This integral will have you on the floor!!

[@kirbo722]

Wow. This is truly one of the best integrals I've ever seen. Floor function? Telescoping series? AND the Basil Problem too???? 10/10!!!!! :))))

[@drpeyam]

Glad you liked it!

[@cycklist]

Basil is a herb 😊. It's Basel, the city in Switzerland.

[@kirbo722]

@@cycklist haha opsie :o

[@rainerzufall42]

@@cycklist Isn't Basil that guy from "Fawlty Towers" (John Cleese)? LOL

[@GeneralSamov]

​@@rainerzufall42 It's the guy who the Moscow cathedral is named after.

[@AstroTibs]

Would have been nice to show how close the final answer is to 1. The answer, for anyone wondering, is about 0.82.

[@orenfivel6247]

Michael penn would love this video due to floor func

[@mathunt1130]

You could have said 1>1/2>1/3>1/4>...>1/N, and then split up your domain as [1,1/2)U[1/2,1/3)U[1/3,1/4).... and then noted that if n

[@decare696]

true, tho the substitution is easier if you're doing this in your head.

[@rainerzufall42]

I appreciate the substitution! Reduces any doubt about the value of the result, btw. Technically, you're right.

[@Bruh-bk6yo]

As far as I understood, you want to use the properties of square parentheses and turn the integral into a discrete sum. Such as if 1/(n+1)

[@DaveJ6515]

@@Bruh-bk6yo in that case, you need Leonard Euler :-)

[@Bruh-bk6yo]

@@DaveJ6515 1/(2n(n+1)²)=A/2n+B/(n+1)+C/(n+1)² An²+2An+A+2Bn²+2Bn+2Cn=1 (A+2B)n²+2(A+B+C)n+A=1 A=1 B=-1/2 C=-1/2 1/(2n)-1/(2(n+1))-1/2(n+1)²=1/2(1/n-1/(n+1)-1/(n+1)²)=1/2(1-π²/6+1)=1/2(2-π²/6)=1-π²/12 Well, I'll be damned, this guy is right...

[@jmcsquared18]

Even though this is an easy calculation, it's intuitively quite incredible that π² ends up in an integral involving nothing but x and the floor of 1/x. You'd think that e would maybe show up since the integral of 1/x is a natural logarithm. Just goes to show how interconnected all of mathematics is, familiar things oftentimes show up in surprising places.

[@hugohugo37]

By the way, great to see you back! I use your PDE videos in a PDE class I'm teaching. Keep up the good work!

[@drpeyam]

Awesome! Thank you!

[@homerthompson416]

The floor integrals are always fun. Kind of surprises me that Mathematica can't solve these in closed form.

[@jeisondurango1118]

8:12 I almost thought he was gonna say "and that's a good place to stop" 🤣

[@martinprince8253]

Math jokes bring me to my limit

[@drpeyam]

Hahaha 😂

[@skilz8098]

When you remove all limitations division by 0 is possible but it's not what you'd expect!

[@solarwinds5114]

And my limit DNE

[@solarwinds5114]

That isn't a derivative joke

[@TymexComputing]

9.86 / 12 is almost 1 for engineering purposes, its more than a standard deviation ratio, but ceiling of that number is exactly 1 :)

[@ekadria-bo4962]

Now my heart is on floor because of great joy of floor integral 🎉

[@drpeyam]

Same!!

[@Pengochan]

So the end result is about 0.822, i.e. smaller than 1, which makes sense, as the floor function makes the integrand smaller in the original integral.

[@hherrera007]

Master Peyam, you are on another level! :)

[@drpeyam]

Awwww I’m dr peyam for a reason :)

[@yurenchu]

From the thumbnail, I thought that the square brackets [...] represented the _rounding_ function rather than the floor function. (i.e. rounding to the nearest integer instead of rounding down to the greatest integer smaller than the argument.) In that case, the value of the integral is quite close to 1 , namely (π² - 6)/4 ≈ 0.967401100 .

[@pkundrat]

Yup. Floor "bracket" has horizontal line on the bottom, ceiling on the top. en.wikipedia.org/wiki/Floor_and_ceiling_functions

[@hassanalihusseini1717]

Great video. I liked how so many problems came together.

[@sriramfavouritesongs32]

Excellent problem, solved with a clever substitution! Well done, Bravo This was a satisfying integral!

