BEAUTIFUL Floor Function Integral (With Zeta Function)
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- čas přidán 23. 05. 2024
- I hope this has been interesting! This is one of my fav integrals in a while so wanted to share. Please comment with any questions or suggestions for new topics. Subscribe to stay updated.
~ Thanks for watching all.
@maths_505 Thanks for taking a look at my solution to this too!
#maths #mathematics #integrals #zetafunction #euler #funproblems #proofs #functions #physics #sums #series #limits #whiteboard #math505
Yo nice!
Subbed
NO WAY! Thank you so much bro I'm the biggest fan.
Hell yeah brother
Alternatively split the integral into two straight away which makes the summations simpler:
Integral[ x=1 to inf: (x-floor(x))/x^4 ]
= Integral[ x=1 to inf: x^(-3) ] - Sum[ k=1 to inf: Integral[ x=k to k+1: k.x^(-4) ] ]
= [ x=1 to inf: (-1/2)x^(-2) ] - Sum[ k=1 to inf: k.[ x=k to k+1: (-1/3)x^(-3) ] ]
= 1/2 + (1/3).Sum[ k=1 to inf: k.(1/(k+1)^3 - 1/k^3) ]
= 1/2 + (1/3).(Sum[ k=2 to inf: (k-1)/k^3 ] - Sum[ k=1 to inf: 1/k^2 ])
Since k-1=0 for k=1:
= 1/2 + (1/3).(Sum[ k=1 to inf: (k-1)/k^3 ] - Sum[ k=1 to inf: 1/k^2 ])
= 1/2 + (1/3).(Sum[ k=1 to inf: 1/k^2 ] - Sum[ k=1 to inf: 1/k^3 ] - Sum[ k=1 to inf: 1/k^2 ])
= 1/2 - (1/3).Sum[ k=1 to inf: k^3 ]
= 1/2 - zeta(3)/3
That’s fire bro!! But could you maybe add timestamps or something when you make those tiny mistakes? (Like forgetting 1/2 while integrating) We all make them but it really throws me off sometimes and I fail to concentrate on the real shit. Maybe it‘s a me problem but a small remark would be greatly appreciated!! Anyways Banger video :)
Yeah absolutely, I'll do that in the future. Thanks so much for the feedback and I'm so glad you enjoyed watching!
Such a cool integral. Really liked your explanation
Thank you so much! Im really glad.
2:35 video start
Banging integral bro🔥
hell mf yeah that is a beaautiful integral. so satisfying. just the right length incredibly simple
Right!! So nice how once you make that first jump it all just simplifies in the perfect way.
@@OscgrMaths i so, so love those little moments in higher math. the "leaps of faith", a name very dramatic but I feel like it's fully deserving of that moniker because as you said, it just clicks into place in the end. the elegance of maths astounded me from day one of my life and has not stopped since
@@theseusswore Exactly the same for me - that's why I wanted to start this channel in the first place! So glad there's such a great community for maths online, I feel really lucky.
@@OscgrMaths I've always wanted to be part of math communities but the fact that I'm too awkward to try and get into groups and just an overall lack of groups which would be willing to accept people with intermediate math knowledge is why im not. that's why it's so beautiful when I find channels like yours and others with passion for the subject oozing out of each video. 3b1b is my all-time favorite, the community is absolutely magnificent
Just found your channel. Gold content
Thanks so much for the comment! I'm so glad you like it.
Sick integral!
I suppose it could be split up into two integrals, the sum of two integrals, the second will have the floor function, that can be split into an infinite series over integer-length intervals, etc...an integral over each interval...
@OscgrMaths Amazing content 🔥 .....showed up on M's feed :)
Nice one. :) I repeated the calculation for an arbitrary natural exponent m instead of the exponent 4 in the denominator, and after some really nice cancellations, I got the quite simple result 1/(m-2) - zeta(m-1)/(m-1), completely agreeing with your result for m = 4.
That's such a great idea - generalising problems like this is always so fun. I'll try and work it through myself as well. Thanks for the comment!
