       Re: M.U.C.

• To: mathgroup at smc.vnet.net
• Subject: [mg20627] Re: M.U.C.
• From: "David Bailey" <db at salford-software.com>
• Date: Thu, 4 Nov 1999 02:13:37 -0500
• Organization: University of Salford, Salford, Manchester, UK
• References: <7vm3io\$q82@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Roscoe <roscoe at hazardcounty.org> wrote in message
news:7vm3io\$q82 at smc.vnet.net...
> How can Mathematica solve the following Linear Operator for a and b?
> (E^(-2 x)((-3) + 10 b - x + 6 a(1 + 5 x)) == 0
>
> By hand it would be :
> (-3+10b+6a)=0
> (-1+30a)=0
>
> a=1/30
> b=whatever
>
> I need mathematica to group the coeeficients of x^0 and x^1, set them
equal
> to 0 and solve.
> I am trying to avoid doing the gruoping by hand.  Thanks.

You can use the Coefficient function to isolate the coefficients of powers
of x in a polynomial. However, your LHS is not actually a polynomial because
you have a multiplier of E^(-2 x) (although it can be divided out). A simple
solution is to generate two simultaneous equations with x set to two
distinct values - e.g. 0 and 1. If you do that you do not need to wory about
making the LHS a polynomial:

eqn = (E^(-2 x)((-3) + 10 b - x + 6 a(1 + 5 x)) == 0)

In:=
Solve[{eqn /. x -> 0, eqn /. x -> 1}, {a, b}]

Out=
{{a->1/30,b->7/25}}

David Bailey
Salford Software

```

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