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Jonask Group Title

Test for convergence

  • one year ago
  • one year ago

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  1. Jonask Group Title
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    \[\int_0^1\frac{dx}{\sqrt{2x-x^2}}\]

    • one year ago
  2. Jonask Group Title
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    improper intergral

    • one year ago
  3. Spacelimbus Group Title
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    Not sure how much that helps but: \[\Large -x^2+2x=-(x^2-2x-1+1)=-(x-1)^2+1 \]

    • one year ago
  4. Spacelimbus Group Title
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    \[\Large \int\frac{dx}{\sqrt{1-(x-1)^2}} \]

    • one year ago
  5. satellite73 Group Title
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    i think you are not going to find a nice closed form for this one, although you can if you try very hard if you are not trying to actually compute the integral, but only say whether or not it converges, then it is much easier

    • one year ago
  6. Jonask Group Title
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    yes we dont really want to compute just to say wherether it conv or not

    • one year ago
  7. satellite73 Group Title
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    you can assert that on the interval \((0,1)\) you have \[2x-x^2>x\] so \[\sqrt{2x-x^2}>\sqrt{x}\] and therefore \[\frac{1}{\sqrt{2x-x^2}}<\frac{1}{\sqrt{x}}\]

    • one year ago
  8. satellite73 Group Title
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    since \[\int_0^1\frac{dx}{\sqrt{x}}=2\] your integral is less than or equal to 2

    • one year ago
  9. satellite73 Group Title
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    i am pretty sure that is what you want to do, because the problem was cooked up so that on \((0,1)\) you have \(2x-x^2>x\)

    • one year ago
  10. satellite73 Group Title
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    \[2x-x^2>x\] \[x-x^2>0\] \[x(1-x)>0\] \[0<x<1\]

    • one year ago
  11. Jonask Group Title
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    so the intergrand becomes infinite when x=0 and x=2 but 2 is of no importance since its not in the interval

    • one year ago
  12. satellite73 Group Title
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    wolfram tells me the integral is in fact \(\frac{\pi}{2}\) which is certainly less than 2, but the closed form is a big fat mess and i can't imagine trying to find it

    • one year ago
  13. Spacelimbus Group Title
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    I thought that this problem relates to the integral test. It should be possible to compute it, or did I mess up with the algebra?

    • one year ago
  14. satellite73 Group Title
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    integral test is usually used to determine if an infinite series converges, and you use \[\int_a^{\infty}f(x)dx\]

    • one year ago
  15. satellite73 Group Title
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    this is a comparison test, compare the improper integral to another improper integral that you know converges

    • one year ago
  16. Jonask Group Title
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    yes this is what the book says thank you bot @Spacelimbus and @satellite73

    • one year ago
  17. Spacelimbus Group Title
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    Oh I see that now @satellite73 , I just thought because by computing it you also get the value of 90° or pi/2. But I never encountered comparing the integrals :-) Interesting problem anyway.

    • one year ago
  18. satellite73 Group Title
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    i.e. since \(\frac{1}{\sqrt{2x-x^2}}<\frac{1}{\sqrt{x}}\) you have \[\int_0^1\frac{dx}{\sqrt{2x-x^2}}\leq\int_0^1\frac{dx}{\sqrt{x}}=2\]

    • one year ago
  19. satellite73 Group Title
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    @Spacelimbus it it clear why it is \(\frac{\pi}{2}\) ? it is not clear to me

    • one year ago
  20. satellite73 Group Title
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    somehow it gives a quarter of the circle, but i don't see it

    • one year ago
  21. Jonask Group Title
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    well i see a primitive of the form\[ \sin^{-1}x\]

    • one year ago
  22. Spacelimbus Group Title
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    Well, as I said, I was just naively computing it - I am pretty sure however that your way is more accurate. \[\Large \int \frac{1}{\sqrt{1-(x-1)^2}}dx \] because: \[\Large -x^2+2x=-(x-1)^2+1 \] Substitute u=sinalpha and then you get a simple integral \[\Large \int 1 d\alpha \]

    • one year ago
  23. Spacelimbus Group Title
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    \[\Large \left.\sin^{-1}(x-1)\right|^1_0 \]

    • one year ago
  24. Jonask Group Title
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    interesting

    • one year ago
  25. electrokid Group Title
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    what if you set x-1=sin theta

    • one year ago
  26. electrokid Group Title
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    \(x-1=\sin\theta\)

    • one year ago
  27. Spacelimbus Group Title
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    that's what I just did @electrokid

    • one year ago
  28. electrokid Group Title
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    it'd lead to the same result as @spacelimbus

    • one year ago
  29. electrokid Group Title
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    @Spacelimbus lol yeah

    • one year ago
  30. electrokid Group Title
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    our brains were on parallel dimension there!

    • one year ago
  31. Spacelimbus Group Title
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    Hehe, yes, I believe we got the same idea.

    • one year ago
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