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antisymmetric-declare-exp-log-solve

IP: (%K1 *(%PBase+(PB...

declare (IP, antisymm...

solve(IP=%LimIP, PBra...

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antisymmetric-declare

h((0, 0),(1, 1),(2, 2...

declare(h,antisymmetr...

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antisymmetric

? antisymmetric;

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antisymmetric-declare

h((a, a),(b, b),(c, c...

declare(h,antisymmetr...

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antisymmetric

h((1, 2),(1, 4),(1, 3...

? antisymmetric;

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antisymmetric

? antisymmetric;

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antisymmetric

h((1, 2),(1, 4),(1, 3...

? antisymmetric;

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antisymmetric-declare

h((a, a),(b, b),(c, c...

declare(h,antisymmetr...

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antisymmetric

h((1, 2),(1, 4),(1, 3...

? antisymmetric;

Calculate

antisymmetric

h((1, 2),(1, 4),(1, 3...

? antisymmetric;

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antisymmetric

Run Example
(%i1)? antisymmetric;

 -- Declaration: antisymmetric
     If `declare(h,antisymmetric)' is done, this tells the simplifier
     that `h' is antisymmetric.  E.g. `h(x,z,y)' will simplify to `-
     h(x, y, z)'.  That is, it will give (-1)^n times the result given
     by `symmetric' or `commutative', where n is the number of
     interchanges of two arguments necessary to convert it to that form.


(%o1)                                true
(%i2) 
Run Example
h((x, x),(y, z),(z, y));
(%o1)                             h(x, z, y)
(%i2) declare(h,antisymmetric);
(%o2)                                done
(%i3) 
Run Example
h((0, 0),(1, 1),(2, 2),(3, 3),(4, 4),(1, 0),(0, 2),(1, 2),(3, 2));
(%o1)                    h(0, 1, 2, 3, 4, 0, 2, 2, 2)
(%i2) declare(h,antisymmetric);
(%o2)                                done
(%i3) 

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