Re: Using InterpolateRoot Function in Mathematica
- To: mathgroup at smc.vnet.net
- Subject: [mg41139] Re: [mg41127] Using InterpolateRoot Function in Mathematica
- From: Bobby Treat <drmajorbob+MathGroup3528 at mailblocks.com>
- Date: Mon, 5 May 2003 02:42:51 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
First, why write
f(L) = Integrate[sqrt(1+z^2) dx] - L0 (Integration limits are from 0 to
L)
if you mean
f[L_]:= Integrate[sqrt(1+z^2),{z,0,L}] - L0
??
This leaves me wondering if you entered that, or something else. The
possible errors you MIGHT have made are endless, so it would really
help if you just showed us the statement you entered.
Secondly, what is L0? Another unknown? A parameter?
Bobby
-----Original Message-----
From: Haritha Yalamanchili <haritha12 at attbi.com>
To: mathgroup at smc.vnet.net
Subject: [mg41139] [mg41127] Using InterpolateRoot Function in Mathematica
Hi,
I am trying to use Mathematica to find a value of "L" that satisfies the
equation
Integrate[sqrt(1+z^2) dx] - L0 = 0 (Integration limits are from 0 to L)
where,
z= -2 x(L C1 - 3 C2)/L^2 - 3 x^2(-L C1 + 2 C2)/L^3
In order to find the value of L that satisfies the above equation, I
have
setup the problem in Mathematica as shown below. Can some one help to
verify
if the problem is setup properly of if Mathematica is capable of
finding a
root for such functions.
******
f(L) = Integrate[sqrt(1+z^2) dx] - L0 (Integration limits are from 0 to
L)
InterpolateRoot[ f(L),{L,0,L0} ]
*******
(L0=13, C1=0.12, C2=0.25)
Value of L is close to L0 and hence, L0 can be used as the initial guess
value.
Thank You and Best Regards
Prasad