[@MetaBuddha]

You might have just helped me with an important proof of mine. Thank you 🙏

[@MrAlvaro412]

actually it gets me on the floor ... all above the scary background sound you made for this special video

[@gabest4]

I checked out the visual representation of this function, it's like the pictogram of a factory building, you can almost calculate the integral manually. (a 1x1 square with increasing smaller missing triangles on the top)

[@Chariotuber]

I miss your videos Dr. Peyam!!! this integral indeed did have me on the floor, amazing!

[@drpeyam]

Thank you :3

[@richardmullins44]

Thanks for this!!!! (I am aged 79 and am writing in the middle of the night after waking up from a dream ("nightmare") where I was fired from a job). Open AI initially gave a guess. It said "A quick numerical evaluation gives an approximate result of 0.25". However, Open AI is very good at improving its attempts. Say to it: "A man who works the problem on youtube says the answer is (pi squared)/12". ChatGPT said: "The result of (pi squared)/2 is indeed correct for the integral in question .... ​ "This is a special case where the sum of the infinite series directly corresponds to a result involving (pi squared)/12 , as derived in some advanced calculus and number theory problems". (I would have liked to post what open ai said verbatim, but I have not found a way to clip the text - it produces gibberish). ​I wonder if open ai can explain to me the Lebesgue integral, which I have never understand. I even suspected that the Lebesgue integral was a hoax to prop up "the new maths". ​

[@byronvega8298]

Truly a jaw dropping integral

[@ArthurvanH0udt]

Very Confusing!! It must be me, but I do not see a floor function, I see square paranthesis!!

[@TheWeeb-j5g]

They're the same Just different notations to depict the same function

[@ArthurvanH0udt]

@@TheWeeb-j5gso what does ceiling function then look like. As far as I know the floor function closes at the bottom only and the ceiling function at the top only.

[@jyotsanabenpanchal7271]

In some places floor fun. Is written as [ ] and in some places it's written like you are saying.​@@ArthurvanH0udt

[@TheWeeb-j5g]

@@ArthurvanH0udt ah your point is valid but I haven't seen any second type of notation for the ceiling function other than just being closed at top. Maybe a secondary notation exists for it or maybe not

[@xinpingdonohoe3978]

How do you do round(x)? Is it a left floor bracket and a right ceiling bracket?

[@michaelz2270]

Number theorists like these integrals. If you don't mind the use of delta functions... if you're integrating [x]/ x^3 from 1 to infinity you can integrate by parts, integrating the 1/x^3 and differentiating the [x], and you get 1/2 + the integral from 1 to infinity of 1/(2x^2) [sum n >= 2 delta(x - n)] = 1/2 + [sum n >= 2] (1/2 n^2) = pi^2/12.

[@awesometronic]

I was wondering where you disappeared to! I hope you keep making math videos; you're one of my favorites!

[@MrRyanroberson1]

i kind of expected euler-masceroni to appear since the *difference* between floor(1/x) and (1/x) in the range 0 to 1 is a reflection of the classical definition of the constant

[@xinpingdonohoe3978]

That's actually an interesting point. Now I'm expecting γ to show up somehow, and I've already seen the answer.

[@nicogehren6566]

very nice approach

[@AA-le9ls]

Shouldn't the first question be if the integral exists? The integrand is not continuous, so it's not clear if it's Riemann integrable or not.

[@nolanr1400]

Correct. You first have to prove that the integral exists

[@VintageGearFreak]

When you split the integral to get rid of the floor function was a real aha moment for me

[@trogdorbu]

I like how every scene cut is accompanied by a very ominous sounding stool being dragged along the floor 😄

[@worldnotworld]

Astonishing! All my life I've disliked the floor function for its brutal castration of the reals into the integers -- all my life until these last few years, with cool problems and cooler solutions like this popping up. I was trying desperately to get the floor of 1/x with 0

[@the_magisterate]

I haven't taken a math class since college, but I when I saw that sum I was like "this looks like a telescoping series". I actually yelped when I got it right haha still got that dog in me

[@kaustubhpandey1395]

Brilliant!

[@Fircasice]

If that is your notation for floor then what does your notation for ceiling look like?

[@ryaneakins7269]

I was thinking the same thing.

