The value of $\bigl\lfloor x\bigr\rfloor+\bigl\lfloor x^2\bigr\rfloor+\bigl\lfloor x^3\bigr\rfloor+\bigl\lfloor x^4\bigr\rfloor$

Question

There was a multiple choices saying:

Find the value of $\bigl\lfloor x\bigr\rfloor+\bigl\lfloor x^2\bigr\rfloor+\bigl\lfloor x^3\bigr\rfloor+\bigl\lfloor x^4\bigr\rfloor$ knowing that $x^2+x<0$

The right answer is $-2$.

For solving this Maths test, I solved $x^2+x<0$ which is $(-1,0)$ and then find the value of each term of the sum.

My question is: Is there any approach more formal than I did? Thank you for your time.

• $\lfloor x\rfloor$ is floor$(x)$

Answer 1

As $x^2+x<0, (x-0)(x-(-1))<0$

Now the product two terms is negative, so

either ($x-0>0$ and $x-(-1)<0$) or ($x-0<0$ and $x-(-1)>0$).

If $x>0$ and $x<-1\implies -1>x>0\implies -1>0$, which is clearly impossible.

If $x<0$ and $x>-1$, $-1<x<0$

$\implies -1<x^{2m+1}<0\implies \bigl\lfloor x^{2m+1}\bigr\rfloor=-1$

$\implies 0<x^{2m}<1\implies \bigl\lfloor x^{2m}\bigr\rfloor=0$.

So, $\bigl\lfloor x\bigr\rfloor+\bigl\lfloor x^2\bigr\rfloor+\bigl\lfloor x^3\bigr\rfloor+\bigl\lfloor x^4\bigr\rfloor=-1+0-1+0=-2$

In general, if $(x-a)(x-b)<0$ where $a<b$,

either $x<a$ and $x>b\implies a>x>b\implies a>b$, but $a<b$(given)

or $x>a$ and $x<b\implies a<x<b$ Here $a=-1,b=0$

Answer 2

$$x^2+x<0\Longrightarrow x(x+1)<0\Longleftrightarrow -1<x<0\Longrightarrow \left\{\begin{array}{}\;\;\; 0 <x^n<1&\,n\,\,\text{ is even}\\-1<x^n<0&\,n\,\,\text{ is odd}\end{array}\right. $$

Thus, passing to the floor function under the above condition:

$$\sum_{n=1}^4\lfloor x^n\rfloor=-1+0-1+0=-2$$

Of course, the above is just what you did slightly more fleshed out.