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TuringTest Group Title

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

  • one year ago
  • one year ago

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  1. TuringTest Group Title
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    or \(x\to-\infty^-\) wither, right?

    • one year ago
  2. TuringTest Group Title
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    either*

    • one year ago
  3. mathslover Group Title
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    When the variable is x, and it takes on only positive values, then it becomes positively infinite. We write \[x\to\infty^+\]

    • one year ago
  4. satellite73 Group Title
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    you gotta start really really far out i guess

    • one year ago
  5. mathslover Group Title
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    http://www.themathpage.com/acalc/infinity.htm

    • one year ago
  6. TuringTest Group Title
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    @mathslover no, I think you are thinking of \(x\to+\infty\)

    • one year ago
  7. mathslover Group Title
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    oh ... yes .

    • one year ago
  8. Hero Group Title
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    I'm still trying to figure out this: \[\int\limits \frac{x \dot\ \sin x}{1 + \cos^2x} dx\]

    • one year ago
  9. TuringTest Group Title
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    I mean, infinity is not a number even how do you approach a non-number from a higher number no number can be higher than a non-number seems like asking if 7>purple

    • one year ago
  10. TuringTest Group Title
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    @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

    • one year ago
  11. UnkleRhaukus Group Title
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    \(∞^\pm\) aren't even in the Extended real number line

    • one year ago
  12. TuringTest Group Title
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    good point^

    • one year ago
  13. Hero Group Title
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    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.

    • one year ago
  14. TuringTest Group Title
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    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

    • one year ago
  15. Hero Group Title
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    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

    • one year ago
  16. TuringTest Group Title
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    me too :) @zarkon actually taught me how to solve this, but he made use of the bounds heavily

    • one year ago
  17. Hero Group Title
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    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.

    • one year ago
  18. TuringTest Group Title
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    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

    • one year ago
  19. Hero Group Title
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    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

    • one year ago
  20. Hero Group Title
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    There are many indefinite integrals with solutions that are not finite.

    • one year ago
  21. TuringTest Group Title
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    true... I am stretching my capacities here admittedly

    • one year ago
  22. Hero Group Title
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    Well, if by finite you mean a specific value.

    • one year ago
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