Solving a Quadratic Floor Equation
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- Опубліковано 5 вер 2024
- This video is about a floor equation
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Conjecture:
floor(x**a) = floor(x)**a
for values of x,a where
0
Interesting ceiling/floor problems. Btw sqrt(2) at the end can be avoided, we have (n-2)^2
Nice!
@@SyberMath also I wanted to tell you that 2-sqrt(2)~0.5858 not 0.6
Very good and satisfying solution as well as the problem 😁
Thank you!
@@SyberMath 3 IS ALSO A SOLUTION!!!!
Great video! I had another method where I said floor(x) =m therefore floor(x^2) = m^2 +n for non negative integers m,n. It follows n =0 or 1 and we can solve for x in each case. It was great to see another way of thinking from you! :)
I don't think it follows that n = 0, 1. What does follow is that 0 =< n =< 2·m
@@angelmendez-rivera351 if you plug it back into the original equation you get m^2 + n - 4m +3 = 0.
This gives (m - 2)^2 + n-1=0.
Or (m-2)^2=1-n. Because LHS is a perfect square, RHS>=0 giving n=0 or 1 as only integer solutions.
@@renatsamur9099 Yes, but that follows from floor(x) = m AND the given equation, not floor(x) = m alone. This needs to be clarified, because otherwise, your comment gives the impression that n = 0, 1 is a property of floor functions, especially to those who are not the most familiar with the subject.
@@angelmendez-rivera351 I was giving a brief explanation of how I solved the question. If anyone needed clarification they could have asked. I’ll be sure to mind that next time though. :)
Thanks for sharing!
Amazing, just loved it, ❤️☺️☺️☺️☺️,thank you very much
My pleasure 😊
@@SyberMath Wait, if the floor function always gives an integer right..the nearest integer less than the x value..Why can't you plug in 3 and 1 since those are answers are integers..then x can be 3. some decimal or 1. some decimal then the floor of x squared could still be 3 squared or 1 squared as long as the decimal part of x isnt so high that squared gives you the next integer up like 10 or 2..
@@leif1075 x=1 and x=3 definitely work but for greater numbers there is an upper bound. For example if x is greater than sqrt(2), then it will not work!
The best way to solve this is on the floor not on the roof!!
😁
We can also use floor (x^2)=x^2-{x^2}; floor(x)=x-{x}. We have then (x^2-4x+3)+(4{x}-{x^2})=0. But (4{x}-{x^2} )is defined on interval
Thank u nice problem !!
You're welcome!
Thank you for your beautiful Floor Eqn
Np. Thank you for watching
Marvelous solution, thank you very much sir.
You are most welcome
When you have found that x is between n and n + 1 and x^2 is between n^2 and (n+1)^2, you just have to use substitutation : floor value of x is u so the floor value of x^2 is u^2. Now, you just have to use discriminante : delta = - 4^2 - 4*3 = 4 so u1 = 1 and u2 = 3. Of course, x needs to be (or not to be 😂) between 1 and 2 or between 3 and 4. No?
Thank you! Please shoot a video about gcd properties
I will take a look at it! Sounds like a good idea
@@SyberMath Thank you!
@@SyberMath Yes, please..
Excelente video, me salió a la primera 🥹, un saludo desde Perú.
Thank you for showing that. If I want to explain it to a 9th grade good student, I will draw a nice y=x^2+3 graph, and y=4x graph. Show that they intersect at 1 and 3. Make lines parellal to x axis , length 1, from (0,0) , (1,4) , (2,8) , (3,12) , (4,16) , (5,20). Lets start with the line from (1,4). The parabulla changes to a line from x=1 to x=2^0.5 . So these lines are "together". Next two lines that are "together" are from x=5^0.5 to 6^0.5 ... you got the idea. At x=15^0.5 and on, they do not overlap any more.
Nice!
i am upvoting every time i post a computer solution, like now:
*𝚁𝚎𝚍𝚞𝚌𝚎[𝙵𝚕𝚘𝚘𝚛[𝚡^𝟸] - 𝟺 𝙵𝚕𝚘𝚘𝚛[𝚡] + 𝟹 == 0, 𝚡, 𝚁𝚎𝚊𝚕𝚜]*
this way i know when my post was deleted when i revisit the video thanks! :P
btw please doht feel annoyed when i post computer solution. i actually post the working input only. it is interesting to see how we can formulate the problem in code. i doht post the output.
hope you doht mind!
i wish the question was (floorx)^2-4floorx+3 so we would factorize it like (floorx-3)(floorx-1)=0 that would have made the answer x (0,1),(3,4) but this intersting aswell and harder also
we know frac(x) can be up to 1, so x^2 can be up to (round(x)+1)^2, and at least round(x)^2
round(x)^2-4round(x)+3
Nice video!!!
Thanks for the visit!
Wow this one was tougher than expected actually:
\x^2/-4\x/+3=0
A: x=n+f with n integer and f in R, 0>=f
Thank you. Very good video.
Please can you look at this one
|X²|- |X|²=100,
No solutions because |X²| = |X|²
waiting.and eager!!!
Liked it?
Good video
Thanks
Yeap! I solved the problem 💖☺💖
Thanks a Zettabyte, Boss!! 🌿🌿🌿
Np. Nice work!
Square is always positive so square can also be taken outside mod just solve the quadratic u will get 4 solutions ±3,±1
👍
(x-3)(x-1)
good!
Thanks!
x=3
Best chnl,,,
Thanks, Mahesh!
8:49 beacaus this a continous function If its not then we have to proof why its always positive
If im wrong pleaz tel me beacaus im not sure !!!!!!!!
It's monotonically non-decreasing, and it's only discontinuous at countably many points.
@@angelmendez-rivera351 thank u!!!
@@tonyhaddad1394 Well, actually, there is a better, simpler explanation as to why it's always positive: n^2 - 2·n + 4 = n^2 - 2·n + 1 + 3 = (n - 1)^2 + 3 > 0, and since (n - 1)^2 >= 0 is trivially true, (n - 1)^2 + 3 >= 3 is always true, for any n. But 3 > 0 is trivial, so there is no information to be gained from this.
@@angelmendez-rivera351 yes bro thank u that a better proof 👍👍👍
Great discussion! Thanks to you both!
Today was the turn of geometry. Isn't it? 😟😟😟 Well. Today's equation was also interesting. Thankyou BTW.🙃🙃🙃
Ashish, I announced in yesterday's video that I will be doing fewer geometry puzzles and more algebra, number theory, and trigonometry problems. I will focus more on the quality of the puzzles!
@@SyberMath ohh I see. I was in a bit hurry yesterday, so I have left the vedio just after the solution so I didn't notice it. Fine! 😁😁😁
Thankyou for your response. 😄😄😄
Fuvest 2024
(x-3)(x-1)
👍