Well done to all math wizards here - I wouldn't even know where to begin on most of these haha. I'm still stumped by the integral with the floor of the floor of x - would love to see an explanation to the solution of that one
Let I be our integral. We can write every real number x as k + t for some integer k and some real 0 < t < 1. Then, we may change our integration into a summation w.r.t. k of integrals w.r.t. t. This gives us the following expression: Sum_(k=1)^9 (Int_0^1 (Floor(k(k+t) dt) We start from k = 1 because for any x such that k = 0, the corresponding integral is zero. Therefore: I = Sum_(k=1)^9 (Int_0^1 (Floor(k^2 + kt) dt) Since k^2 is an integer, we can separate Floor(k^2 + kt) into k^2 + Floor(kt). Thus: I = Sum_(k=1)^9 (Int_0^1 (k^2 + Floor(kt)) dt) By linearity of integration, we have: = Sum_(k=1)^9 (k^2 Int_0^1 (1 dt) + Int_0^1 (Floor(kt) dt)) = Sum_(k=1)^9 (k^2 + Int_0^1 (Floor(kt) dt)) Now, Floor(kt) depends on t, which ranges from 0 to 1. Note that the interval [0, 1] may be decomposed into a union of k disjoint intervals: (0,1) = [0, 1/k) ∪ [1/k, 2/k) ∪ [2/k, 3/k) ∪ ... ∪ [(k-1)/k, 1) We can ignore [0, 1/k) because in that interval, the floor is zero, and therefore, so does the integrand. Thus, we may substitute Floor(kt) = m for some integer 0 < m < k and write the remaining integral as a summation from m = 1 to k-1 of Floor(kt) times the width of each subinterval in the decomposition above, which is 1/k: = Sum_(k=1)^9 (k^2 + Sum_(m=1)^(k-1) (m/k)) = Sum_(k=1)^9 (k^2 + 1/k Sum_(m=1)^(k-1) (m)) By the linearity of summation, we have: = Sum_(k=1)^9 (k^2) + Sum_(k=1)^9 (1/k Sum_(m=1)^(k-1) (m)) Note that the sum of the first n integers is n(n+1)/2, and the sum of the first n integer squares is n(n+1)(2n+1)/6. You may prove this via induction, via telescoping series, or whatever method you come up with. Using these two theorems, we should get: = 9(9 + 1)(18 + 1)/6 + Sum_(k=1)^9 (1/k k(k - 1)/2) = 15*19 + Sum_(k=1)^9 ((k - 1)/2) = 15*19 + 1/2 Sum_(k=1)^9 (k - 1) (distributive property) Now, note that summing the first 9 (k-1)'s (starting from k=1) is the same as summing the first 8 k's. Thus: = 15*19 + 1/2 Sum_(k=1)^8 (k) = 15*19 + 1/2 8(8+1)/2 = 15*19 + 2*9 = 15*19 + 18 = 15*19 + 15 + 3 = 15*(19 + 1) + 3 = 15*20 + 3 = 300 + 3 = 303 Hope this helps.
22:23 - You can see the critical mistake on the board: substitution to 1-x exchanges integration limits, so we shouldn't add minus sign at the beginning of the integral
Also appreciate the man with the purple bucket hat writing his method in the first round! I'm not a math genius, but I do like watching these, and I could actually understand what was happening :)
1:11:03 board 1 is incorrect , the denominator of last 2 terms , they should be whole square and whole cube respectively , but the he wrote 2 cube and 2 square which is obviously different
Well done to all math wizards here - I wouldn't even know where to begin on most of these haha. I'm still stumped by the integral with the floor of the floor of x - would love to see an explanation to the solution of that one
Let I be our integral.
We can write every real number x as k + t for some integer k and some real 0 < t < 1. Then, we may change our integration into a summation w.r.t. k of integrals w.r.t. t. This gives us the following expression:
Sum_(k=1)^9 (Int_0^1 (Floor(k(k+t) dt)
We start from k = 1 because for any x such that k = 0, the corresponding integral is zero.
