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Can $|-x^2 | < 1 $ imply that $-1<x<1$? My steps are as follows? $$| -x^2| < 1 $$ $$-1<(-x^2)< 1 $$ $$-1<(-x^2)< 1 $$ $$-1<x^2< 1 $$ $$\sqrt{-1}<x< \sqrt 1 $$

I'm actually looking for the radius of convergence for the power series of $\frac{1}{1-x^2}$:

$$\frac{1}{1-x^2}=\sum\limits_{n=0}^\infty (-x^2)^n \hspace{10mm}\text{for} \,|-x^2|<1$$ This is derived from the equation $$\frac{1}{1-x}=\sum\limits_{n=0}^\infty x^n \hspace{10mm}\text{for} \,|x|<1$$ According to my textbook, the power series $$\sum\limits_{n=0}^\infty (-x^2)^n \hspace{10mm}\text{for} \,|-x^2|<1$$ is 'for the interval (-1,1)' which means that $|-x^2 | < 1 $ implies that $-1<x<1$. However, that implication does not make sense to me.

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    $\begingroup$ Whoa! $x\gt\sqrt{-1}$? That doesn't fly. $\endgroup$ Mar 17, 2013 at 23:31
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    $\begingroup$ Recall that $|x|=-x$ if $x<0$. And $-x^2<0$. (if $x$ is real number.) $\endgroup$
    – Hanul Jeon
    Mar 17, 2013 at 23:36
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    $\begingroup$ Lol, I get it. I just missed the step $|-x^2|=|x^2|$ $\endgroup$
    – raindrop
    Mar 17, 2013 at 23:43
  • $\begingroup$ hum, $\frac{1}{1-x^2}=\sum\limits_{n=0}^\infty (x^2)^n$ not $\sum\limits_{n=0}^\infty (-x^2)^n$ $\endgroup$
    – wece
    Mar 18, 2013 at 0:15

4 Answers 4

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$$|(-x)^2|<1$$ $$|x^2|<1$$ $$|x|^2<1$$ $$|x|<\sqrt{1}$$ $$|x|<1$$ $$-1<x<1$$ Seem more correct to me. (The $\sqrt{-1}$ isn't really well define).

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You want to prove that $|-x^2|<1$ implies $|x|<1$. Note this is the same as $|x^2|<1$ implies $|x|<1$.

Let's prove the contrapositive. If $|x|\geq 1$ then $|x^2|\geq 1$. If $x=1,-1$ this is clear. Since $|x|\geq 0$, the argument for $x>1$ will also serve for $x<1$. Thus, let $x>1$. Then $x=|x|>1$. Multiplying through $|x|$, we find that $$x^2=|x|^2=|x^2|>|x|\geq 1$$ So our claim follows. If $|x|\geq 1$, then $|x^2|\geq 1$.

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The conclusion is correct. However, the steps you took to reach it aren't. The $\sqrt{-1}$ at the end should ring an alarm bell.

A better approach would be to notice that $f(x) = x^2$ is a strictly increasing function on $[0, \infty)$ and strictly decreasing on $(-\infty, 0]$. We know that $f(-1) = f(1) = 1$. Thus, if $x^2 < 1$, then $0 \le x < 1$ or $-1 < x \le 0$. Put together to get $-1 < x < 1$.

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Be careful with the last line. $\sqrt{-1}$ is not a real number so it doesn't make much sense to say $x$ is greater than it. You can, however, deduce from $x^2\geq 0$ for all real numbers, that $0\leq x^2<1$ which implies that $-1<x<1$ (you can see this from the graph of $y=x^2$. For a more rigorous proof, consider the cases $x<0$, $x=0$ and $x>0$ seperately.).

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