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Jonask

  • one year ago

Test for convergence

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  1. Jonask
    • one year ago
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    \[\int_0^1\frac{dx}{\sqrt{2x-x^2}}\]

  2. Jonask
    • one year ago
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    improper intergral

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

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

  5. satellite73
    • one year ago
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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

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

  7. satellite73
    • one year ago
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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 \[\frac{1}{\sqrt{2x-x^2}}<\frac{1}{\sqrt{x}}\]

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

  9. satellite73
    • one year ago
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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 \(2x-x^2>x\)

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

  11. Jonask
    • one year ago
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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

  12. satellite73
    • one year ago
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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

  13. Spacelimbus
    • one year ago
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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?

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

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

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

  17. Spacelimbus
    • one year ago
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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.

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

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

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

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

  22. Spacelimbus
    • one year ago
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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 \[\Large \int 1 d\alpha \]

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

  24. Jonask
    • one year ago
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    interesting

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

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

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

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

  29. electrokid
    • one year ago
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    @Spacelimbus lol yeah

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

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

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