Diskussion:Pfeilschreibweise

aus Wikipedia, der freien Enzyklopädie

Folgendes steht noch in der englischen Version. Ich hab allerdings keine Zeit mehr das noch zu übersetzen. Bitte gern entfernen, wenn erledigt.

--Till P. 01:10, 28. Feb. 2008 (CET)

[Bearbeiten] Pfeilnotation in der Exponentialschreibweise

Versucht man $a \uparrow \uparrow b$ in gewöhnlicher Exponentialschreibweise zu schreiben, erhält man einen Turm von Exponenten. Wird b zu groß um diesen Turm von as zu schreiben, muss man auf Punkte ausweichen, die mit Hilfe einer Zahl angeben, wie hoch der Turm sein müsste.

requires a row of such power towers, separated by braces: there are b power towers, including the last with height 1, hence simply the number a. If b is too large  to write all these power towers, we use dots to indicate a row of them, and for the number of power towers a "cross-brace" (the number of braces is one less).  requires a row of such rows of power towers; there are b rows of power towers, including the last, which consists of only one "power tower" of height 1, so is simply the number a. If b is too large to write all these rows, we use a "cross-cross-brace" with this number b next to it (the number of cross-braces is one less). And so on.

Since the power notation is in direction "/", the braces are too. A row of them could be written in perpendicular direction "\", and the cross-brace too. A row of cross-braces could then extend in the direction "/", with a cross-cross-brace too, etc.

 Example:

*For 46 there are six power towers, including the last with height 1, hence simply the number 4; writing out the fifth power tower we have only five:

Using the left-superscript notation for tetration we have one "level of braces" less:  requires a "tetration tower" in the direction "\", and a brace with the number b next to it, indicating the height of the tetration tower.  requires a row of such tetration towers, separated by braces: there are b tetration towers, including the last with height 1, hence simply the number a.  If b is too large to write all these tetration towers, we use a "cross-brace" with this number b next to it. And so on.

Beispiele:

Das vorstehende Beispiel wird zu:

$^{^{^{^{^{4}4}4}4}4}4$

Für die vierte Ackermann Zahl $4 \uparrow \uparrow \uparrow \uparrow 4$ sind es vier Tetrationstürme, inklusive dem Letzten, der acht Mal die 1 enthält, also einfach 4 ergibt. Schreibt man den dritten Tetrationsturm aus, ergibt es lediglich drei:

$\begin{matrix} \underbrace{\begin{matrix} \underbrace{^{^{^{^{^{^{^{4}.}.}.}4}4}4}4} \\ ^{^{^{^{^{^{^{4}.}.}.}4}4}4}4 \end{matrix}} \\ ^{^{^{^4}4}4}4 \end{matrix}$

including the last with height 1, hence simply the number 4; writing out the third tetration tower we have only three:

[Bearbeiten] Verallgemeinerung

Einige Zahlen sind so groß, dass die Pfeilschreibweise unübersichtlich wird. Es wird dann der n-Pfeiloperator $\uparrow^n$ nützlich. Auch um eine variable Anzahl von Pfeilen darstellen zu können, kann sie genutzt werden, ebenso für den Hyper-Operator.

Einige Zahlen sind jedoch selbst für diese Notation noch zu groß, wie beispielsweise die Graham-Zahl.

Some numbers are so large that even that notation is not sufficient. Graham's number is an example. The Conway chained arrow notation can then be used: a chain of three elements is equivalent with the other notations, but a chain of four or more is even more powerful.

$\begin{matrix} a\uparrow^n b & = & \mbox{hyper}(a,n+2,b) & = & a\to b\to n \\ \mbox{(Knuth)} & & & & \mbox{(Conway)} \end{matrix}$

It is generally suggested that Knuth's arrow should be used for relatively smaller magnitude numbers, and the chained arrow or hyper operators for larger ones.

Right associativity is also natural because we can rewrite the iterated arrow expression $a\uparrow^n\cdots\uparrow^na$ that appears in the expansion of $a \uparrow^{n + 1}b$ as $a\uparrow^n\cdots\uparrow^na\uparrow^n1$ , so that all the $a$ s appear as left operands of arrow operators. This is significant since the arrow operators are not commutative.

Writing $(a\uparrow ^m)^b$ for the bth functional power of the function $f(n)=a\uparrow ^m n$ we have $a\uparrow ^n b = (a\uparrow ^{n-1})^b 1$ .

The definition could be extrapolated one step, starting with $a\uparrow^n b= ab$ if n = 0, because exponentiation is repeated multiplication starting with 1. Extrapolating one step more, writing multiplication as repeated addition, is not as straightforward because multiplication is repeated addition starting with 0 instead of 1. "Extrapolating" again one step more, writing addition of n as repeated addition of 1, requires starting with the number a. Compare the Vorlage:Ml, where the starting values for addition and multiplication are also separately specified.

[Bearbeiten] Tables of values

Computing $2\uparrow^m n$ can be restated in terms of an infinite table. We place the numbers 2 $n$ in the top row, and fill the left column with values 2. To determine a number in the table, take the number immediately to the left, then look up the required number in the previous row, at the position given by the number just taken.

