Re: FindRoot for the determinant of a matrix with a varying size
- To: mathgroup at smc.vnet.net
- Subject: [mg59722] Re: [mg59720] FindRoot for the determinant of a matrix with a varying size
- From: Curtis Osterhoudt <gardyloo at mail.wsu.edu>
- Date: Fri, 19 Aug 2005 04:31:45 -0400 (EDT)
- References: <200508180417.AAA08644@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Hi,
My linear algebra is rusty, but the value which your f[x] returns is
always 1/x (for x > 3, anyway). Thus, the solution to f[x] == 0.1 is
simply x = 10.
That probably isn't really what you were looking for, but in this
case, that's the solution.
Regards,
Curtis O.
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?
>
>Thanks,
>
>Wonseok Shin
>
>
>
>
--
PGP Key ID: 0x235FDED1
Please avoid sending me Word or PowerPoint attachments.
http://www.gnu.org/philosophy/no-word-attachments.html
- References:
- FindRoot for the determinant of a matrix with a varying size
- From: "Wonseok Shin" <wssaca@gmail.com>
- FindRoot for the determinant of a matrix with a varying size