An interesting integral with the floor function.
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- Опубликовано: 8 апр 2020
- We present an interesting integral involving the floor function.
Playlist: • Interesting Integrals
www.michael-penn.net
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www.randolphcollege.edu/mathem...
I've never seen anyone write a summation sign with such ferocity
You opened my eyes to new and crazy things!! And that camera angle switch at 6:55 was fresh!
Thanks, I found another tripod at the thrift store a few months ago and finally put it to good use!
This channel should be called "theFloorFuncionGuy"
Indeed. Kinda surprising calculus is helpful on something so agresssively discontinuous.
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).
Yes, that is a much easier way.
Yes, from the Taylor series of ln(x+1)
@@virajagr no directly from the integral
For me I was like "I wonder if it's going to be ln(2)" without really knowing why I thought it 😂
Its crazy that the steps you used were so bizarre and unconventional but it still got you to the same answer! Absolutely brilliant!
Love how simple the final expression is. Btw the audio seems to clip a lot more than in previous videos.
I'd guess the recording level was too low and he had to up the gain in post
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.
Also if you remember the Taylor series of ln(1+x) and vigilant enough, you could easy avoid like last 5 steps.
I love the way you make your videos. Keep it up, I learned a lot from you :)
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Love this and the use towards the end of the anti-derivative is icing on a cake!
This is amazingly mind blowing stuff, man! Encore, encore, encore...please.
pk tu fais le français là mon reuf?
You make it sound like a fine poetry and that is awesome!
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.
Great Michael, it is a very nice problem, never seen any problem close to this.
"And that finishes this video". Have I got the right Michael Penn?
I've never seen these tricks before, thank you!
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!
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Very nice integral evaluation. i love watching your videos a lot.
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Very interesting. The function graph would also be welcome. Congratulations on the channel.
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That's brilliant one! Especially with geometric series representation.
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 !!!
I think this is a "famous" problem. I have seen it before, but I don't really know where.
@@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)
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.
I think You are using famous math sum, but you resolve sum in the great wonderful manner. Thank you
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)
I love the deadpan "great", keep it up!
He's just containing all his excitement inside of him :3
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.
It's about teaching, not about finishing first.
رائع جدا كالعادة.
أحب هذه التكاملات الغريبة
Gracias por tan interesante ejercicio.
Hey, interesting problem. Thanks. You've got a tiny mistake 2:55. You've forgotten du in your substitution.
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THAT'S SO GREAT!
awesome problem!
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
amazing!!!
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.
Thanks for giving me new insight...........its brilliant...hve subscribed ur channel...
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Hey can you make a video on the 4th isomorphism theorem and composition series. Thanks a ton !
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.
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.
@@tracyh5751 Thank you! I was really confused about that step.
Tracy H which equals 1/2 right?
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
@@Hiltok Very nice. Although you wouldn't need sum 1/n^2 = pi^2/6, just sum 1/(n^2+n) = 1.
I get -1+2ln2 for the case of the ceiling function instead of the floor function. Thank you.
Increíble resolución.
amazing
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
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!
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.
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Dude you are awesome
Amazing！！！
You are brilliant.
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awesome
this is so cool
Loveeeed❤❤❤❤❤
(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.
The correct answer, 1 - 2 ln 2, is negative because the integrand is -1 on the entire interval 1/2
Great.
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).
10:58 "because this is absolutely convergent"
Is it? If you consider the absolute sum you just get 1/n which diverges.
That’s my bad, it’s actually because of the dominated convergence theorem.
@@MichaelPennMath because of alternating series test also
@@harlock7521 That's just a convergence test. It doesn't tell you whether you can interchange the integral and the sum.
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?).
Yeah, somehow the audio seems off indeed. But please keep the great maths coming :)
I think your mic is clipping for some reason. might require a replacement
Nice new integral form. Here is an extension: Do the result (I have not) for [1/x^n] inside the integral. May be interesting.
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At the end, you should substitute x=1 first to the expression 1-2lnx, right?
The 1 is outside of the evaluation. Notice that a few steps before we had 1-2integral.
How does the geometric series converge at the right endpoint of the interval x=1? Shouldn’t |x| < 1?
that was cool
Fancy new camera angle 👌
Thanks!
Great
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.
He did say something to justify this, something about the harmonic series, but I didn't catch it.
@@dansheppard2965 He said that it is absolutely convergent, which is not true.
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 😁
Wtf? You didn't fail even once?
@@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.
@@DeepDeepEast Oh that makes sense. Are you working in the industry now?
Can you actually separate the sum if the series doesn't converge absolutely?
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.
I forget the problem every time he turns around and I see his back muscles.
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?
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.
mannnnnnnnnnnnnnnnnnnnn this is amazingggg
The power n of -1 is either 1 or -1, why their sum is not an integer?
Didn't you forget the du when doing the u-substitution?
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?
The range of x in the integral is from 0 to 1 so we only consider x in that range.
Can you explain 10:00 why we did that?
I broke up into subintervals without the substitution and got the same sum without having to integrate 1/u², just 1 and -1.
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 .
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.
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
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.
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!
I would write it as 1-ln(4) but very good video’
In Wolframalpha
∫ (-1)^floor(1/x) dx from 0 to 1
I got 0.50... which not equal to 1-2ln(2)
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
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.
feel like I need a cigarette after that.
i lost it at 9:21 when that 2 appeared
I got a little bit confused with 2
Never mind.
Math is crazy!
nice problem
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.
Well, in all honesty a substitution is never really necessary, it just makes things easier to see.
Maple Calculator evaluates this definite integral as 1, without the 2 ln(2) term
If the floor is lava, how do we integrate it?
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.
Oh, it was simple. Adding two convergent series converges to the sum of each series, looking RHS to LHS.
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?
the bounds are only for the latter term, not for the 1
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)
GhostyOcean Mr5nan Thanks for the tips! I make small but obvious mistakes at times
I got Ln(4) - 1
Czy ty zauważyłeś, że w ostatnich sekundach logo twojego kanału zasłania ostateczny wynik?
Really nice talk
👍👍
For the same integral but the roof function instead of floor, the result is the opposite
It's dx=-du/u^2
Why are you adding x around 10:12?
To get an integral and then switch the order of summation and integration to solve the problem
Michael does that a lot
Missed the usual "that's a good place to stop"
Good place to stop: ∅
I got immediately lost at 10:00
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.
What a body 👍 , Sirrrrrr 😉 .
Reordering a conditionally convergent series? Mh...
nice video but why does he scream?
The answer is approximately -0.338629436111989061883446424291635, which I mean, come on,, obviously! ;)