Kinda surprising calculus is helpful on something so agresssively discontinuous.
@ireallydontknow32994 жыл бұрын
That was cool. I'm happy that I was able to solve this on my own, but only because I was able to leave the answer in function of the sum S = 1 - 1/2 + 1/3 - 1/4 + ..., which I happened to know was ln(2).
@alessandropizzotti9324 жыл бұрын
Yes, that is a much easier way.
@virajagr3 жыл бұрын
Yes, from the Taylor series of ln(x+1)
@shrirammaiya98673 жыл бұрын
@@virajagr no directly from the integral
@lyrimetacurl02 жыл бұрын
For me I was like "I wonder if it's going to be ln(2)" without really knowing why I thought it 😂
@technoguyx4 жыл бұрын
Love how simple the final expression is. Btw the audio seems to clip a lot more than in previous videos.
@FadkinsDiet3 жыл бұрын
I'd guess the recording level was too low and he had to up the gain in post
@gradecracker4 жыл бұрын
Its crazy that the steps you used were so bizarre and unconventional but it still got you to the same answer! Absolutely brilliant!
@abushahid11504 жыл бұрын
Also if you remember the Taylor series of ln(1+x) and vigilant enough, you could easy avoid like last 5 steps.
@Nomnomlick4 жыл бұрын
I noticed very quickly that the integral from 1/n to 1/n+1 was just 1/n-1/n+1 with alternating signs starting with negative. So you get -(1-1/2) + (1/2-1/3) - (1/3-1/4) + ... Getting rid of the paranthesis and grouping terms with common denominator you get: -1 + 2/2 - 2/3 + 2/4 - 2/5. Substracting and adding 1, factoring out -2 you get 1-2AlternateHarmonic. Found ln2 by finding that the Maclaurin series of ln(1+x) fit the alternating harmonic series pretty well.
@wannabeactuary013 жыл бұрын
Love this and the use towards the end of the anti-derivative is icing on a cake!
@willyh.r.12164 жыл бұрын
This is amazingly mind blowing stuff, man! Encore, encore, encore...please.
@norajcarnaj92072 жыл бұрын
pk tu fais le français là mon reuf?
@1089S2 жыл бұрын
You make it sound like a fine poetry and that is awesome!
@el_variable4 жыл бұрын
I love the way you make your videos. Keep it up, I learned a lot from you :)
@adamjennifer64373 жыл бұрын
Just watch this impressive Math channel kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@peterdecupis82962 жыл бұрын
Very cool work! As a matter of fact, the function is well integrable in Peano sense, because the discontinuity of the floor function occurs only in a numerable set of isolated points; then we can correctly replace the improper integral with a series whose terms are definite integrals relevant to adjacent unit intervals [k, k+1), for any positive index k. Then the final evaluation of the series is correctly performed by exploiting the convergence condition of alternate sign series with vanishing terms; a quicker approach to the final evaluation is based on the Taylor-McLaurin series expansion of the function f(x)=ln(1+x) for x=1, i.e. f(1)=ln2, that is immediately recognized as the alternate harmonic series.
@PatrickOfTav2 жыл бұрын
"And that finishes this video". Have I got the right Michael Penn?
@suniltshegaonkar78093 жыл бұрын
Great Michael, it is a very nice problem, never seen any problem close to this.
@stasiawright3774 жыл бұрын
I've never seen these tricks before, thank you!
@spideramazon50324 жыл бұрын
Very nice integral evaluation. i love watching your videos a lot.
@adamjennifer64373 жыл бұрын
Just watch this impressive Math channel kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@mathunt11303 жыл бұрын
The substitution was key here. It made things very clear to see and is clearly a very good technique for these types of problems. Finding the series by integration was also a neat trick.
@fivestar58552 жыл бұрын
That's brilliant one! Especially with geometric series representation.
@AltinoSantos4 жыл бұрын
Very interesting. The function graph would also be welcome. Congratulations on the channel.
@adamjennifer64373 жыл бұрын
Just watch this impressive Math channel kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@wolfmanjacksaid4 жыл бұрын
I love the deadpan "great", keep it up!
@ByteOfCake3 жыл бұрын
He's just containing all his excitement inside of him :3
@minwithoutintroduction Жыл бұрын
رائع جدا كالعادة. أحب هذه التكاملات الغريبة
@alexpavlov67543 жыл бұрын
I think You are using famous math sum, but you resolve sum in the great wonderful manner. Thank you
@Salvador964 Жыл бұрын
Gracias por tan interesante ejercicio.
