Re: solve question

• To: mathgroup at smc.vnet.net
• Subject: [mg90995] Re: solve question
• From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
• Date: Sat, 2 Aug 2008 03:27:02 -0400 (EDT)
• Organization: The Open University, Milton Keynes, UK
• References: <g6uchj\$pdf\$1@smc.vnet.net>

```Francisco Gutierrez wrote:

> Frieds: I have a question, as always I imagine it is quite simple.
>
> I have made several functions. I give the following example:
>
> newsubstraction[a_List, b_List]:=Module[{k=((b[[2]]+a[[2]])-(b[[1]]+a[[=
> 1]]))/2},{middle[a]-middle[b]-k, middle[a]-middle[b]+k}]
>
> This function takes two lists of the form {a1,b1} and {a2,b2} and returns a list of identical form.  Now, suppose I have an equation with an unknown (of the form {u1,u2}). My question is: how to solve it?
> I would like something of the sort:
> IMAGINEDSOLVE[newsubstraction[firstlist, UNKNOWN]==secondlist, UNKNOWN]
> I would like this IMAGINEDSOLVE to work for newsubstraction and other functions of the same type. Thanks for any help.

Something along the lines,

mySolve[l1_, l2_, uk_] :=
Solve[Thread[newsub[l1, uk] == newsub[l2, uk]], uk]

should do it.

Regards,
-- Jean-Marc

```

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