Re: Recursive FindRoot with initial values as a list

*To*: mathgroup at smc.vnet.net*Subject*: [mg77047] Re: Recursive FindRoot with initial values as a list*From*: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>*Date*: Fri, 1 Jun 2007 02:48:59 -0400 (EDT)*Organization*: The Open University, Milton Keynes, UK*References*: <f3m4qv$fgl$1@smc.vnet.net>

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... > > ~R.G Hi, One possible way of doing what you are looking for is the use a functional construct such as NestList and a pure function (that why you can see #1 and &). For instance, In[1]:= f[a_, b_] = Integrate[Sin[x] + 3, {x, a, b}]; With[{Initialvalue = 0, n = 5}, Rest[NestList[FindRoot[f[#1, unknown] == 2, {unknown, Initialvalue}][[1, 2]] & , 0, n]]] Out[1]= {0.607102, 1.13938, 1.64271, 2.15069, 2.69881} Cheers, Jean-Marc