@JonathonV4 жыл бұрын
That was incredible! Very cool integral and algebra/calculus substitutions. I also really liked your camera angle switch at 6:55 so I could see the board. You should keep the camera there! And for some more unsolicited but hopefully still welcome advice: I suggest moving your “please like and subscribe” thing to the end of the video so that your promo thumbnails for other videos don't block your grand reveal! I had to scrub back and forth through the video so I could see the solution on the board. Anyway, cool problem!
@adamjennifer64373 жыл бұрын
Just watch this impressive Math channel kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@nailabenali74884 жыл бұрын
I've just seen this on my book this morning and yet here I am watching your video because it's much more interesting, maybe if I start working on summable families you may do a video about it too xD! Thank you a lot !!!
@MichaelPennMath4 жыл бұрын
I think this is a "famous" problem. I have seen it before, but I don't really know where.
@user-yt8xc8zw6v4 жыл бұрын
@@MichaelPennMath This is work of Dr Ovidiu Furdui , you can find it in his research papers or even in his book entitled by (Limits, Series, and Fractional Part Integrals)
@Someone-cr8cj3 жыл бұрын
THAT'S SO GREAT!
@phee41742 жыл бұрын
another nice property of this integral is that if you replace 1/x with x^s you get something that's essentially the alternating Riemann zeta function (though it's evaluated at something like 1/s instead of s, and is multiplied by and has added to it a factor of 1/x^2)
@operationmike85534 жыл бұрын
You are brilliant.
@adamjennifer64373 жыл бұрын
Just watch this impressive Math channel kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@zapahaha3 жыл бұрын
amazing!!!
@user-fh5rm2ef4n4 жыл бұрын
Hey, interesting problem. Thanks. You've got a tiny mistake 2:55. You've forgotten du in your substitution.
@adamjennifer64373 жыл бұрын
Just watch this impressive Math channel kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@Walczyk3 жыл бұрын
awesome problem!
@anshusingh14934 жыл бұрын
Thanks for giving me new insight...........its brilliant...hve subscribed ur channel...
@adamjennifer64373 жыл бұрын
Just watch this impressive Math channel kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@erickherrerapena89814 жыл бұрын
Increíble resolución.
@douglasmagowan27094 жыл бұрын
By minute 7, we had an expression that was obvious to anyone who knew the Taylor series of ln (1+x). But rather than saying that, or deriving the Taylor series, the expanation becomes circuitious over its last 5 minutes.
@zwz.zdenek4 жыл бұрын
It's about teaching, not about finishing first.
@br758574 жыл бұрын
Dude you are awesome
@laventin43324 жыл бұрын
Hey can you make a video on the 4th isomorphism theorem and composition series. Thanks a ton !
@trevorsong43454 жыл бұрын
Amazing!!!
@ningliu36483 жыл бұрын
hi Penn, I got confused about some of the steps. Since you only have the property of conditional convergence, you are not allowed to change the order of summation to guarantee you reach the same number. What you did in your video, you change the order of the summation.
@dansheppard29652 жыл бұрын
He did say something to justify this, something about the harmonic series, but I didn't catch it.
@feitao13 Жыл бұрын
@@dansheppard2965 He said that it is absolutely convergent, which is not true.
@pikkutonttu26974 жыл бұрын
I was about to try the substitition method, but I backed off as the wikipedia article on integration by substitution gives a theorem, which assumes that the integrand is a continuous function on the interval of integration. I suppose the assumption can be relieved at least to piecewise continuity, since your initial and subsequent methods give the same result.
@mryip062 жыл бұрын
amazing
@Camll Жыл бұрын
Loveeeed❤❤❤❤❤
@lason914 жыл бұрын
awesome
@antoniopalacios81604 жыл бұрын
I get -1+2ln2 for the case of the ceiling function instead of the floor function. Thank you.
@TheMichaelmorad Жыл бұрын
You could just use a little bit of reasoning: Between 1/(n+1) and 1/n the function evaluates at (-1)ⁿ which means that this expression is the sun where 1
@cycklist4 жыл бұрын
Fancy new camera angle 👌
@MichaelPennMath4 жыл бұрын
Thanks!
