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There is an amazing and counterintuitive theorem:

For all $n$, there exists a $k$ such that the $k$-th term of the Goodstein sequence $G_k(n)=0$. In other words, every Goodstein sequence converges to $0$.

How can I find $N$ such $G_{N}(n)=0$?

for instance if $n=100$

$G_0(100)=2^{2^{2}+2}+2^{2^{2}+1}+2^2=100$

$G_1(100)=3^{3^{3}+3}+3^{3^{3}+1}+3^3-1=228767924549636$

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How to find $k$ such $G_{k}(100)=0$?

vito
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2 Answers2

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You don't.

Let $g$ be the "Goodstein terminating" function: $g(n)=k$ iff $G_k(n)=0$ and $G_{k-1}(n)\not=0$. Then $g$ grows insanely fast: if you've heard of the Ackermann function, it's of a similar species. In fact, new notation and concepts had to be invented to even talk about how fast such functions grow! Look up "fast-growing hierarchies" (or see https://en.wikipedia.org/wiki/Goodstein%27s_theorem#Sequence_length_as_a_function_of_the_starting_value).

In fact, $g$ grows so fast, that when I say $$\mbox{$g(100)$ has more digits than there are atoms in the known universe,}$$ that's an understatement.

Peter
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Noah Schweber
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  • Probably greater than the number of plank volumes.... You can sometimes calculate some properties of the number though like its right most didget but even that would take a while. – shai horowitz Jul 01 '16 at 20:51
  • @shaihorowitz Really? I don't see a way to even calculate $g(100)$ mod 2 more efficiently than just calculating $g(100)$ itself - there isn't an obvious role parity plays in the Goodstein sequence construction (or am I missing something?). – Noah Schweber Jul 01 '16 at 20:56
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    $g(4)=6.895\times10^{121210694}$ see here – vito Jul 01 '16 at 21:20
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    @vito Yes - and $g(100)$ is unimaginably larger than that. – Noah Schweber Jul 01 '16 at 21:53
  • @Noah Schweber Since $100 = 2^{2^2 + 2} + 2^{2^2+1} + 2$, $g(100) = H_{\omega^{\omega^\omega+\omega} + \omega^{\omega^\omega+1} + \omega}(3)-3 = F_{\omega^\omega+\omega}(F_{\omega^\omega+1}(F_1(3)))-3$, where $H$ and $F$ are the Hardy hierarchy and fast-growing hierarchy respectively. This will eventually resolve to $F_1(n)-3 = 2n-3$ for some natural number $n$, so $g(100)$ is odd. In fact $g(n)$ is odd for any $n>0$ by the same reasoning. – Deedlit Jul 02 '16 at 21:09
  • Actually, we don't need the formulas; for $n \ge 2$, the number will start off at least as big as the base, so at some point the number will equal the base. Call this number $n$. Then the next step will have number $n$ and base $n+1$, and then $n$ steps later we will have number $0$ and base $2n+1$. Since the base started off as $2$, the process took $2n-1$ steps. – Deedlit Jul 02 '16 at 21:34
  • @Deedlit Could you check my answer? I do believe I've done the approximation correctly. (expanding out all the smaller parts so that the result is a tad more accurate) – Simply Beautiful Art Jul 04 '17 at 00:55
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We happen to have an approximation to this sequence using a thing called the Hardy Hierarchy. Here, we happen to have

$$k_{100}\approx H_{\omega^{\omega^\omega+\omega}+\omega^{\omega^\omega+1}}(2^{402653184}402653184-1)$$

This may be noted by observing the behavior of the Goodstein sequence:

$$n^ka_k+n^{k-1}a_{k-1}+\dots+na_1+a_0$$

If $a_0>0$, then this behaves like a successor ordinal:

$$n^ka_k+n^{k-1}a_{k-1}+\dots+na_1+a_0\\\to(n+1)^ka_k+(n+1)^{k-1}a_{k-1}+\dots+(n+1)a_1+a_0-1$$

The Hardy Hierarchy works similarly:

$$H_{\alpha+1}(n)=H_\alpha(n+1)$$

As you can see, the base in both goes up by one, while the overall number goes down by one.

If $a_p=0$ for $p\in[0,j]$, then this behaves similarly to the limit ordinal:

$$n^ka_k+n^{k-1}a_{k-1}+\dots+n^{j+1}a_j\\\small\to(n+1)^ka_k+(n+1)^{k-1}a_{k-1}+\dots+(n+1)^{j+1}(a_j-1)+(n+1)^jn+\dots+(n+1)n+n$$

For the Hardy Hierarchy:

$$H_{\omega^ka_k+\omega^{k-1}a_{k-1}+\dots+\omega^ja_j}(n)=H_{\omega^ka_k+\omega^{k-1}a_{k-1}+\dots+\omega^j(a_j-1)+\omega^j(n-1)+\dots+\omega(n-1)+n}(n)$$

This is the same with the very small exception that the base (the number in the parenthesis) does not go up. However, this is extremely close to the Goodstein sequence...

