Re: Closed Form solution too much to hope for?
- To: mathgroup at smc.vnet.net
- Subject: [mg66845] Re: Closed Form solution too much to hope for?
- From: "Valeri Astanoff" <astanoff at yahoo.fr>
- Date: Thu, 1 Jun 2006 06:55:00 -0400 (EDT)
- References: <e5augm$no4$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Assuming mean(x) is about .75, here is an empirical formula
that might be useful to you:
In[1]:=x[c_,d_,n_]:=Sqrt[c](d-1)(2d-1)+c^(1/n)d(2d-1)-
(2(d-1)d(-2n+Sqrt[2]Sqrt[36c+n(9+2n+9Log[4/3])-9(1+Log[2])]))/3
In[2]:=x[c,0,n]
Out[2]=Sqrt[c]
In[3]:=x[c,1,n]
Out[3]=c^(1/n)
In[4]:=x[.5,.6,9]
Out[4]=0.88615
In[5]:=FindRoot[d x^n+(1-d)x^2==c /. c -> .5 /. d -> .6 /. n ->
9,{x,.5}]
Out[5]={x -> 0.880133}
hth
V.Astanoff