Another lame question : defining function by a result.
Subject: Another lame question : defining function by a result.
From: Viktor T. Toth
Date: Wed, 22 Dec 2004 08:35:19 -0500
Try define(f(x),%o2); it'll work.
The reason is simple: when you define a function with :=, the RHS is NOT
evaluated. In other words, your function definition will contain the SYMBOL
%o2, not its value. When you use define(), the second argument IS evaluted:
(%i5) define(f(x),%o2);
2 x + 1
2 ATAN(-------)
LOG(x + x + 1) SQRT(3) LOG(x - 1)
(%o5) f(x) := - --------------- - ------------- + ----------
6 SQRT(3) 3
Compare this with your line %o3.
Viktor
-----Original Message-----
From: maxima-admin@math.utexas.edu [mailto:maxima-admin at math] On
Behalf Of Jordan Tuzsuzov
Sent: Wednesday, December 22, 2004 8:08 AM
To: maxima@math.utexas.edu
Subject: Another lame question : defining function by a result.
Hello,
It is just about the expectance of work. I want to
define a function f(x) by a result of previous
calculation, related to x. Example :
(%i1) 1 / ( x^3 - 1 );
1
(%o1) ------
3
x - 1
(%i2) integrate(%,x);
2 x + 1
2 ATAN(-------)
LOG(x + x + 1) SQRT(3) LOG(x - 1)
(%o2) - --------------- - ------------- +
----------
6 SQRT(3) 3
(%i3) f(x) := %o2;
(%o3) f(x) := %o2
(%i4) f(10);
2 x + 1
2 ATAN(-------)
LOG(x + x + 1) SQRT(3) LOG(x - 1)
(%o4) - --------------- - ------------- +
----------
6 SQRT(3) 3
The x is not substituted by 10 in call %i4. Why ?
Thanx.
Jordan
__________________________________
Do you Yahoo!?
Take Yahoo! Mail with you! Get it on your mobile phone.
http://mobile.yahoo.com/maildemo
_______________________________________________
Maxima mailing list
Maxima@www.math.utexas.edu
http://www.math.utexas.edu/mailman/listinfo/maxima