I have never heard of the floor function before in my math classes. It's something new I learned in this area. You explained very clearly to me, with great examples. Keep on going.
Floor function is very important in data science and quant development (algo-trading). Veeeeeeery important!! Thanks again bro. From the various MathTubers that I follow, you've become my personal favorite during these last few months. You're in the level of the best of the best!
Thank you sir! I have tackled definite integrals using geometry (not calculus) before. Never have I tackled an indefinite integral using those techniques. I wondered if we were going to use the piecewise definition of the floor function to solve.
Nice job. For me, it is easier to see that we have the areas 1+2+3+4+... minus the half the upper square (triangle) with area 1/2, if we compare with f(x)=x. Then we have to sum them up.
So... Problem. Suppose I want to integrate floor(x) dx from 1 to 1.5. Well, that's super easy, as it's equal to 1 everywhere in that range. So, it's 0.5. Now, let's use your magic formula. It's (x^2 - x)/2 evaluated at 1.5, then same evaluated at 1, and we take the difference. Well, plugging in x = 1, we get 0. Plugging in x = 1.5, we get (2.25 - 1.5)/2 = 0.75/2 = 0.375. So, we get the integral evaluating to 0.375. So... 0.375 = 0.5. Either your magic formula is busted, or math is broken. The secret is, your formula is busted whenever one of the end points isn't an integer. To fix it, try the following: The integral of floor(x) dx = floor(x) ( floor(x) - 1 ) / 2 + floor(x)(x - floor(x)) + c.
So the integral of the ceiling function would be the integral of the floor function plus the number of integers because the ceiling function is the floor function plus 1 except where x = an exact integer.
Almost, but the adaption you have to make is pretty obvious. You calculate up to the the floor of the upper bound (=n) with the gaussian formular and then add width*n, where the width is the difference between the upper bound and the floor. You can generalize this also with a real lower bound which is just as straight forward. Another note: this is actually a way less weird integral than it seems as in fact the integral is constructed over "simple functions" and the floor function is an example of such a function.
@@PrimeNewtonswhat about when you integrate from 0.5 to 1. The area should be 0, but the formula gives (1²-1)/2 - (0.5²-0.5)/2 = 0.125. Or did I do something wrong there?
How do you prove that the integral of floor(x) is (x^2-x)/2+C? For some reason, when I looked up through Wolfram Alpha and Symbolab, it doesn't have in terms of standard mathematical functions. I don't see where it's coming from.
this only works for integer x, I'm pretty sure. Easy way to see that is that within each step of the staircase, the area should increase linearly, not with x²
I'm retired from Slovakia and I love the mathematical examples posted on youtube. I find this valuable, I have never solved such an integrated one.
I have never heard of the floor function before in my math classes. It's something new I learned in this area.
You explained very clearly to me, with great examples. Keep on going.
Glad it was helpful!
I've seen it back in school, but our teacher never explained what it was, since it wouldn't be on exams
Your voice is great and your teaching is so great ❤❤❤
Floor function is very important in data science and quant development (algo-trading). Veeeeeeery important!!
Thanks again bro. From the various MathTubers that I follow, you've become my personal favorite during these last few months.
You're in the level of the best of the best!
You are great sir 🎉
When I watch your videos.. I get so interested in MATHS..
Thank you for your valuable contributions..
It's the ceiling. Basically.
Very good explanation mate!
Nice job...
Thanks alot... learn alot...❤❤❤.
And if it's possible,make a video for ceil function too...the integral of ceil function...
Will upload soon
@@PrimeNewtons thanks...❤️❤️❤️.
Thank you sir! I have tackled definite integrals using geometry (not calculus) before. Never have I tackled an indefinite integral using those techniques. I wondered if we were going to use the piecewise definition of the floor function to solve.
Very nice one!
Very good. Thanks Sir
Great explanation!
