integrate with the FLOOR and CEILING | MIT Integration Bee 2023
ฝัง
- เผยแพร่เมื่อ 2 ธ.ค. 2024
- "CAVENDISH YOU SCOUNDREL! How was it that you were able to come by an integral from the the current year MIT Integration Bee?" ask Muffleford. "Why I obtained the information from a friend of mine", Cavendish began to sweat heavily and Meffleford could see it and gave Cavendish a look of skepticism. Heading off his inquiry, Cavendish quickly added, "...a friend...his...his name is...Chalk."
"CHALK?!", exclaimed Muffleford. "THE Chalk? The...the. one that....", "yes!", Cavendish interrupted. "Yes. the very same...I was desperate and needed a hit of Calculus and I got this integral". Okay, stop reading this nonsense, watch the video all the way through and comment "CAVENDISH YOU SCOUNDREL!"
🌟Support the channel🌟
Patreon: / michaelpennmath
Merch: teespring.com/...
My amazon shop: www.amazon.com...
🟢 Discord: / discord
🌟my other channels🌟
mathmajor: / @mathmajor
pennpav podcast: / @thepennpavpodcast7878
🌟My Links🌟
Personal Website: www.michael-pen...
Instagram: / melp2718
Twitter: / michaelpennmath
Randolph College Math: www.randolphcol...
Research Gate profile: www.researchga...
Google Scholar profile: scholar.google...
🌟How I make Thumbnails🌟
Canva: partner.canva....
Color Pallet: coolors.co/?re...
🌟Suggest a problem🌟
forms.gle/ea7P...
10:55 The first integral from 0 to 1 is indeed one half as we take the antiderivative x²/2 but the second one is just a constant multiplied by the length of the interval, so it should be 1/2 + n instead of 1/2 + n/2
Yes, and @ 12:23 the summand should be n*(n+1)*(2*n+1) making the summand @ 13:00
n*(n-1)*(2*n-1). Then, multiplying that out and using the formulas for sums of consecutive first, second and third powers of natural numbers, you will see that the final answer is still the same.
Please pin this comment.
Yep, obviously with or without change of variable, the integral evaluates to 1/2 + n. And hence we should end up with the same summand. Not sure how he didn't realise he clearly ended up with a different sum and result!
Yes correct
Get well soon Michael!
By expanding the sum to powers of n^3, n^2 and n, the answer is 99990000/4 = 24997500. The general result summing from 0 to N is N^2(N^2-1)/4, slightly smaller than n^4/4
Once we got to n(n+1)(2n+1)/2 it’s simpler to multiply the terms into polynomial and apply summation formulas to n^3, n^2, n terms
No way he's integrating the whole house
This simplifies a lot if you take the sum from 1 to 100 instead of from 0 to 99. Everything works out nicely in even powers of 10.
0*0.5*1 + 1*1.5*2 + 2*2.5*3 + ... is a polynomial of degree 4 because it's the partial sums of a polynomial of degree 3. You can take the first 4 numbers, (0, 3, 15, 42), compute their differences (3, 12, 27), again (9, 15) and again (6). Now the sum we want is the combinatorial numbers multiplied by the first numbers of those rows, like this: 0*n + 3*n*(n-1)/2 + 9*n*(n-1)*(n-2)/6 + 6*n*(n-1)*(n-2)*(n-3)/24. Substitute n=100 and you get 24,997,500.
That result at the end isn't the same though (I get 12,582,075 when I evaluate it algorithmically). As others have pointed out, the integral from t=0 to t=1 of n dt evaluates to n, not n/2, hence the result being off by about a factor of 2.
Yes, I noticed that as well. It just goes to show that if we know the end, we're bound to believe it.
I've been thinking about his video a lot. I love mixing the discrete with the continuous. This is so very cool and I'm so glad you're presenting this.
