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solveexplicit

? solveexplicit ;

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solveexplicit-sqrt

solveexplicit;

solve([x=sqrt(x+1)+y]...

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solveexplicit-sqrt

solveexplicit;

solve([x=sqrt(x+1)+y]...

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solveexplicit-tanh

? solveexplicit;

srd:(1/96000);

t1(x):= tanh(x);

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solveexplicit-tanh

? solveexplicit;

srd:(1/96000);

t1(x):= tanh(x);

Calculate

solveexplicit-sqrt

[solveexplicit:true];

solve([x=sqrt(x+1)+y]...

Calculate

solveexplicit

? solveexplicit ;

Calculate

solveexplicit-sqrt

[solveexplicit:true];

solve([x=sqrt(x+1)+y]...

Calculate

solveexplicit-sqrt

solveexplicit;

solve([x=sqrt(x+1)+y]...

Calculate

solveexplicit-sqrt

[solveexplicit:true];

solve([x=sqrt(x+1)+y]...

Calculate

solveexplicit

Run Example
(%i1)[solveexplicit:true];
(%o1)                               [true]
(%i2) [solvedecomposes:true];
(%o2)                               [true]
(%i3) solve([x=sqrt(x+1)+y],x);
(%o3)                                 []
(%i4) 
Run Example
solveexplicit:true;
(%o1)                                true
(%i2) assume(r>
0);
(%o2)                               [r > 0]
(%i3) f1:(x-vx)^2+vy^2=r^2;
                                     2     2    2
(%o3)                        (x - vx)  + vy  = r
(%i4) f2:vx^2+(y-vy)^2=r^2;
                                     2     2    2
(%o4)                        (y - vy)  + vx  = r
(%i5) f3:solve(f1,vx);
                              2     2                  2     2
(%o5)         [vx = x - sqrt(r  - vy ), vx = x + sqrt(r  - vy )]
(%i6) f4:subst(rhs(f3[1]),vx,f2);
                            2              2     2  2    2
(%o6)               (y - vy)  + (x - sqrt(r  - vy ))  = r
(%i7)  	solve(f4,vy);
(%o7)                                 []
(%i8) 
Run Example
? solveexplicit ;

 -- Option variable: solveexplicit
     Default value: `false'

     When `solveexplicit' is `true', inhibits `solve' from returning
     implicit solutions, that is, solutions of the form `F(x) = 0'
     where `F' is some function.


(%o1)                                true
(%i2) 

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