P3C タキシング
前回はアイゼンシュタイン級数の数学的な定義を分解して以下のようなMaximaで計算できる形で定義してみました。
(%i59) Eis(z,k,M,N);
$$ \tag{%o59} \sum_{n=1}^{N}{\sum_{m=1}^{M}{\frac{1}{\left(m\,z+n\right)^{k}}}}+\sum_{n=1}^{N}{\sum_{m=1}^{M}{\frac{1}{\left(m\,z-n\right)^{k}}}}+\sum_{n=1}^{N}{\sum_{m=1}^{M}{\frac{1}{\left(n-m\,z\right)^{k}}}}+\sum_{n=1}^{N}{\sum_{m=1}^{M}{\frac{1}{\left(-m\,z-n\right)^{k}}}}+\frac{\sum_{m=1}^{M}{\frac{1}{m^{k}}}}{z^{k}}+\frac{\sum_{m=1}^{M}{\frac{1}{m^{k}}}}{\left(-z\right)^{k}}+\frac{\sum_{n=1}^{N}{\frac{1}{n^{k}}}}{\left(-1\right)^{k}}+\sum_{n=1}^{N}{\frac{1}{n^{k}}} $$
この後ろの方の項にゼータ関数の姿が見えます。実は第１項〜第４項にもゼータ関数が隠れています。今回はそちらを見てみます。
(%i60) F:first(%);
$$ \tag{%o60} \sum_{n=1}^{N}{\sum_{m=1}^{M}{\frac{1}{\left(m\,z+n\right)^{k}}}} $$
この２重和の内側の式でm, nは全ての自然数を渡るのですが、m, nの最大公約数をdとして、\( m=m_{d}\,d \), \( n=n_{d}\,d \) とすると \( m_{d} \), \( n_{d} \)は互いに素で、
(%i61) 1/(m*z+n)^k=1/(d^k*(m[d]*z+n[d])^k);
$$ \tag{%o61} \frac{1}{\left(m\,z+n\right)^{k}}=\frac{1}{d^{k}\,\left(m_{d}\,z+n_{d}\right)^{k}} $$
というように因数分解できます。
M=7, N=10として70の項について実際にこれをやってみます。
(%i62) powerdisp:true$
(%i63) a:F,M:7,N:10,nouns;
$$ \tag{%o63} \frac{1}{\left(1+z\right)^{k}}+\frac{1}{\left(2+z\right)^{k}}+\frac{1}{\left(3+z\right)^{k}}+\frac{1}{\left(4+z\right)^{k}}+\frac{1}{\left(5+z\right)^{k}}+\frac{1}{\left(6+z\right)^{k}}+\frac{1}{\left(7+z\right)^{k}}+\frac{1}{\left(8+z\right)^{k}}+\frac{1}{\left(9+z\right)^{k}}+\frac{1}{\left(10+z\right)^{k}}+\frac{1}{\left(1+2\,z\right)^{k}}+\frac{1}{\left(2+2\,z\right)^{k}}+\frac{1}{\left(3+2\,z\right)^{k}}+\frac{1}{\left(4+2\,z\right)^{k}}+\frac{1}{\left(5+2\,z\right)^{k}}+\frac{1}{\left(6+2\,z\right)^{k}}+\frac{1}{\left(7+2\,z\right)^{k}}+\frac{1}{\left(8+2\,z\right)^{k}}+\frac{1}{\left(9+2\,z\right)^{k}}+\frac{1}{\left(10+2\,z\right)^{k}}+\frac{1}{\left(1+3\,z\right)^{k}}+\frac{1}{\left(2+3\,z\right)^{k}}+\frac{1}{\left(3+3\,z\right)^{k}}+\frac{1}{\left(4+3\,z\right)^{k}}+\frac{1}{\left(5+3\,z\right)^{k}}+\frac{1}{\left(6+3\,z\right)^{k}}+\frac{1}{\left(7+3\,z\right)^{k}}+\frac{1}{\left(8+3\,z\right)^{k}}+\frac{1}{\