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Converge or diverge arctan [please move to calculus section]

by rcmango
Tags: arctan, calculus, converge, diverge
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rcmango
#1
Jan16-08, 08:37 PM
P: 238
1. The problem statement, all variables and given/known data


Sorry wrong section!! should be in calculus!!

does this problem converge or diverge?

2. Relevant equations



3. The attempt at a solution

Not sure where to start, i've never seen a series arctan problem before, is there a way to switch it around to look like i want it too, maybe with sin or cos. please help.
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foxjwill
#2
Jan16-08, 08:51 PM
P: 355
Is you're question whether [tex]\lim_{x\rightarrow \infty}\arctan{x}[/tex] exists? My first instinct is to try for a series representation, but I'm not sure.

Edit:
Come to think of it, by definition, the range of [tex]\arctan{x}[/tex] is [tex](-\frac{\pi}{2},\frac{\pi}{2})[/tex] while its domain is [tex](-\infty, \infty)[/tex], so, not only does [tex]\lim_{x\rightarrow \infty}\arctan{x}[/tex] exist, it equals [tex]\frac{\pi}{2}[/tex].
rcmango
#3
Jan16-08, 10:39 PM
P: 238
Okay, so I know the limit is pi/2, so is there a series test i can use to prove this?

foxjwill
#4
Jan16-08, 10:57 PM
P: 355
Converge or diverge arctan [please move to calculus section]

I don't know (>_<), but it's kind of not something that needs to be proven since it's part of the definition.
rcmango
#5
Jan16-08, 11:32 PM
P: 238
So if i show the definition as work, would you believe that to be enough? thanks for all the help.
rcmango
#6
Jan16-08, 11:37 PM
P: 238
also, could i show this using the squeeze theorem with bounds -pi/2 and upper bound pi/2
rcmango
#7
Jan20-08, 07:23 PM
P: 238
anyone else who can comment on whats been said here please?
Feldoh
#8
Jan20-08, 07:32 PM
P: 1,345
Quote Quote by rcmango View Post
Okay, so I know the limit is pi/2, so is there a series test i can use to prove this?
You could.

arctan(x) = x - x^3/3 + x^5/5 - x^7/7 + x^9/9 - ... (-1 < x < 1) is the maclaurin series for it.


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