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swissgirl Group Title

Calculus: Find K

  • 2 years ago
  • 2 years ago

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  1. swissgirl Group Title
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    |dw:1353910548451:dw|

    • 2 years ago
  2. swissgirl Group Title
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    You need to use integration by parts

    • 2 years ago
  3. hartnn Group Title
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    put u= -y^2/2 then.

    • 2 years ago
  4. ivanmlerner Group Title
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    Is it:\[\int\limits\limits_{-\infty}^{\infty}ke^{-\frac{y^2}{2}}dy=1\]?

    • 2 years ago
  5. swissgirl Group Title
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    Yes

    • 2 years ago
  6. hartnn Group Title
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    u= -y^2/2 find dy =... ?

    • 2 years ago
  7. swissgirl Group Title
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    I used u=y^2

    • 2 years ago
  8. swissgirl Group Title
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    but maybe its easier with u=-y^2/2

    • 2 years ago
  9. swissgirl Group Title
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    Its not smart to have that negative there

    • 2 years ago
  10. swissgirl Group Title
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    Do you agree?

    • 2 years ago
  11. swissgirl Group Title
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    @hartnn ?

    • 2 years ago
  12. hartnn Group Title
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    oh, sorry.... yeah, its easier with e^u constants can be rearranged easily...

    • 2 years ago
  13. ivanmlerner Group Title
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    I got the wrong answer when evaluating the integral, and I won't do it again now. But this function is a gaussian, and it is expected that the constant k in this case has the value 1/sqrt(2pi)=sqrt(2pi)/2pi. That comes from the statistic theory.

    • 2 years ago
  14. swissgirl Group Title
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    ohhhhh

    • 2 years ago
  15. swissgirl Group Title
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    Yes the answer is 1/ sqrt(2*pi) I just so dont follow how they got that answer.

    • 2 years ago
  16. hartnn Group Title
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    did u try that substitution and then integration by parts ?

    • 2 years ago
  17. swissgirl Group Title
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    I was getting weird answers. I have not really touched calculus for over a yr so i cld be messing up

    • 2 years ago
  18. hartnn Group Title
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    \(\huge\int \frac{e^{-u}u^{-1/2}}{\sqrt2}du\) when 2u=y^2 yes, its messy...

    • 2 years ago
  19. swissgirl Group Title
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    How would you continue from here?

    • 2 years ago
  20. swissgirl Group Title
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    Can you show me?

    • 2 years ago
  21. swissgirl Group Title
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    ok dont show me

    • 2 years ago
  22. hartnn Group Title
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    u need to know gamma function to solve that ... i am so sorry, my internet are giving me connection problems....

    • 2 years ago
  23. swissgirl Group Title
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    hmmm I know the gamma distribution function

    • 2 years ago
  24. swissgirl Group Title
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    ohhhh I see it now lol

    • 2 years ago
  25. hartnn Group Title
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    \( \Gamma(z) = \int_0^\infty e^{-t} t^{z-1} dt \)

    • 2 years ago
  26. swissgirl Group Title
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    Yesssssss

    • 2 years ago
  27. hartnn Group Title
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    now , you'll get it easily.

    • 2 years ago
  28. hartnn Group Title
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    ask if u don't get...

    • 2 years ago
  29. swissgirl Group Title
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    Thanks let me try to work it out then. Thanks :)))))

    • 2 years ago
  30. swissgirl Group Title
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    what?

    • 2 years ago
  31. hartnn Group Title
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    \(\Gamma{1/2} = \sqrt{\pi}\)

    • 2 years ago
  32. hartnn Group Title
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    you'll need it.

    • 2 years ago
  33. swissgirl Group Title
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    ohhh coool never knew that

    • 2 years ago
  34. swissgirl Group Title
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    I feel like i am missing the basics in math :(

    • 2 years ago
  35. hartnn Group Title
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    http://en.wikipedia.org/wiki/Gamma_function

    • 2 years ago
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