[@samsonblack]

ceil(x) = -[-x]

[@gabritex14]

While I agree that it's not the best notation (I was tought that was the notation for the integer part of a number, not the floor), at least for x>0 the behaviour is the same as the floor function

[@Fircasice]

@@gabritex14 I guess that's a good point, indeed.

[@NeedBetterLoginName]

Every time KZbin recommends one of your videos, I can't help but wonder what kind of accent you have. I've never heard anything like it. (Great video, btw)

[@82rah]

Fantastic!

[@misterdubity3073]

Good one. But this also had the broken chalk on the floor.

[@Ðęłẽțëď]

The [1/x] looks like a round function though, if it is a round function it is still easy because then it will be 0 to 1 of x*floor(1/x)+1/2 dx and then just bring the 1/2 out so it will be 0 to 1 of x*floor(1/x) + 0 to 1 of 1/2 and then the answer should be pi^2/12+1/2 (and also i think it will be better to say 0+ to 1 instead of 0 to 1 because 1/0)

[@Biggyweezer69]

I assume by round function you mean the function that rounds the number to the nearest integer? If so the solution is pi^2/4 - 3/2. There is actually a really nice way to show this using the identity round(x) = floor(2x)-floor(x) and the integral solved in the video: int from 0 to 1 of x*round(1/x) dx = int from 0 to 1 of x*(floor(2/x)-floor(x)) dx = int from 0 to 1 of x*floor(2/x) dx - pi^2/12 = (by u substitution, let 2/x = u) 4 * int from 0 to 1/2 of u*floor(1/u) du - pi^2/12 = (by int 0 to 1/2 = int 0 to 1 - int 1/2 to 1) 4 * (pi^2/12 - int from 1/2 to 1 of u*floor(1/u) du) - pi^2/12 = (floor(1/u) on interval 1/2 to 1 is equal to 1) 4 * (pi^2/12 - int from 1/2 to 1 of u du) - pi^2/12 = (simplification) pi^2/4 - 3/2

[@KurtHaag]

You really took this to the ceiling.

[@drpeyam]

Hahahaha 😂

[@colinbaker6404]

I'm not clear what the [1/x in square brackets floor means]...? Can someone explain.... I would have just cancelled the x!

[@victorsago]

What looks like square brackets is not really square brackets, but the floor function. The function floor(x) means just the biggest integer that is less than or equal to x. So, floor(2.5)=2, floor(3.1)=3, floor(100.00000001)=100, and so on. For the negative numbers it also holds: floor(-4.2)=-5, and so on. Of course, the floor of an integer is just that integer: floor(5)=5, and floor(-3)=-3. Floor function looks like square brackets with the top part missing. In the video he's just using the square bracket, but that's an old notation going back to Gauss. If you look it up in Wikipedia, you'll see. Hope this helps.

[@joshthompson5267]

The floor function of an argument gives the greatest integer less than or equal to the argument itself. For example, [1.2] = 1, and [-1.2] = -2 - It's for this reason why you could cancel if and only if the domain of x is a subset of the integers (so that x = [x]). This definitely isn't true here, as an integral should really be defined on a continuous interval (of which the integers are not comprised).

[@davidt939]

​@@victorsago This is confusing in XXIth century

[@skilz8098]

As victorsago pointed out, think of it in terms of software, programming as in conversion between floating points and integer types that pertains to the truncation of values. Consider this: float PI = 3.14... int CloseToPi = (int)(PI). printf( %d, CloseToPi) The output will be 3. If we were to take this int type and cast it back to a floating-point type, it would still be 3 but it would print 3.0 due to the previous truncation. The floor function is similar to this. They're not exactly the same, but it's an easy way to see what the floor function is actually doing. The ceiling function is the opposite as it is closer to a rounding (UP) type function. The main difference between the floor function and truncation due to type conversions is when you start observing the results within their negative domains. The ranges aren't the same.

[@robertanderson1043]

@@victorsago Then why not use the half bracket floor notation that everybody else uses? Seems intentionally misleading for clicks.

[@hectorpinheirobatista1740]

By the writing, this is not floor of 1/x !

[@drpeyam]

It is actually

[@__-1234]

Cool to see there is now a version in English, I'm curious to know where you found this, the first time I saw it was in French as it was given at the prestigious Mines-Ponts competitive exam.

[@drpeyam]

I found it on Instagram actually, someone wrote about it!

