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Calculus1 23 Online
OpenStudy (anonymous):

test for convergence

OpenStudy (anonymous):

OpenStudy (anonymous):

can you check my work?

OpenStudy (anonymous):

one of the simplest techniques is that check whether the derivative of the function is always decreasing or not.

OpenStudy (anonymous):

so you have to check dy/dx is decreasing, so for that find whether d^(2)y/dx^(2) <0

OpenStudy (anonymous):

we werent taught that method in class os im not sure if we're allowed to use it in our exam

OpenStudy (anonymous):

@us probably not

OpenStudy (anonymous):

any help?

OpenStudy (anonymous):

Is sec(x) convergent ?? i don't think so. It's periodic function. I may be wrong or i am interpreting something wrong.

OpenStudy (anonymous):

it approaches 1 from what i see but i may be wrong

terenzreignz (terenzreignz):

I'm sorry, but why did you decide to break it at x = 1?

terenzreignz (terenzreignz):

I don't think secant behaves oddly around x = 1

OpenStudy (anonymous):

there is no need to break it at 1 as it is continuous at 1 also. btw i got what approach you were having. See, you were integrating this function but you should integrate it from -infinity to infinity and if you get some finite value for your function then it is convergent. and you should also check where the function looses its continuity and where it is differentiable. You are trying to sum up all the values of the function over the interval, if it gives a finite value, then i will be convergent. I may be wrong.

OpenStudy (anonymous):

it doesnt, but i saw in an example in my book they say you usually break it around there

OpenStudy (anonymous):

from what i see it's only discontinuous at pi/2

OpenStudy (anonymous):

yes you should break it at pi/2 because it reaches infinity and so its not convergent. that's is what i was trying to hint at.

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