How to implement split-complex numbers?
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For those who do not know, the split-complex numbers are an analogue to the complex numbers where J is defined such that $J^2=1$ but $Jnepm1$, so they are all of the form $a+bJ$.
By using TagSetDelayed, I tried to define the split-complex numbers as so:
J /: J^2 := 1
If I then type J^2, I get the output 1.
However, if I type J^3, I just get the output J^3. I would like to instead get the output $J$, since $J^3=J^2J=1J=J$.
Is there a better way to implement this number system?
programming algebraic-manipulation algebra
asked Jan 22 at 22:18
volcanrbvolcanrb
1734
$endgroup$
add a comment |
$begingroup$
For those who do not know, the split-complex numbers are an analogue to the complex numbers where J is defined such that $J^2=1$ but $Jnepm1$, so they are all of the form $a+bJ$.
By using TagSetDelayed, I tried to define the split-complex numbers as so:
J /: J^2 := 1
If I then type J^2, I get the output 1.
However, if I type J^3, I just get the output J^3. I would like to instead get the output $J$, since $J^3=J^2J=1J=J$.
Is there a better way to implement this number system?
programming algebraic-manipulation algebra
asked Jan 22 at 22:18
volcanrbvolcanrb
1734
$endgroup$
2
$begingroup$
Might be easier to do this using 2x2 matrix representations.
$endgroup$
– Daniel Lichtblau
Jan 22 at 22:28
$begingroup$
Ummmm.... isn't $J = -1$?
$endgroup$
– David G. Stork
Jan 22 at 23:28
$begingroup$
Sorry, I forgot to specify that $Jne -1$ either, I've edited the question to fix that
$endgroup$
– volcanrb
Jan 22 at 23:46
add a comment |
$begingroup$
For those who do not know, the split-complex numbers are an analogue to the complex numbers where J is defined such that $J^2=1$ but $Jnepm1$, so they are all of the form $a+bJ$.
By using TagSetDelayed, I tried to define the split-complex numbers as so:
J /: J^2 := 1
If I then type J^2, I get the output 1.
However, if I type J^3, I just get the output J^3. I would like to instead get the output $J$, since $J^3=J^2J=1J=J$.
Is there a better way to implement this number system?
programming algebraic-manipulation algebra
asked Jan 22 at 22:18
volcanrbvolcanrb
1734
$endgroup$
For those who do not know, the split-complex numbers are an analogue to the complex numbers where J is defined such that $J^2=1$ but $Jnepm1$, so they are all of the form $a+bJ$.
By using TagSetDelayed, I tried to define the split-complex numbers as so:
J /: J^2 := 1
If I then type J^2, I get the output 1.
However, if I type J^3, I just get the output J^3. I would like to instead get the output $J$, since $J^3=J^2J=1J=J$.
Is there a better way to implement this number system?
programming algebraic-manipulation algebra
programming algebraic-manipulation algebra
asked Jan 22 at 22:18
volcanrbvolcanrb
1734
asked Jan 22 at 22:18
volcanrbvolcanrb
1734
edited Jan 22 at 23:46
volcanrb
asked Jan 22 at 22:18
volcanrbvolcanrb
1734
asked Jan 22 at 22:18
volcanrbvolcanrb
1734
asked Jan 22 at 22:18
volcanrbvolcanrb
1734
1734
2
$begingroup$
Might be easier to do this using 2x2 matrix representations.
$endgroup$
– Daniel Lichtblau
Jan 22 at 22:28
$begingroup$
Ummmm.... isn't $J = -1$?
$endgroup$
– David G. Stork
Jan 22 at 23:28
$begingroup$
Sorry, I forgot to specify that $Jne -1$ either, I've edited the question to fix that
$endgroup$
– volcanrb
Jan 22 at 23:46
add a comment |
2
$begingroup$
Might be easier to do this using 2x2 matrix representations.
$endgroup$
– Daniel Lichtblau
Jan 22 at 22:28
$begingroup$
Ummmm.... isn't $J = -1$?
$endgroup$
– David G. Stork
Jan 22 at 23:28
$begingroup$
Sorry, I forgot to specify that $Jne -1$ either, I've edited the question to fix that
$endgroup$
– volcanrb
Jan 22 at 23:46
2
2
$begingroup$
Might be easier to do this using 2x2 matrix representations.
$endgroup$
– Daniel Lichtblau
Jan 22 at 22:28
$begingroup$
Might be easier to do this using 2x2 matrix representations.
$endgroup$
– Daniel Lichtblau
Jan 22 at 22:28
$begingroup$
Ummmm.... isn't $J = -1$?
$endgroup$
– David G. Stork
Jan 22 at 23:28
$begingroup$
Ummmm.... isn't $J = -1$?
$endgroup$
– David G. Stork
Jan 22 at 23:28
$begingroup$
Sorry, I forgot to specify that $Jne -1$ either, I've edited the question to fix that
$endgroup$
– volcanrb
Jan 22 at 23:46
$begingroup$
Sorry, I forgot to specify that $Jne -1$ either, I've edited the question to fix that
$endgroup$
– volcanrb
Jan 22 at 23:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Try this:
J /: Power[J, p_Integer?OddQ] := J
J /: Power[J, p_Integer?EvenQ] := 1
J^Range[-10, 10]
{1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1}
answered Jan 22 at 22:21
Henrik SchumacherHenrik Schumacher
51.6k469146
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Try this:
J /: Power[J, p_Integer?OddQ] := J
J /: Power[J, p_Integer?EvenQ] := 1
J^Range[-10, 10]
{1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1}
answered Jan 22 at 22:21
Henrik SchumacherHenrik Schumacher
51.6k469146
$endgroup$
add a comment |
$begingroup$
Try this:
J /: Power[J, p_Integer?OddQ] := J
J /: Power[J, p_Integer?EvenQ] := 1
J^Range[-10, 10]
{1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1}
answered Jan 22 at 22:21
Henrik SchumacherHenrik Schumacher
51.6k469146
$endgroup$
add a comment |
$begingroup$
Try this:
J /: Power[J, p_Integer?OddQ] := J
J /: Power[J, p_Integer?EvenQ] := 1
J^Range[-10, 10]
{1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1}
answered Jan 22 at 22:21
Henrik SchumacherHenrik Schumacher
51.6k469146
$endgroup$
Try this:
J /: Power[J, p_Integer?OddQ] := J
J /: Power[J, p_Integer?EvenQ] := 1
J^Range[-10, 10]
{1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1}
answered Jan 22 at 22:21
Henrik SchumacherHenrik Schumacher
51.6k469146
answered Jan 22 at 22:21
Henrik SchumacherHenrik Schumacher
51.6k469146
answered Jan 22 at 22:21
Henrik SchumacherHenrik Schumacher
51.6k469146
answered Jan 22 at 22:21
Henrik SchumacherHenrik Schumacher
51.6k469146
51.6k469146
add a comment |
add a comment |
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2
$begingroup$
Might be easier to do this using 2x2 matrix representations.
$endgroup$
– Daniel Lichtblau
Jan 22 at 22:28
$begingroup$
Ummmm.... isn't $J = -1$?
$endgroup$
– David G. Stork
Jan 22 at 23:28
$begingroup$
Sorry, I forgot to specify that $Jne -1$ either, I've edited the question to fix that
$endgroup$
– volcanrb
Jan 22 at 23:46