left(9+3\,z\right)^{k}}+\frac{1}{\left(10+3\,z\right)^{k}}+\frac{1}{\left(1+4\,z\right)^{k}}+\frac{1}{\left(2+4\,z\right)^{k}}+\frac{1}{\left(3+4\,z\right)^{k}}+\frac{1}{\left(4+4\,z\right)^{k}}+\frac{1}{\left(5+4\,z\right)^{k}}+\frac{1}{\left(6+4\,z\right)^{k}}+\frac{1}{\left(7+4\,z\right)^{k}}+\frac{1}{\left(8+4\,z\right)^{k}}+\frac{1}{\left(9+4\,z\right)^{k}}+\frac{1}{\left(10+4\,z\right)^{k}}+\frac{1}{\left(1+5\,z\right)^{k}}+\frac{1}{\left(2+5\,z\right)^{k}}+\frac{1}{\left(3+5\,z\right)^{k}}+\frac{1}{\left(4+5\,z\right)^{k}}+\frac{1}{\left(5+5\,z\right)^{k}}+\frac{1}{\left(6+5\,z\right)^{k}}+\frac{1}{\left(7+5\,z\right)^{k}}+\frac{1}{\left(8+5\,z\right)^{k}}+\frac{1}{\left(9+5\,z\right)^{k}}+\frac{1}{\left(10+5\,z\right)^{k}}+\frac{1}{\left(1+6\,z\right)^{k}}+\frac{1}{\left(2+6\,z\right)^{k}}+\frac{1}{\left(3+6\,z\right)^{k}}+\frac{1}{\left(4+6\,z\right)^{k}}+\frac{1}{\left(5+6\,z\right)^{k}}+\frac{1}{\left(6+6\,z\right)^{k}}+\frac{1}{\left(7+6\,z\right)^{k}}+\frac{1}{\left(8+6\,z\right)^{k}}+\frac{1}{\left(9+6\,z\right)^{k}}+\frac{1}{\left(10+6\,z\right)^{k}}+\frac{1}{\left(1+7\,z\right)^{k}}+\frac{1}{\left(2+7\,z\right)^{k}}+\frac{1}{\left(3+7\,z\right)^{k}}+\frac{1}{\left(4+7\,z\right)^{k}}+\frac{1}{\left(5+7\,z\right)^{k}}+\frac{1}{\left(6+7\,z\right)^{k}}+\frac{1}{\left(7+7\,z\right)^{k}}+\frac{1}{\left(8+7\,z\right)^{k}}+\frac{1}{\left(9+7\,z\right)^{k}}+\frac{1}{\left(10+7\,z\right)^{k}} $$
各項を因数分解してみます。以下に現れる各項の分母のカッコの中のzの係数と定数は互いに素になっています。カッコの外には\( 2^{k} \), \( 3^{k} \), \( 4^{k} \), などが現れているのが見て取れます。
(%i64) b:map(factor,a);
$$ \tag{%o64} \frac{1}{\left(1+z\right)^{k}}+\frac{1}{2^{k}\,\left(1+z\right)^{k}}+\frac{1}{3^{k}\,\left(1+z\right)^{k}}+\frac{1}{4^{k}\,\left(1+z\right)^{k}}+\frac{1}{5^{k}\,\left(1+z\right)^{k}}+\frac{1}{6^{k}\,\left(1+z\right)^{k}}+\frac{1}{7^{k}\,\left(1+z\right)^{k}}+\frac{1}{\left(2+z\right)^{k}}+\frac{1}{2^{k}\,\left(2+z\right)^{k}}+\frac{1}{3^{k}\,\left(2+z\right)^{k}}+\frac{1}{4^{k}\,\left(2+z\right)^{k}}+\frac{1}{5^{k}\,\left(2+z\right)^{k}}+\frac{1}{\left(3+z