Hello,

I've been having trouble on the following question:

We let g(x) = ⌊x⌊x⌋⌋ for x >= 0.

i. Find all x >= 0 so that g(x) = 1.

ii. Find all x >= 0 so that g(x) = 3.

iii. Find all x >= 0 so that g(x) = 5.

iv. Find the number of possible values of g(x) for 0 <= x <= 10.

I started on i. using the properties of a floor function and got 1 <= x⌊x⌋ < 2, but I have no idea what to do next. Please help as soon as possible.
Thanks in advance.

Let's take a look at a few values of x.

g(.5) = ⌊.5 ⌊.5⌋⌋ = ⌊.5*0⌋ = ⌊0⌋ = 0

g(1.9) = ⌊1.9⌊1.9⌋⌋ = ⌊1.9*1⌋ = ⌊1.9⌋ = 1

g(2.4) = ⌊2.4⌊2.4⌋⌋ = ⌊2.4*2⌋ = ⌊4.8⌋ = 4
g(2.7) = ⌊2.7⌊2.7⌋⌋ = ⌊2.7*2⌋ = ⌊5.4⌋ = 5
so, for any 0<z<1, g(2+z) = ⌊(2+z)⌊2+z⌋⌋ = ⌊(2+z)*2⌋ = ⌊4+2z⌋
So, g(2+z) = 4 for z < 0.5
g(2+z) = 5 for z >= 0.5

So, I'd guess that for any integer k>1,
g(k+z) = k^2 for z < 1/k
k^2+1 for 1/k <= z < 2/k
...
k^2+(k-1) for (k-1)/k <= z < 1

I haven't checked it thoroughly, but see what you can do with that. It does seem to indicate that there is no solution for g(x) = 3

Thank you for your response. I have solved (a), (b) and (c). Yes, there is no solution for g(x)=3.

Please help on part iv, I’ve no idea where to start.

To solve the problem, we need to understand the properties of the floor function in order to analyze the equation g(x) = ⌊x⌊x⌋⌋.

i. To find all x >= 0 so that g(x) = 1, we need to substitute g(x) with 1 in the original equation and find the corresponding value of x.

1 = ⌊x⌊x⌋⌋

Since the floor function returns the largest integer less than or equal to the input, the expression ⌊x⌊x⌋⌋ will always be an integer.

For g(x) = 1, we need to find x such that 1 <= x⌊x⌋ < 2.

The first step is to consider the range of values for x⌊x⌋. Notice that x can be any non-negative real number, and since ⌊x⌋ will also be a non-negative integer, x⌊x⌋ will take integer values.

Since we know 1 <= x⌊x⌋ < 2, we can start exploring the possible values of x. One approach is to consider different cases based on the value of ⌊x⌋.

Case 1: ⌊x⌋ = 0
If ⌊x⌋ = 0, then x⌊x⌋ = 0, and 1 <= 0 < 2, which is not possible.

Case 2: ⌊x⌋ = 1
If ⌊x⌋ = 1, then x⌊x⌋ = x, and 1 <= x < 2. This means the values of x in this case are in the range [1, 2).

Case 3: ⌊x⌋ = 2
If ⌊x⌋ = 2, then x⌊x⌋ = 2x, and 1 <= 2x < 2.
Simplifying this inequality, we find 1/2 <= x < 1.

Case 4: ⌊x⌋ = 3
If ⌊x⌋ = 3, then x⌊x⌋ = 3x, and 1 <= 3x < 2.
Simplifying this inequality, we find 1/3 <= x < 2/3.

We can see that as the values of ⌊x⌋ increase, the range of possible values for x decreases. Continuing this pattern, for ⌊x⌋ = 4, we would have 1/4 <= x < 1/2, and so on.

To summarize, for g(x) = 1, the possible values of x are in the following intervals:
[1, 2) ∪ [1/2, 1) ∪ [1/3, 2/3) ∪ [1/4, 1/2) ∪ ...

ii. Following a similar approach, we can solve for g(x) = 3.

3 = ⌊x⌊x⌋⌋

For g(x) = 3, we need to find x such that 3 <= x⌊x⌋ < 4.

We can again consider different cases based on the value of ⌊x⌋.

Case 1: ⌊x⌋ = 0
Not possible, as x⌊x⌋ = 0 < 3.

Case 2: ⌊x⌋ = 1
Not possible, as x⌊x⌋ = x < 3.

Case 3: ⌊x⌋ = 2
Not possible, as x⌊x⌋ = 2x < 3.

Case 4: ⌊x⌋ = 3
If ⌊x⌋ = 3, then x⌊x⌋ = 3x, and 3 <= 3x < 4.
Simplifying this inequality, we find 1 <= x < 4/3.

For ⌊x⌋ = 4, we would have 3/4 <= x < 2/3, and so on.

To summarize, for g(x) = 3, the possible values of x are in the following intervals:
[1, 4/3) ∪ [3/4, 2/3) ∪ [1/2, 3/5) ∪ [3/5, 4/7) ∪ ...

iii. Following the same method, we can solve for g(x) = 5.

5 = ⌊x⌊x⌋⌋

For g(x) = 5, we need to find x such that 5 <= x⌊x⌋ < 6.

After considering the different cases and intervals, you can find the possible values of x for g(x) = 5.

iv. To find the number of possible values of g(x) for 0 <= x <= 10, you would need to analyze the intervals obtained for each value of g(x) (1, 3, 5). Count the number of intervals that intersect with the range [0, 10] and include the endpoints if they are a valid solution.