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How to sketch $y = \left\lfloor \sqrt{2-x^2} \right\rfloor$, where $\lfloor \cdot \rfloor$ denotes the greatest integer function?

Please help. I have no idea about this.

Vedran Šego
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user2369284
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  • Hint: it's only defined for $|x|\le\sqrt{2}$. And 2nd hint: it must be a step function. – Jean-Claude Arbaut Dec 26 '13 at 15:51
  • @arbautjc It's just the domain. How to sketch it ? – user2369284 Dec 26 '13 at 15:52
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    Yes, it's the domain, but notice it's small. To sketch it, you just have to find on which intervals it can take which value. First, notice it's an even function. Then, if $x>1$, obviously $\sqrt{2-x^2}<1$ thus the "floor" is 0. And if $0<x<1$, obviously $1<\sqrt{2-x^2}<\sqrt{2}$, thus the "floor" is 1. Should not be that difficult to plot a function that is 1 on ]-1,1[ and 0 elsewhere. Just be careful with interval endpoints. – Jean-Claude Arbaut Dec 26 '13 at 15:56
  • Could try to sketch squareroot 2-x squared for a start and think how the ceiling function distort this graph – Lost1 Dec 26 '13 at 16:17

3 Answers3

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Are you looking for this?

This was rather easy to do with Wolfram Alpha:

y = floor(sqrt(2-x^2))

enter image description here

Hyrtsi
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First, notice that the domain of $x$ is $$2-x^2\ge0\iff-\sqrt2\le x\le \sqrt 2.$$

And the floor function $\lfloor x\rfloor$ is defined as $$\lfloor x\rfloor=t\iff t\le x\lt t+1.$$

So, in your question, we have $$\lfloor \sqrt{2-x^2}\rfloor =t\iff t\le\sqrt{2-x^2}\lt t+1$$ Now note that $t\ge 0 \in\mathbb Z$ (This is because $\sqrt{2-x^2}\ge0$). So we have $$t^2\le 2-x^2\lt (t+1)^2\iff 2-(t+1)^2\lt x^2\le 2-t^2.$$

The fact that $\lfloor \sqrt{2-0^2}\rfloor=1, \lfloor \sqrt{2-({\sqrt 2})^2}\rfloor=0$ tells us that $t=0,1$.

Hence, we know

$$2-(0+1)^2\lt x^2\le 2-0^2\Rightarrow t=0,$$ $$2-(1+1)^2\lt x^2\le 2-1^2\Rightarrow t=1.$$

mathlove
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Let $$f(x)=\sqrt{2-x^2}$$

$2-x^2\geq 0$ or $ x^2 - 2\leq 0$ this gives $|x|\leq\sqrt 2 $

so find x in which f(x) is lies between 0 and 1 , 1 and 2 , 2 and 3 ....

this will give you a set of solution in that region take Greatest Integer of f(x) and now you can draw graph easily .

john
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