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π^π^π^πは整数か？(2次までの超Taylor編)

199

はじめに

少し前のことですが、このような動画を見ました。
その時に思ったのは、「これ、 $x^{x^{x^{x}}}$ の $x = π$ におけるTaylor展開使えば求まるのでは」ということでした。
実際にこの計算を行おうとすると、 $x^{x^{x^{x}}}$ の微分を求めなければなりません。 $(x^{x^{x^{x}}})^{'} = x^{x^{x^{x}}} (x^{x^{x} - 1} + x^{x^{x}} \log x (x^{x - 1} + x^{x} \log x (1 + \log x)))$ ( Wolfram Mathematica を用いて計算)より、 $x = π$ の時の微分係数が $π^{π^{π^{π}}}$ よりも大きくなるはずなので計算が頓挫したのですが、これは次の考えにつながります。
「では超Taylor展開を使えばいいのでは?」
つい先日Y.K.氏によりそれが見つけられたとのことなので[1]、それを使ってひたすらに計算をしていこうと思います。

材料

$a_{n}$ : $π$ の小数第 $n$ 桁目まで( $a_{0} = 3$ )
$b_{n} := a_{n} + 10^{- n}$
$e_{n}$ : $e$ の小数第 $n$ 桁目まで( $e_{0} = 2$ )
$\begin{aligned} (x^{x^{x^{x}}})^{‘} & = \frac{x \cdot x^{x^{x^{x}}} (x^{x^{x} - 1} + x^{x^{x}} \log x (x^{x - 1} + x^{x} \log x (1 + \log x)))}{x^{x^{x^{x}}}} \\ = x^{x^{x}} (1 + x^{x} \log x \cdot (1 + x \log x \cdot (1 + \log x))) \end{aligned}$
$(x^{x^{x^{x}}})^{‘ ‘} = \frac{x^{x} (2 + \log x \cdot (x^{x} (1 + x \log x \cdot (1 + \log x))^{2} + x (4 + \log x \cdot (6 + x + \log x \cdot (1 + 2 x + x \log x)))))}{1 + x^{x} \log x \cdot (1 + x \log x \cdot (1 + \log x))}$
2024/12/17追記：超導関数は Wolfram Mathematica を用いて計算

手法

$x = a_{n}$ における $k$ 次の $f (x) = x^{x^{x^{x}}}$ の超Taylor級数 $T_{k} (x)$ を考え、比 $\frac{T_{k} (b_{n})}{T_{k} (a_{n})}$ が $1 + 10^{⌊ - \log_{10} T_{k} (a_{n}) ⌋}$ よりも小さくなるような最大の $n$ を考える。
$n$ が現在計算されている $π$ の桁数よりも小さいなら、現時点で計算は可能となるため、計算する最大限の努力をする。