Therefore:
I = Sum_(k=1)^9 (Int_0^1 (Floor(k^2 + kt) dt)
Since k^2 is an integer, we can separate Floor(k^2 + kt) into k^2 + Floor(kt). Thus:
I = Sum_(k=1)^9 (Int_0^1 (k^2 + Floor(kt)) dt)
By linearity of integration, we have:
= Sum_(k=1)^9 (k^2 Int_0^1 (1 dt) + Int_0^1 (Floor(kt) dt))
= Sum_(k=1)^9 (k^2 + Int_0^1 (Floor(kt) dt))
Now, Floor(kt) depends on t, which ranges from 0 to 1. Note that the interval [0, 1]
may be decomposed into a union of k disjoint intervals:
(0,1) = [0, 1/k) ∪ [1/k, 2/k) ∪ [2/k, 3/k) ∪ ... ∪ [(k-1)/k, 1)
We can ignore [0, 1/k) because in that interval, the floor is zero, and therefore, so does the integrand.
Thus, we may substitute Floor(kt) = m for some integer 0 < m < k and write the remaining integral as a summation from m = 1 to k-1 of Floor(kt) times the width of each subinterval in the decomposition above, which is 1/k:
= Sum_(k=1)^9 (k^2 + Sum_(m=1)^(k-1) (m/k))
= Sum_(k=1)^9 (k^2 + 1/k Sum_(m=1)^(k-1) (m))
By the linearity of summation, we have:
= Sum_(k=1)^9 (k^2) + Sum_(k=1)^9 (1/k Sum_(m=1)^(k-1) (m))
Note that the sum of the first n integers is n(n+1)/2, and the sum of the first n integer squares is n(n+1)(2n+1)/6. You may prove this via induction, via telescoping series, or whatever method you come up with. Using these two theorems, we should get:
= 9(9 + 1)(18 + 1)/6 + Sum_(k=1)^9 (1/k k(k - 1)/2)
= 15*19 + Sum_(k=1)^9 ((k - 1)/2)
= 15*19 + 1/2 Sum_(k=1)^9 (k - 1) (distributive property)
Now, note that summing the first 9 (k-1)'s (starting from k=1) is the same as summing the first 8 k's. Thus:
= 15*19 + 1/2 Sum_(k=1)^8 (k)
= 15*19 + 1/2 8(8+1)/2
= 15*19 + 2*9
= 15*19 + 18
= 15*19 + 15 + 3
= 15*(19 + 1) + 3
= 15*20 + 3
= 300 + 3
= 303
Hope this helps.
I'm surprised Problem 16 stumped so many people. A simple integration by parts would've led to the correct answer.
22:23 - You can see the critical mistake on the board: substitution to 1-x exchanges integration limits, so we shouldn't add minus sign at the beginning of the integral
i got (1/2025) - (1/1013) + (1/2027) it is right?
Feels good to be Able to solve some of them, especially with the time limit! Hope that one day I might be able to do it alongside of them :)
Second problem can also be solve by taking x to the power. 4047 from denominator
33:38 this guy's smile made my day. So happy to see him :D
Next time would it be possible to keep the integral on the screen while showing the contestants solving them?
Appreciate the subtitles :D
Also appreciate the man with the purple bucket hat writing his method in the first round!
I'm not a math genius, but I do like watching these, and I could actually understand what was happening :)
Sorry perple bucket hat man, I don't know your name 🥲
Chirag falor jee advanced air1 2020❤
dang that's luke from mathcounts 2018 he was famous in that vid
1:11:03 board 1 is incorrect , the denominator of last 2 terms , they should be whole square and whole cube respectively , but the he wrote 2 cube and 2 square which is obviously different
I don't understand. Answer is same as correct...
Wait how did they graph this function? Can someone explain? 13:05
ahh so as its in log to the base 43 so from 1 till 43 the function gonna give a value of 0, from 43 to 43 squared its gonna be 2 and so on.
Asian guys are smarter and hard working
This is fun
Most are chinese and indians
And the indian scored the highest❤
he's chirag falor, jee advanced AIR 1