\right)^{k}}+\frac{1}{2^{k}\,\left(3+z\right)^{k}}+\frac{1}{3^{k}\,\left(3+z\right)^{k}}+\frac{1}{\left(4+z\right)^{k}}+\frac{1}{2^{k}\,\left(4+z\right)^{k}}+\frac{1}{\left(5+z\right)^{k}}+\frac{1}{2^{k}\,\left(5+z\right)^{k}}+\frac{1}{\left(6+z\right)^{k}}+\frac{1}{\left(7+z\right)^{k}}+\frac{1}{\left(8+z\right)^{k}}+\frac{1}{\left(9+z\right)^{k}}+\frac{1}{\left(10+z\right)^{k}}+\frac{1}{\left(1+2\,z\right)^{k}}+\frac{1}{2^{k}\,\left(1+2\,z\right)^{k}}+\frac{1}{3^{k}\,\left(1+2\,z\right)^{k}}+\frac{1}{\left(3+2\,z\right)^{k}}+\frac{1}{2^{k}\,\left(3+2\,z\right)^{k}}+\frac{1}{3^{k}\,\left(3+2\,z\right)^{k}}+\frac{1}{\left(5+2\,z\right)^{k}}+\frac{1}{2^{k}\,\left(5+2\,z\right)^{k}}+\frac{1}{\left(7+2\,z\right)^{k}}+\frac{1}{\left(9+2\,z\right)^{k}}+\frac{1}{\left(1+3\,z\right)^{k}}+\frac{1}{2^{k}\,\left(1+3\,z\right)^{k}}+\frac{1}{\left(2+3\,z\right)^{k}}+\frac{1}{2^{k}\,\left(2+3\,z\right)^{k}}+\frac{1}{\left(4+3\,z\right)^{k}}+\frac{1}{2^{k}\,\left(4+3\,z\right)^{k}}+\frac{1}{\left(5+3\,z\right)^{k}}+\frac{1}{2^{k}\,\left(5+3\,z\right)^{k}}+\frac{1}{\left(7+3\,z\right)^{k}}+\frac{1}{\left(8+3\,z\right)^{k}}+\frac{1}{\left(10+3\,z\right)^{k}}+\frac{1}{\left(1+4\,z\right)^{k}}+\frac{1}{\left(3+4\,z\right)^{k}}+\frac{1}{\left(5+4\,z\right)^{k}}+\frac{1}{\left(7+4\,z\right)^{k}}+\frac{1}{\left(9+4\,z\right)^{k}}+\frac{1}{\left(1+5\,z\right)^{k}}+\frac{1}{\left(2+5\,z\right)^{k}}+\frac{1}{\left(3+5\,z\right)^{k}}+\frac{1}{\left(4+5\,z\right)^{k}}+\frac{1}{\left(6+5\,z\right)^{k}}+\frac{1}{\left(7+5\,z\right)^{k}}+\frac{1}{\left(8+5\,z\right)^{k}}+\frac{1}{\left(9+5\,z\right)^{k}}+\frac{1}{\left(1+6\,z\right)^{k}}+\frac{1}{\left(5+6\,z\right)^{k}}+\frac{1}{\left(7+6\,z\right)^{k}}+\frac{1}{\left(1+7\,z\right)^{k}}+\frac{1}{\left(2+7\,z\right)^{k}}+\frac{1}{\left(3+7\,z\right)^{k}}+\frac{1}{\left(4+7\,z\right)^{k}}+\frac{1}{\left(5+7\,z\right)^{k}}+\frac{1}{\left(6+7\,z\right)^{k}}+\frac{1}{\left(8+7\,z\right)^{k}}+\frac{1}{\left(9+7\,z\right)^{k}}+\frac{1}{\left(10+7\,z\right)^{k}} $$
(%i65) powerdisp:false$