Values of $2\uparrow^m n$ = hyper(2, m + 2, n) = 2 → n → m
m\n	1	2	3	4	5	6	7	formula
0	2	4	6	8	10	12	14	$2 n$
1	2	4	8	16	32	64	128	$2 n$
2	2	4	16	65536	$2^{65536}\approx 2.0 \times 10^{19,729}$	$2^{2^{65536}}\approx 10^{6.0 \times 10^{19,728}}$	$2^{2^{2^{65536}}}\approx 10^{10^{6.0 \times 10^{19,728}}}$	$2\uparrow\uparrow n$
3	2	4	65536	$\begin{matrix} \underbrace{2_{}^{2^{{}^{.\,^{.\,^{.\,^2}}}}}} \\ 65536\mbox{ copies of }2 \end{matrix}\approx (10\uparrow)^{65531}(6.0 \times 10^{19,728})$				$2\uparrow\uparrow\uparrow n$
4	2	4	$\begin{matrix} \underbrace{2_{}^{2^{{}^{.\,^{.\,^{.\,^2}}}}}}\\ 65536\mbox{ copies of }2 \end{matrix}$					$2\uparrow\uparrow\uparrow\uparrow n$

Note: $(10\uparrow)^k$ denotes a functional power of the function $f (n) = 10 n$ (the function also expressed by the suffix -plex as in googolplex).

The table is the same as that of the Ackermann function, except for a shift in $m$ and $n$ , and an addition of 3 to all values.

Computing $3\uparrow^m n$

We place the numbers 3 $n$ in the top row, and fill the left column with values 3. To determine a number in the table, take the number immediately to the left, then look up the required number in the previous row, at the position given by the number just taken.

Values of $3\uparrow^m n$ = hyper(3, m + 2, n) = 3 → n → m
m\n	1	2	3	4	5	formula
0	3	6	9	12	15	$3 n$
1	3	9	27	81	243	$3 n$
2	3	27	7,625,597,484,987	$3 7,625,597,484,987$		$3\uparrow\uparrow n$
3	3	7,625,597,484,987	$\begin{matrix} \underbrace{3_{}^{3^{{}^{.\,^{.\,^{.\,^3}}}}}}\\ 7,625,597,484,987\mbox{ copies of }3 \end{matrix}$			$3\uparrow\uparrow\uparrow n$
4	3	$\begin{matrix} \underbrace{3_{}^{3^{{}^{.\,^{.\,^{.\,^3}}}}}}\\ 7,625,597,484,987\mbox{ copies of }3 \end{matrix}$				$3\uparrow\uparrow\uparrow\uparrow n$

Computing $10\uparrow^m n$

We place the numbers 10 $n$ in the top row, and fill the left column with values 10. To determine a number in the table, take the number immediately to the left, then look up the required number in the previous row, at the position given by the number just taken.

Values of $10\uparrow^m n$ = hyper(10, m + 2, n) = 10 → n → m
m\n	1	2	3	4	5	formula
0	10	20	30	40	50	$10 n$
1	10	100	1,000	10,000	100,000	$10 n$
2	10	10,000,000,000	$10 10,000,000,000$	$10^{10^{10,000,000,000}}$	$10^{10^{10^{10,000,000,000}}}$	$10\uparrow\uparrow n$
3	10	$\begin{matrix} \underbrace{10_{}^{10^{{}^{.\,^{.\,^{.\,^{10}}}}}}}\\ 10\mbox{ copies of }10 \end{matrix}$	$\begin{matrix} \underbrace{10_{}^{10^{{}^{.\,^{.\,^{.\,^{10}}}}}}}\\ 10^{10}\mbox{ copies of }10 \end{matrix}$	$\begin{matrix} \underbrace{10_{}^{10^{{}^{.\,^{.\,^{.\,^{10}}}}}}}\\ 10^{10^{10}}\mbox{ copies of }10 \end{matrix}$		$10\uparrow\uparrow\uparrow n$
4	10	$\begin{matrix} \underbrace{^{^{^{^{^{10}.}.}.}10}10}\\ 10\mbox{ copies of }10 \end{matrix}$	$\begin{matrix} \underbrace{^{^{^{^{^{10}.}.}.}10}10}\\ 10^{10}\mbox{ copies of }10 \end{matrix}$			$10\uparrow\uparrow\uparrow\uparrow n$

Note that for 2 ≤ n ≤ 9 the numerical order of the numbers $10\uparrow^m n$ is the lexicographical order with m as the most significant number, so for the numbers of these 8 columns the numerical order is simply line-by-line. The same applies for the numbers in the 97 columns with 3 ≤ n ≤ 99, and if we start from m = 1 even for 3 ≤ n ≤ 9,999,999,999.

See also ebooksgratis.com: no banners, no cookies, totally FREE.

Diskussion:Pfeilschreibweise

aus Wikipedia, der freien Enzyklopädie

[Bearbeiten] Pfeilnotation in der Exponentialschreibweise

[Bearbeiten] Verallgemeinerung

[Bearbeiten] Tables of values

Views

Navigation

Mitmachen

Suche