So I figured this out this evening while studying for my DiffEq final exam tomorrow. My math friends all checked it and found no errors, so I might as well publish it.
Plug it into your calculator and see.
aka ln(-1)/i
anybody want to see some of the cool stuff Ive figured out?
Looks good, although technically it isn't an approximation.
I spent 2.5 hours once finding rational approximations for square roots of prime numbers; I got to 37 and have about a third of them memorized.
But ln(i^2)=ln(-1)=meaningless in terms of traditional arithmetic, and so is only meaningful in terms of the original equation because it cannot be approximated.
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aka ln(-1)/I
Which is less useful because it yields an ERR: NONREAL ANS on the calc.
Quote:
anybody want to see some of the cool stuff Ive figured out?
Sure.
jpez wrote:
But ln(I^2)=ln(-1)=meaningless in terms of traditional arithmetic, and so is only meaningful in terms of the original equation because it cannot be approximated.
We're now in the realm of imaginary numbers. Traditional arithmetic has been left in the dust. Now there are only a few undefined functions:
x/0
∞-∞
0*∞
∞^0
∞/∞
anything involving limits towards +/-∞ in occillating functions (such as lim(sin(x),x,∞))
elfprince13 wrote:
aka ln(-1)/I
anybody want to see some of the cool stuff Ive figured out?
No, that doesn't work. That's why this is so cool
wait, how does that not work? plug that into your calc you will get pi.
rivereye wrote:
wait, how does that not work? plug that into your calc you will get pi.
ln(i^2)/i works, but ln(-1)/i does not.
KermMartian wrote:
rivereye wrote:
wait, how does that not work? plug that into your calc you will get pi.
ln(I^2)/I works, but ln(-1)/I does not.
Probaly because the first simplifies to 2ln(i)/i instead of just bruting ln(-1) out.
I bet you're right about that. Now I just need to find a good approximation of ln(i)/i and I've got pi/2
I just typed in ln(-1)/i and got 3.14159264, so how does that not work?
Note:My calc was in a+bi mode.
Dude, your calculator was in a+bi mode. Kerm's method get's around that and allows you to make the calculation in real mode, which your method does not.
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Please don't necropost.
But it's a productive post (or at least he's asking a question); he's fine 
I guess it was, but I thought through the topic it was solved for some odd reason, my bad.
calc84maniac wrote:
I just typed in ln(-1)/i and got 3.14159264, so how does that not work?
Note:My calc was in a+bi mode.
Yeah, which causes problems if you're in real mode. My method works in any mode.
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