Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

TuringTest

  • 2 years ago

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

  • This Question is Closed
  1. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    or \(x\to-\infty^-\) wither, right?

  2. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    either*

  3. mathslover
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    When the variable is x, and it takes on only positive values, then it becomes positively infinite. We write \[x\to\infty^+\]

  4. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you gotta start really really far out i guess

  5. mathslover
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    http://www.themathpage.com/acalc/infinity.htm

  6. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @mathslover no, I think you are thinking of \(x\to+\infty\)

  7. mathslover
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh ... yes .

  8. Hero
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm still trying to figure out this: \[\int\limits \frac{x \dot\ \sin x}{1 + \cos^2x} dx\]

  9. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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

  10. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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

  11. UnkleRhaukus
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \(∞^\pm\) aren't even in the Extended real number line

  12. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    good point^

  13. Hero
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  14. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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

  15. Hero
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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

  16. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    me too :) @zarkon actually taught me how to solve this, but he made use of the bounds heavily

  17. Hero
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  18. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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

  19. Hero
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 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

  20. Hero
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    There are many indefinite integrals with solutions that are not finite.

  21. TuringTest
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    true... I am stretching my capacities here admittedly

  22. Hero
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Well, if by finite you mean a specific value.

  23. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.