swissgirl
Calculus: Find K
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swissgirl
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|dw:1353910548451:dw|
swissgirl
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You need to use integration by parts
hartnn
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put u= -y^2/2 then.
ivanmlerner
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Is it:\[\int\limits\limits_{-\infty}^{\infty}ke^{-\frac{y^2}{2}}dy=1\]?
swissgirl
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Yes
hartnn
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u= -y^2/2
find dy =... ?
swissgirl
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I used u=y^2
swissgirl
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but maybe its easier with u=-y^2/2
swissgirl
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Its not smart to have that negative there
swissgirl
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Do you agree?
swissgirl
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@hartnn ?
hartnn
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oh, sorry.... yeah, its easier with e^u
constants can be rearranged easily...
ivanmlerner
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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.
swissgirl
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ohhhhh
swissgirl
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Yes the answer is 1/ sqrt(2*pi)
I just so dont follow how they got that answer.
hartnn
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did u try that substitution and then integration by parts ?
swissgirl
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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
hartnn
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\(\huge\int \frac{e^{-u}u^{-1/2}}{\sqrt2}du\)
when 2u=y^2
yes, its messy...
swissgirl
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How would you continue from here?
swissgirl
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Can you show me?
swissgirl
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ok dont show me
hartnn
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u need to know gamma function to solve that ...
i am so sorry, my internet are giving me connection problems....
swissgirl
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hmmm I know the gamma distribution function
swissgirl
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ohhhh I see it now lol
hartnn
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\( \Gamma(z) = \int_0^\infty e^{-t} t^{z-1} dt \)
swissgirl
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Yesssssss
hartnn
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now , you'll get it easily.
hartnn
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ask if u don't get...
swissgirl
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Thanks let me try to work it out then. Thanks :)))))
swissgirl
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what?
hartnn
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\(\Gamma{1/2} = \sqrt{\pi}\)
hartnn
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you'll need it.
swissgirl
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ohhh coool never knew that
swissgirl
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I feel like i am missing the basics in math :(