Thanks for this video
This one was great, the rational is flawless. Ive always call thai function the ladder function
Nice job. For me, it is easier to see that we have the areas 1+2+3+4+... minus the half the upper square (triangle) with area 1/2, if we compare with f(x)=x. Then we have to sum them up.
POV:it’s 2am, and you like math
So... Problem. Suppose I want to integrate floor(x) dx from 1 to 1.5. Well, that's super easy, as it's equal to 1 everywhere in that range. So, it's 0.5.
Now, let's use your magic formula. It's (x^2 - x)/2 evaluated at 1.5, then same evaluated at 1, and we take the difference. Well, plugging in x = 1, we get 0. Plugging in x = 1.5, we get (2.25 - 1.5)/2 = 0.75/2 = 0.375. So, we get the integral evaluating to 0.375.
So... 0.375 = 0.5. Either your magic formula is busted, or math is broken.
The secret is, your formula is busted whenever one of the end points isn't an integer. To fix it, try the following:
The integral of floor(x) dx = floor(x) ( floor(x) - 1 ) / 2 + floor(x)(x - floor(x)) + c.
Loved that. Ceiling and floor were not in my math vocabulary.
It is just y=x triangle (x*x/2) with n small triangles (0.5) substracted.
So S(floor(x)) = x^2/2 - x/2
So the integral of the ceiling function would be the integral of the floor function plus the number of integers because the ceiling function is the floor function plus 1 except where x = an exact integer.
So (x(x + 1))/2 would actually be the ceiling, right?
Sir How we get the area under a discontinuos function ?
Well done sir 🎉🎉🎉🎉. ( What about the ceiling function)
probably just x instead of x - 1 because you're not deleting the first rectangle
Does this still work if the bounds of the integral are not integers?
Absolutely, just understand that the beginning and end bases won't be equal to 1
Almost, but the adaption you have to make is pretty obvious. You calculate up to the the floor of the upper bound (=n) with the gaussian formular and then add width*n, where the width is the difference between the upper bound and the floor. You can generalize this also with a real lower bound which is just as straight forward.
Another note: this is actually a way less weird integral than it seems as in fact the integral is constructed over "simple functions" and the floor function is an example of such a function.
Thanks
The next video should be about the ceiling function
got a general formula int(a,b)floor(x)=na+{n(n-1)/2} where a and b are positive integers and b=a+n
Strictly speaking the integral does not exist because the function is not continuous, no?
it'll be more simple: sum of all integers from the lower limit( here 1) to just less than the upper limit(here 5) so it's sum of 1 to 4 & it's 10
Have done a lot of questions like this while preparing for jee
Where was floor function
GIF is what I saw
@@raja2850 Bhai ji usko hi floor function kehte hai 😃
@@SakshamMahajanB Aaj pehli baar floor suna 😁
In the very last step, is it the integral of x dx or integral of the floor of x dx?
Like a ladder function, sir
Does this formula works only when integral interval is integers?
I think it works for all real x. Remember, you can cat the Rectangles to tiny straps not just 1
@@PrimeNewtons will arifmetic sequence work after that strap? I don't think so
I may need to think about it.
@@PrimeNewtonswhat about when you integrate from 0.5 to 1. The area should be 0, but the formula gives (1²-1)/2 - (0.5²-0.5)/2 = 0.125.
Or did I do something wrong there?
The floor function should have been called 'staircase' function, because that it how it behaves! 😀
How do you prove that the integral of floor(x) is (x^2-x)/2+C? For some reason, when I looked up through Wolfram Alpha and Symbolab, it doesn't have in terms of standard mathematical functions. I don't see where it's coming from.
this only works for integer x, I'm pretty sure. Easy way to see that is that within each step of the staircase, the area should increase linearly, not with x²
The way I remember the sum is: The average of the number and its square.
Sir give us viedos on parabola, ellipse and hyperbola and graph of functions.
I already have those
@@PrimeNewtons Please pin the link.
😭 "promosm"