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swissgirl

  • 2 years ago

Calculus: Find K

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  1. swissgirl
    • 2 years ago
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    |dw:1353910548451:dw|

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

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

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

  5. swissgirl
    • 2 years ago
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    Yes

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

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

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

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

  10. swissgirl
    • 2 years ago
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    Do you agree?

  11. swissgirl
    • 2 years ago
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    @hartnn ?

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

  13. ivanmlerner
    • 2 years ago
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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.

  14. swissgirl
    • 2 years ago
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    ohhhhh

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

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

  17. swissgirl
    • 2 years ago
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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

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

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

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

  21. swissgirl
    • 2 years ago
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    ok dont show me

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

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

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

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

  26. swissgirl
    • 2 years ago
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    Yesssssss

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

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

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

  30. swissgirl
    • 2 years ago
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    what?

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

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

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

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

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

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