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Jonask

  • 2 years ago

Test for convergence

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

  2. Jonask
    • 2 years ago
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    improper intergral

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

  5. satellite73
    • 2 years 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
    • 2 years ago
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    yes we dont really want to compute just to say wherether it conv or not

  7. satellite73
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    \[2x-x^2>x\] \[x-x^2>0\] \[x(1-x)>0\] \[0<x<1\]

  11. Jonask
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    this is a comparison test, compare the improper integral to another improper integral that you know converges

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

  17. Spacelimbus
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    @Spacelimbus it it clear why it is \(\frac{\pi}{2}\) ? it is not clear to me

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

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

  22. Spacelimbus
    • 2 years 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
    • 2 years ago
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    \[\Large \left.\sin^{-1}(x-1)\right|^1_0 \]

  24. Jonask
    • 2 years ago
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    interesting

  25. electrokid
    • 2 years ago
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    what if you set x-1=sin theta

  26. electrokid
    • 2 years ago
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    \(x-1=\sin\theta\)

  27. Spacelimbus
    • 2 years ago
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    that's what I just did @electrokid

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

  29. electrokid
    • 2 years ago
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    @Spacelimbus lol yeah

  30. electrokid
    • 2 years ago
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    our brains were on parallel dimension there!

  31. Spacelimbus
    • 2 years ago
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    Hehe, yes, I believe we got the same idea.

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