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TuringTest
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quick question: there is no meaning in the concept of approaching infinity from the right, correct?
\(x\to\infty^+\)
 2 years ago
 2 years ago
TuringTest Group Title
quick question: there is no meaning in the concept of approaching infinity from the right, correct? \(x\to\infty^+\)
 2 years ago
 2 years ago

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TuringTest Group TitleBest ResponseYou've already chosen the best response.2
or \(x\to\infty^\) wither, right?
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
When the variable is x, and it takes on only positive values, then it becomes positively infinite. We write \[x\to\infty^+\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
you gotta start really really far out i guess
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
http://www.themathpage.com/acalc/infinity.htm
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
@mathslover no, I think you are thinking of \(x\to+\infty\)
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
oh ... yes .
 2 years ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
I'm still trying to figure out this: \[\int\limits \frac{x \dot\ \sin x}{1 + \cos^2x} dx\]
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
I mean, infinity is not a number even how do you approach a nonnumber from a higher number no number can be higher than a nonnumber seems like asking if 7>purple
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
@Hero it has been solved, and I will be happy to give you hints should you desire let me note that that integral is unsolvalble (at least by me) as an indefinite integral, you must put the bounds
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
\(∞^\pm\) aren't even in the Extended real number line
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
good point^
 2 years ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
I want to solve it without the bounds. I already saw the solution with bounds. Since the function is continuous I will assume an indefinite integral exists.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
Good luck with that! I don't know how to solve it without making use of the interval 0 to pi that it is supposed to be
 2 years ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
Apparently no one or no thing does. Not even mathematica, geogebra, maple or TI. But not to worry, I know one person who for sure will be able to tell if it is possible. If it is integrable this guy will know. I kinda miss @JamesJ
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
me too :) @zarkon actually taught me how to solve this, but he made use of the bounds heavily
 2 years ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
I contacted TI about this today and they wanted me to provide them with the solution to this so that they could use it to update their algorithm.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
wow, far out I remember one thinbg was that we proved that\[\int_0^\infty{x\sin x\over1+\cos^2x}dx\]does not have a finite value, though I forget how we proved it. that suggests to me that no closed form of the indefinite integral exists, but I could be wrong
 2 years ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
Hmmm interesting. It may not have a finite value, but that doesn't mean it can't be solved. Last time I checked, if you take the integral of something and get sin(x) as the result....well sin(x) isn't "finite" either, but it still exists as a solution
 2 years ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
There are many indefinite integrals with solutions that are not finite.
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
true... I am stretching my capacities here admittedly
 2 years ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
Well, if by finite you mean a specific value.
 2 years ago
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