Re: Trouble with FindRoot
- To: mathgroup at smc.vnet.net
- Subject: [mg79246] Re: Trouble with FindRoot
- From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
- Date: Sun, 22 Jul 2007 04:17:45 -0400 (EDT)
- Organization: The Open University, Milton Keynes, UK
- References: <f7sh9g$t29$1@smc.vnet.net>
chuck009 wrote:
>> Sooraj R wrote:
>
>>> Now for the initial guess for second row, I need to
>>> use the first row as the initial guess, and so on.
>>> How do I do this?
>
> Jean-Marc, Looks like your code is not doing this. I modified it to do so:
Hi,
I plea guilty: I forgot to reply to the second part of the OP's query.
Regards,
Jean-Marc
> eh = Table[Eh, {10}, {16}];
>
> m = 3.85857975 10^-31;
> charge = 1.602 10^-19;
> hbar = 1.054571 10^-34;
> LH = {88.83663979522105,
> 93.83693172861489, 98.83720928295747, 103.83746882256844, \
> 108.83770913229637, 113.837930338503, 118.83813328307562, 123.83831915913095, \
> 128.83848929997706, 133.8386450587284, 138.83878774185274, 143.8389185748323, \
> 148.8390386868427, 153.83914910653547, 158.83925076412058, \
> 163.83934449683832};
> F = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} 10^6;
> ZeroField = Table[0, {16}];
> For[i = 1, i < 17, i++, ZeroField[[i]] = (Pi hbar)^2/(2 m charge (LH[[i]]
> 10^-10)^2)]
>
> (*Added Print function so ZeroField is displayed*)
> Print[ZeroField // MatrixForm];
> (* now do the first row *)
>
> For[i = 1, i < 17, i++, eh[[1, i]] = Eh /. FindRoot[AiryAi[-((2*
> m)/(charge*hbar*F[[j]])^2)^(1/3) (Eh
> charge + charge*F[[j]]*(LH[[i]] 10^-10)/2)] AiryBi[-((2*m)/(charge*hbar*
> F[[j]])^2)^(1/3) (Eh charge - charge*F[[j]]*(LH[[i]]
> 10^-10)/2)] == AiryAi[-((2*m)/(charge*hbar*F[[j]])^2)^(1/
> 3) (Eh charge - charge*F[[j]]*(
> LH[[i]] 10^-10)/2)] AiryBi[-((2*m)/(charge*
> hbar*F[[j]])^2)^(
> 1/3) (Eh charge + charge*F[[j]]*(LH[[i]]
> 10^-10)/2)], {Eh, ZeroField[[i]]}]]
>
>
> (* now do succeeding
> rows by using the
> row above each as the starting values for \
> FindRoot *)
>
> For[j = 2, j < 11, j++, For[i = 1, i < 17,
> i++,(*Here we are : the symbol Eh located at (j, i) is going to be replace \
> by the value returned by FindRoot by applying the replacement operator /. to \
> the transformation rule of the form {Eh -> somenumber}*)eh[[j,
> i]] = Eh /. FindRoot[AiryAi[-((2*m)/(
> charge*hbar*F[[j]])^2)^(1/3) (Eh charge + charge*F[[j]]*(
> LH[[i]] 10^-10)/2)] AiryBi[-((2*m)/(charge*hbar*F[[j]])^2)^(
> 1/3) (Eh charge - charge*F[[j]]*(LH[[i]] 10^-10)/2)] == \
> AiryAi[-((2*m)/(charge*hbar*F[[j]])^2)^(1/3) (Eh charge - \
> charge*F[[j]]*(LH[[i]] 10^-10)/2)] \
> AiryBi[-((2*m)/(charge*hbar*F[[j]])^2)^(1/3) (Eh
> charge + charge*F[[j]]*(LH[[i]] 10^-10)/2)], {Eh, eh[[j - 1, i]]}]]]
> eh // MatrixForm
>