You can also look at the greatest common divisor (4 in this case) and plug in values in each interval of size 1/gcd
@wesleydeng71 Жыл бұрын
Since ⌊x⌋ is an integer, it can be moved outside of other floor functions and also is ⌊2x⌋ which means, y = ⌊4x⌋ - ⌊2x⌋ - ⌊x⌋ which is easier to analyze.
@DrBarker Жыл бұрын
This is an interesting point - it would be fun to further explore/prove how we can manipulate expressions with the floor function, e.g. when does ⌊ab⌋ = ⌊⌊a⌋⌊b⌋⌋, or ⌊a + b⌋ = ⌊⌊a⌋ + ⌊b⌋⌋.
@JuniorTheCuber Жыл бұрын
I don't believe that holds for -ve x, say u have x = -1.5. Then floor(2x+floor(x)) = floor(-3-2)=-5 But floor(2x)+floor(x) = -4. I believe in general u can only split, for some integer n, floor(n+x) into n+floor(x) only for positive values of x
@charlottedarroch Жыл бұрын
@@JuniorTheCuber If x = -1.5, then floor(2x) = -3 and floor(x) = -2. So floor(2x)+floor(x) = -5. If you work with the piecewise definition of the floor function, you'll see that floor(x+n) = floor(x)+n for all integers n.
@JuniorTheCuber Жыл бұрын
@@charlottedarroch floor(1-0.5)
@Cosmonaut_Sloth Жыл бұрын
@@JuniorTheCuber … gives 0, since 1+floor(-0.5) = 1-1 = 0 = floor(0.5) = floor(1-0.5). If your argument here is that n is -0.5, note that -0.5 is a rational, but not an integer (as Daniel pointed out that it must be), so this doesn’t contradict anything.
@composerlmythomorphic2635 Жыл бұрын
Suppose x=a+r where a is the integral part. y=[4a+4r+[2a+2r+a]]=[4a+4r-[3a+2r]] For rr>1/2, y=[a+4r-1]. Then consider the case r
@5ucur Жыл бұрын
Listening to this Desmos makes an interesting tune: rising, falling, rising again but longer, and so on! Especially when zoomed in a bit or the playback is slowed down a bit.
@txikitofandango Жыл бұрын
Very interesting and beautiful problem here. The reverse problem would also have been interesting, if you are given a graph with some strange but repeating pattern of steps, can you construct a rule for it using the floor function? Also, it would look cool as a polar graph, you could scale theta by a multiple of pi, or not
@DrBarker Жыл бұрын
The reverse problem sounds like it should be possible to solve, as long as our graph is made up of straight lines. It might be easiest to construct such a function as a sum of indicator functions, then work out a general way of converting from indicator functions to floor functions.
@hylianpikachu37 Жыл бұрын
I found it easier to simplify by noticing that ⌊4x - ⌊2x + ⌊x⌋⌋⌋ = ⌊4x - (⌊2x⌋ + ⌊x⌋)⌋ = ⌊4x⌋ - ⌊2x⌋ - ⌊x⌋ since the integer values can be "pulled out" of the floor functions. From there, it is easier to analyze this function on the four intervals [n, n + 1/4), [n + 1/4, n + 1/2), [n + 1/2, n + 3/4), and [n + 3/4, n + 1) to get the behaviour.
@MrRyanroberson1 Жыл бұрын
4:36 i have somewhat of a simpler method. notice the floor function is blind to integer steps: floor(x+1) = floor(x)+1 always. notice also the floor function always returns integers... so in floor(4x-floor(2x+floor(x))) floor(2x+floor(x)) = floor(2x) + floor(x) and floor(4x - floor(2x) - floor(x)) = floor(4x) - floor(2x) - floor(x) the overall slope should be approximately 1 since 4x-2x-x = x, and because the common period is also 1 that means you only need to look at the range [0,1) to know how this all behaves
@lexinwonderland5741 Жыл бұрын
Would love to see somebody make a graphing calculator (desmos/geogebra) app for this problem with various parameters!!
@kummer45 Жыл бұрын
Outstanding video. I need a lot of training with floor and ceiling functions.
@MathOrient Жыл бұрын
A very distinctive problem 👍
@jamiewalker329 Жыл бұрын
I would have just written x = floor(x) + x* where x* is the fractional part. You then get y = floor(4floor(x) + 4x* - floor(3floor(x) + 2x*)) Then this will give one case where 0
@osneket5713 Жыл бұрын
Since I couldn't be bothered to get a piece of paper I tried to solve it entirely in my head. Quite a fun challange actually. I belive the graph can be simplified to: y = [x - 0.25] (The square brackets are floor brackets). My math might be completely off though. I haven't seen the solution as of writing this.