Great job this was very cool integral. Can you make a video on solving the zeta function of 3?
The floor function truncates for positive inputs, but negatives are not a simple truncation. This is because the definition of the floor function is "the the largest integer less than or equal to x". (ex. floor(2.5)=2 , floor(3)=3 , floor(-1.5)=-2)
Yeah absolutely, it works differently for negatives so that the ceiling of x is always greater than the floor of x. Thanks for the comment!
Nice
Great job! I will share with my students!
Great! Really glad you enjoyed.
this video cured my stupidity.
i know know yet another thing without being able to explain it.
You'll get there! If you have any questions then feel free to ask.
Brilliant. Thanks.
No problem at all - glad you enjoyed!
hell yeah i’m glad i was recommended this
Thanks so much! Really glad you enjoyed.
Genial
Awesome explanation
Thanks so much!
Amazing explanation!👍👍👍
Thanks so much! Im glad you enjoyed.
Rewrite problem as
sum n=1 to inf (Integ n to n+1 (x-n)dx/x^4)
Rewrite as:
sum n=1 to inf (Integ n to n+1 (1/x^3-n/x^4)dx)
Integrate
sum n=1 to inf (n/(3x^3)-1/(2x^2)) | n to n+1)
replace x with n+1 and n and subtract the first with the second:
sum n=1 to inf n/(3(n+1)^3)-1/(2(n+1)^2)-n/(3n^3)+1/(2n^2))
sum n=1 to inf 2n/(6(n+1)^3)-3/(6(n+1)^2)-2/(6n^2)+3/(6n^2))
sum n=1 to inf 1/(6n^2))+2n/(6(n+1)^3)-3/(6(n+1)^2)
sum n=1 to inf 1/(6n^2))+(2n+2-2)/(6(n+1)^3)-3/(6(n+1)^2)
sum n=1 to inf 1/(6n^2))+2(n+1)/(6(n+1)^3)-2/(6(n+1)^3)-3/(6(n+1)^2)
sum n=1 to inf 1/(6n^2))+2/(6(n+1)^2)-2/(6(n+1)^3)-3/(6(n+1)^2)
sum n=1 to inf 1/(6n^2))-2/(6(n+1)^3)-1/(6(n+1)^2)
the terms 1/(6(n+1)^2) and 1/(6n^2)) will cancel each other out with the exception of 1/(6n^2)) when n=1 so:
1/6 -sum n=1 to inf 1/(3(n+1)^3)
replace sum with zeta function:
1/6 +(1-z(3))/3
Super clean, loved the format and the solution! Subbed
Thank you so much! Glad you enjoyed.
Clean? The writing is sloppy.
@@OscgrMaths It's not clean. The writing is doctor's writing.
Nice one. Thx!
No problem! Glad you enjoyed.
Very nice! I love it when the "clunkiness" of truncation somehow yields elegant results. (You might have pointed out to those not familiar with the floor function that _x-floor(x)_ truncates left of the decimal point, i.e. eliminates the integer part; most people will notice, but it's a fun fact!)
Yeah I probably should have clarified that... Glad you enjoyed anyway!
❤❤❤
nice
5:38 : where’d the 2 go? Or, why was there a 2 in the denominator to cancel with?
Edit: 6:39 oh, ok
Sorry my bad for forgetting it - next time i'll try and use time stamps or edit it out.
Hey man, great video, just wanted to add:
Would we not need to account for the discontinuities that occur on each segment of the floor function? I.e. for the integral from 1 to 2 of the floor function, you’d need to write it as an improper integral with an upper bound approach two in the limit? Ik this doesn’t affect the result, just an important step
You're absolutely right, I probably should have done that in order to be more rigorous. Really glad you enjoyed the video! Next time I'll definitely mention things like that if they come up.
Very nice
Thank you!
Nice video, liked iRL, liked in video, subbed, saved
Great! Thanks so much, really glad you enjoyed it.
subbed
Thanks so much!
new sub 😊
Thanks for subbing!