@TheFinav4 жыл бұрын
Great.
@filipsperl2 жыл бұрын
this is so cool
@vbcool834 жыл бұрын
Got the answer as 1-2ln(2) - let's see if it turns out right! Very strange that this crazy power integral involving floors related to logs!
@danielhan20074 жыл бұрын
because you can solve this problem geometrically, i.e., alternating sums of area of rectangles with lengths of (1-1/2), (1/2-1/3), ... and height of 1 and -1.
@adamjennifer64373 жыл бұрын
Just watch this impressive Math channel kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@cicik57 Жыл бұрын
okay, you can move easier and straightvorward way. Basic idea with whole part or fraction part is to split integral on parts where this prackets give the same value. for example, you can split intevral 0..1 on : I(1/2 ..1) of -1 dx + I(1/3 ..1/2) of 1 dx + I(1/4 ..1/3) of -1 dx +...= -x evaluated in (1, 1/2) +x | (1/2..1/3) - x(1/3, 1/4)... = -1 + 1/2 + 1/2 -1/3-1/3 + 1/4 + 1/4 = the sign change harmonic series is well known and is ln2, so -1 + 2(1-ln2) = 1-2ln2
@rsactuary87634 жыл бұрын
I forget the problem every time he turns around and I see his back muscles.
@jeremy.N4 жыл бұрын
lol, I would have done the same thing. except that I would have made at least 3 mistakes with plus and minus. Great video, although your audio is a little much, maybe you can invest into a new micro (a ball like bprp?).
@Boe17713 жыл бұрын
Yeah, somehow the audio seems off indeed. But please keep the great maths coming :)
@sword71634 жыл бұрын
that was cool
@JoJoModding4 жыл бұрын
10:58 "because this is absolutely convergent" Is it? If you consider the absolute sum you just get 1/n which diverges.
@MichaelPennMath4 жыл бұрын
That’s my bad, it’s actually because of the dominated convergence theorem.
@harlock75214 жыл бұрын
@@MichaelPennMath because of alternating series test also
@martinepstein98263 жыл бұрын
@@harlock7521 That's just a convergence test. It doesn't tell you whether you can interchange the integral and the sum.
@mushroomsteve4 жыл бұрын
I got stuck at 10:55 when you said "this is absolutely convergent". It looks like the series in question is just a different form of sum(0,inf,(-1)^n/(n+1)), which is an alternating harmonic series, and thus not absolutely convergent.
@tracyh57514 жыл бұрын
You are right. Absolute convergence does not work here. Instead, he can use the dominated convergence theorem along with the alternating series partial sum estimate to justify the interchange.
@mushroomsteve4 жыл бұрын
@@tracyh5751 Thank you! I was really confused about that step.
@jordiplotnikovpous48444 жыл бұрын
Tracy H which equals 1/2 right?
@Hiltok4 жыл бұрын
Sum = 1 - 1/2 + 1/3 + 1/4 - 1/5 ... = [1 + 1/2] + [1/3 - 1/4] + ... # this is the 'alternating partial sum' Tracy H mentions = Sum 1/(n^2+n) for n=1,2,3,... < Sum 1/n^2 = finite (= pi^2/6) # this is the dominated convergence
@alephnull40444 жыл бұрын
@@Hiltok Very nice. Although you wouldn't need sum 1/n^2 = pi^2/6, just sum 1/(n^2+n) = 1.
@DeepDeepEast4 жыл бұрын
Great Content. , I should have watched those when I began my math bachelor. Now I am finished soon and I still lack basic stuff in calculus 😁
@pbj41843 жыл бұрын
Wtf? You didn't fail even once?
@DeepDeepEast3 жыл бұрын
@@pbj4184 No never failed an exam. But also didn't receive good grades sometimes and after I learned for an exam I forgot a lot of things. Or didn't dig deeper into details. The stuff I find interesting I will remember for ever.
@pbj41843 жыл бұрын
@@DeepDeepEast Oh that makes sense. Are you working in the industry now?
@liyi-hua21113 жыл бұрын
(i didn't learn really well at math which sadly was my major.) would you mind to tell me: 1. how can "sum((-1)^n*(1/n-1/n+1)),n=1 to inf" split into two parts? from what i recall, alternating series can converge to different result by proper rearrangement. Won't splitting things up make a different result? 2. Considering "integral (-1)^floor(1/x), from 0 to 1" as an alternating series formed by area pieces, integration seems to be no order to converge to a certain answer, or are there any conventions or theorems for this kind of integration? it would be grateful if you could reply to my questions.