[@sokiuwu]

Imagine waking up after 2 years of coma and this is all you see for the first 8 minutes

[@drpeyam]

Hahahaha it would be worth it!!

[@CornishMiner]

Beautiful.

[@randirbox]

This is interesting. However, you have an infinite sequence with each term close to 1/k at 6:07. This one is only conditionally convergent, so regrouping its terms can give us any result. That should be no problem as terms of the sequence could be represented as { 2 over k(k + 1) } - { 1 over k (k + 1)^2 } if I'm not wrong.

[@Zevoxian]

We haven’t used the limit yet so as is it’s a finite sum that we can rearrange

[@fridgepuff1722]

Thank you dr peyam

[@francescobedinijacobini]

The tricky part is the notation. To me the floor of a number is n(x), not [x].

[@josephtorrescardona8819]

What if we use the ceil function instead? Would it have any relationship with this result? 8:17

[@drpeyam]

You can use that ceiling = floor + 1 so it would add 1/2 to the integral I believe

[@leif1075]

Thanks for sharing.

[@Pilot_engineer_19]

Well, my scientific calculator says it is one. Go argue with a machine.

[@douglaszare1215]

Is there an essentially different way to do this integral, which would then give an alternative summation of 1/k^2?

[@quinnculver]

The graph puts in a lot of work. It shows that another (better?) approach---as @mathunt1130 indicated below---is to split up on intervals from 1/(n+1) to 1/n, where the function is simply n*x. That quickly yields the series at 4:54.

[@imnotarobot6927]

Cool! I can very much recommend the integral sin(floor(1/x)) I had a lot of fun with it

[@drpeyam]

Ooooh I like this!!

[@adumont]

@@drpeyam oh I hope tp see a video sin(floor(1/x)) is sick!!!!

[@imnotarobot6927]

@@drpeyam I just uploaded a rough hand-wavey outline on how to tackle this integral. Few years back I recorded a proper step by step video, but never uploaded it. It gets extremely messy at the end, so maybe it doesn't make for the best video, but it is still a lot of fun to do :) edit: taken down for a while for privacy reasons

[@personxy7443]

Nice explanation!COOL T-shirt

[@drpeyam]

Thank you :3

[@hansisbrucker813]

Can someone explains how we got step 4 with the integral from ∞ to 1? I have rewinded that part for over 5 times and I still don't get it 🤔

[@joshthompson5267]

The interval of the integral in x was from 0 to 1, but since it is a substitution, that of the integral in u=1/x should be defined by corresponding bounds dependent on x. Considering the value of u where x = 1, the upper bound can be found by u = 1/x = 1/1 = 1. I'm somewhat informal with this, but since the interval for x is never negative, the lower bound can be though of as x-> 0⁺ (meaning very tiny positive), so u = 1/x -> +∞. Hence, the interval for intergration in u can be written as ∞ to 1. The rest of the integral comes from reconfiguring the expression in the first line by writing x as 1/u, 1/x as 1/(1/u) = u, and dx = -1/u^2 du as worked out in the third step, which can be done by differentiation by the power rule. Hope this helps.

[@doctorb9264]

Excellent problem and solution. I subscribed.

[@drpeyam]

Thanks so much :3

[@dank.]

Amazing!

[@PetrovichJLA]

This is so beautiful I can't even figure out how can people don't like maths

[@drpeyam]

Agreeeeed!!!

[@ralfbodemann1542]

I solved it without the initial u-substitution. I got the correct result, but the path to the solution was less elegant than yours.

[@martinstu8400]

2:15 smart move

[@sciencelover-c2j]

Is there another way to solve this integral????

[@whiteshadow8520]

Why can’t you cancel the x initially? Does it depend on where it comes from?

[@donmoore7785]

Because what is in the brackets is not equal to 1/x. It is floor(1/x). Those are brackets, not parentheses.

[@parthasm]

Quite interesting i follow your videos regularly and share your wisdom with my only math student - my son!

[@drpeyam]

Awwwww thank you!!!

[@egillandersson1780]

Very nice !

