Math Help Forum - View Single Post

shawsend · #9 December 2nd, 2008, 07:31 AM

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 [math] 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.