Re: Starting values with FindRoot
- To: mathgroup at smc.vnet.net
- Subject: [mg20649] Re: [mg20620] Starting values with FindRoot
- From: BobHanlon at aol.com
- Date: Sun, 7 Nov 1999 02:09:54 -0500
- Sender: owner-wri-mathgroup at wolfram.com
Benoit,
eq = Table[z[i]^2 == 2 * i, {i, 2}];
x0 = Table[{z[i], i + .5}, {i, 2}];
FindRoot[eq, Evaluate[Sequence @@ x0]]
{z[1] -> 1.4142135623746899, z[2] -> 2.0000000929222947}
Bob Hanlon
In a message dated 11/4/1999 7:18:45 AM, benoit at ecn.ulaval.ca writes:
>I am fairly new to Mathematica. I have to solve numerically a large set
>
>of nonlinear equations. My question is: How can I set the starting
>values without having to explicitly name every variable? Here is an
>example of what I mean: Suppose I try to solve numerically the following
>
>two equations:
>eq = Table[z[i]\^2 == 2 i, {i, 2}]
>
>It would be nice if starting values could be set as a List. Example:
>
>x0 = Table[{z[i], i + .5}, {i, 2}]
>
>and to solve the equations with the following statement:
>
>FindRoot[eq,x0]
>
>Unfortunetly, it does not work. Mathematica does not seem to accept
>starting values as lists. The problem appears to be the outer {}. In
>place of x0 above, Mathematica looks for something like
>{z[1],1.5},{z[2],2.5}, while x0={ {z[1],1.5},{z[2],2.5} }. There is an
>
>extra set of {}.
>
>I have found a solution by using a chain of FullForm[], ToString[],
>StringDrop[] and ToExpression[]. Here is my solution:
>
>ToExpression[ToString[FindRoot] <> "[" <> ToString[FullForm[eq]] <> ","
>
><>
> StringDrop[StringDrop[ToString[x0], 1], -1] <> "]"]
>
>Althought this works OK, it is nevertheless a very roundabout way of
>doing a simple thing. Anyone knows a more elegant way of doing this?
>