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

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$(x)$
-
Your approach can be done completely formally. – Alex Becker Sep 8 '12 at 6:00
There is nothing casual about your approach. for example, if $-1\lt x\lt 0$ then $0\lt x^2\lt 1$ so $[x^2]=0$. The other are done the same way. – André Nicolas Sep 8 '12 at 6:03
Babak, I don't know what you mean by formal. I can't imagine any approach that would be better in any significant way. – Gerry Myerson Sep 8 '12 at 6:11

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 $a=-1,b=0$

-
Your solution of the inequality is, I think, incorrect: the system "or", as you wrote, exists for a quadratic inequality of this kind iff the inmequality sign was exactly the opposite one. In the present case, the inequality's solution is directly $\,-1<x<0\,$, which is a system "and" solution. All this is pretty clear if we draw the parabola $\,x^2+x=x(x+1)\,$ and locate its roots. – DonAntonio Sep 8 '12 at 7:01
@DonAntonio, could you please explain "the inequality sign was exactly the opposite one"? – lab bhattacharjee Sep 8 '12 at 7:49
I mean that the inequality $\,(x-a)(x-b)>0\,\,,\,\,a<b\,$ has precisely the solution $\,x<a\,\,or\,\,x>b\,$ , which is an "or" system, whereas the opposite inequality $\,(x-a)(x-b)<0\,\,,\,\,a<b\,$ has the solution $\,a<x<b\,$ , which is an "and" system. – DonAntonio Sep 8 '12 at 8:04
@Don: you're focusing too much on form: the end result are the same. One common method to solving inequalities is to solve for the signs, then use the signs to solve for the variables. In this particular case, it means you get a spurious sign solution you have to consider and rule out. – Hurkyl Sep 8 '12 at 8:18
@Hurkyl, I can't tell you. I've taught this stuff to high school kids and refreshed it to university ones, and we try to be formal and precise. The very beautiful interaction between algebra and geometry in this particular case makes it very easy both to prove and then to solve this kind of inequalities, and that's why I stressed what I did. That the solution "is the same" does not, imho, justify imprecisions of mistakes, because later we might face another cases where we can be misled. – DonAntonio Sep 8 '12 at 8:33

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

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.

-
Thanks Don for your attempt here. Thanks for your time. – Babak S. Sep 8 '12 at 9:14
Any time, @BabakSorouh. – DonAntonio Sep 8 '12 at 9:18