This integral will have you on the floor 🤣🤣
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- Опубликовано: 11 апр 2023
- Integral of 1 over integer part of 1/x from 0 to 1. I calculate a nice integral with a floor function by using Riemann sums and using telescoping series, it’s a must see for calculus lovers. The answer involves the Euler Mascheroni constant
Integral fractional part of 1/x
• Integral of the fracti...
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I didn't recognise this "floor" notation and thought you were asking me to find the continuous integral of x dx between 0 and 1.
Interesting problem with an interesting result, though.
Glad I got it right. Euler's solution to the Basel problem is a fun topic by itself. I love it when continuous and discrete domains are mixed. Reminds me of signal processing math.
WTF happened to the floor.
Floor is the piecewise function that let him break it into a bunch of simple integrals.
I was making a pokemon talk reference.
@@tyruskarmesin5418 can you give more info on this please? I'm new to these functions. I looked it up on wiki but it described it in a very strange way and I don't understand
@@yogi30051972 So for each interval of x that he integrates 1/x over, (1/2 to 1 and 1/3 to 1/2 ect), the floor of 1/x for x being in that interval is just a constant. For example 1/x where x is a value somewhere between 1/3 and 1/2 means 1/x is between 2 and 3 so the floor of 1/x is just 2 (decimal part removed). Which allows him to integrate the floor of 1/x as just 2 over that interval. If you add all those pieces together you get the whole integral from 0 to 1.
memory unlocked
I love your enthusiasm!
Any math problem will have me on the floor in a fetal position
Great video as always! I am pretty sure that the last integral, of the frational part of {1/x} from 0 to 1, is actually positive and equal to 1-gamma.
I transfered the integral to a pretty weird sum, and as wolfram alpha said, the sum equals 1-gamma.
No it had me on the "ceiling"!
Great video! Vear clear explanation, and a cool final result.
If you want a challenge, you could try:
Integral from 0 to infinity of {e^x} - 1/2
i.e fractional part of e^x, subtract 1/2
Very cool solution - thanks 👍
Great video!
I accidently discovered what a floor function was trying to find the wave equation of a triangular traveling wave. It scared me and so does this integral.
Fun math, fun video and fun guy.
very good video, it could help me to solve an integral with matrices or with linear algebra, greetings from Mexico
The calculation you did in the video uses the floor function, which is different from the image shown in the thumbnail
The thumbnail shows the floor function
@@drpeyam the Thumbnail shows square brackets
@@a.b.c.d.e... Greatest integer function
No it doesn’t. The thumbnail shows square brackets [ ]
It’s another notation for the floor function…
The last integral is positive: 1-γ.
Hi PROF nice to see U back☺
Very cool!
I literally just saw this problem on Brilliant yesterday and thought... "Huh!? Where do I even begin??" Thanks.
Thank you so much ❤
I felt smart when I figured this out without watching the video, then I felt even smarter after reading the comments lol!
bruh look at the thumbnail its not the floor f unction there, so people solved it on the thumbnail
Always looks like there is something mysterious when we manage integer numbers and suddenly appears Pi
Perfect❤❤ i love your show
That right-hand expression collapses like a cheap suit.
I must be missing something. Isn't [1/x]expn(-1) just x? So that would be Integral(x) from 0 to 1. And wouldn't that just be 1/2?
No it’s the floor function
The integral of the integer part does not necessarily need to be an integer because the integer part is just in the integrand correct and it becomes an infinite series as we saw what you did. - Good to see Euler Mascheroni constant again!
My face in the floor when doctor peyam make a video ❤😊
Awwwwww thank youuuu!!!
I derived the series at 1:47 using integral substitution. If we let u=1/x, then dx = 1/u du (note I didn’t mix up x and u here, the two formulations are equivalent).
From there it’s a relatively straightforward forward sum of integrals.
Doubly cool!
how is that second sum telescoping? Did you mistake it for the sum of (1/n - 1/(n+1)) or did I miss something?
It is telescoping, you can use partial fractions
Aaah I see, the numerator is 1 + n - n.
Beautiful result out of ugly integral 😂
After reading the comment section I am convinced that a lot of people, even in the math community, don't know what the floor function is.
I know right?? I’m so shook 😂
@@drpeyam thumbnail doesnt CONTAIN it
I've used it a lot over 20+ years in MATLAB / C / C++ where we just see it as floor() but have never seen this "partial square bracket" notation.
@@jameeztherandomguy5418 Yes that's why I thought the problem was to integrate x dx between 0 an 1, which of course is trivial. Then on the white board it looks different, if you spot it.
People do know what the floor function is. The problem is the thumbnail did not show the floor function.
Amazing 🥰🥰🥰🥰
Pi squared? No way!
Pi are round.
Cornbread are square.
Pie a la mode ~
Potato casserole squared
your thumbnail shows nearest integer, not floor
No that’s the notation for floor
@@drpeyam no its not, you put in brackets
What about studying the triples as [2,3,5] with one even and two odds numbers such that 2*3+5=11 (prime); 2*5+3=13 (prime); 3*5+2=17 (prime) ? Never heard about this topic.
Seems quite interesting excluding symmetries. Another example could be [3,7,10] ... and so on and so forth...
