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Numerical Integration

  1. Aug 8, 2008 #1
    Hello

    I have the following question in numerical integration in higher dimension. Any help/suggestion would be welcome.

    The integral is ( I am using maple notation ):

    int( int( int( int( int( int( f(x,y,z), z=-infinity..w), w=-infinity..infinity),
    y=-infinity..v), v=-infinity..infinity),
    x=-infinity..u), u=-infinity..infinity);

    The integral is convergent. Indeed f(x,y,z) = K*exp(-( x^2+(y-x)^2+(z-y)^2) )/2), where K is a given constant.

    Thanks in advance

    Kind regards
    hpriye
     
  2. jcsd
  3. Aug 8, 2008 #2

    Borek

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    Staff: Mentor

    And you are asking about... ?

    Could be I am missing something obvious, it happens, but I see only statements, no questions.
     
  4. Aug 8, 2008 #3

    HallsofIvy

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    What is your question? Surely it is not just "how do you do a numerical integration"! That is much too broad.
     
  5. Aug 8, 2008 #4
    I am sorry, I thought it obvious that I was asking for a numerical method for evaluating the 6 dimension integral.

    Many thanks in advance
    hpriye
     
  6. Aug 8, 2008 #5

    arildno

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    Dearly Missed

    Hmm..I would first restrict the infinity to a radial variable, i.e, switch to spherical coordinates and see what simplifications, if any, that would give me.
     
  7. Aug 8, 2008 #6
    Ok. I tried the following transformation:

    x = r*cos(theta)*cos(phi)
    y = r*cos(theta)*(cos(phi) + sin(phi))
    z = r*cos(theta)*(cos(phi) + sin(phi)) + r*sin(theta)

    and the Jacobian is -r^2*cos(theta). This means I am retaining the variables u,v,w as they are.

    Now the question: what are the limits for the six variables: r, theta, phi, u, v, w?

    Another question: Is there a transformation for removing u, v, w from being present in the limits?

    Kind regards and many thanks
    hpriye


     
  8. Aug 8, 2008 #7

    tiny-tim

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    Hi hpriye! :smile:

    The integral is over all x and u with -∞ < x < u < ∞, so:

    [tex]\int_{-\infty}^{\infty} \int_{-\infty}^u f(x,y,x) dx du\ =\ \int_{-\infty}^{\infty} f(x,y,x)\left(\int_x^{\infty} du\right) dx\ =\ \infty[/tex]

    or have I misunderstood the question? :confused:
     
  9. Aug 8, 2008 #8
    More than misunderstanding I would say misreading :(

    hpriye


     
  10. Aug 8, 2008 #9
    i would use monte carlo integration. how about you write out your integral in tex so we can all understand it
     
  11. Aug 8, 2008 #10

    tiny-tim

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    Nope … can't be misreading …

    'cos I don't read maple! :biggrin:

    … is it canadian? :confused:

    … i know tex is texan … :rolleyes:
     
  12. Aug 9, 2008 #11
    ∫_(-∞)^∞▒∫_(-∞)^u▒∫_(-∞)^∞▒∫_(-∞)^v▒∫_(-∞)^∞▒∫_(-∞)^w▒〖f(x,y,z) dz dw dy dv dx du〗

    f(x,y,z)=1/〖√2π〗^3 e^(-1/2(x^2+(y-x)^2+(z-y)^2))-1/√6 e^(-x^2/2-y^2/4-z^2/6)

    I have tried to give the MsWord 2007 format. Hope it is readable! Underscore stands for subscript and ^ stands for superscript

    Kind regards
    hpriye
     
  13. Aug 9, 2008 #12

    tiny-tim

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    in texas we say …

    Hi hpriye! :smile:

    If you mean [tex]\int_{-\infty}^{\infty}\int_{-\infty}^u\int_{-\infty}^{\infty}\int_{-\infty}^v\int_{-\infty}^{\infty}\int_{-\infty}^w f(x,y,z)\,dz dw dy dv dx du[/tex]

    [tex]f(x,y,z)\ =\ \frac{1}{(2\pi)^3} e^{-1/2(x^2+(y-x)^2+(z-y)^2)}\ -\ \frac{1}{\sqrt{6}}e^{-(\frac{1}{2}x^2\,-\,\frac{1}{4}y^2\,-\,\frac{1}{6}z^2)}[/tex]

    it's:

    [noparse][tex]\int_{-\infty}^{\infty}\int_{-\infty}^u\int_{-\infty}^{\infty}\int_{-\infty}^v\int_{-\infty}^{\infty}\int_{-\infty}^w f(x,y,z)\,dz dw dy dv dx du[/tex]

    [tex]f(x,y,z)\ =\ \frac{1}{(2\pi)^3} e^{-1/2(x^2+(y-x)^2+(z-y)^2)}\ -\ \frac{1}{\sqrt{6}}e^{-(\frac{1}{2}x^2\,-\,\frac{1}{4}y^2\,-\,\frac{1}{6}z^2)}[/tex][/noparse]
     
  14. Aug 9, 2008 #13
    Re: in texas we say …

    Yes, exactly. But I observed square root symbol is missing, even in the original typesetting. It should be (sqrt(2*pi))^3. Thank you very much!

