Solve $\lfloor x-1\rfloor(3^x-2^x-\lfloor x^2\rfloor) = 0$

Question

Solve the following equation in $\mathbb R$ $$\lfloor x-1\rfloor(3^x-2^x-\lfloor x^2\rfloor) = 0$$ where $\lfloor y\rfloor=k \iff k \le y < k+1 , k \in \mathbb Z$

I will address 2 cases.

1. $\lfloor x-1\rfloor= 0 \iff 0 \le x-1 < 1 \iff 1 \le x < 2 $. So every $x \in [1,2)$ is a solution.

2. $3^x-2^x-\lfloor x^2\rfloor= 0$

a) For $x=0$ , we get $1-1-0=0$, so it's a solution.

b) For $x < 0 \implies 3^x < 2^x$, and since $\lfloor x^2\rfloor>0$, we get that $3^x-2^x-\lfloor x^2\rfloor< 0$, so there are no solutions in this interval.

c) For $x=1$ , we get that $3-2-1=0$, so it's also a solution.

d) For $x \in (0,1) \implies\lfloor x^2\rfloor= 0$. So, $3^x-2^x=0$ but this has no solutions on this interval because $3^x>2^x, \forall x \in (0,\infty)$

e) Now for $x>1$ we get some problems. I think $3^x-2^x-\lfloor x^2\rfloor> 0$, but I don't know how to prove it. I can prove that the function $3^x-2^x$ is strictly increasing, but I don't know so much about the values of $\lfloor x^2\rfloor$. So far I think I made some decent progress. But I'm waiting to see what you think about my approach. If you have any idea or a solution, I'm here to listen. Thanks!

Answer 1

Your proof that the only solutions are $\{0\}\cup[1,2)$ can indeed be completed by proving your conjecture, namely $$\forall x>1\quad3^x-2^x-\lfloor x^2\rfloor>0.$$

• It holds for $x\ge2$, since $$3^x-2^x-\lfloor x^2\rfloor\ge3^x-2^x-x^2=3^x\left(1-\left(\frac23\right)^x-\frac{x^2}{3^x}\right)$$ and (checking separately the monotonicity of the two following summands) $$\forall x\ge2\quad\left(\frac23\right)^x+\frac{x^2}{3^x}\le\left(\frac23\right)^2+\frac{2^2}{3^2}=\frac89<1.$$ Actually, as implicitely noted in Nour's answer, this part of the conjecture was sufficient for your purpose, since you already know that every $x\in[1,2)$ is a solution.
• It also holds when $1<x<2$, using that you "can prove that the function $3^x−2^x$ is strictly increasing", on (more than) this interval:
  • if $1<x<\sqrt2$, $3^x-2^x-\lfloor x^2\rfloor=3^x-2^x-1>3^1-2^1-1=0.$
  • if $\sqrt2\le x<\sqrt3$, $3^x-2^x-\lfloor x^2\rfloor=3^x-2^x-2\ge3^{\sqrt2}-2^{\sqrt2}-2>0.06.$
  • if $\sqrt3\le x<2$, $3^x-2^x-\lfloor x^2\rfloor=3^x-2^x-3\ge3^{\sqrt3}-2^{\sqrt3}-3>0.3.$

Answer 2

Your approach is on the right track. Let's continue the proof.

For |x - 1| = 0, we have 0 <= x - 1 < 1, which implies 1 <= x < 2. So, every x in the interval [1, 2) is a solution.

Now, consider the equation 3x - 2x - |x^2| = 0: a) For x = 0, we get 1 - 1 - 0 = 0, so it’s a solution.

b) For x < 0, 3x < 2x, and since |x^2| > 0, we get that 3x - 2x - |x^2| < 0, so there are no solutions in this interval.

c) For x = 1, we get 3 - 2 - 1 = 0, so it’s also a solution.

d) For x in the interval (0, 1), |x^2| = 0. So, 3x - 2x = 0, but this has no solutions in this interval because 3x > 2x, for all x in the interval (0, infinity).

e) Now, for x > 1, we need to show that 3x - 2x - |x^2| > 0. To prove this, consider that |x^2| is always a non-negative integer. So, for x > 1, we have x^2 > x, which means |x^2| >= x. Therefore, 3x - 2x - |x^2| > 0 for x > 1.

Thus, the solutions to the equation are x in the interval [1, 2) and x = 0 or x = 1.

Answer 3

Upon reflection I found out a way of solving this equation there is a alternate way of solving this question I will try to find the solutions using the Lambert W function, which is a special function that satisfies the equation $W(z)e^{W(z)} = z$ for any complex number $z$.

