Re: Recursive FindRoot with initial values as a list
- To: mathgroup at smc.vnet.net
- Subject: [mg77048] Re: [mg77023] Recursive FindRoot with initial values as a list
- From: Sseziwa Mukasa <mukasa at jeol.com>
- Date: Fri, 1 Jun 2007 02:49:30 -0400 (EDT)
- References: <200705310932.FAA15745@smc.vnet.net>
On May 31, 2007, at 5:32 AM, R.G wrote:
> Greetings...
> Say I have a function defined as:
> f[a_, b_]:= Integrate[Sin[x] + 3, {x, a, b}]
>
> I can find b with given f[a,b]=2 and initial value=0:
> In[1]:= Initialvalue = 0;
> s1 = FindRoot[f[0, unknown] == 2, {unknown,Initialvalue}];
> x1 = s1[[ 1, 2]]
>
> Now, using the x1 value obtained from FindRoot, I can find x2:
> s2 = FindRoot[f[x1, unknown] == 2, {unknown,Initialvalue}];
> x2 = s2[[ 1, 2]]
>
> If I have initialvalue in a list form:
> InitialValues={initialvalue1,initialvalue2,...,initialvalueN}
>
> in which each si should get InitialValues[[i]], in general form:
>
> \!\(s\_i =
> FindRoot[{f[x\_\(i - 1\), unknown] \[Equal] 2}, {unknown,
> InitialValues[\([i]\)]}]\)
>
> where
>
> \!\(x\_i = s\_i[\([1, 2]\)]\)
>
> How do I code it in Mathematica please? Thanking in advance...
You have not defined x0 but from your example I assume it's 0. The
rest is just an application of FoldList
Rest[FoldList[FindRoot[f[#,unknown]==2,{unknown,#2}][[1,2]]&,0(*This
is x0*),InitialValues]]
This also works and avoids the awkward use of Part
Rest[FoldList[unknown/.FindRoot[f[#,unknown]==2,{unknown,#2}]&,0
(*This is x0*),InitialValues]]
Regards,
Ssezi