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last edited 5 years ago by test int |
Edit detail for SandBoxReduce revision 27 of 36
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Editor: crashcut
Time: 2008/01/31 02:49:23 GMT-8 |
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added:
From crashcut Thu Jan 31 02:49:20 -0800 2008
From: crashcut
Date: Thu, 31 Jan 2008 02:49:20 -0800
Subject:
Message-ID: <20080131024920-0800@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 |
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int(sqrt(1-sin(x)*cos(x)),x); | reduce |
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example from my daughter's college calc --pbwagner, Mon, 10 Sep 2007 12:58:25 -0500 reply
int(log(log(x)),x); | reduce |
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axiomintegrate(log(log(x)),x)
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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 |
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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 |
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}); | reduce |
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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}); | reduce |
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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}); | reduce |
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