ある行列式の計算

ある行列式の計算

Muntzの定理を証明するのに必要となった公式の計算をする。近似理論では誤差を行列式で評価できる。式が横に長くなるのでスマホで見る場合は横にして見てほしい。

$${{{\Delta }_{n}}=\left| \begin{matrix}\frac{1}{{{a}_{1}}+{{b}_{1}}} & \frac{1}{{{a}_{1}}{{b}_{2}}}&\cdots & \frac{1}{{{a}_{1}}+{{b}_{n}}} \\\frac{1}{{{a}_{2}}+{{b}_{1}}} & \frac{1}{{{a}_{2}}{{b}_{2}}}&\cdots & \frac{1}{{{a}_{2}}+{{b}_{n}}} \\\vdots & \vdots & \ddots & \vdots \\\frac{1}{{{a}_{n}}+{{b}_{1}}} & \frac{1}{{{a}_{n}}+{{b}_{2}}} & \cdots & \frac{1}{{{a}_{n}}+{{b}_{n}}} \\\end{matrix} \right|}$$

$${=\frac{\prod\limits_{i=1}^{n-1}{\left( {{a}_{n}}-{{a}_{i}} \right)\left( {{b}_{n}}-{{b}_{i}} \right)}}{\left( {{a}_{n}}+{{b}_{n}} \right)\prod\limits_{i=1}^{n-1}{\left( {{a}_{n}}+{{b}_{i}} \right)\left( {{b}_{n}}+{{a}_{i}} \right)}}{{\Delta }_{n-1}}}$$
がなりたつ。
 
証明：
 
それぞれの行から一番下の行を引く

$${{{\Delta }_{n}}=\left| \begin{matrix}\frac{1}{{{a}_{1}}+{{b}_{1}}}-\frac{1}{{{a}_{n}}+{{b}_{1}}} & \frac{1}{{{a}_{1}}+{{b}_{2}}}-\frac{1}{{{a}_{n}}+{{b}_{2}}} & \cdots & \frac{1}{{{a}_{1}}+{{b}_{n}}}-\frac{1}{{{a}_{n}}+{{b}_{n}}} \\\frac{1}{{{a}_{2}}+{{b}_{1}}}-\frac{1}{{{a}_{n}}+{{b}_{1}}} & \frac{1}{{{a}_{2}}+{{b}_{2}}}-\frac{1}{{{a}_{n}}+{{b}_{2}}} & \cdots & \frac{1}{{{a}_{2}}+{{b}_{n}}}-\frac{1}{{{a}_{n}}+{{b}_{n}}} \\\vdots & \vdots & \ddots & \vdots \\\frac{1}{{{a}_{n}}+{{b}_{1}}} & \frac{1}{{{a}_{n}}{{b}_{2}}}&\cdots & \frac{1}{{{a}_{n}}+{{b}_{n}}} \\\end{matrix} \right|}$$

$${{{\Delta }_{n}}=\left| \begin{matrix}\frac{{{a}_{n}}-{{a}_{1}}}{{{a}_{1}}+{{b}_{1}}}\frac{1}{{{a}_{n}}+{{b}_{1}}} & \frac{{{a}_{n}}-{{a}_{1}}}{{{a}_{1}}+{{b}_{2}}}\frac{1}{{{a}_{n}}+{{b}_{2}}} & \cdots & \frac{{{a}_{n}}-{{a}_{1}}}{{{a}_{1}}+{{b}_{n}}}-\frac{1}{{{a}_{n}}+{{b}_{n}}} \\\frac{{{a}_{n}}-{{a}_{2}}}{{{a}_{2}}+{{b}_{1}}}\frac{1}{{{a}_{n}}+{{b}_{1}}} & \frac{{{a}_{n}}-{{a}_{2}}}{{{a}_{2}}+{{b}_{2}}}\frac{1}{{{a}_{n}}+{{b}_{2}}} & \cdots & \frac{{{a}_{n}}-{{a}_{2}}}{{{a}_{2}}+{{b}_{n}}}-\frac{1}{{{a}_{n}}+{{b}_{n}}} \\\vdots & \vdots & \ddots & \vdots \\\frac{1}{{{a}_{n}}+{{b}_{1}}} & \frac{1}{{{a}_{n}}+{{b}_{2}}} & \cdots & \frac{1}{{{a}_{n}}+{{b}_{n}}} \\\end{matrix} \right|}$$

共通項をくくりだす。

$${{{\Delta }_{n}}=\frac{\left( {{a}_{n}}-{{a}_{1}} \right)\left( {{a}_{n}}-{{a}_{2}} \right)\cdots \left( {{a}_{n}}-{{a}_{n-1}} \right)}{\left( {{a}_{n}}+{{b}_{1}} \right)\left( {{a}_{n}}+{{b}_{2}}\right)\cdots \left( {{a}_{n}}+{{b}_{n}} \right)}\left| \begin{matrix}\frac{1}{{{a}_{1}}+{{b}_{1}}} & \frac{1}{{{a}_{1}}+{{b}_{2}}} & \cdots & \frac{1}{{{a}_{1}}+{{b}_{n}}} \\\frac{1}{{{a}_{2}}+{{b}_{1}}} & \frac{1}{{{a}_{2}}+{{b}_{2}}} & \cdots & \frac{1}{{{a}_{2}}+{{b}_{n}}} \\\vdots & \vdots & \ddots & \vdots \\1 & 1 & \cdots & 1 \\\end{matrix} \right|}$$

