In the case of "floor of sin(1/x)", a convenient way to split up the interval (0,1) should be the following: (1,1/π),(1/π,1/2π),(1/2π,1/3π), etc. So, the interval (1,1/π) and those of the forns of the form (1/2nπ,1/(2n+1)π), n ≥ 1, would have value equal to zero for the floor of sin(1/x) while the intervals of the form (1/(2n-1)π,1/2nπ), n ≥ 1, would have value equal to (-1) for the floor of sin(x) and value (-1/π)*[(1/(2n-1))-(1/2n)], n ≥ 1, for the integral of the interval. So, the result of the integrals for the intervals with non-zero value would be: (-1/π)*(1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... ). In short, this is (-1/π)*(Ln(2)). @drpeyam, please let me know whether my reasoning is correct.
at around 01:40 I went a different direction and replaced sin(n) with the imaginary part of exp(i*n). After some crazy complex journey I arrived at the same result, but I got the cosine version of this problem "for free" at the end, by just taking the real part
Nooo. The thumbnail is wrong. Solved it myself, proving the integral of the floor of sin (1/x) is 0. (It's non zero in a numerable subset of the interval, so the lebesgue integral is 0). But opened the video ans the integral is different :(
@csilval18 but can't you just substitute 1/x = u then break up the u domain such that the floor of sin(u) either become 0 or -1 then evaluate the resulting series by some prefered method (series is a result of the 1/u^2 that you get)
@@csilval18 If I understand correctly, the thumbnail originally said int_0^1 floor(sin(1/x)) dx. This is Riemann integrable since it has only countably many discontinuities.
That's not the floor symbol, it's an undecided mixture of floor and ceiling!
In some textbooks they use this notation.
In the case of "floor of sin(1/x)", a convenient way to split up the interval (0,1) should be the following: (1,1/π),(1/π,1/2π),(1/2π,1/3π), etc. So, the interval (1,1/π) and those of the forns of the form (1/2nπ,1/(2n+1)π), n ≥ 1, would have value equal to zero for the floor of sin(1/x) while the intervals of the form (1/(2n-1)π,1/2nπ), n ≥ 1, would have value equal to (-1) for the floor of sin(x) and value (-1/π)*[(1/(2n-1))-(1/2n)], n ≥ 1, for the integral of the interval. So, the result of the integrals for the intervals with non-zero value would be: (-1/π)*(1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... ). In short, this is (-1/π)*(Ln(2)). @drpeyam, please let me know whether my reasoning is correct.
at around 01:40 I went a different direction and replaced sin(n) with the imaginary part of exp(i*n). After some crazy complex journey I arrived at the same result, but I got the cosine version of this problem "for free" at the end, by just taking the real part
I love floor sins.
floor of sin or sin of floor?
Oops hahaha 😂
Nooo. The thumbnail is wrong.
Solved it myself, proving the integral of the floor of sin (1/x) is 0. (It's non zero in a numerable subset of the interval, so the lebesgue integral is 0). But opened the video ans the integral is different :(
Sorry :/
Interesting problem ,well presented !
Is the answer to the integral in the thumbnail (-ln 2)/pi ?
No, it's 0. It's not Reimman integrable pretty sure. But Lebesgue integration gives you 0
@csilval18 but can't you just substitute 1/x = u then break up the u domain such that the floor of sin(u) either become 0 or -1 then evaluate the resulting series by some prefered method (series is a result of the 1/u^2 that you get)
@@csilval18 If I understand correctly, the thumbnail originally said int_0^1 floor(sin(1/x)) dx. This is Riemann integrable since it has only countably many discontinuities.
Which video does he calculate the cos sum?
The one in the description is just for the sin
@@Happy_Abe u can see blackpenredpen
Where’s the floor? DJ!
omg, yay!
The summations are making it sigma
Second
wtf