Beautiful illustration of the change of the order of summation
What's the point of doing the integral substitution? Evaluating it directly you still get (2n+1)/2 as the integral and you can split it up the same way as n+1/2 and continue as in the second method.
I'm pretty sure that Michael is contractually obligated to do at least one integral substitution per video, no matter if necessary or not. That scoundrel Cavendish is behind it.
The point is that instead of an integral with a range depending on n we just get an integral over the range 0 to 1, which greatly simplifies the rest of the steps.
It is interesting that the solution is exactly 2,500 = 100^2/4 less than the integral of x^3 over the same domain (which is 25,000,000=100^4/4).
If you let the upper bound of the integral be K, then the result nicely becomes: (K^4-K^2)/4, so the pattern above holds that the integral with the floor and ceiling is always K^2/4 less than the integral where the floor and ceiling are ignored (just replaced with x).
10:56 I'm think integral from 0 to 1 of n*dt = n, not n/2
Final result is (N^4-N^2)/4 for N=100
(N^4)/4 is the primitive of N^3 what is going on here?
@@JacobPlat I have (sum from n=0 to N-1 of (2n^3+3n^2+n)/2) = (sum from n=1 to N of (2(n-1)^3+3(n-1)^2+(n-1))/2) = (sum from n=1 to N of (2n^3-3n^2+n)/2), using formulas for sum of first N powers of k, the result for me is (N^4-N^2)/4
@@juanixzx I see clever!
Always good for a hit of calculus.
One can use rising factorials (see Wiki) to do it faster. First, put n and (n+1) out of the integral, then the remaining x yields the area of the trapezoid with bases n and (n+1) and height 1. So we arrive at the sum of f(n)=n*(n+1/2)*(n+1) over n=0,...,99. Note that f(n)=n*(n+1)*(n+2)-(3/2)*n*(n+1). Hence to find F(n) such that F(n+1)-F(n)=f(n) one may set F(n)=(1/4)*(n-1)*n*(n+1)*(n+2)-(1/2)*(n-1)*n*(n+1)=1/4*(n-1)*n^2*(n+1)=(n^2-1)*n^2/4. Now the integral is F(100)-F(0)=9999*2500-0=24,997,500.
Very interesting and great vedios you do!
2:25, did anyone else notice the counting error?
CAVENDISH YOU SCOUNDREL! I was just thinking about if switching summations like what is done here was possible, and that geometric representation is just what I needed to understand it. Very cool!
What I did was split the integral into smaller integrals over the intervals [n-1 n) = I_n and summed them up from n=1 to n=N>1 (N being Natural).
Since in those intervals floor(I_n) = n-1 and ceil(I_n)= n, I can evaluate them and treat them as constants, so I pull them out of the integral, integrate the remaining x over the interval and then just simplifly the resulting polynomial of n until I get the sum from 1 to N of 1/2*(2n^3-3n^2+n) which is
(N(N+1)/2)^2 - N(N+1)(2N+1)/4 + N(N+1)/4.
With N=100 this is equal to 24,997,500
Love to see it!
Ahh I remember I saw a similar integral of x*floor(x). I felt so proud of myself after solving it entirely by hand
Both methods are the same. No matter if you integrate with x or t, you finally get n * (n + 1) * (2n + 1) under sum.
Good explanation
Thanks mc penn
CAVENDISH YOU SCOUNDREL!
Isn't the integral from 0 to 1 of n with respect to t simply n? An antiderivate is nt, which gives n when evaluated at the endpoints.
But its the integral of nx(n+1) wrt x.
@@colinjava8447 no, it isn't. He substituted x = t+n, so there's no longer any x dependence.
Yes, he made a mistake: the integral from 0 to 1 of t + n is 1/2 + n. And then, we end up with the same sum of general term n(n+1)(2n+1)/2. Not sure how he didn't realise he ended up with a clearly different sum/result... 🤷♂
For more integrals and solutions. There's a new book "MIT Integration Bee: Solutions of Qualifying Tests from 2010 to 2023"
£. Integral xdx r*r+1. Limits of each step wise interval is r and r+1
May I ask, what is the purpose, that changing the single summation to a double summation, changing the order of summation, then change back to two single summation again? Since in my point of view, the original single summation can already be evaluated, isn't it?
