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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$(x)$
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    Babak, I don't know what you mean by formal. I can't imagine any approach that would be better in any significant way.2012-09-08

2 Answers 2

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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

$\implies -1

$\implies 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,

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

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

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    I'm not erasing my first comments in order for others to see what's been going on here, yet I will upvote the answer instead. +12012-09-08
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$x^2+x<0\Longrightarrow x(x+1)<0\Longleftrightarrow -1

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.

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    Any time, @BabakSorouh.2012-09-08