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Math Help - Regression of series equation

  1. #1
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    Regression of series equation

    Hello to everbody,
    I wave an equation with datas and want to regress in non linear form with plot.
    Would you help me to solve the equation bellow by any softwares, please?

    The data, the equation and parameters are given bellow. I want to fit the data to non-linear equation to find D and k values.
    I just need help.
    Your sincerely,,

    Equation:
    M(t)=Mo - (Mo - Ms)*Pi*8*R*Ms*D [Sum((k*R^2 + Pi^2*D n^2)*k*R^2*t -Pi^2*R^2*D n^2 - (Pi^2*R^2*D n^2*
    Exp[-t*(k*R^2 + D*Pi^2 n^2)/R^2])/(k*R^2 + n^2*D*Pi^2)^2]

    [Sum] goas from 1 to 1000

    Ms=140
    Mo=11.58
    R=0.004

    Variables and Data
    t M(t)
    0 11.58
    30 72.61
    60 97.89
    90 106.05
    120 109.63
    150 117.02
    180 119.57
    210 125.99
    240 129.56
    270 132.85
    Last edited by ayildirim; November 30th 2008 at 11:44 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ayildirim View Post
    Hello to everbody,
    I wave an equation with datas and want to regress in non linear form with plot.
    Would you help me to solve the equation bellow by any softwares, please?

    The data, the equation and parameters are given bellow. I want to fit the data to non-linear equation to find a and k values.
    I just need help.
    Your sincerely,,

    Equation:
    M(t)=Mo - (Mo - Ms)*Pi*8*R*Ms*D [Sum((k*R^2 + Pi^2*D n^2)*k*R^2*t -Pi^2*R^2*D n^2 - (Pi^2*R^2*D n^2*
    Exp[-t*(k*R^2 + D*Pi^2 n^2)/R^2])/(k*R^2 + n^2*D*Pi^2)^2]

    [Sum] goas from 1 to 1000

    Ms=140
    Mo=11.58
    R=0.004

    Variables and Data
    t M(t)
    0 11.58
    30 72.61
    60 97.89
    90 106.05
    120 109.63
    150 117.02
    180 119.57
    210 125.99
    240 129.56
    270 132.85
    Try using the solver shipped with Excel.

    CB
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  3. #3
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    Quote Originally Posted by ayildirim View Post

    Equation:
    M(t)=Mo - (Mo - Ms)*Pi*8*R*Ms*D [Sum((k*R^2 + Pi^2*D n^2)*k*R^2*t -Pi^2*R^2*D n^2 - (Pi^2*R^2*D n^2*
    Exp[-t*(k*R^2 + D*Pi^2 n^2)/R^2])/(k*R^2 + n^2*D*Pi^2)^2]

    [Sum] goas from 1 to 1000

    Ms=140
    Mo=11.58
    R=0.004

    Variables and Data
    t M(t)
    0 11.58
    30 72.61
    60 97.89
    90 106.05
    120 109.63
    150 117.02
    180 119.57
    210 125.99
    240 129.56
    270 132.85
    You got undefined D and I don't see any a in there.
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  4. #4
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    Here D and k are constans to be evaluated.

    Equation:
    M(t)=Mo - (Mo - Ms)*Pi*8*R*Ms*D [Sum((k*R^2 + Pi^2*D n^2)*k*R^2*t -Pi^2*R^2*D n^2 - (Pi^2*R^2*D n^2*
    Exp[-t*(k*R^2 + D*Pi^2 n^2)/R^2])/(k*R^2 + n^2*D*Pi^2)^2]

    [Sum] goas from 1 to 1000

    Ms=140
    Mo=11.58
    R=0.004
    Variables and Data
    t M(t)
    0 11.58
    30 72.61
    60 97.89
    90 106.05
    120 109.63
    150 117.02
    180 119.57
    210 125.99
    240 129.56
    270 132.85
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  5. #5
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    Hey is this it:

    M(t)=Mo-C_1\sum _{n=0}^{1000} k R^2 t \left(\pi ^2 d n^2+k R^2\right)-\frac{\pi ^2 D n^2 R^2 \exp \left(-\frac{t \left(\pi ^2 D n^2+k R^2\right)}{R^2}\right)}{\left(\pi ^2 D n^2+k R^2\right)^2}-\pi ^2 D n^2 R^2<br />

    C_1=\pi  8 D \text{Ms} R (\text{Mo}-\text{Ms})

    Ms=140,\; Mo=11.58,\;R=0.004

    with D and k to be determined by the fit?
    Last edited by shawsend; December 1st 2008 at 03:15 AM.
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  6. #6
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    regression

    My question on attacment.
    Attached Files Attached Files
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  7. #7
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    Quote Originally Posted by ayildirim View Post
    My question on attacment.
    Ayildirim, that's harder to read than my equation above and also it looks to have another variable M_s or something other. Is the equation I posted the correct one and if not, what corrections have to be made.
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  8. #8
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    In the equation that you send me, the division is only related to exp term. This is wrong. The division have to be for over all terms in Sum function.
    Also the limits of sum function must be from 1 to 1000.
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  9. #9
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    Ok, this is what I did then:

    Using:

    M(t)=Mo-C_1\sum _{n=1}^{1000} \frac{ k R^2 t \left(\pi ^2 d n^2+k R^2\right)-\pi ^2 d n^2 R^2 \exp \left(-\frac{t \left(\pi ^2 d n^2+k R^2\right)}{R^2}\right)-\pi ^2 d n^2+k R^2}{(\pi^2 d n^2+k R^2)^2}

    (changed D to d since D means derivative in Mathematica)

    C_1=\pi  8 d \text{Ms} R (\text{Mo}-\text{Ms})

    Ms=140,\; Mo=11.58,\;R=0.004

    I wrote the following Mathematica code to try and fit the data to your expression. Unfortunately, Mathematica returns a negative value for d which quickly causes overflow with the sum (even with smaller upper limits on the sum overflow occurs):

    Code:
    vals={{0,11.58},{30,72.61},{60,97.89},
    {90,106.05},{120,109.63},{150,117.02},
    {180,119.57},{210,125.99},{240,129.56},
    {270,132.85}}; 
    lp = ListPlot[vals, PlotRange -> {{0, 200}, {0, 300}}]
    coef = FindFit[vals, M[t], {d, k}, t]
    p1 = Plot[f[t] /. coef, {t, 0, 3}]
    Mathematica returns:

    {d -> -1.84547*10^8, k -> 122.325}

    Oh yeah, if the equation is still not correct, click on it. A small window will appear with the LaTex code I wrote to generate the math code. You can understand it. \frac means fraction in french brackets, power is ^, so forth. with the [tex] brackets starting and ending it. Try and cut and paste it into a new post and edit the code as you wish even if it comes out messy. It's a start and you should try posting in Latex so everything is neat, well defined and explicit so people will follow it easily.
    Last edited by shawsend; December 2nd 2008 at 05:50 AM.
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  10. #10
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    regression of series equation

    After Mo in the general equation (+) sign can be changed to (-) sign.
    like this

    M(t)=Mo+C1*.........................
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  11. #11
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    Quote Originally Posted by ayildirim View Post
    After Mo in the general equation (+) sign can be changed to (-) sign.
    like this

    M(t)=Mo+C1*.........................
    Hey ayildirim. That doesn't help. Still getting either underflow or overflow depending on how many terms I use for the sum.
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