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MathGroup Archive 2007

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RE: Re: Hold and Equal

  • To: mathgroup at smc.vnet.net
  • Subject: [mg73760] RE: [mg73739] Re: [mg73715] Hold and Equal
  • From: "Virgilio, Vincent - SSD" <Vincent.Virgilio at itt.com>
  • Date: Wed, 28 Feb 2007 04:23:09 -0500 (EST)

Mr. Woll,

step[] has an interesting syntactic style. Did you intentionally group
"&" with the following argument, rather than with the preceding Function
body? That reminds me of the alternative practices between C and C++,
which either groups "*" with the declarator (C) or the type specifier
(C++).

Vince Virgilio

-----Original Message-----
From: Carl Woll [mailto:carlw at wolfram.com]
Sent: Tuesday, February 27, 2007 5:45 AM
To: mathgroup at smc.vnet.net
Subject: [mg73760] [mg73739] Re: [mg73715] Hold and Equal

Murray Eisenberg wrote:

>How can I produce in an Output cell (under program control) an
>expression like the following,
>
>   (a+b)^2 = a^2+ 2 a b + b^2
>
>where instead of the usual Equal (==) I get a Set (=), as in
>traditional math notation?  I want to input the unexpanded (a+b)^2 and
>have the expansion done automatically.
>
>Of course, I can try something like the following:
>
>   (a+b)^2 == Expand[(a+b)^2])
>
>So how do I convert the == to =?  Of course
>
>   ((a + b)^2 == Expand[(a + b)^2]) /. Equal -> Set
>
>gives a Set::write error.  And
>
>   (Hold[(a + b)^2 == Expand[(a + b)^2]]) /. Equal -> Set
>
>doesn't actually evaluate the Expand part and leaves the "Hold"
wrapper.
> 
>
Murray,

How about using HoldForm?

step[x_] := HoldForm[x = #] &[Expand[x]]

step[(a+b)^2]
(a+b)^2=a^2+2 b a+b^2

Carl Woll
Wolfram Research

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