## Callisto 3 years ago In the domain [0, 100], how many solutions are there for the equation $x^2- \lfloor x \rfloor x = 81.25$ How to start?

1. anonymous

Instead of thinking of it as one function defined on [1,100], maybe think of it as 100 different function defined on the intervals [n,n+1] as n goes from 0 to 99.

2. anonymous

That way you can minimize the frustration the floor function brings to the table.

3. anonymous

er, i guess the interval should really be [n,n+1)

4. anonymous

Then on each interval, the question is really:$x^2+nx-81.25=0$where x ranges from [n,n+1). Not sure if this leads to a solution, but it might be a better way of thinking of the problem.

5. Callisto

Domain = [0, 100] :S

6. anonymous

wait, the function$x^2+\lfloor x \rfloor x$is strictly increasing, isnt it?

7. anonymous

then it can have at most one solution.

8. Callisto

That means it has 0 :'(

9. anonymous

strictly increasing on the interval [0,100] anyways.

10. anonymous

well, im not 100% sure on the strictly increasing part, im a little fuzzy on what happens when you jump from one unit interval to the other.

11. anonymous

no, its strictly increasing for sure.

12. Callisto

Sorry!!! Wait... a mistake in the question posted here!!

13. anonymous

ah ok, that makes more sense.

14. Callisto

I'm sorry!!!! :(

15. anonymous

if you look at the function:$x^2-\lfloor x \rfloor x$restricted to an interval {n,n+1) (open on the right), note that the values start at 0 when x = n, and as x approaches n+1, the values also approach n+1:|dw:1357103977725:dw|

16. anonymous

So the equation $x^2-\lfloor x \rfloor x=81.25$only has a solution when $x\le 81$the interval {81,82) is going to be the first time the parabola piece gets as large as 81.25.

17. anonymous

less than >.> i meant greater than lol.

18. anonymous

$x\ge 81.25$

19. anonymous

>.<$x\ge 81$

20. anonymous

So there are 19 solutions, one in each of the intervals [81, 82), [82,83),....., [99,100)

21. Callisto

22. anonymous

The idea is that the floor function on large intervals in a pain, but on unit intervals, its just a the constant function. an interesting problem indeed :)

23. Callisto

I'll remember it! Thanks again!!

24. shubhamsrg

x-[x] = {x} right.. ? {x} = fractional part of x so if we simplify a little, we see we have x*{x} = 81.25 does this help ?

25. shubhamsrg

x= 81.25 /{x} x can go max to 100 so fractional part can be minimum 0.8125 and max 0.999999...

26. shubhamsrg

just substitute in new values of {x} to get new values of x so according to me,,infinite solutions..

27. shubhamsrg

ohh wait,,did i make a mistake in logic,,i am getting a feeling,,!??

28. shubhamsrg

@joemath314159 @mukushla @calculusfunctions

29. anonymous

if you look at the function:$\frac{81.25}{\{x\}}$ since the fractional part of x can only go from 0 to 1, the only time that will have a solution is when x is bigger than 81.

30. anonymous

if you look at domain, say, [79, 80), this equation:$x=\frac{81.25}{\{x\}}$wont have a solution. since the smallest$\frac{81.25}{\{x\}}$can be is 81.25

31. shubhamsrg

maybe this is where am flawed: since we are dealing with fractional part separately, x , cannot have a fractional part..!! like for example if we set {x} = 0.9, x = 90.2777.. hence contradiction --> 2 different fractional parts..

32. shubhamsrg

i see,,only solutions will be when x is an integer..so that the contradiction with fractional part doesnt come into play and since x has to be greater than 81.25 as @joemath314159 rightly pointed, x can take values 82,83...100 -->19 solutions..

33. shubhamsrg

i mean x can not take those values,, but there'll be a solution each in [82,83) and so on.. so yes,,19 solutions..

34. anonymous

problem says how many solutions so its done...and this is just for fun :) we can find exact solutions:$[x]+\{x\}=\frac{81.25}{\{x\}}$$[x]=\frac{81.25}{\{x\}}-\{x\}$now we know that $[x]=\{81,82,...,99\}$so for example for the solution that lies on the interval [81,82) we have$81=\frac{81.25}{\{x\}}-\{x\}$this is a quadratic for $$\{x\}$$

35. shubhamsrg

yep..i understand better now..thank you..