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Re: Differencing two equations

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  • Subject: [mg131485] Re: Differencing two equations
  • From: "djmpark" <djmpark at>
  • Date: Mon, 5 Aug 2013 23:09:10 -0400 (EDT)
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  • References: <> <kfae2p$m4o$> <6814623.17519.1375697295577.JavaMail.root@m06>

Inner is also sometimes a useful construct for this.

eqn1 = a1 == a2;
eqn2 = b1 == b2; 

Inner[Plus, eqn1, eqn2, Equal] 

a1 + b1 == a2 + b2 

Inner[Plus, eqn1, Minus /@ eqn2, Equal] 

a1 - b1 == a2 - b2 

Inner[Plus, eqn1, 3 # & /@ eqn2, Equal] 

a1 + 3 b1 == a2 + 3 b2 

David Park
djmpark at 

From: David Bailey [mailto:dave at] 
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
 >> 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

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