How can I implement Hyperbolic numbers in Mathematica? How could I express their polar form? How could I calculate their power?
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6$\begingroup$ How to write a good question! $\endgroup$– halirutan ♦Jan 25, 2013 at 15:20
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2$\begingroup$ Depends on what exactly you want to do with them. For the "standard" case x+j*y with j^2=1, could use symmetric 2x2 matrices. $\endgroup$– Daniel LichtblauJan 26, 2013 at 22:33
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2 Answers
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ScalarQ[c_] := NumericQ[c] && ! MatchQ[Head[c], Complex | Hyperbolic];
Hyperbolic /: Hyperbolic[a_, 0] := a;
Hyperbolic /: c_?ScalarQ + Hyperbolic[a_, b_] := Hyperbolic[c + a, b];
Hyperbolic /: Hyperbolic[a_, b_] + Hyperbolic[c_, d_] :=
Hyperbolic[a + c, b + d];
Hyperbolic /: c_?ScalarQ*Hyperbolic[a_, b_] /; ScalarQ[c] :=
Hyperbolic[c a, c b];
Hyperbolic /: Hyperbolic[a_, b_]*Hyperbolic[c_, d_] :=
Hyperbolic[a c + b*d, b c + a d];
Hyperbolic /: Power[Hyperbolic[a_, b_], 0] := 1;
Hyperbolic /: Power[Hyperbolic[a_, b_], -1] :=
Hyperbolic[a/(a^2 - b^2), -b/(a^2 - b^2)];
Hyperbolic /: Power[Hyperbolic[a_, b_], n_Integer?Positive] :=
Hyperbolic[a, b] Power[Hyperbolic[a, b], n - 1];
Hyperbolic /: Power[Hyperbolic[a_, b_], n_Integer?Negative] :=
1/Power[Hyperbolic[a, b], -n];
Hyperbolic /: Power[Hyperbolic[a_, b_], x_?ScalarQ] :=
Which[a^2 - b^2 > 1,
Hyperbolic[(Sqrt[Abs[a^2 - b^2]])^x*
Cosh[x*ArcTan[b/a]], (Sqrt[Abs[a^2 - b^2]])^x*
Sinh[x*ArcTan[b/a]]], a^2 - b^2 < -1,
Hyperbolic[(Sqrt[Abs[a^2 - b^2]])^x*
Sinh[x*ArcTan[b/a]], (Sqrt[Abs[a^2 - b^2]])^x*
Cosh[x*ArcTan[b/a]]], a^2 - b^2 == 0, 0];
Hyperbolic /:
f_[Hyperbolic[a_, b_]] /; MemberQ[Attributes[f], NumericFunction] :=
Hyperbolic[f[a], b f'[a]];
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Add this line to the top of the notebook:
$Post = #/.J->{-1,1}/.{x_,y_}->(x+y)/2+J(y-x)/2&;
Use like this:
In:= Log[J]
Out:= (I Pi)/2 - (I Pi J)/2
In:= (3J+5)^3
Out:= 260 + 252 J
In:= I^J
Out:= I J
