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Re: FindRoot for the determinant of a matrix with a varying size

  • To: mathgroup at smc.vnet.net
  • Subject: [mg59736] Re: [mg59720] FindRoot for the determinant of a matrix with a varying size
  • From: Sseziwa Mukasa <mukasa at jeol.com>
  • Date: Fri, 19 Aug 2005 04:32:04 -0400 (EDT)
  • References: <200508180417.AAA08644@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On Aug 18, 2005, at 12:17 AM, Wonseok Shin wrote:

> Hello everyone,
>
> I am a user of Mathematica 5.1 for Mac .
> I defined the function using the determinant of a matrix of a varying
> size.  Even though this function is  well-behaving, it seems that
> FindRoot cannot deal this function.  Please look at the following  
> code:
>
> -------------------------------------------------
> In[1]:=
> f[x_] := Det[Table[Exp[(i - j)/x]/x , {i, 2, 5, x}, {j, 2, 5, x}]]
>
> In[2]:=
> Plot[f[x], {x, 3, 30}]
> -------------------------------------------------
>
> By running the above Plot command, you can see clearly that the
> function f is very smooth in the interval 3< x < 30, and f[x] == 0.1
> has a solution in 5 < x < 15.
>
> But I've failed to find a solution of f[x] == 0.1 using FindRoot:
>
> -------------------------------------------------
> In[3]:=
> FindRoot[f[x] == 0.1, {x, 5}]
>
> Table::iterb : Iterator {i, 2, 5, x} does not have appropriate bounds.
> -------------------------------------------------
>
> Is there any workaround for this problem?

The iterator in your definition of x produces more than one element  
only in the case 1 <= x <= 3, furthermore iterators must be specified  
over ranges of integers so a real valued x cannot be used.  Since you  
suspect your solution lies between 5 and 15 (actually since for x > 3  
the expression for f[x] simplifies to the one element matrix {{1/x}}  
the solution can be determined trivially) I would remove the  
dependence on x in the iterators of your expression.

Regards,

Ssezi



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