@housamkak6464 жыл бұрын
mannnnnnnnnnnnnnnnnnnnn this is amazingggg
@lesprivatrizal4 жыл бұрын
Great
@hypernova43343 жыл бұрын
How does the geometric series converge at the right endpoint of the interval x=1? Shouldn’t |x| < 1?
@WmTyndale4 жыл бұрын
Nice new integral form. Here is an extension: Do the result (I have not) for [1/x^n] inside the integral. May be interesting.
@adamjennifer64373 жыл бұрын
Just watch this impressive Math channel kzfaq.info/love/ZDkxpcvd-T1uR65Feuj5Yg
@Omcsesz4 жыл бұрын
At the end, you should substitute x=1 first to the expression 1-2lnx, right?
@MichaelPennMath4 жыл бұрын
The 1 is outside of the evaluation. Notice that a few steps before we had 1-2integral.
@derekcresswell73523 жыл бұрын
I'm a tad confused by one step. Why is turning floor(u) to a constant valid? Wouldn't u take on the value of n + 1 at a single point in the integration?
@feitao13 Жыл бұрын
Changing the value of a function at one single point would not affect the integration, i.e., integral of f = integral of g over some interval, if f = g except at one point.
@stewartzayat75264 жыл бұрын
Yay! I arrived at the same answer. I wanted to check if I was correct on wolfram alpha, but wolfram got the wrong answer lol
@MichaelPennMath4 жыл бұрын
wolfram alpha and mathematica have a hard time with this one. If you set the lower bound to 0.01 instead of 0 it will approximate it correctly.
@tokajileo59284 жыл бұрын
i lost it at 9:21 when that 2 appeared
@ethanchandler39342 жыл бұрын
I would write it as 1-ln(4) but very good video’
@AlfonsoNeilJimenezCasallas4 жыл бұрын
Math is crazy!
@lluisllacer72954 жыл бұрын
How do you know that the sequence $\sum_{n=1}^\infty \int_n^{n+1}(-1)^u {1\over u^2}du$ converges to the same number as $\int_1^\infty (-1)^u{1\over 1}du$ does? I mean, the first one is a subsequence from the second. This doesn't mean that if de subsequence converges the main sequence converges too or, in case both converge, share de same limit.
@fulla14 жыл бұрын
Oh, I hate it so much, when people write a summation sign in a hurry (or from an uncomfortable angle) that it looks like the thing at 6:27. But great video, though!
@bautibunge7372 жыл бұрын
Can you actually separate the sum if the series doesn't converge absolutely?
@dominiquelaurain64274 жыл бұрын
But (-1)^(n+1)/(n+1) equals -1 (not 1) when n = 0 as you say @8:51 ...but parenthesis is then not zero later (you mistakenly substract 1 in parenthesis and add 1 outside). Check out the missing "-" on your formula @9:21 (and remove your trick adding and substracting 1).
@davidblauyoutube Жыл бұрын
The correct answer, 1 - 2 ln 2, is negative because the integrand is -1 on the entire interval 1/2
@tubamazouz Жыл бұрын
👍👍
@SlidellRobotics3 жыл бұрын
I broke up into subintervals without the substitution and got the same sum without having to integrate 1/u², just 1 and -1.
@JonahFoley4 жыл бұрын
nice problem
@MichaelJamesActually Жыл бұрын
feel like I need a cigarette after that.
@smiley_10004 жыл бұрын
Didn't you forget the du when doing the u-substitution?
@jesusandrade1378 Жыл бұрын
Maple Calculator evaluates this definite integral as 1, without the 2 ln(2) term
@abdallahal-dalleh64533 жыл бұрын
Can you explain 10:00 why we did that?
@bookworm84143 жыл бұрын
The power n of -1 is either 1 or -1, why their sum is not an integer?
@romanpavelko59944 жыл бұрын
But inserting x^(n + 1) to the series means that for that series to be convergent x should be -1 < x < 1. How to prove that x is between 0 and 1 only?
@user-sk5nl6vj3e4 жыл бұрын
The range of x in the integral is from 0 to 1 so we only consider x in that range.
