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Re: Differencing two equations

  • To: mathgroup at smc.vnet.net
  • Subject: [mg131484] Re: Differencing two equations
  • From: Bob Hanlon <hanlonr357 at gmail.com>
  • Date: Mon, 5 Aug 2013 23:08:50 -0400 (EDT)
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  • References: <20130210082414.C3A5F6937@smc.vnet.net>

If you want to ensure that the inputs are equalities (and of the same
length) just include the constraints in the function definition.


equationOperate::usage =
  "equationOperate[equation1, equation2, operation] performs a
listable operation (e.g., Plus, Subtract, Times, Divide) on the input
equations. For example, equationOperate[a == b, c == d, Subtract] produces
the equation a - c == b - d";


equationOperate[eqn1_Equal, eqn2_Equal, op_] :=
  Equal @@ op @@ List @@@ {eqn1, eqn2} /;
   Length[eqn1] == Length[eqn2];


SetAttributes[f, Listable]


equationOperate[a == r, b == s, #] & /@
 {Subtract, Plus, Times, Divide,
  Power, Power[#2, #1] &, f}


{a - b == r - s, a + b == r + s, a*b == r*s,
   a/b == r/s, a^b == r^s, b^a == s^r,
   f[a, b] == f[r, s]}


equationOperate[a == r == x, b == s == y, #] & /@
 {Subtract, Plus, Times,
  Divide, Power, Power[#2, #1] &, f}


{a - b == r - s == x - y, a + b == r + s == x + y,
   a*b == r*s == x*y, a/b == r/s == x/y,
   a^b == r^s == x^y, b^a == s^r == y^x,
   f[a, b] == f[r, s] == f[x, y]}



Bob Hanlon


On Mon, Aug 5, 2013 at 6:04 AM, David Bailey <dave at removedbailey.co.uk>wrote:

> On 11/02/2013 09:36, Murray Eisenberg wrote:> Yeah, Mathematica
> stubbornly refuses t
>  >
>  >> I'm brand new to Mathematica, so I apologize for the naive questions...
>  >>
>  >> I'm trying to figure out how to difference two equations.  Basically
> if I have:
>  >> a==r
>  >> b==s
>  >>
>  >> I'd like to get:
>  >> a-b == r-s
>  >>
>  >> What I'm getting is more like (a==r) - (b==s).  I'm not sure how
> that's a useful result, but is there a function to do what I'm looking for?
>  >>
>  >> A quick search of the archives seem to bring up ways of doing this
> from using transformation rules to swap heads to unlocking the Equals
> operator and hacking its behavior.  I'd like to avoid doing that kind of
> rewiring for a simple operation, and I'd like to keep the syntax clean.
>  >>
>  >> The Core Language documentation makes a big point of how everything
> is basically a list with different heads.  In this case, what I'm trying
> to do would work if it were treated as a list ({a,b}-{r,s} returns
> {a-b,r-s}) but doesn't work under Equal.
>  >>
>  >> Thanks for any suggestions.
>  >
>
> I think it helps to manipulate algebraic expressions in a way that makes
> it easy to understand when you come back to a notebook sometime later!
> So my approach would be to define a function to subtract equations, so
> that it is pretty obvious what is going on!
>
> In[16]:= subtract[a1_ == a2_, b1_ == b2_] := (a1 - b1) == (a2 - b2)
>
> In[17]:= subtract[x + 3 y == 6, x - y == 2]
>
> Out[17]= 4 y == 4
>
> This also has the advantage that if you accidentally supply an argument
> that is not an equality, you will not get a misleading answer!
>
> A slightly more general function might take a third argument that would
> act as a multiplier of the second equality:
>
> In[20]:= combine[a1_ == a2_, b1_ == b2_,
>    k_] := (a1 + k b1) == (a2 + k b2)
>
> In[21]:= combine[x + 3 y == 6, x - y == 2, -1]
>
> Out[21]= 4 y == 4
>
> In[23]:= combine[x + 3 y == 6, x - y == 2, 3] // Simplify
>
> Out[23]= x == 3
>
> (You could also include the Simplify as part of your combine function)
>
> David Bailey
> http://www.dbaileyconsultancy.co.uk
>
>
>


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