2n^3 + 3n^2 + n (assuming you leave the fractions outside), then it's 3 known sums (first n integers, squares, cubes) -- so yes, you can. it is probably quite a bit more arithmetic, though, so if you figure out the tricks quickly then saving the time on the arithmetic is the "point" I suppose
I guess he wanted to see how long he could make his method, because he did mention that the first method is the longer one.
CAVENDISH YOU SCOUNDREL
I really need to work on memorizing these sequence sum identities. There's just so many.
"CAVENDISH, YOU SCOUNDREL!"
Doesn't the integral of n*dt on the interval 0 to 1 evaluate out to 'n' NOT 'n/2'??? Or am I crazy??
It should be n. That's why he got 1/2 * n(n+1)^2 in his second sum instead of the proper 1/2 * n(n+1)(2n+1) that he got using the first method. Basically the (n+1)/2 should have been n+1/2 = (2n+1)/2.
Another approach... do an integration by parts, integrating x to x^2/2 and differentiating the floor times ceiling to 2n times the delta function at n, from n =1 to 99. The result is 99*100 * (100)^2/2 - 0 - the sum of 2n times n^2/2 from 0 to 99. The latter sum is the sum of n^3 from 1 to 99, or 99^2 * 100^2/4, so the final answer is 99*100 * (100)^2/2 - 99^2 * 100^2/4 = 24,997,500.
I just split it up into 100 integrals like he did, but without the error integrating \int_0^1 n dt as being n/2 instead of n. (the indefinite integral is nt, not nt/2, as n is a constant with respect to t.) but I also got 24997500 so I think you're right
Hi,
Seems like you have a "chat dans la gorge" 🤧Get well soon Michael!
I saw the problem and gave up finding an exact answer, so I just used Trapezoidal Rule with 100 equal subintervals...
Nice problem.
Had to remind myself how to derive all of the "sum of the first n mth powers" formulae.
After that it was a doddle :)
This channel always has such interesting math problems.
Man, it seems like we're going to integrate an entire house very soon 😄💀💀
I got the first bit right, but bogged down on the sum of the cubic polynomial.
i started solution that integral from integer bounds : n to n+1 = integral ( n*x*(n+1)) = 1/2 *n*(n+1) * ( (n+1)² - n²) = 1/2 *n(n+1)(2*n+1) , and then you make the summ of this expression from n = 0 to n = 99, what is for sure possible and are just technical steps, is this solution correct?
much easier is to see telescoping sums in sum(n * (n+1) * ((n+1)² - n²))
reindex the first sum and you will have sum(n⁴-n³) - sum(n⁴+n³)all n⁴ cancel in the process except 100⁴, then you are left with 2 sums of cubes
what ended up happening to the intern guy?
I don't get it; after just 5 minutes you've obtained a sum of " polynomial (n)." It's even only a third degree, and you already PUT formulas for sigma n^k up to k = 3 on the board in the first minute. Why not simply plug in the formulas at this point? Why does this video continue for nine more minutes??
An interesting addition to this problem could be replacing the 'x' with {x} where {.} is the fractional part function defined as {x}= x - floor(x).
Edit: grammar
Reading these comments I see no one can agree on the answer...
u sound sick; hope ur ok !
Dude, rest that voice. It really sounds like you're struggling. Allergies? Cold? Either way, take care of yourself.
I think you made that harder than it was. Just expand the cubic and sum each piece individually. You ended up doing that at the end with your fancy double sum.
what happened to your voice?
He had a cold I guess
Definitely sounds like it.
CAVENDISH YOU SCOUNDREL!
CAVENDISH YOU SCOUNDREL!