[@lukandrate9866]

WTH I just made up and solved this same integral like ten minutes ago when solving a similar one. Such a coincidence

[@MultiNeurons]

Wow!! What a surprising integral

[@joeeeee8738]

You should GRAPH the integral at end

[@NeilDesmond]

Those aren't the correct grouping symbols for expressing the floor of 1/x. The correct grouping symbols look like a larger uppercase "L" with a very short horizontal line on the left side, and a horizontally oriented mirror image of that on the right side, sort of like this: |_ 1/x _|.

[@drpeyam]

[1/x] is equally valid

[@guitaristxcore]

Wonderful video, lovely math problem

[@Happy_Abe]

I don’t think it’s always true that we can go from an indefinite integral to the limit of the integral from o to N in integers N. I think it works here since the functions always positive so if the limit exists in N it will exits in general, but for say an oscillating function like a sin it may have an integer limit but not a general limit.

[@drpeyam]

Yes that’s why I was saying that if the improper integral has a limit then you can say it works for integers

[@Happy_Abe]

@@drpeyam gotcha, right! And even if the improper integral doesn’t converge it wouldn’t in integers either since the function is positive

[@drpeyam]

No if the integral doesn’t converge then it might converge if n is an integer, think sin(pi x)

[@Happy_Abe]

@@drpeyam but sin(pix) is not positive. For a positive function f, the integral of f in integers converge iff the improper integral converges. I thought so at least

[@blackpenredpen]

Did u drop the marker bc u didn’t want to break ur mic? 😂

[@blackpenredpen]

I watched until the end tho not “just the end”

[@GeoffryGifari]

so the key to dealing with floor function in the integrand is splitting the full interval into unit intervals?

[@archangecamilien1879]

Perhaps it is 0, etc...if you think about it as an area under a curve, floor(1/x) is zero between 0 and 1, and only 1 at the point 1, so the area under the curve would be 0 anyway...if I'm not mistaken, etc...

[@victorsago]

@@archangecamilien1879 floor(1/x) is only 0 if x>1. If 0

[@martinepstein9826]

@@archangecamilien1879 If x is between 0 and 1 then floor(1/x) is never 0.

[@archangecamilien1879]

Oh my, lol...ok, I am incorrect....

[@archangecamilien1879]

So I guess it'll be an infinite sum of integral, etc...each one over an integer interval...

[@pizza8725]

I though that was the round function

[@usethechenlu4096]

and they said this function was beneath you

[@jyotsanabenpanchal7271]

Very Nice!

[@gwpiaser]

thank you

[@elhombre9171]

You should explain why x [1/x] is not reducible to 1 in the first place... however... instead, all of a sudden, then you change 0 for infinity... then a sum... etc... A complete nightmare!

[@drpeyam]

[1/x] is the floor of 1/x

[@elhombre9171]

@@drpeyam floor? the floor on which 1/x walks? what are you talking about?

[@jamesshavrnoch8665]

But if you subtract the log cosine of the invers Finochia sequence... you get pie...

[@henrikholst7490]

I was a bit worried about the infinite number of discontinuites .. should I be? Can we exclude a countable set of points on the real line as long as the function is bounded? Perhaps I should be more worried about the anti derivative.... 😂

[@alextrebek8293]

so why can't you cancel the x's?

[@slavinojunepri7648]

Every step well explained. Keep up the great work. Be well 🙏

[@drpeyam]

Thank you!!

[@leif1075]

​​@@drpeyamThanks but I don't see why anyone wpuld.think of doing a substitutjon in tbjs case wluld you agree..so why do thst offf tje bat..I think you can you solve without it like rewrite x as the inegwr part I plus tje decimal part so the floor would just be 1/I plus the decimal and work from there?

[@leif1075]

​@@drpeyamHope you can respond. Thanks very much.

[@jeremylongstreet7613]

What am I missing that you can't cancel the 'x'? i put this integral into the AI and questioned it on the cancel of 'x'. It says it is fine to cancel the 'x'.

[@drpeyam]

Here [1/x] is the floor of 1/x, AI cannot be always trusted

[@v.r.kildaire4063]

use floor brackets pls

[@drpeyam]

I am using floor brackets

[@eamonnwalker4512]

@@drpeyam Judging from the other comments, I am not alone in thinking ⌊x⌋ is the more common modern notation for floor(x). Without context, a large portion of your audience is likely to interpret [1/x] as being equivalent to (1/x) instead of ⌊1/x⌋.

