|
|
|
last edited 6 years ago by test int |
Edit detail for SandBoxReduce revision 23 of 36
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 | ||
|
Editor: crashcut
Time: 2008/01/31 02:41:31 GMT-8 |
||
| Note: | ||
added:
From crashcut Thu Jan 31 02:41:31 -0800 2008
From: crashcut
Date: Thu, 31 Jan 2008 02:41:31 -0800
Subject:
Message-ID: <20080131024131-0800@axiom-wiki.newsynthesis.org>
\begin{reduce}
solve({y=-1/l*(1-(l*x+c1)^2)^(1/2)+c2, c1=((-3*l^2-l+8)^(1/2)/4-l, c2=-3((-3*l^2-l+8)^(1/2)+7*l)/12*l}, {y});
\end{reduce}
Try Reduce calculations here. For example:
\begin{reduce}
solve({z=x*a+2},{z,x});
int(sqrt(1-sin(x)*cos(x)),x);
\end{reduce}
solve({z=x*a+2},{z,x}); | reduce |
 |
int(sqrt(1-sin(x)*cos(x)),x); | reduce |
 |
example from my daughter's college calc --pbwagner, Mon, 10 Sep 2007 12:58:25 -0500 reply
int(log(log(x)),x); | reduce |
 |
axiomintegrate(log(log(x)),x)
 | (1) |
Type: Union(Expression Integer,...)
solve({c=-1/m*(1-(m*a+c1)^2)^(1/2)+c2, b=-1/m*(1-c1^2)^(1/2)+c2}, {c1,c2}); | reduce |
 |
solve({y=-1/l(1-(lx+c1)^2)^(1/2)+c2, c1=((-3l^2-l+8)^(1/2)/4-l, c2=-3((-3l^2-l+8)^(1/2)+7l)/12l}, {y});
solve({y=-1/l*(1-(l*x+c1)^2)^(1/2)+c2, c1=((-3*l^2-l+8)^(1/2)/4-l, c2=-3((-3*l^2-l+8)^(1/2)+7*l)/12*l},
{y});;)}y{,solve({y=-1/l*(1-(l*x+c1)^2)^(1/2)+c2,c1=((-3*l^2-l+8)^(1/2)/4-l,c2=-
3((-3*l^2-l+8)^(1/2)+7*l)/12*l | reduce |
$ | reduce |
$$} ***** Too few right parentheses ***** Continuing with parsing only ... | reduce |