Locate the discontinuities

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September 17th 2008, 08:11 PM
silencecloak

Locate the discontinuities

Can someone tell me if my logic behind this question is correct

$ y=ln(\tan^2(x)) $

the answer is

$ n\pi/2 $

where 'n' is any integer

logic: since arctangent is the inverse of tangent and

$ -\pi/2 \le arctan \le \pi/2 $

multiplying these boundaries by tan would result in a discontinuity
September 17th 2008, 08:33 PM
Jhevon

Quote:

Originally Posted by silencecloak

Can someone tell me if my logic behind this question is correct

$ y=ln(\tan^2(x)) $

the answer is

$ n\pi/2 $

where 'n' is any integer

logic: since arctangent is the inverse of tangent and

$ -\pi/2 \le arctan \le \pi/2 $

multiplying these boundaries by tan would result in a discontinuity

your answer is (partially) correct. your logic makes no sense to me though. what does arctan(x) have to do with anything?
September 17th 2008, 08:47 PM
Jhevon

Quote:

Originally Posted by silencecloak

Can someone tell me if my logic behind this question is correct

$ y=ln(\tan^2(x)) $

the answer is

$ n\pi/2 $

where 'n' is any integer

logic: since arctangent is the inverse of tangent and

$ -\pi/2 \le arctan \le \pi/2 $

multiplying these boundaries by tan would result in a discontinuity

here is a hint:

the domain of the logarithm is the set of positive real numbers. if what is being logged is zero or negative, then the log is undefined.

tangent is discontinuous where cos(x) = 0 and it is zero where sin(x) = 0
September 17th 2008, 09:19 PM
silencecloak

Quote:

Originally Posted by Jhevon

here is a hint:

the domain of the logarithm is the set of positive real numbers. if what is being logged is zero or negative, then the log is undefined.

tangent is discontinuous where cos(x) = 0 and it is zero where sin(x) = 0

so where does the $\pi/2$ for the answer come from

that's why i was tying arctan in :/

EDIT:

oh wait...we are playing with the unit circle now huh?

cos(90) = 0

but how do you know that the tan is discontinuous where cos(x) = 0 and is zero where sin(x) = 0
September 18th 2008, 05:36 AM
Aryth

We know that because of the identity using $\tan{x}$ . We know that:

$\tan{x} = \frac{\sin{x}}{\cos{x}}$

So, when $\cos{x} = 0$

$\tan{x} = \frac{\sin{x}}{0}$

Which is undefined and therefore discontinuous, and when $\sin{x} = 0$

$\tan{x} = \frac{0}{\cos{x}}$

Which is zero for all nonzero results of $\cos{x}$ , so the logarithm is therefore discontinuous because it is not defined at zero.