How do you read the --> in z-->z²+c

Chillheimer

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Fractal Schemer

Posts: 972

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How do you read the --> in z-->z²+c

« on: April 25, 2016, 12:34:33 PM »

Hi!
I'm making a short movie about fractals and have to read the m-set formula.
do you say z equals z²+c?
or does the arrow have its own word?
Thx in advance!


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claude

Fractal Bachius

Posts: 563

Re: How do you read the --> in z-->z²+c

« Reply #1 on: April 25, 2016, 01:06:00 PM »

"z equals z squared plus c" is wrong (it defines a quadratic equation with two solutions).

"z maps to z squared plus c" is probably what I would say, or modifying or leaving out the "maps" depending on context ("repeated iterations of the mapping 'z to z squared plus c'" could work), although the LaTeX symbol is slightly different: $\mapsto$ \mapsto, vs $\to$ \to

alternatively subscripts with iteration can use an explicit equals: "z n plus one equals z n squared plus c"


« Last Edit: April 25, 2016, 01:41:29 PM by claude, Reason: fix typos (thanks quaz0r) »	Logged

kram1032

Fractal Senior

Posts: 1863

Re: How do you read the --> in z-->z²+c

« Reply #2 on: April 25, 2016, 01:21:15 PM »

"is mapped to" or "is replaced by" or "is substituted for"
The difference between
$\left(1\right) \ \ \ z \to z^2+c$
and
$\left(2\right) \ \ \ z=z^2+c$
is that $\left(1\right)$ refers to the M-set at any given iteration step - it's valid for all of them, whereas $\left(2\right)$ talks about the limit set. Only in the limit of infinitely many iterations is it actually true that $z=z^2+c$ .

Although claude is right that usually $z=z^2+c$ would be considered a quadratic equation with two solutions. It only works as I described if you consider z to be the entire Julia- or Mandelbrot set, rather than a single point which only holds true for a few trivial cases (namely the above mentioned solutions of the quadratic equation).


« Last Edit: April 25, 2016, 01:45:08 PM by kram1032 »	Logged

quaz0r

Fractal Molossus

Posts: 652

Re: How do you read the --> in z-->z²+c

« Reply #3 on: April 25, 2016, 01:34:43 PM »

all of that is rather awkward for casually speaking the words... couldnt one say Zn+1 equals Zn^2 + c, or for that matter, just say z equals z^2 + c, the fact that you are referring to how each iteration is calculated either being implied or explicitly stated, as opposed to deffining a quadractic euqation (claude hasnt had his coffee yet)

i think the casual listener probably won't assume you are wrongly defining a quadratic equation and feel all horribly offended and led astray..


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kram1032

Fractal Senior

Posts: 1863

Re: How do you read the --> in z-->z²+c

« Reply #4 on: April 25, 2016, 01:49:10 PM »

In how far is saying "z maps to z squared plus c" any more or less awkward than saying "z equals z squared plus c" and how is it any more awkward than explicitly mentioning the iteration variable "z n plus one equals z n squared plus c"? Why mention variables that are only used internally for housekeeping? Explicitly mentioning the iteration variables is both awkward and ambiguous.
"z equals z squared plus c" will rarely but occasionally be ambiguous.
"z maps to z squared plus c" never is ambiguous.


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Chillheimer

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Fractal Schemer

Posts: 972

Just another fractal being floating by..

Re: How do you read the --> in z-->z²+c

« Reply #5 on: April 26, 2016, 10:45:35 AM »

Thx guys, "maps to" is exactly what I was looking for.
Falls jemand die deutsche Übersetzung kennt, die wüsste ich auch gerne..


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hobold

Fractal Bachius

Posts: 573

Re: How do you read the --> in z-->z²+c

« Reply #6 on: April 26, 2016, 11:34:43 AM »

Quote from: Chillheimer on April 26, 2016, 10:45:35 AM

Thx guys, "maps to" is exactly what I was looking for.
Falls jemand die deutsche Übersetzung kennt, die wüsste ich auch gerne..

"z wird abgebildet auf z² + c"


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