I’m not a number theorist
(1/x)-¹ = x
But not if there is a floor in the middle.
on 0:29 you start with the lower bound of the first integration with 1/2 not 0 why?!
Because for x being 0 to 1/2 (floor(1/x))^-1 will be zero
Because if x belongs to (1/2, 1], floor(1/x) = 1. The next interval that results in the same operator being 2 is (1/3, 1/2] and so on until (countable or discrete) infinity. Hope that gives you clarity.
actually he goes till that , its just that he has started with 1/2 to 1 then 1/3 to 1/2 soo eventually it will reach 0 and be included (which we will take care of in the summation)
I don't understand this at all. I thought the answer would just be 1/2??
The answer 1/2 is true if the expression in the brackets was just 1/x. But it isn't. Look carefully, 1/x isn't surrounded by square brackets (although it looks like one). It is actually a floor function applied on 1/x. Just google the symbol for floor function and you'll understand.
@@sanjayg1728 Thanks, man.
@@sanjayg1728it's square brackets in the thumbnail lmao
@@NKY5223 Yeah, that's right.
@@sanjayg1728No, it is brackets because it's the round not the floor function.
Hello! What is the convolution of two cauchy distribution?
No idea
@@drpeyam I tought you could help me.
@@drpeyam😂😂😂your lying
I'm from India ❤ I like your video
Never seen someone so excited to see pi^2/6 as an answer. P.S. where could i find the solution to getting pi^2/6
The videoxof 3blue1brown about it is nice
Google "Basel Problem"
fourier series
@@BridgeBum holy hölle!
Fourier series
great
i initially got confused with the floor functions, but then i realised it is actually the least integer function (thats what we call it in india)
and your ceiling function would be Greatest Integer Function
It's the other way around. Floor is the greatest integer function
1/2
Nope
When you talk about π²/6, you showed a picture of π/6!
Maybe...that was actually π²/6? Please tell me!
😂😂
What.
Can't that be a beta function😅😂
PROF why do U write the N capital ? Thanks)
My n are illegible
@@drpeyam😂😂😂😂eh
Answer:1/2
No
@@drpeyam Well per the problem presented on the thumbnail, yes. On the whiteboard however, you're right, no.
@@thegrandmuftiofwakandathe thumbnail is dependent on the country
@@wakeupthewublins69 I can only comment on what I can see, and what I can see is ambiguity.
Isn’t the answer 1/2
No because this is the floor(1/x)^(-1) not (1/x)^(-1)
@@minecraftxd4996 you just said the same function
@@mokshamahey how is floor(1/x)^(-1) equivalent to (1/x)^(-1)?
@@mokshamahey Keyword: floor.
I see
If we substitute p = 1/x
==> dx = - p^-2 dp
Also
If x = 0, then p range = infinity
and x = 1 then p range = 1
Now we integrate (- p^-3) dp within the p range infinity and 1
Is it workable ?
and once evaluated I get the result
1/2 + C !
No that doesn’t work
@@drpeyamtrue may that be, simply stating the falsehood of a solution without providing correction or guidance is both disheartening to someone who was excited about a problem you posed, as well as being unhelpful in any potential aspiration for improvement.
Okay but why do a substitution like that? If you interpret the task like you just did, you can just use the power rule (x^-1)^-1 = x^(-1 * -1) = x, and then the integral becomes int_0^1 x dx which is 1/2 x² from the bounds 0 to 1, so 1/2.
But all of this isn't reading the problem correctly: There is a floor function around the 1/x, which doesn't let you do any operations like this. (floor 1/x is not the same as 1/floor x)
@manny4148 Ok
@@drpeyam why does your head look like a strawberry?
Fun video!
Isn't it 0.5 bro?
Its just x^2/2 and integral of that is 1/2
Nope
cool
aint no way bro used an integral to find a triangle
It’s not x
@@drpeyam
bro got the supreme clickbait. Are you trying to get pepole to say its 1/2 because you didnt show the floor function on your thumbnail???
I showed the floor function in the thumbnail
@@drpeyam
brackets can mean anything. That is not commonly accepted notation -- especially since you showed it correctly in the video
sems like clickbait
@@drpeyam brackets are not floor, floor notation misses the top part of bracktes
Why isnt (x^-1)^-1 just x?
Because this square brackets means floor function. The Floor function returns the smallest integer not greater than the argument. This aspect makes the question harder and, of course, more interesting.
🙏🌺🙏
i! ???
As a first year calc student, (1/x)^-1 is literally just x. What are you even doing? 😂
This isn't (1/x)^-1, this is floor(1/x)^(-1). They are two completely different functions.
@@minecraftxd4996 Woah, I didn't even realise that's the floor function. Damn.
Thumbnail is wrong
?
@@drpeyam The thumbnail doesn't show a floor function, just normal square brackets
That is the notation for floor…
@@tomasbeltran04050 The square brackets are another common notation for the floor function, introduced by Gauss.
@@tomasbeltran04050 this notation is widely used in asian Countries and it's also called Gif (Greatest Integer Function).
🫣👻
no way, no hope, pi always pops up!!! :))))))
1/2
1/2
No?
@@drpeyam I suppose that they misinterpreted the floor function [1/x] as parantheses (1/x). Great video!