[VA] SRC #009 - Pi Day 2021 Special

[J-F Garnier]

(03-16-2021 06:36 PM)Valentin Albillo Wrote:  

J-F Garnier Wrote:For point (c), well ... it's a nice identity

If you say so ... this means you think that  2 + 2 = 4   is a "nice identity" too ?

By identity, I meant that the expression is independent from the 'e' value, and always produces \(\pi\), quite similar to
arctan(1/x) = \(\pi\)/2 - arctan(x) (for x>0).
I'm ready to admit that identity not the correct word, if you think so.

Quote:

J-F Garnier Wrote:I don't think we can say that the equation e^(i π) + 1 = 0 can be used to get π from e. If you try to get π from this expression, you will just end with π = acos(-1).

Well, you can isolate \(\pi\) in the equation and you get   \(\pi\) = log_e(-1) / i, which apart from constants -1 and i features a logarithm base e as a fundamental part of it, and which your Emu71+Math ROM readily evaluates as:

      >LOG((-1,0))/(0,1)

            (3.14159265359, 0),     i.e.: \(\pi\)

and I see no cosine in that evaluation.

Right but how, practically, do you compute a complex logarithm ?
log(z) = log(abs(z)) + i.arg(z) [1]
for z=(-1,0) , log(abs(z)) = 0 and arg(z)=atan2(0,-1) [ANGLE(-1,0) in HP-71's BASIC ]
So, we get \(\pi\) = log_e(-1) / i = atan2(0,-1)
No 'e' value involved anywhere.
[1] This is the method used to compute log(z) in the HP-71's Math ROM, and all RPN/RPL descendants, and very likely the 15C too.

Quote:

J-F Garnier Wrote:I don't know -and don't think there is - any relation that can be used to get π from e.

Paraphrasing Hamlet:

      "There are more things in heaven and earth, Jean-François, than are dreamt of in your philosophy"

In other words, you bet there are.

Looking forward to see examples, if included in your final comments !

J-F