newRPL - build 1255 released! [updated to 1299]
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10-12-2018, 02:26 PM
Post: #302
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RE: newRPL - build 1089 released! [update:build 1111]
(10-11-2018 03:36 PM)okkama Wrote: The result is not wrong considering is numeric, not symbolic. Unless the case i,-i is treated as a special case, you won't get an exact result. If you want Z^-1 it's best to use INV(Z) which uses multiplication by the conjugate, and can give you an exact result for exact arguments. The general case Z^a is used when you use ^, which has no choice but to use transcendental functions: Z^a = exp(ln(Z^a)) = exp(a * ln(Z)) = exp( a * (1-i*π/2) ) ** I replaced already Z=(0,-1) above as r*exp(i*Arg) = 1*exp(i*-π/2) = exp( a - a*π/2 i ) = exp(a) * exp(i*(-a*π/2)) = exp(a) * [ cos(-a*π/2) + i * sin(-a*π/2) ] Now we can see the problem: cos(π/2) can't ever be zero unless you can exactly represent π/2, which is numerically impossible, so it returns what you saw: a tiny number in the order of 10^-32 (because you used the default 32 digits, the result will vary at different precisions). Unlike many other calculators, newRPL will "by design" NEVER try to cheat or manipulate results in any way. |
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