@osneket5713 Жыл бұрын
Turns out my math was wrong. But at least my answer matches the correct one for 75% of all x values
@mrigayu Жыл бұрын
What a fun task! I wonder what patterns analogous to the one in your video would look like with logarithms, sinusoids, or even polynomial functions of higher order.
@AndrejPanjkov Жыл бұрын
square wave: y = 1- [x - [x/2 + [x/2]]]
@rob876 Жыл бұрын
I really enjoy your videos. I never miss one.
@DrBarker Жыл бұрын
I'm glad you're enjoying them!
@alexandermorozov2248 Жыл бұрын
Great, interesting! Good work! 👍
@NoNameAtAll2 Жыл бұрын
axis scaling slightly confused me, as I expected roughly y=1*x growth it is indeed so, just not as straightforward as 45 degrees :)
@kianushmaleki Жыл бұрын
Lovely
@joefarrow1599 Жыл бұрын
Is there an algorithmic way to take functions of this form (kind of roughly 'floor linear') and return your plotting algorithm of, "go to the right 1/4 go up one, go to the right 1/2 go down one etc."?
@maxvangulik1988 Жыл бұрын
Is there a graph of the limit case for this? (That is to say, would the sum of floor((-2^n)x) converge to a continuous function of x?)
@maxvangulik1988 Жыл бұрын
The next partial sum would be floor(8x+floor(4x-floor(2x+floor(x)))), then floor(16x-floor(8x+floor(4x-floor(2x+floor(x))))), etc
@maxvangulik1988 Жыл бұрын
I have a gut feeling that approaches a line
@DrBarker Жыл бұрын
I'd need to think quite carefully about this. Given that the graph jumps up and down, I wouldn't be surprised if we end up with quite a complicated fractal graph.
@TaladrisKpop Жыл бұрын
The problem becomes much simpler when remarking that f(x+1)=f(x)+1
@ilafya Жыл бұрын
It will be very easyer if you use the periodicity of the fonction wwuch is equal to one
@ngc-fo5te Жыл бұрын
Yes.
@General12th Жыл бұрын
Hi Dr. Barker! Is there anything the _ceiling_ function could add here? Or is taking the ceiling function usually the same as taking the floor function, but everything is shifted vertically?
@DrBarker Жыл бұрын
They're very similar, the only big difference is that the discontinuities are the opposite way around. So the ceiling function could be useful if we want our function to be left-continuous, rather than right-continuous.
@kappascopezz5122 Жыл бұрын
An easy identity to relate them is ceil(x) = -floor(-x), so they're essentially both doing the same thing. Though it's easy to make the mistake of thinking that ceil(x) = floor(x) + 1, even though as Dr Barker pointed out, doing that would shift each integer to be mapped to one greater than itself, rather than keeping all the integers the same.
@nathanisbored Жыл бұрын
i like the haircut
@padraiggluck2980 Жыл бұрын
I have Desmos 🙂
@hypebeastuchiha9229 Жыл бұрын
2:18 THE ONE PIECE IS REAL
@tomctutor Жыл бұрын
I notice a couple of things: There is periodicity associated with these mixed linear-floor combinations (n:1 hence no inverse function); a clearer example is simply x-⇂x⇃which is a sawtooth waveform sitting along the x-axis of magnitude unity... ... ╱╱╱╱ ... Then ax - ⇂x⇃ = (a-1)x+( x-⇂x⇃) which is simply a sloped y=mx line plus the sawtooth earlier. Finally cx - ⇂bx⇃ ≡ bax - ⇂bx⇃ = b(ax - ⇂x⇃) which is just more vertical scaling. I do admit, once you start nesting mixed functions like these, it can get very awkward to visualize. For DrB's example just copy the command, plot floor(4x-floor(2x+floor(x))) into WolframAlpha. 😗
@BraileySims-qr1yx Жыл бұрын
Denote 'floor', ⌊x⌋, by Fx ≡ F(x), and let {x} := x -⌊x⌋ denote the fractional part of x. So, x = Fx + {x} u := F(2x + Fx) = F(2(Fx + {x}) + Fx) = F(3Fx + 2{x}) = 3Fx, if {x} < 1/2, = 3Fx + 1 otherwise So, v := 4x - u = 4(Fx + {x}) - 3Fx = Fx + 4{x}, if {x} < 1/2, = Fx - 1 + 4{x}, otherwise. And, finally, y = F(4x - F(2x + Fx)) = F(v) = Fx, if {x} < 1/4, = Fx + 1, if 1/4 =< {x} < 1/2, = Fx + 1, if 1/2 =< {x} < 3/4, = Fx + 2, if 3/4 =< {x} < 1.