9 replies to this topic

[James33]

For those that don't know, the floor function which I will denote by [x] is the largest integer less than or equal to x so for example [6]=6, [8.7]=8 and [-4.3]=5. Solving equations using this function can be very interesting and there are often loads of ways to do it.

An easier one to start off:

Solve -6x^2+[x^4]+19=0

This one is harder;

Solve [x^3]-x^2+6[x]=23

[CaptainEd]

I'm more accustomed to [-4.3] = (-5)
Is that what you mean? Or do you really intend to take the absolute value of the floor?

[James33]

I'm more accustomed to [-4.3] = (-5)
Is that what you mean? Or do you really intend to take the absolute value of the floor?

Yes that is what I meant.

[WitchOfSecrets]

For the first:

Spoiler for solved

19 is a small number and fourth powers grow quickly. So I try assuming that [x] = 2 and thus [x^4]=16. This gives an answer of x = sqrt(35/6), or about 2.4. Negative also works.

Edited by WitchOfSecrets, 10 June 2012 - 12:16 AM.

[superprismatic]

Spoiler for for the first

It seems that [x⁴]=6x²-19 which means 6x²-19≤x⁴<6x²-18 and so -19≤x⁴-6x²<-18 thus -10≤x⁴-6x²+9<-9 which is -10≤(x²-3)²<-9 which is impossible since (x²-3)²>0 in the reals where the function [ ] is defined. So, there is no solution.

[WitchOfSecrets]

Whoops. I did [x]^4, not [x^4]. I think SP is right.

[CaptainEd]

Spoiler for an answer for second problem

Noting that x^2 must be integer, I decided x = sqrt(y), for some integer y.
plotting the function for y between 0 and 10, I saw the inflection points and traced it down to y = 7, so

x = sqrt(7)

Edited by CaptainEd, 10 June 2012 - 04:18 AM.

[James33]

Superprismatic is correct for the first, CaptainEd is correct for the second but there are much better ways of doing these than looking at the graph, also you have to prove there are no other solutions.

[CaptainEd]

Spoiler for Not proud about brute forcing, but there is little more to say

First, the two expressions with floor are both therefore integers.
This implies that x^2 must also be integer.
Hence the restriction that x = +/- sqrt(y) for some integer y.
Notice, the unfloored variant of the function is never more than 7 away from the functional value itself.
By definition of floor,
[x^3] <= x^3 < 1+[x^3] and
6[x] <= 6x < 6+6[x]
So [x^3] - x^2 + 6[x] - 23 <=
x^3 - x^2 + 6x - 23 <
[x^3]+1 - x^2 + 6[x]+6 - 23 = 7

When we plug into the unfloored function the value sqrt(10) for x,
we get 10*sqrt(10) - 10 + 6*sqrt(10) - 23 = 16*sqrt(10) - 33, roughly 15.
This is too high, even higher than our buffer of 7, and
Looking at the first derivative 3x^2 -2x + 6 is positive, and so is the second derivative (6x-2).
So no solutions are above sqrt(10).

------- Similarly, plugging in 0 for x, we get -23, which is too low.
First derivative is 6, indicating that the function will diverge further with decreasing x, and second derivative is negative and more negative with decreasing x.

So no solutions are below 0. So there are no solutions outside the set of x = sqrt(y), for integer y in [0,9]. An exhaustive search through the integers from 0 through 9 shows that only x=sqrt(7) is a solution.

I'd like to add that this whole area is bizarre, and yet it does seem (you've said so, and SP has shown an example) that one could do some sort of analysis for problems in this area. Thank you, James33, for a glimpse into a weird world!

Edited by CaptainEd, 11 June 2012 - 10:24 PM.

[James33]

Spoiler for Not proud about brute forcing, but there is little more to say

First, the two expressions with floor are both therefore integers.
This implies that x^2 must also be integer.
Hence the restriction that x = +/- sqrt(y) for some integer y.
Notice, the unfloored variant of the function is never more than 7 away from the functional value itself.
By definition of floor,
[x^3] <= x^3 < 1+[x^3] and
6[x] <= 6x < 6+6[x]
So [x^3] - x^2 + 6[x] - 23 <=
x^3 - x^2 + 6x - 23 <
[x^3]+1 - x^2 + 6[x]+6 - 23 = 7

When we plug into the unfloored function the value sqrt(10) for x,
we get 10*sqrt(10) - 10 + 6*sqrt(10) - 23 = 16*sqrt(10) - 33, roughly 15.
This is too high, even higher than our buffer of 7, and
Looking at the first derivative 3x^2 -2x + 6 is positive, and so is the second derivative (6x-2).
So no solutions are above sqrt(10).

------- Similarly, plugging in 0 for x, we get -23, which is too low.
First derivative is 6, indicating that the function will diverge further with decreasing x, and second derivative is negative and more negative with decreasing x.

So no solutions are below 0. So there are no solutions outside the set of x = sqrt(y), for integer y in [0,9]. An exhaustive search through the integers from 0 through 9 shows that only x=sqrt(7) is a solution.

I'd like to add that this whole area is bizarre, and yet it does seem (you've said so, and SP has shown an example) that one could do some sort of analysis for problems in this area. Thank you, James33, for a glimpse into a weird world!

Thats very similar to the way I was expecting it to be solved, a lot of these problems cannot be solved perfectly analyticly but, as you did, can reduce the possibilities of x to a small, finite number of options which can be eliminated without much trouble.