    Kind regards
    hpriye

     
  15. Aug 9, 2008 #14

    tiny-tim

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    Re: in texas we say …

    hmm … so it's:

    [tex]\int_{-\infty}^{\infty}\int_{-\infty}^u\int_{-\infty}^{\infty}\int_{-\infty}^v\int_{-\infty}^{\infty}\int_{-\infty}^w f(x,y,z)\,dz dw dy dv dx du[/tex]

    [tex]f(x,y,z)\ =\ \frac{1}{(2\pi)^\frac{3}{2}} e^{-1/2(x^2+(y-x)^2+(z-y)^2)}\ -\ \frac{1}{\sqrt{6}}e^{-(\frac{1}{2}x^2\,-\,\frac{1}{4}y^2\,-\,\frac{1}{6}z^2)}[/tex]

    with an extra:

    [noparse][tex]\frac{3}{2}[/tex][/noparse] :wink:
     
  16. Aug 9, 2008 #15
    Hi hypre.

    As you already know this task is not so simple to solve because of multi-integration. First it is very useful and not so painful see whether first integral exists: :smile:

    [tex] \Theta(u,x,v,y,w)=\int_{-\infty}^{w} f(u,x,v,y,w,z) dz = \frac{1}{2}e^{-\frac{x^2}{2}-\frac{y^3}{4}}\;\sqrt{\pi} \left(1+Erf(\frac{w}{\sqrt{6}}) \right) +
    \frac{1}{4\pi}\; e^{-x^2+xy-\frac{y^2}{2}} \left(1+Erf(\frac{w-y}{\sqrt{2}}) \right)
    [/tex]

    Integrals would exist if [tex]u,v,w \in (-\infty,0) [/tex] otherwise integrals diverge !
    You can see that if above integral split on two integrals
    [tex] \int_{-\infty}^{0} \Theta(u,x,v,y,w)\; dw = 0.586571 + O(10^{-6}) [/tex]
    [tex]\int_{0}^{\infty} \Theta(u,x,v,y,w)\; dw \rightarrow \bf{\infty} [/tex]

    Trick how to see whether this exist to plot it. Put [tex] x=y=1 [/tex] as you already know [tex] x \in (-\infty,\infty)[/tex] and [tex]y \in (-\infty,\infty)[/tex]. I made also graph so see what is the shape of function and is attached to this answer.

    If you don't understand just send me email and I will try to explain.

    Enjoy :smile:
    MrSnoopy
     

    Attached Files:

  17. Aug 9, 2008 #16
    Sorry I forgot to mention numerical solution [tex]= 0.586571 + ...[/tex] comes form [tex]x=y=1[/tex]. It is only to see that first has numerical solution other diverge.
     
  18. Aug 12, 2008 #17
    Thankyou very much for your efforts!

    One important aspect of this integral is this: It belongs to a class of integrals which apparantly diverge, but not in the real sense.

    For example define two functions
    [tex] f(x)\ =\ x^2 + e^{-x^2} \ [/tex] and
    [tex] g(x)\ =\ x^2 + e^{-2 x^2} \ [/tex]
    then individually the two integrals:
    [tex] \int_{-\infty}^\infty f(x)\ dx [/tex] and [tex] \int_{-\infty}^\infty g(x)\ dx [/tex] diverge but [tex] \int_{-\infty}^\infty (f(x)-g(x))\ dx = \sqrt{\pi} - \sqrt{\frac{\pi}{2}} [/tex]

    Similarly, the six-dimensional integral has another 3 dimension integral in the denominator (which I have not posted or mentioned earlier) such that the ratio is a finite number, in a limiting sense. I know this is a very very complicated integral, which naturally arises while solving problems in financial mathematics, so what I am interested in is, a numerical method to handle such integrals involving limiting case:

    [tex] \lim_{a \rightarrow \infty} \left(\frac{\int_{-a}^a \int_{-a}^u \int_{-a}^a \int_{-a}^v \int_{-a}^a \int_{-a}^w \nu(x,y,z) \ dw\ dz\ dv\ dy\ dx\ du}{\int_{-a}^a \int_{-a}^a \int_{-a}^a \delta(x,y,z) \ dz\ dy\ dx} \right) [/tex]

    Kind regards and thanks to one and all.
    hpriye
     
    Last edited: Aug 12, 2008
  19. Aug 12, 2008 #18
    No problem.

    I would solved it by Simpsons method, has good accuracy at low number of steps. Don't forget functions are very quick :smile:
     
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