Nice, i have another one, similar but more challenging for you. I= int from 0 to 1 ( x * floor(1/x) ) dx. Good luck.
I'll take a look at this later! Thanks so much.
@@OscgrMaths I give you the final result, it is pi^2/12
It is a bit more challenging than yours because of the inverse, but same principle otherwise, you have to chop it into pieces.
Just write s+1 in place of 4 and solve, you basically get an alternate definition of the zeta function, but with no actual usefulness I guess😭😭😭😭😭
That is interesting!! Always good to have multiple ways to write important functions like the zeta function. Great idea.
@@OscgrMaths Thanks😇😇
bro i came for the math but you're beautiful no diddy
This integral looks familiar from somewhere.... 🤔
Now you can cheat and see the solution 😁!
@@OscgrMaths but then I won’t get three stars ☹️(seriously tho that was actually a very satisfying solution in the end- that first step was genius)
Wouldn’t a u-sub with u=-ln(x) work quickly?
Hey, that sounds really interesting but I can't work out how to start it - what were were your steps?
i like your rubber/pen holder
I tried to do the same but with x³ in the denominator, but I ended up with Σ(n=1,∞)(1/(2n³+4n²+2n)), and when I computed this on a calculator it seemed to equal 1-(π²/12), do you know how to evaluate this sum ?
Yes I just did it now - factorise the bottom part and then use partial fractions decomposition. It should split into three fractions. From there you can rearrange and manipulate the sums. If you need any more hints just message!
@@OscgrMaths the problem I have is that I don't know how to not end up with the harmonic series, since the degree of the denominator is odd. I can easily isolate 1/(n+1)² or 1/n² and do a substitution but then I'd have another sum that doesn't converge (namely 1/n or 1/n+1)
@@rasmodii You should end up with by partial fraction decomposition
1/2n -1/2(n+1) - 1/2(n+1)^2
These first two harmonic series will cancel out for all terms other than the first
1/2 + 1/4 + 1/6 ... - 1/4 - 1/6 ...
so we are left with just 1/2. We can use the trick of substitution that was used in the video to show that 1/(n+1)^2 is the same as pi^2/6-1, so
= 1/2 - 1/2(pi^2/6 -1)
=1-pi^2/12
@@OscgrMaths I'm incredibly silly I didn't realize it had to have the form A/n+B/(n+1)+C/(n+1)², thank you so much and keep up the amazing content!
@@rasmodii Ah no problem!!! Easy mistake to make. Glad you're enjoying - always feel free to comment!
question :
does x - floor(x) >= 0 ?? for all x
Yep, it’s pretty easy to see if you graph it. It looks like a sawtooth wave that stays between y=0 and y=1
You're absolutely right, since x >= floor(x), by subtracting floor(x) on both sides we get what you said:
x-floor(x)>=0
But also, correct me if i'm wrong, I think we can also bound it from above as well since the greatest difference between x and floor x would be if x were approaching the integer above floor(x)
(eg... if x=3.999999 and floor(x) =3 )
Given that this is a difference of just below 1, this means that
1>x-floor(x)>=0 for all x
Is floor(2)=1?
Great question, floor(2)=2 because it's at this point that we take a step up to the next integer!
The camera is at too much of a steep angle and writing is sloppy.
i simplified into two integrals first
Yeah that's also a great way to do it - probably makes things easier in the long run!
Horrible camera angle
Well... you need some proper software. Not a whiteboard at an angle. And don't explain 1/3 + 1/6 = 1/2, I mean, who is your audience?
Sorry, I meant that part to come off as sarcastic. I've been thinking of looking into using manim - do you know any good softwares I could look at?
@@OscgrMathsI like the whiteboard! Really enjoyed the video. If you do want to move to software though, 3Blue1Brown made his own python library called Manim, for making Maths videos.
@@conkeydong2095 Thanks so much! To be honest I quite like the whiteboard too. I still might take a look at manim though.