計算

まず、 $f (x)$ の1次の超Taylor級数を計算する。
$\begin{aligned} T_{1} (x) & = f (a_{n}) {(\frac{x}{a_{n}})}^{f^{‘} (a_{n})} \\ = {a_{n}}^{{a_{n}}^{{a_{n}}^{a_{n}}}} {(\frac{x}{a_{n}})}^{{a_{n}}^{{a_{n}}^{a_{n}}} (1 + {a_{n}}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n})))} \end{aligned}$ [1]
よって比 $\frac{T_{1} (b_{n})}{T_{1} (a_{n})}$ は、
$\begin{aligned} \frac{T_{1} (b_{n})}{T_{1} (a_{n})} & = \frac{{a_{n}}^{{a_{n}}^{{a_{n}}^{a_{n}}}} {(\frac{b_{n}}{a_{n}})}^{{a_{n}}^{{a_{n}}^{a_{n}}} (1 + {a_{n}}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n})))}}{{a_{n}}^{{a_{n}}^{{a_{n}}^{a_{n}}}} {(\frac{a_{n}}{a_{n}})}^{{a_{n}}^{{a_{n}}^{a_{n}}} (1 + {a_{n}}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n})))}} \\ = {(\frac{b_{n}}{a_{n}})}^{{a_{n}}^{{a_{n}}^{a_{n}}} (1 + {a_{n}}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n})))} \\ = {(1 + \frac{10^{- n}}{a_{n}})}^{{a_{n}}^{{a_{n}}^{a_{n}}} (1 + {a_{n}}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n})))} \\ > 1 + \frac{10^{- n}}{a_{n}} \cdot {a_{n}}^{{a_{n}}^{a_{n}}} (1 + {a_{n}}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))) \end{aligned}$
と表される。
そのため、
$\begin{aligned} \frac{T_{1} (b_{n})}{T_{1} (a_{n})} < 1 + 10^{⌊ - \log_{10} T_{1} (a_{n}) ⌋} \\ \Rightarrow 1 + \frac{10^{- n}}{a_{n}} \cdot {a_{n}}^{{a_{n}}^{a_{n}}} (1 + {a_{n}}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))) < 1 + 10^{- \log_{10} T_{1} (a_{n})} \\ \Leftrightarrow \frac{10^{- n}}{a_{n}} \cdot {a_{n}}^{{a_{n}}^{a_{n}}} (1 + {a_{n}}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))) < \frac{1}{T_{1} (a_{n})} \\ \Leftrightarrow {a_{n}}^{{a_{n}}^{{a_{n}}^{a_{n}}}} \cdot a_{n}^{{a_{n}}^{a_{n}} - 1} (1 + {a_{n}}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))) < 10^{n} \end{aligned}$
以上より、 $n$ は、 ${a_{m}}^{{a_{m}}^{{a_{m}}^{a_{m}}}} < 10^{m}$ を満たす最小の $m$ より大きく、現在計算されている円周率の桁数よりも大きくなってしまうため、現状では計算できない。
同様に2次の超Taylor級数でも計算する。
$\begin{aligned} T_{2} (x) & = f (a_{n}) e^{\frac{f^{‘} (a_{n})}{f^{‘ ‘} (a_{n})} ({(\frac{x}{a_{n}})}^{f^{‘ ‘} (a_{n})} - 1)} \\ = {a_{n}}^{{a_{n}}^{{a_{n}}^{a_{n}}}} \cdot e^{\frac{{a_{n}}^{{a_{n}}^{a_{n}}} (1 + a_{n}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n})))^{2}}{{a_{n}}^{a_{n}} (2 + \log a_{n} \cdot ({a_{n}}^{a_{n}} (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))^{2} + a_{n} (4 + \log a_{n} \cdot (6 + a_{n} + \log a_{n} \cdot (1 + 2 a_{n} + a_{n} \log a_{n})))))} ({(\frac{x}{a_{n}})}^{\frac{1 + a_{n}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))}{{a_{n}}^{a_{n}} (2 + \log a_{n} \cdot ({a_{n}}^{a_{n}} (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))^{2} + a_{n} (4 + \log a_{n} \cdot (6 + a_{n} + \log a_{n} \cdot (1 + 2 a_{n} + a_{n} \log a_{n})))))}} - 1)} \end{aligned}$ [1]