[@rainerzufall42]

@@eamonnwalker4512 No doubt, that it should be something like ⌊x⌋ = floor(x). [x] looks like a mixture of floor(x) and ceil(x). Could also be round(x)! As ceil(x) isn't used very often in mathematics, Gauss' old "Gaussklammer" notation will be preferably used for floor(x), but with an international audience, I would never expect the viewers to use the term "Gaussklammer" (in English). You certainly didn't, I am sure, you said "floor" instead of "Gaussklammer"! Thus ⌊x⌋, not [x].

[@rainerzufall42]

@@drpeyam Wikipedia(en): "In mathematics, the floor function is [...] denoted ⌊x⌋ or floor(x)." Furthermore: "Similarly, the ceiling function [...], denoted ⌈x⌉ or ceil(x)." (citation: Graham, Knuth, & Patashnik, Ch. 3.1)

[@rainerzufall42]

@@drpeyam There's even a paragraph about this topic: "Carl Friedrich Gauss introduced the square bracket notation [x] in his third proof of quadratic reciprocity (1808).[3] This remained the standard[4] in mathematics until Kenneth E. Iverson introduced, in his 1962 book A Programming Language, the names "floor" and "ceiling" and the corresponding notations ⌊x⌋ and ⌈x⌉. (Iverson used square brackets for a different purpose, the Iverson bracket notation.)" That means, you're not totally off, but be assured, that this square bracket notation is very "special" or at least outdated.

[@skilz8098]

Isn't there an association between the Floor and Ceiling functions with the Taylor Series or the Harmonic Series?

[@carultch]

The Harmonic series would be integral floor(1/x) dx from 1 to n.

[@skilz8098]

@@carultch I thought so... It's just been so long since I've done something on those lines. I did vaguely remember or come across something like this a few years back when I started to freshen up on the Fourier Series for implementing FFTs and their Inverses as well as ODEs such as Runge Kutta or RK3 & RK4 algorithms.

[@Lolwutdesu9000]

It would be nice to draw the floor function notation correctly, at least.

[@drpeyam]

It’s correct notation

[@reidflemingworldstoughestm1394]

What manner of sorcery is this?

[@drpeyam]

Peyamagic 🧙

[@alejrandom6592]

3:22 *falling noises*

[@Bethos1247-Arne]

so it is half of the sum of the recipokes of quadratics?

[@drpeyam]

Yes exactly!!

[@bradyoder1694]

What about the integral of the same function from -1 to 0?

[@yurenchu]

{ ∫ x*floor(1/x) dx , from x=-1 to x=0 } = = { ∫ x*floor(1/x) dx , from x=-1 to x=-½ } + { ∫ x*floor(1/x) dx , from x=-½ to x=-⅓ } + { ∫ x*floor(1/x) dx , from x=-⅓ to x=-¼ } + ... = Σ { ∫ x*floor(1/x) dx , from x=-1/k to x=-1/(k+1) } from k=1 to k=+inf ... note: -1/k < x < -1/(k+1) ==> -(k+1) < 1/x < -k ==> floor(1/x) = -(k+1) ... = Σ { ∫ x*(-(k+1)) dx , from x=-1/k to x=-1/(k+1) } from k=1 to k=+inf = Σ -(k+1) * { ∫ x dx , from x=-1/k to x=-1/(k+1) } from k=1 to k=+inf = Σ -(k+1) * { ½x² | from x=-1/k to x=-1/(k+1) } from k=1 to k=+inf = -½ * { Σ (k+1) * ( 1/(k+1)² - 1/k² ) from k=1 to k=+inf } = ½ * { Σ (k+1) * ( 1/k² - 1/(k+1)² ) from k=1 to k=+inf } = ½ * { (1+1)*(1/1² - 1/(1+1)²) + (2+1)*(1/2² - 1/(2+1)²) + (3+1)*(1/3² - 1/(3+1)²) + ... } = ½ * { (2)*(1/1² - 1/2²) + (3)*(1/2² - 1/3²) + (4)*(1/3² - 1/4²) + ... } = ½ * { 2/1² - 2/2² + 3/2² - 3/3² + 4/3² - 4/4²) + ... } = ½ * { 2/1² + 1/2² + 1/3² + 1/4² + ... } = ½ * { 1 + 1/1² + 1/2² + 1/3² + 1/4² + ... } = ½ + ½ * { 1/1² + 1/2² + 1/3² + 1/4² + ... } ... Basel problem: {1/1² + 1/2² + 1/3² + 1/4² + ... } = π²/6 ... = ½ + ½ * { π²/6 } = ½ + π²/12 = (π²+6)/12 Note also: { ∫ x*floor(1/x) dx , from x=-1 to x=0 } = ... substitute u = -x , dx = -du , 1/x = -1/u , floor(1/x) = floor(-1/u) = -ceil(1/u) ... = { ∫ (-u)*(-ceil(1/u)) (-du) , from u=+1 to u=0 } = { ∫ u*ceil(1/u) du , from u=0 to u=1 } = ½ + { ∫ u*floor(1/u) du , from u=0 to u=1 } Note: generally (for non-integer z), ceil(z) = 1+floor(z) , so ∫ u*ceil(1/u) du = = ∫ u*(1+floor(1/u)) du = ∫ u du + ∫ u*floor(1/u) du = [½u²] + ∫ u*floor(1/u) du

