4 ms·

That's not his argument; he says defining Θ = e^2πi is not useful because e^2πi = 1, so Θ and Θ^x would also be 1. That's why he defined Θ = e^2π instead, so th

by twiss 3y ago

That's not his argument; he says defining Θ = e^2πi is not useful because e^2πi = 1, so Θ and Θ^x would also be 1. That's why he defined Θ = e^2π instead, so that Θ^x (or possibly Θ^ix) is a useful operation.

cycomanic 3y ago
he writes: > where ln(e^2πi)=ln(1)=0, that is incorrect because the logarithm of a complex number is multi-valued. He even cites the correct source on wikipedia, but his argument is incorrect (and because the exponent is 2πi he actually would get a meaningful results I believe).
- cycomanic 3y ago
  just to expand on this the issue is not that e^2πi = 1 but that (e^z)^x =/= e^(zx) when z is complex, because in general z^w . instead it would be e^(x ln(e^z)), where the logarithm of a complex number is a multivalued function. We can compute the principle value of ln(e^2πi)=ln(1) + i(2π + 2πk) (where k is an integer) so therefore (e^2πi)^x = e^(xln(e^2πi)) = e^(x2π(k+1)i)
- mgunyho 3y ago
  Hmm, this indeed seems to be the case! I think I was confused by ln(1), since 1 is a real number, but the multi-valued complex logarithm of 1 is indeed k 2pi (and now I see the appropriate notation should be "log" instead of "ln"). Perhaps the whole thing could indeed be reduced to "1^x" with an asterisk that 1^x means complex exponentiation. I'll have to update this section in the post.
  - cycomanic 3y ago
    See my other reply down below. That said I agree with the final conclusion that including the i is not a good idea, although I'm a bit on the fence on if the whole thing (I'm convinced by the tau=2pi arguments on the other hand). I think this really is just a redefinition of sin and cosine (and consequentially the exponential). My feeling is that we now have to deal with different derivative operators for periodic and non-periodic functions and a lot of other weird disconnects, e.g. we don't have a relation between a cycle and the radius (diameter, circumference ...) anymore. I'm not convinced that the (minor IMO) conveniences that gives us is worth it.

by twiss 3y ago

cycomanic 3y ago

he writes: > where ln(e^2πi)=ln(1)=0, that is incorrect because the logarithm of a complex number is multi-valued. He even cites the correct source on wikipedia, but his argument is incorrect (and because the exponent is 2πi he actually would get a meaningful results I believe).

cycomanic 3y ago
just to expand on this the issue is not that e^2πi = 1 but that (e^z)^x =/= e^(zx) when z is complex, because in general z^w . instead it would be e^(x ln(e^z)), where the logarithm of a complex number is a multivalued function. We can compute the principle value of ln(e^2πi)=ln(1) + i(2π + 2πk) (where k is an integer) so therefore (e^2πi)^x = e^(xln(e^2πi)) = e^(x2π(k+1)i)
mgunyho 3y ago
Hmm, this indeed seems to be the case! I think I was confused by ln(1), since 1 is a real number, but the multi-valued complex logarithm of 1 is indeed k 2pi (and now I see the appropriate notation should be "log" instead of "ln"). Perhaps the whole thing could indeed be reduced to "1^x" with an asterisk that 1^x means complex exponentiation. I'll have to update this section in the post.
- cycomanic 3y ago
  See my other reply down below. That said I agree with the final conclusion that including the i is not a good idea, although I'm a bit on the fence on if the whole thing (I'm convinced by the tau=2pi arguments on the other hand). I think this really is just a redefinition of sin and cosine (and consequentially the exponential). My feeling is that we now have to deal with different derivative operators for periodic and non-periodic functions and a lot of other weird disconnects, e.g. we don't have a relation between a cycle and the radius (diameter, circumference ...) anymore. I'm not convinced that the (minor IMO) conveniences that gives us is worth it.