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Re: Series Simplification - how to truncate the result ?
- To: mathgroup at smc.vnet.net
- Subject: [mg127660] Re: Series Simplification - how to truncate the result ?
- From: "Harvey P. Dale" <hpd1 at nyu.edu>
- Date: Mon, 13 Aug 2012 03:17:15 -0400 (EDT)
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- Delivered-to: l-mathgroup@wolfram.com
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- References: <20120812054957.D8F4E67FE@smc.vnet.net>
Rahul:
Look up the Take command.
Best,
Harvey
-----Original Message-----
From: Rahul Chakraborty [mailto:rahul.6sept at gmail.com]
Sent: Sunday, August 12, 2012 1:50 AM
To: mathgroup at smc.vnet.net
Subject: [mg127660] Series Simplification - how to truncate the result ?
Hi everybody,
While solving a series i'm getting a result too large.Kindly let me
know as how to get it truncated ?
Code as follows:
(* Formula: Subscript[F, e](x,t) = 1/2 (Subscript[C, 0]
Subscript[d, 0])/(Subscript[d, 0]-x)^2*Subscript[V, 1]^2(t) - 1/2
(Subscript[C, 0] Subscript[d, 0])/(Subscript[d, 0]+x)^2*Subscript[V,
2]^2 *)
Clear[x,t,f];
c=4.06*(10^-4);
V[t]=Sqrt[2]*Subscript[V, p]*Sin[2Subscript[\[Omega], 0]t]; =
Subscript[V,
1][t] = 1+(c*V[t]);
r1 = Series[Subscript[V, 1][t],{1,(c*V[t]),2}]; Subscript[V, 2] = 1; =
z =
(Subscript[d, 0]-x)^2;
y = Series[z,{Subscript[d, 0],0,2}]; (*
Taylor expansion at x=0 *)
z1 = (Subscript[d, 0]+x)^2;
y1 = Series[z1,{Subscript[d, 0],0,2}]; (* Taylor
expansion at x=0 *)
a = 1/2*((Subscript[C, 0]*Subscript[d, 0])/y)*r1 //Simplify;
b = 1/2*((Subscript[C, 0]*Subscript[d, 0])/y1)*(Subscript[V, 2]^2)
//Simplify;
Subscript[F, e][x,t] =a-b
Sincerely,
rc
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