[@trewq398]

Wow, and where is the circle in this problem?

[@yurenchu]

Nowhere. It's a _squared_ pi(e) in the outcome, presumably because of a lot of squares, but no circle. ▪️ ▫️ ◾ ◽ ◼ ◻ ⬛ ⬜ 🔲 🔳 ⏹

[@milind006]

Am I the only one bothered by this, or did the symbol for floor change, or I never knew the symbol for floor? In my mind the symbol for the floor function does not have the top horizontal lines of the square brackets, so they look like walls with floors. The way it’s written in the video, they just look like square brackets/boxes whatever your preferred term is.

[@drpeyam]

Both notations are acceptable, we use [x] to denote floor x

[@milind006]

Thanks for responding.

[@richardmullins44]

what would ramanjujan have said?

[@drpeyam]

He would have loved it :)

[@LorenzoWTartari]

Very cool shirt

[@drpeyam]

Thank you!! 🐉

[@gregodify]

I am having a hard time figuring out why dx=-1/U^2dU. Can someone explain, please?

[@joshthompson5267]

Expressing 1/U as U^-1, you can then use the power rule [d/dx x^n = nx^(n-1)] to find the differential by differentiating x with respect to U :) I hope that's helpful!

[@joshthompson5267]

Oh, and more fundamentally, it can be shown by the formal definition of a derivative [d/dx f(x) = lim_(h->0) (f(x+h)-f(x))/h] D = d/dU 1/U D = lim_(h->0) (1/(U+h) - 1/U)/h D = lim_(h->0) (U-(U+h))/(U(U+h))/h D = lim_(h->0) -h/(U(U+h))/h D = lim_(h->0) -1/(U(U+h)) D = -1/U^2 If you want anything else please ask!

[@gregodify]

@@joshthompson5267 Thank you! I knew it was something small that I was missing.

[@士-x7e]

Nice shirt! Dragon?

[@drpeyam]

🐉

[@magilviamax8346]

Shouldn't the integral just be undefined cause x cannot assume a value of zero at the lower bound by definition?

[@johannesrehnstrom1058]

Mathematicians have tricks to deal with that issue. For example, if a function is undefined at only a single point, it will not affect the integral, so it can be ignored. To be a bit more rigorous, an integral can be defined in terms of a limit, if the function is undefined at the boundaries. For example the integral from 0 to 1 of 1/x can be defined as the limit of (integral from a to 1 of 1/x) as a goes to 0.

[@magilviamax8346]

@@johannesrehnstrom1058 Exactly, it's missing the limit. I think it's necessary to include it in the solution as it could change the result.

[@alexcastrotello5165]

Why 1/12 appears on a lot of places? Strange

[@jyotsanabenpanchal7271]

Would love to see video on stolz cesaro Theorem.❤❤

[@drpeyam]

What’s that?

[@jyotsanabenpanchal7271]

@@drpeyam umm well I don't know so that's why I told you sir to do that😅 Limits me use hota hai uska sir

[@tzonic8655]

Who else thought there was a hair on their screen?

[@jaimecastells4283]

I guess I'm ignorant. Why can't the integrand be algebraically simplified?

[@jaimecastells4283]

In fact, the explanation provided uses algebraic manipulation of the integrand. I'm confused.

[@drpeyam]

Because [1/x] is the floor of 1/x and not just 1/x