newRPL: The complexity of complex mode

[Claudio L.]

(08-24-2016 04:06 PM)Vtile Wrote:  Hold on, this complex infinities goes partially beyond my knowledge (i don't understand the directed infinity), but (+∞,3) did take my attention.

PS. Now when I have spend too much time front of my computer rewriting this post, I think that the directed infinities are (lossy) polar representations.

It all boils down calculating the Theta with the some rules of the complex plane (or trigonometric sections of the circle).

You are getting the hang of it quite fast. Directed infinity is a point in the complex plane that's infinitely far from zero, going in a very specific direction. Therefore, I think the most appropriate representation is a polar complex number. A cartesian representation doesn't express the direction very well, should be something like this: \(\infty\times\)(0,3) but this is not parser-friendly, or screen friendly.

(08-24-2016 04:06 PM)Vtile Wrote:  This can be seen when we transform complex number \(Z=(\infty,3)=\infty + j3 \) to polar notation of
\(\left | Z \right | \angle \theta ^{\circ}\) = \(\sqrt{a^2+b^2}\angle ATan(\frac{b}{a})^{\circ}=\sqrt{a^2+b^2}\angle ACot(\frac{a}{b})^{\circ}\), with the values we get
\(\left | Z \right | = \sqrt{a^2+b^2} = \infty\) , when a (.. or b) approaches infinity
\(\theta = ArcTan(\frac{b}{a}) = ArcTan(\frac{0}{\infty}) = ArcCot(\frac{\infty}{0}) = 0\) => \(\infty \angle 0^{\circ}\) and because we were dealing with positive infinity of real part the theta is indeed 0 degrees (or should it be \(\theta = 2^{-\infty}\) hehe).

Now suddenly we can not return back since..
\( \begin{align*}
& a = Cos(\theta^{\circ})*\left | Z \right | = Cos(0^{\circ})*\left | \infty \right | = 1 *\infty =\infty\\
& b = Sin(\theta^{\circ})*\left | Z \right |= Sin(0^{\circ})*\left | \infty \right | = 0 *\infty = 0 (\neq 3)
\end{align*}\)

Yes, that's why I call it a "malformed" infinity. I think cartesian complex numbers involving infinity should not be allowed, or worst case allowed when typed by the user, but immediately converted to polar by the system.

(08-24-2016 04:06 PM)Vtile Wrote:  So how I understand this "directed infinity" is that it is "partially continuous" and we should keep attention in which sector in im / re plane it stays. Then it can be expressed with the "nonlossy" rectangular (complex plane) form.

I don't know why you call rectangular representation "nonlossy". You actually lose more information that way than in polar mode.
(Inf,3) is fine, but what if it's at let's say 30 degrees? (Inf,Inf) doesn't tell you much, you'd have to have a full representation Inf*(\(\sqrt{3}\over{2}\),\(1\over2\) ), which now needs an infinity object + a complex number.
The polar representation allows you to have directed infinity represented with magnitude and argument, just like any normal complex number, which makes it easier to handle.

(08-24-2016 04:06 PM)Vtile Wrote:  

Quote:(0,Inf) = Malformed, should be (Inf ∡90°)
(0,-Inf) = Malformed, should be (Inf ∡-90°)

Malformed why?

I think I covered this above: I think directed infinities should only be represented in polar form, therefore I call these forms "malformed".

(08-24-2016 04:06 PM)Vtile Wrote:  

Quote:(Inf,Inf) = Undirected infinity?

actually if a=inf and b=inf and inf=inf then it would be inf in angle of 45deg +n*90deg ???

Actually, inf=inf is not true. Two infinite quantities are not necessarily the same. For example, f(X)=2X as X tends to infinity is an infinity twice as large as g(X)=X when X tends to infinity (not sure if it makes sense with words), so there's Infinities that are greater than other infinities.
You can check by doing f(X)/g(X) when X tends to infinity: the limit is 2 proving that one infinity is bigger than the other.

This all means we don't know the direction of (inf,inf), could be anything therefore it's undirected infinity.

EDIT: Even though I sound like I knew this, I confess I learned this after the first post...

(08-24-2016 04:06 PM)Vtile Wrote:  PPS. what comes to the question if NewRPL is simple calculator the answer is NO, it is not a calculator, but a mathematical environment. Sorry it is first and foremost your project at this moment so it is your decision in the end.

Actually, this is the paragraph that makes more sense of all the posts in this thread. I think you are right, and I'm inclining to implement all these details into newRPL even if it takes a lot of work.