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Edit detail for SandBoxReduce revision 18 of 36

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Editor: crashcut
Time: 2008/01/30 07:41:15 GMT-8
Note:

added:

From crashcut Wed Jan 30 07:41:15 -0800 2008
From: crashcut
Date: Wed, 30 Jan 2008 07:41:15 -0800
Subject: 
Message-ID: <20080130074115-0800@axiom-wiki.newsynthesis.org>

\begin{reduce}
solve({c=-1/l*(1-(l*a+c1)^2)^(1/2)=c2, b=-1/l*(1-c1^2)^(1/2)+c2}, {c1,c2});
\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/l*sqrt(1-(l*a+c1)^2)+c2, b=-1/l*sqrt(1-c1^2)+c2}, {c1, c2})
reduce

solve({c-b=-1/l*sqrt(1-(l*a+c1)^2)+1/l*sqrt(1-c1^2)}, {c1, c2})
reduce

solve({c-b=-1/l*(1-(l*a+c1)^2)^(1/2)+1/l*(1-c1^2)^(1/2)}, {c1, c2})
reduce

solve({c-b=-1/l*(1-(l*a+c1)^2)^(1/2)+1/l*(1-c1^2)^(1/2)}, {c1})
reduce

solve(c-b=-1/l*(1-(l*a+c1)^2)^(1/2)+1/l*(1-c1^2)^(1/2), c1)
reduce

solve(c-b=-1/l*(1-(l*a+c1)^2)^(1/2)+1/l*(1-c1^2)^(1/2), c1);
reduce
LatexWiki Image 

solve({c=-1/l*(1-(l*a+c1)^2)^(1/2)=c2, b=-1/l*(1-c1^2)^(1/2)+c2}, {c1,c2});
2  2                2
- sqrt( - a *l  - 2*a*c1*l - c1  + 1)
***** c=---------------------------------------- invalid as scalar
l
***** Continuing with parsing only ... 
reduce