Register to reply

Converge or diverge arctan [please move to calculus section]

by rcmango
Tags: arctan, calculus, converge, diverge
Share this thread:
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.
Phys.Org News Partner Science news on Phys.org
NASA team lays plans to observe new worlds
IHEP in China has ambitions for Higgs factory
Spinach could lead to alternative energy more powerful than Popeye
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.


Register to reply

Related Discussions
Does ln(n)sin(n) converge or diverge? Calculus & Beyond Homework 9
Converge or diverge Calculus 14
Converge or diverge? Calculus 4
Converge or diverge? Calculus 3
To converge and to diverge General Physics 3