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richyw

  • 3 years ago

improper integral. sorry for all of these. my textbook thinks that "converges" is a suitable explanation. \[\int^{\infty}_0 \tan\frac{1}{x}\text{d}x\]

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  1. richyw
    • 3 years ago
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    I need to say if it converges or diverges.

  2. abb0t
    • 3 years ago
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    Take the integral as you would from [0, R]

  3. anonymous
    • 3 years ago
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    can you change variables to \(u=\frac{1}{x}\) and then flip the limits of integration?

  4. abb0t
    • 3 years ago
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    Then take the limit of the function as R->∞

  5. richyw
    • 3 years ago
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    is this sufficient to say it converges? \[\lim_{x \to \infty} \tan\frac{1}{x}=0\]

  6. anonymous
    • 3 years ago
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    how?

  7. anonymous
    • 3 years ago
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    thing thing has an infinite number of singularities

  8. electrokid
    • 3 years ago
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    how about the substitution \[x = \tan^{-1}u\]

  9. richyw
    • 3 years ago
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    oh right.

  10. electrokid
    • 3 years ago
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    or \[x=\cot^{-1}(u)\]

  11. anonymous
    • 3 years ago
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    the problem isn't just that the path is infinite,

  12. anonymous
    • 3 years ago
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    it is undefined infinitely often

  13. klimenkov
    • 3 years ago
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    What about \(\tan\frac1x\sim\frac1x,\, x\rightarrow\infty\) ?

  14. electrokid
    • 3 years ago
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    yep!! convergence....

  15. anonymous
    • 3 years ago
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    i think there is a mistake here there is no way this thing converges

  16. electrokid
    • 3 years ago
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    \[\lim_{x\to0}\tan(1/x)=\infty\]

  17. richyw
    • 3 years ago
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    oh wait a second, it does diverge.

  18. anonymous
    • 3 years ago
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    graph it and see what it looks like, then tell me it converges...

  19. electrokid
    • 3 years ago
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    @satellite73 can we say that as \(x\to0\), it is infinite or is it undefined?

  20. richyw
    • 3 years ago
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    but the limits iof integration are supposed to be from 1 to infty. so ignore the zero

  21. anonymous
    • 3 years ago
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    makes no difference, the problem is not the infinite path tangent is undefined infinitely often

  22. klimenkov
    • 3 years ago
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    Use that \(\tan\frac1x\sim\frac1x,\, x\rightarrow\infty\) and immediately say that it diverges.

  23. electrokid
    • 3 years ago
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    @richyw it also has infinite "vertical asymptotes" and hence it is discontinuous over the interval and cannot be integrated!

  24. electrokid
    • 3 years ago
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    @satellite73 so, you are saying that the function cannot exist to begin with?

  25. richyw
    • 3 years ago
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    @klimenkov is that just the first term of the taylor expansion at infinity?

  26. klimenkov
    • 3 years ago
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    If you want Taylor series - yes. But from \(\tan x\sim x,\,x\rightarrow0\) implies \(\tan\frac1x\sim\frac1x,\, x\rightarrow\infty\).

  27. richyw
    • 3 years ago
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    how am I supposed to know that \(\tan x \sim x,\;x\to 0\) ?

  28. klimenkov
    • 3 years ago
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    Compute\[\lim_{x\rightarrow0}\frac{\tan x}{x}\]

  29. richyw
    • 3 years ago
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    AWSOME. ok this is what I was struggling with last night.

  30. richyw
    • 3 years ago
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    why does doing that show that they behave the same?

  31. richyw
    • 3 years ago
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    that step is what really confuses me!

  32. richyw
    • 3 years ago
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    same step was used in the second line of this. I don't understand why I do this!

  33. klimenkov
    • 3 years ago
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    Try to check this theorem in your calculus book.

  34. richyw
    • 3 years ago
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    believe me I have tried. I even emailed my prof asking where the theorem is. does it have a name?

  35. klimenkov
    • 3 years ago
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    Ok. Did you understand why \(\tan\frac1x\sim\frac1x,\, x\rightarrow\infty\) ?

  36. richyw
    • 3 years ago
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    no, because I don't understand why \(\tan x \sim x,\;x\to 0\)

  37. klimenkov
    • 3 years ago
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    Do you know that \[\lim_{x\rightarrow0}\frac{\sin x}{x}=1\]

  38. richyw
    • 3 years ago
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    yes

  39. klimenkov
    • 3 years ago
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    Now \[\lim_{x\rightarrow0}\frac{\tan x}{x}=\lim_{x\rightarrow0}\frac{\frac{\sin x}{\cos x}}{x}=\lim_{x\rightarrow0}\frac{\sin x}{x}=1\]because \(\lim_{x\rightarrow0}\cos x=1\). Got it?

  40. richyw
    • 3 years ago
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    yes, I can compute the limit.

  41. richyw
    • 3 years ago
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    but I don't understand its significance

  42. klimenkov
    • 3 years ago
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    Ah. Ok. If \(f(x)\sim g(x),\,x\rightarrow a\), it means that \(f(x)\) behaves like \(g(x)\) at the point \(x=a\).

  43. richyw
    • 3 years ago
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    right. do you know why?

  44. richyw
    • 3 years ago
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    I mean why the limit thing, not that definition

  45. klimenkov
    • 3 years ago
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    Or \[\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=1\]

  46. richyw
    • 3 years ago
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    AHHH that's so obvious!

  47. klimenkov
    • 3 years ago
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    It is equivalent definitions.

  48. richyw
    • 3 years ago
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    omg. thank you so much.

  49. richyw
    • 3 years ago
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    I feel so stupid now, but if you looked at my questions history I spent like 3 hours last night trying to get this explanation!

  50. richyw
    • 3 years ago
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    thank you so much!

  51. klimenkov
    • 3 years ago
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    Don't be shy to ask. It is normal not to see obvious things because obvious is a relative concept. I wish you a lot of luck in learning math. Love it.

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