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Callisto
In the domain [0, 100], how many solutions are there for the equation \[x^2- \lfloor x \rfloor x = 81.25\] How to start?
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.
That way you can minimize the frustration the floor function brings to the table.
er, i guess the interval should really be [n,n+1)
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.
wait, the function\[x^2+\lfloor x \rfloor x\]is strictly increasing, isnt it?
then it can have at most one solution.
That means it has 0 :'(
strictly increasing on the interval [0,100] anyways.
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.
no, its strictly increasing for sure.
Sorry!!! Wait... a mistake in the question posted here!!
ah ok, that makes more sense.
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|
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.
less than >.> i meant greater than lol.
So there are 19 solutions, one in each of the intervals [81, 82), [82,83),....., [99,100)
Thanks! I've never thought about this way to look at the problem.. :(
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 :)
I'll remember it! Thanks again!!
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 ?
x= 81.25 /{x} x can go max to 100 so fractional part can be minimum 0.8125 and max 0.999999...
just substitute in new values of {x} to get new values of x so according to me,,infinite solutions..
ohh wait,,did i make a mistake in logic,,i am getting a feeling,,!??
@joemath314159 @mukushla @calculusfunctions
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.
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
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..
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..
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..
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 \(\{x\}\)
yep..i understand better now..thank you..