Differing Definitions of arcsec(x) Lead to Confusion over Signs
Date: 12/03/2005 at 21:36:52 From: Richard Subject: sec(arcsec x) Why does sec(arcsec x) equal |x|? Similarly, why does csc(arccsc x) equal |x|? I was trying to prove the derivatives of arcsec and arccsc, but when I checked the answer my book had |x| where I had x. I'm trying to figure out what went wrong. Do you have any idea? Here is my work: d/dx (arcsec x) If y = arcsec x, then arcsec y = x = 1/((sec(arcsec x))tan(arcsec x)) = 1/(x * tan y) sec(arcsec x)= x... right? = 1/(x * sqrt(sec^2 y - 1)) 1 + tan^2 y = sec^2 y = 1/(x * sqrt (x^2 - 1) Since there is |x| instead of x in the correct anwswer, I assume that sec(arcsec x)= |x|. But why?
Date: 12/03/2005 at 22:57:51 From: Doctor Peterson Subject: Re: sec(arcsec x) Hi, Richard. Certainly it is not true that sec(arcsec(x)) = |x|; take a specific negative value of x and verify that the result is negative. It's true that there is an absolute value in the derivative, but it comes from a different step than taking sec(arcsec(x)). Let me edit what you wrote to show what I think you meant, correcting a typo and a derivative you didn't write: d/dx (arcsec x): If y = arcsec x, then sec y = x dy/dx = 1/(dx/dy) = 1/((sec(arcsec x))tan(arcsec x)) = 1/(x * tan y) sec(arcsec x)= x... right? = 1/(x * sqrt(sec^2 y - 1)) 1 + tan^2 y = sec^2 y = 1/(x * sqrt (x^2 -1) The problem is not in sec(arcsec x), but in tan(arcsec x)! While it's true that sec(arcsec x) = x for all x in the domain, the sign of tan(y) has to be considered. Assuming your arcsec is defined to have a range of [0, pi], excluding pi/2, its tangent is positive in [0, pi/2) and negative in (pi/2, pi]. You used sqrt(sec^2 y - 1), which is always positive! Remember that there are two square roots, and you have to choose the appropriate one. You want to take the positive root when x = sec y happens to be positive, and the negative root when x is negative, so we can write the last two lines this way, using the "sgn" function that gives the sign of a number (+1 or -1): = 1/(x * sgn(x) * sqrt(sec^2 y - 1)) = 1/(|x| * sqrt (x^2 -1) Here, x times its own sign gives it absolute value (always positive). Does that help? These sign issues can be very tricky, and I can see why this one mystified you. The moral of the story: when you take a square root, ALWAYS check which sign to use under which conditions. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 12/05/2005 at 15:55:08 From: Richard Subject: sec(arcsec x) Hello! Ok, I almost understand it now. But I don't get what the "sgn" means and why you have to multiply that by x. Also, this whole arcsec function has me confused. I just realized I have never even seen the graph of it...
Date: 12/05/2005 at 17:36:51 From: Doctor Peterson Subject: Re: sec(arcsec x) Hi, Richard. I briefly defined the sgn function; here is a fuller version: sgn(x) = 1 if x>0 0 if x=0 -1 if x<0 So x sgn(x) = x if x>0 0 if x=0 -x if x<0 which is the same as |x|. So what I did was to use sgn to introduce the correct sign for each value of x, and then merging the sgn(x) with x to get |x|. I was just looking at an article about the unpopularity of the arcsec function in calculus courses, which is partly tied to the fact that texts differ on its exact definition--specifically what range to use when x < 0 (College Mathematics Journal, Nov 2005). It is pointed out that if you take pi <= arcsec(x) < 3pi/2 for x <= -1, then the derivative is 1/[x sqrt(x^2 - 1)], but if you define it (as I do) to have pi/2 <= arcsec(x) <= pi for x < -1, then the derivative has the absolute value. If your text asked for the derivative of arcsec(x), I hope it has also shown you its graph somewhere, or at least defined it as they are using it! Here is some information on it: Inverse Secant http://mathworld.wolfram.com/InverseSecant.html - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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