$\begin{aligned} \frac{T_{2} (b_{n})}{T_{2} (a_{n})} & = \frac{{a_{n}}^{{a_{n}}^{{a_{n}}^{a_{n}}}} \cdot e^{\frac{{a_{n}}^{{a_{n}}^{a_{n}}} (1 + a_{n}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n})))^{2}}{{a_{n}}^{a_{n}} (2 + \log a_{n} \cdot ({a_{n}}^{a_{n}} (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))^{2} + a_{n} (4 + \log a_{n} \cdot (6 + a_{n} + \log a_{n} \cdot (1 + 2 a_{n} + a_{n} \log a_{n})))))} ({(\frac{b_{n}}{a_{n}})}^{\frac{1 + a_{n}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))}{{a_{n}}^{a_{n}} (2 + \log a_{n} \cdot ({a_{n}}^{a_{n}} (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))^{2} + a_{n} (4 + \log a_{n} \cdot (6 + a_{n} + \log a_{n} \cdot (1 + 2 a_{n} + a_{n} \log a_{n})))))}} - 1)}}{{a_{n}}^{{a_{n}}^{{a_{n}}^{a_{n}}}} \cdot e^{\frac{{a_{n}}^{{a_{n}}^{a_{n}}} (1 + a_{n}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n})))^{2}}{{a_{n}}^{a_{n}} (2 + \log a_{n} \cdot ({a_{n}}^{a_{n}} (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))^{2} + a_{n} (4 + \log a_{n} \cdot (6 + a_{n} + \log a_{n} \cdot (1 + 2 a_{n} + a_{n} \log a_{n})))))} ({(\frac{a_{n}}{a_{n}})}^{\frac{1 + a_{n}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))}{{a_{n}}^{a_{n}} (2 + \log a_{n} \cdot ({a_{n}}^{a_{n}} (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))^{2} + a_{n} (4 + \log a_{n} \cdot (6 + a_{n} + \log a_{n} \cdot (1 + 2 a_{n} + a_{n} \log a_{n})))))}} - 1)}} \\ = e^{\frac{{a_{n}}^{{a_{n}}^{a_{n}}} (1 + a_{n}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n})))^{2}}{{a_{n}}^{a_{n}} (2 + \log a_{n} \cdot ({a_{n}}^{a_{n}} (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))^{2} + a_{n} (4 + \log a_{n} \cdot (6 + a_{n} + \log a_{n} \cdot (1 + 2 a_{n} + a_{n} \log a_{n})))))} ({(\frac{b_{n}}{a_{n}})}^{\frac{1 + a_{n}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))}{{a_{n}}^{a_{n}} (2 + \log a_{n} \cdot ({a_{n}}^{a_{n}} (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))^{2} + a_{n} (4 + \log a_{n} \cdot (6 + a_{n} + \log a_{n} \cdot (1 + 2 a_{n} + a_{n} \log a_{n})))))}} - 1)} \\ > e^{\frac{{a_{n}}^{{a_{n}}^{a_{n}}} (1 + a_{n}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n})))^{2}}{{a_{n}}^{a_{n}} (2 + \log a_{n} \cdot ({a_{n}}^{a_{n}} (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))^{2} + a_{n} (4 + \log a_{n} \cdot (6 + a_{n} + \log a_{n} \cdot (1 + 2 a_{n} + a_{n} \log a_{n})))))} (\frac{10^{- n}}{a_{n}} \cdot \frac{1 + a_{n}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))}{{a_{n}}^{a_{n}} (2 + \log a_{n} \cdot ({a_{n}}^{a_{n}} (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))^{2} + a_{n} (4 + \log a_{n} \cdot (6 + a_{n} + \log a_{n} \cdot (1 + 2 a_{n} + a_{n} \log a_{n})))))})} \\ = e^{\frac{{a_{n}}^{{a_{n}}^{a_{n}}} (1 + a_{n}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n})))^{3}}{({a_{n}}^{a_{n}} (2 + \log a_{n} \cdot ({a_{n}}^{a_{n}} (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))^{2} + a_{n} (4 + \log a_{n} \cdot (6 + a_{n} + \log a_{n} \cdot (1 + 2 a_{n} + a_{n} \log