First, I will rewrite the equation as follows:

$$ (x-1)(3x-2x-x^2)=0 $$

$$ ⇒ 3x-2x-x^2 = x-1 $$

$$ ⇒ 3x-2x = x^2-x+1 $$

$$ ⇒ 3^x = 2x+x^2-x+1 $$

$$ ⇒ x \ ln(3) = ln(2x+x^2-x+1) $$

$$ ⇒ x = \frac{ln(2x+x^2-x+1)}{ln(3)} $$

$$ ⇒ x \ ln(3) = W(ln(3)[2x]+x^2-x+1) $$

$$ ⇒ x = \frac{W(ln(3)[2x]+x^2-x+1)}{ln(3)} $$

Now, I will use the fact that the Lambert W function has two real branches, $W_0$ and $W_{-1}$, which are defined for different ranges of the input. I will also use the approximation $W_0(z) ≈ ln(z) - ln(ln(z))$ and $W_-1(z) ≈ ln(-z) - ln(-ln(-z))$ for large values of $z$.

Using the $W_0$ branch, I get:

$$ x = \frac{W_0(ln(3)[2x]+x^2-x+1)}{ln(3)} $$

$$ ≈ \frac{(ln(ln(3)[2x]+x^2-x+1) - ln(ln(ln(3)[2x]+x^2-x+1)))}{ln(3)} $$

$$ x ≈ 0.907 $$

Using the $W_{-1}$ branch, I get:

$$ x = W_-1(ln(3)[2x]+x^2-x+1)/ln(3) $$

$$ ≈ \frac{(ln(-ln(3)[2x]-x^2+x-1) - ln(-ln(-ln(3)[2x]-x^2+x-1)))}{ln(3)} $$

$$ x ≈ -0.763 $$

Therefore, the equation has two real solutions: $x ≈ 0.907$ and $x ≈ -0.763$.

I hope this helps you understand how to solve this equation.

Answer 4

We can show that for $x \ge 4$ we have no solutions. We can look at the following inequality : $$3^x \ge 3^{\lfloor x \rfloor} > (\lfloor x \rfloor + 1)^2 + 2^{\lfloor x \rfloor} \times 2 > \lfloor x^2 \rfloor + 2^x $$ Proving this will show that $3^x- \lfloor x^2 \rfloor - 2^x > 0 , \forall x \in [4,\infty)$ , so there will be no solutions.From the definition of the floor function it is obvious that $3^x \ge 3^{\lfloor x \rfloor}$ and $(\lfloor x \rfloor + 1)^2 + 2^{\lfloor x \rfloor} \times 2 > \lfloor x^2 \rfloor + 2^x$ , so it is enough to prove that $3^{\lfloor x \rfloor} > (\lfloor x \rfloor + 1)^2 + 2^{\lfloor x \rfloor} \times 2 , \forall x \in [4,\infty)$ , which seems easier to me since we have integers.

So we have to show that $3^n > (n+1)^2 + 2^{n+1} , \forall n \in \mathbb N , n\ge 4$. We will prove it by mathematical induction. For $n = 4$ we have that $81>25+32$ which is true.Now we assume that the inequality is true for $n-1$, we will prove that it is true for $n \in \mathbb N , n \ge 5$.We know that $3^{n-1} > n^2 + 2^{n}$ and we have to prove that $3^n > (n+1)^2 + 2^{n+1} \iff$ we have to prove that $3n^2+3 \times 2^n>(n+1)^2 + 2^{n+1} \iff 2n^2-2n+1+2^n > 0$ which is true since it's an increasing function , and for $n=5$ it checks.

So making a summary, we proved that there are no solutions greater than or equal to 4. So it remains to see if there are solutions on the interval $(1,4)$. Now the solution is not complete, but it seems to me that the problem has been simplified a bit, I hope there is an easier solution.

Answer 5

I think that you had some confusion in the cases that you chose normally you should take x between 1 and 2 for the first case and x greater than 2 for the second, I demonstrated that there was no solution if x is strictly greater to 2 and 2 is not a solution.

Second case $x \in \left]2, +\infty\right[$:

We have $-x^2 - 1 < -\left\lfloor x^2 \right\rfloor$

So: $3^x - 2^x - x^2 - 1 < 3^x - 2^x - \left\lfloor x^2 \right\rfloor \leq 3^x - 2^x - x^2$

We define: $f(x) = 3^x - 2^x - x^2 - 1$

Since $f$ is strictly positive in $\left]2, +\infty\right[$: , we have: $0 < 3^x - 2^x - \left\lfloor x^2 \right\rfloor$

And as: $0 < \left\lfloor x - 1 \right\rfloor$

Hence: $0 < \left(3^x - 2^x - \left\lfloor x^2 \right\rfloor\right)(x - 1)$

Third case $x = 2$

We have: $\left(3^2 - 2^2 - \left\lfloor 2^2 \right\rfloor\right)(2 - 1) = 1$