@skwbusaidi4 жыл бұрын
In Wolframalpha ∫ (-1)^floor(1/x) dx from 0 to 1 I got 0.50... which not equal to 1-2ln(2)
@MichaelPennMath4 жыл бұрын
This is one that wolframalpha and Mathematica have a hard time approximating. If you give them some help and change the bound of integration to 0.01 to 1 it works: bit.ly/2JQP5Gm
@felipelopes31714 жыл бұрын
In general, you should never put an integral with a singularity in a numerical integration software, because it messes up their methods. You can try splitting it and do a change of variables to remove the singularity.
@aranbrico10054 жыл бұрын
I got a little bit confused with 2
@aranbrico10054 жыл бұрын
Never mind.
@kkalyan55694 жыл бұрын
i have done this as(i.e with 1/x)it is without substitution and ended up with the alternating series of 1-1+1-1+1...... which is of course not the correct one .Someone pls correct me .
@Zero-tq6hv4 жыл бұрын
You've probably just made an error during computation. INT from 0 to 1 (-1)^floor(1/x) dx = Sum from n=1 to inf (INT from 1/(n+1) to 1/n (-1)^floor(1/x) dx) = Sum from n=1 to inf (INT from 1/(n+1) to 1/n (-1)^n dx) = Sum from n=1 to inf (((-1)^n)/n+((-1)^(n+1))/(n+1)) After simplification we get 1-2ln(2). Quite simillar to method used in the video.
@zackmercurys4 жыл бұрын
I think your mic is clipping for some reason. might require a replacement
@khai-hoannguyen-dang90824 жыл бұрын
Really nice talk
@jcfgykjtdk3 жыл бұрын
I got Ln(4) - 1
@HideyukiWatanabe2 жыл бұрын
7:20 Why can you change the order of sum? The series Σ(-1)^n (1/n) and Σ(-1)^n (1/(n+1)) aren't absolutely convergent.
@HideyukiWatanabe2 жыл бұрын
Oh, it was simple. Adding two convergent series converges to the sum of each series, looking RHS to LHS.
@albertreitsma9883 жыл бұрын
Missed the usual "that's a good place to stop"
@matthewryan48443 жыл бұрын
Good place to stop: ∅
@diabl2master4 жыл бұрын
I feel that the substitution was totally unecessary. I arrived at the sum on the right at 7:37 doing essentially exactly the same thing but without the substitution.
@xshortguy4 жыл бұрын
Well, in all honesty a substitution is never really necessary, it just makes things easier to see.
@thecustomer28044 жыл бұрын
Wait hold on. 1 - 2ln(1 + 0) = 1, so plugging in the limits of integration should get you (1 - 2ln(2)) - (1) = -2ln(2) rightt?
@Mr5nan4 жыл бұрын
the bounds are only for the latter term, not for the 1
@GhostyOcean4 жыл бұрын
Some parentheses would've helped, or putting the +1 after the integral. It should be written as 1-(2ln(x+1)|x=1 x=0)
@thecustomer28044 жыл бұрын
GhostyOcean Mr5nan Thanks for the tips! I make small but obvious mistakes at times
@seroujghazarian63433 жыл бұрын
For the same integral but the roof function instead of floor, the result is the opposite
@samiulfahim53843 жыл бұрын
What a body 👍 , Sirrrrrr 😉 .
@mashtonish4 жыл бұрын
I got immediately lost at 10:00
@ByteOfCake3 жыл бұрын
hes essentially just multiplying it by 1. If you use the bound he used you get 1^(n+1)-0^(n+1) which is just 1.
@VUrosov3 жыл бұрын
It's dx=-du/u^2
@jannowak9052 Жыл бұрын
Czy ty zauważyłeś, że w ostatnich sekundach logo twojego kanału zasłania ostateczny wynik?
@muhammadsarimmehdi4 жыл бұрын
Why are you adding x around 10:12?
@pbj41843 жыл бұрын
To get an integral and then switch the order of summation and integration to solve the problem
@pbj41843 жыл бұрын
Michael does that a lot
@tomhase70073 жыл бұрын
Reordering a conditionally convergent series? Mh...
@disgruntledtoons3 жыл бұрын
If the floor is lava, how do we integrate it?
@paulkohl92674 жыл бұрын
The answer is approximately -0.338629436111989061883446424291635, which I mean, come on,, obviously! ;)