Limiting sum q (1 Viewer)

flavurr · Oct 20, 2015

r=-tan^2(theta)

For what values of theta between the interval -pi/2 < theta < pi/2 does the limiting sum exist?

InteGrand · Oct 20, 2015

flavurr said:
r=-tan^2(theta)

For what values of theta between the interval -pi/2 < theta < pi/2 does the limiting sum exist?

$The limiting sum exists if and only if $-1<r<1$. In this case, $r=-\tan^2 \theta$, so $r$ is non-positive, so the limiting sum exists if and only if $r>-1$. So we just need to solve the inequation $-\tan^2 \theta > -1$. We have$

$-\tan^2 \theta > -1 \Longleftrightarrow \tan^2 \theta < 1$

$\Longleftrightarrow -1< \tan \theta < 1$

$$\Longleftrightarrow -\frac{\pi}{4}<\theta <\frac{\pi}{4}$ (answer).$

davidgoes4wce · Oct 20, 2015

is that a squared? or tan (2 theta) ?

flavurr · Oct 20, 2015

InteGrand said:
$The limiting sum exists if and only if $-1<r<1$. In this case, $r=-\tan^2 \theta$, so $r$ is non-positive, so the limiting sum exists if and only if $r>-1$. So we just need to solve the inequation $-\tan^2 \theta > -1$. We have$

$-\tan^2 \theta > -1 \Longleftrightarrow \tan^2 \theta < 1$

$\Longleftrightarrow -1< \tan \theta < 1$

$$\Longleftrightarrow -\frac{\pi}{4}<\theta <\frac{\pi}{4}$ (answer).$

How did you go from tan^2(theta) <1 to -1 < tan(theta) <1 ?

sorry im sounding silly here

flavurr · Oct 20, 2015

davidgoes4wce said:
is that a squared? or tan (2 theta) ?

squared

InteGrand · Oct 21, 2015

flavurr said:
How did you go from tan^2(theta) <1 to -1 < tan(theta) <1 ?

sorry im sounding silly here

$A real number $a$ satisfies $a^2 < 1$ if and only if $-1<a<1$ (see the graph of $y=x^2$ for instance; $x^2$ is below $y=1$ when $-1<x<1$, and $not$ below it otherwise. In other words, $x^2 < 1$ if and only if $-1<x<1$).$

Limiting sum q (1 Viewer)

flavurr

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InteGrand

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davidgoes4wce

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flavurr

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flavurr

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InteGrand

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