Jump to content
BrainDen.com - Brain Teasers
  • 0

Equations with the Floor Function


James33
 Share

Question

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

Link to comment
Share on other sites

9 answers to this question

Recommended Posts

  • 0

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
Link to comment
Share on other sites

  • 0

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.

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Answer this question...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
 Share

  • Recently Browsing   0 members

    • No registered users viewing this page.
×
×
  • Create New...