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MathGroup Archive 2007

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Re: Recursive FindRoot with initial values as a list

  • To: mathgroup at smc.vnet.net
  • Subject: [mg77043] Re: [mg77023] Recursive FindRoot with initial values as a list
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Fri, 1 Jun 2007 02:46:55 -0400 (EDT)
  • Reply-to: hanlonr at cox.net

Since f can be evaluated symbolically, use Set ( = ) rather than SetDelayed ( := ).

f[a_, b_] = Integrate[Sin[t] + 3, {t, a, b}];

n = 5;
x = NestList[unknown /. 
    FindRoot[f[#, unknown] == 2, {unknown, #}] &, 
     0, n]

{0,0.607102,1.13938,1.64271,2.15069,2.69881}


Bob Hanlon

---- "R.G" <gobiithasan at yahoo.com.my> 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
> 
> 
> Send instant messages to your online friends http://uk.messenger.yahoo.com 
> 
> 



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