Re: FindRoot for the determinant of a matrix with a varying size
- To: mathgroup at smc.vnet.net
- Subject: [mg59747] Re: FindRoot for the determinant of a matrix with a varying size
- From: Bill Rowe <readnewsciv at earthlink.net>
- Date: Fri, 19 Aug 2005 04:32:43 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
On 8/18/05 at 12:17 AM, wssaca at gmail.com (Wonseok Shin) wrote: >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? A small change to the definition for f fixes things In[1]:= f[(x_)?NumericQ] := Det[Table[Exp[(i - j)/x]/x, {i, 2, 5, x}, {j, 2, 5, x}]] In[2]:= FindRoot[f[x] == 0.1, {x, 5}] Out[2]= {x -> 10.} -- To reply via email subtract one hundred and four