a_{n}))))))^{2}} \frac{10^{- n}}{a_{n}}} \\ > (1 + 10^{- n})^{\frac{{a_{n}}^{{a_{n}}^{a_{n}}} (1 + a_{n}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n})))^{3}}{a_{n} ({a_{n}}^{a_{n}} (2 + \log a_{n} \cdot ({a_{n}}^{a_{n}} (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))^{2} + a_{n} (4 + \log a_{n} \cdot (6 + a_{n} + \log a_{n} \cdot (1 + 2 a_{n} + a_{n} \log a_{n}))))))^{2}}} \\ > 1 + 10^{- n} \cdot \frac{{a_{n}}^{{a_{n}}^{a_{n}}} (1 + a_{n}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n})))^{3}}{a_{n} ({a_{n}}^{a_{n}} (2 + \log a_{n} \cdot ({a_{n}}^{a_{n}} (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))^{2} + a_{n} (4 + \log a_{n} \cdot (6 + a_{n} + \log a_{n} \cdot (1 + 2 a_{n} + a_{n} \log a_{n}))))))^{2}} \end{aligned}$
$\begin{aligned} \frac{T_{2} (b_{n})}{T_{2} (a_{n})} < 1 + 10^{⌊ - \log_{10} T_{2} (a_{n}) ⌋} \\ \Rightarrow 1 + 10^{- n} \cdot \frac{{a_{n}}^{{a_{n}}^{a_{n}}} (1 + a_{n}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n})))^{3}}{a_{n} ({a_{n}}^{a_{n}} (2 + \log a_{n} \cdot ({a_{n}}^{a_{n}} (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))^{2} + a_{n} (4 + \log a_{n} \cdot (6 + a_{n} + \log a_{n} \cdot (1 + 2 a_{n} + a_{n} \log a_{n}))))))^{2}} < 1 + 10^{- \log_{10} T_{2} (a_{n})} \\ \Leftrightarrow 10^{- n} \cdot \frac{{a_{n}}^{{a_{n}}^{a_{n}}} (1 + a_{n}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n})))^{3}}{a_{n} ({a_{n}}^{a_{n}} (2 + \log a_{n} \cdot ({a_{n}}^{a_{n}} (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))^{2} + a_{n} (4 + \log a_{n} \cdot (6 + a_{n} + \log a_{n} \cdot (1 + 2 a_{n} + a_{n} \log a_{n}))))))^{2}} < \frac{1}{T_{2} (a_{n})} \\ \Leftrightarrow {a_{n}}^{{a_{n}}^{{a_{n}}^{a_{n}}}} \cdot \frac{{a_{n}}^{{a_{n}}^{a_{n}}} (1 + a_{n}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n})))^{3}}{a_{n} ({a_{n}}^{a_{n}} (2 + \log a_{n} \cdot ({a_{n}}^{a_{n}} (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))^{2} + a_{n} (4 + \log a_{n} \cdot (6 + a_{n} + \log a_{n} \cdot (1 + 2 a_{n} + a_{n} \log a_{n}))))))^{2}} < 10^{n} \end{aligned}$
ここで、
$\frac{{a_{n}}^{{a_{n}}^{a_{n}}} (1 + a_{n}^{a_{n}} \log a_{n} \cdot (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n})))^{3}}{a_{n} ({a_{n}}^{a_{n}} (2 + \log a_{n} \cdot ({a_{n}}^{a_{n}} (1 + a_{n} \log a_{n} \cdot (1 + \log a_{n}))^{2} + a_{n} (4 + \log a_{n} \cdot (6 + a_{n} + \log a_{n} \cdot (1 + 2 a_{n} + a_{n} \log a_{n}))))))^{2}}$
の評価をする。
関数
$f (x) = \frac{x^{x^{x}} (1 + x^{x} \log x \cdot (1 + x \log x \cdot (1 + \log x)))^{3}}{x (x^{x} (2 + \log x \cdot (x^{x} (1 + x \log x \cdot (1 + \log x))^{2} + x (4 + \log x \cdot (6 + x + \log x \cdot (1 + 2 x + x \log x))))))^{2}}$
は連続なので、このグラフより、 $a_{n}$ において $f (a_{n})$ は $1$ よりも大きい。
以上より、 $n$ は、 ${a_{m}}^{{a_{m}}^{{a_{m}}^{a_{m}}}} < 10^{m}$ を満たす最小の $m$ より大きく、現在計算されている円周率の桁数よりも大きくなってしまうため、現状では計算できない。
以上より、2次までの超Taylor級数では $π^{π^{π^{π}}}$ が整数であるかどうかは現状わからない。