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Edit detail for SandBoxReduce revision 23 of 36

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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
LatexWiki Image 
int(sqrt(1-sin(x)*cos(x)),x);
reduce
LatexWiki Image 

example from my daughter's college calc --pbwagner, Mon, 10 Sep 2007 12:58:25 -0500 reply
int(log(log(x)),x);
reduce
LatexWiki Image 

axiom
integrate(log(log(x)),x)
LatexWiki Image(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
LatexWiki Image 

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