So this is basically how I got the approximation, after expanding out $g_{2^{402653184}402653184-1}(100)$ (easily done once you've spotted the patterns)

All that being said, this approximation only goes to show that this is much larger than what most people could fathom. In terms of the fast growing hierarchy, which a few people may know better, we have

$$k_{100}\approx f_{\omega^\omega+\omega}(f_{\omega^\omega+1}(2^{402653184}402653184-1)$$

As Deedlit notes, the base being off can be fixed by considering

$$H_{\omega^ka_k+\omega^{k-1}a_{k-1}+\dots+\omega^ja_j}(n+1)$$

Thus, we actually get an exact value:

$$\begin{align}k_{100}&=H_{\omega^{\omega^\omega+\omega}+\omega^{\omega^\omega+1}+\omega^\omega}(3)-3\\&=H_{\omega^{\omega^\omega+\omega}}(H_{\omega^{\omega^\omega+1}}(H_{\omega^\omega}(3)))\\&=f_{\omega^\omega+\omega}\left(f_{\omega^\omega+1}\left(2^{2^{24}\cdot24}\cdot2^{24}\cdot24\right)\right)\end{align}$$

Simply Beautiful Art
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    Actually, the correspondence with the Hardy Hierarchy is not an approximation, it's an equality! As you have observed, when $n$ corresponds to a successor ordinal it's clear that both procedures do the same thing. At limit ordinals, the correspondence is more subtle; the Goodstein procedure corresponds to taking fundamental sequences (at the value of the new base) until you reach a successor ordinal, then subtract one. So for example: If we have a base of 4 and the number is $4^3$, then we take $5^3 - 1 = 5^2+5^2+5^2+5^2+5^2-1=5^2+5^2+5^2+5^2+5+5+5+5+5-1=5^24+54+15-1=5^24+54+14$$. – Deedlit Jul 05 '17 at 01:37
  • @Deedlit You sure its exact? Using the regular fundamental sequences and Hardy Hierarchy? If my understanding is correct, does the exact equality come out if we have: $$H_\alpha(n)=\begin{cases}n,&\alpha=0\H_\beta(n+1),&\alpha=\beta+1\H_{\alpha[n+1]}(n),&\alpha\in\mathbb{Lim}\end{cases}$$ – Simply Beautiful Art Jul 05 '17 at 01:41
  • For the Hardy Hierarchy, this corresponds to $f_{\omega^3}(5) = f_{\omega^2 * 5}(5)=f_{\omega^24+\omega5}(5)=f_{\omega^24+\omega4+5}(5)=f_{\omega^24+\omega4+4}(6)$, so the Goodstein process is equivalent to taking 4 steps of the Goodstein sequence. Notice that the argument for the Hardy Hierarchy is one more than the current base for the Goodstein sequence. – Deedlit Jul 05 '17 at 01:45
  • Hm, don't you mean to start at $f_{\omega^3}(4)$? – Simply Beautiful Art Jul 05 '17 at 01:47
  • Assuming $4^3$ corresponds to $H_{\omega^3}(4)$, we would have, by my definition, $4^3\to5^24+5\cdot4+4$ and $H_{\omega^3}(4)=H_{\omega^25}(4)=H_{\omega^24+\omega5}(4)=H_{\omega^24+\omega4+5}(4)=H_{\omega^24+\omega4+4}(5)$, which is an exact equality. – Simply Beautiful Art Jul 05 '17 at 01:50
  • Also, I shall have to continue this conversation tomorrow. Good night! – Simply Beautiful Art Jul 05 '17 at 01:50
  • No, since the base increments to $5$, when we subtract one we get the effect of $\omega$ being converted to $5$. The exact value should be $k_{100}=H_{\omega^{\omega^\omega+\omega}+\omega^{\omega^\omega+1}+{\omega^\omega}}(3)-3 = H_{\omega^{\omega^\omega+\omega}+\omega^{\omega^\omega+1}}(402653184 \cdot 2^{402653184}) - 3$. – Deedlit Jul 05 '17 at 01:51
  • Ah, I see. Hm, so we would've wanted the base number in $H_\alpha(n)$ to be one larger than the current number, with $\alpha$ the corresponding ordinal to $k$, and this gives us an exact equality... beautiful! – Simply Beautiful Art Jul 05 '17 at 01:52
  • The exact result is in the Wikipedia article and proved in Caicedo's paper cited there. – r.e.s. Jul 05 '17 at 15:57
  • @r.e.s. Thanks. In the Wikipedia article, do you know what FS it uses? It appears that it has $H_{\omega^\alpha}(n)=f_\alpha(n+1)$, which is a bit off the normal FS AFAIK. – Simply Beautiful Art Jul 05 '17 at 17:03
  • Where it's citing Caicedo (2007), presumably the fundamental sequences are those defined there (on p.5). With the hierarchy definitions given in that paper, it follows that $H_{\omega^\alpha}(n)=f_{\alpha}(n+1)-1$ (see Eq.1, p.7). – r.e.s. Jul 07 '17 at 13:35