$${{{\Delta }_{n}}=\frac{\left( {{a}_{n}}-{{a}_{1}} \right)\left( {{a}_{n}}-{{a}_{2}} \right)\cdots \left( {{a}_{n}}-{{a}_{n-1}} \right)}{\left( {{a}_{n}}+{{b}_{1}} \right)\left( {{a}_{n}}+{{b}_{2}}\right)\cdots \left( {{a}_{n}}+{{b}_{n}} \right)}\left| \begin{matrix}\frac{1}{{{a}_{1}}+{{b}_{1}}} & \frac{1}{{{a}_{1}}+{{b}_{2}}} &\cdots & \frac{1}{{{a}_{1}}+{{b}_{n}}} \\\frac{1}{{{a}_{2}}+{{b}_{1}}} &\frac{1}{{{a}_{2}}+{{b}_{2}}} & \cdots & \frac{1}{{{a}_{2}}+{{b}_{n}}} \\\vdots & \vdots & \ddots & \vdots \\\frac{1}{{{a}_{n-1}}+{{b}_{1}}} & \frac{1}{{{a}_{n-1}}+{{b}_{2}}} & \cdots & \frac{1}{{{a}_{n-1}}+{{b}_{n}}} \\1 & 1 & \cdots &1\\\end{matrix} \right|}$$

各列から最後の列をひく
 

$${{{\Delta }_{n}}=\frac{\left( {{a}_{n}}-{{a}_{1}} \right)\left( {{a}_{n}}-{{a}_{2}} \right)\cdots \left( {{a}_{n}}-{{a}_{n-1}} \right)}{\left( {{a}_{n}}+{{b}_{1}} \right)\left( {{a}_{n}}+{{b}_{2}}\right)\cdots \left( {{a}_{n}}+{{b}_{n}} \right)}\left| \begin{matrix}\frac{1}{{{a}_{1}}+{{b}_{1}}}-\frac{1}{{{a}_{1}}+{{b}_{n}}} & \frac{1}{{{a}_{1}}+{{b}_{2}}}-\frac{1}{{{a}_{1}}+{{b}_{n}}} & \cdots & \frac{1}{{{a}_{1}}+{{b}_{n}}} \\\frac{1}{{{a}_{2}}+{{b}_{1}}}-\frac{1}{{{a}_{2}}+{{b}_{n}}} & \frac{1}{{{a}_{2}}+{{b}_{2}}}-\frac{1}{{{a}_{2}}+{{b}_{n}}} & \cdots & \frac{1}{{{a}_{2}}+{{b}_{n}}} \\\vdots & \vdots & \ddots & \vdots \\\frac{1}{{{a}_{n-1}}+{{b}_{1}}}-\frac{1}{{{a}_{n-1}}+{{b}_{n}}} & \frac{1}{{{a}_{n-1}}+{{b}_{2}}}-\frac{1}{{{a}_{n-1}}+{{b}_{n}}} & \cdots & \frac{1}{{{a}_{n-1}}+{{b}_{n}}} \\0 & 0 & \cdots & 1 \\\end{matrix} \right|}$$

$${{{\Delta }_{n}}=\frac{\prod\limits_{i=1}^{n-1}{\left( {{a}_{n}}-{{a}_{i}} \right)}}{\prod\limits_{i=1}^{n}{\left( {{a}_{n}}+{{b}_{i}} \right)}}\left| \begin{matrix}\frac{{{b}_{n}}-{{b}_{1}}}{{{a}_{1}}+{{b}_{1}}}\frac{1}{{{a}_{1}}+{{b}_{n}}} & \frac{{{b}_{n}}-{{b}_{2}}}{{{a}_{1}}+{{b}_{2}}}\frac{1}{{{a}_{1}}+{{b}_{n}}} & \cdots & \frac{1}{{{a}_{1}}+{{b}_{n}}} \\\frac{{{b}_{n}}-{{b}_{1}}}{{{a}_{2}}+{{b}_{1}}}\frac{1}{{{a}_{2}}+{{b}_{n}}} & \frac{{{b}_{n}}-{{b}_{2}}}{{{a}_{2}}+{{b}_{2}}}\frac{1}{{{a}_{2}}+{{b}_{n}}} & \cdots & \frac{1}{{{a}_{2}}+{{b}_{n}}} \\\vdots & \vdots & \ddots & \vdots \\\frac{{{b}_{n}}-{{b}_{1}}}{{{a}_{n-1}}+{{b}_{1}}}\frac{1}{{{a}_{n-1}}+{{b}_{n}}} & \frac{{{b}_{n}}-{{b}_{2}}}{{{a}_{n-1}}+{{b}_{2}}}\frac{1}{{{a}_{n-1}}+{{b}_{n}}} & \cdots & \frac{1}{{{a}_{n-1}}+{{b}_{n}}} \\0 & 0 & \cdots & 1 \\\end{matrix} \right|}$$

共通項をくくりだす

 $${{{\Delta }_{n}}=\frac{\prod\limits_{i=1}^{n-1}{\left( {{a}_{n}}-{{a}_{i}} \right)\left( {{b}_{n}}-{{b}_{i}} \right)}}{\prod\limits_{i=1}^{n}{\left( {{a}_{n}}+{{b}_{i}} \right)\left( {{b}_{n}}+{{a}_{i}} \right)}}\left| \begin{matrix}\frac{1}{{{a}_{1}}+{{b}_{1}}} & \frac{1}{{{a}_{1}}+{{b}_{2}}} & \cdots & \frac{1}{{{a}_{1}}+{{b}_{n-1}}} & 1 \\\frac{1}{{{a}_{2}}+{{b}_{1}}} & \frac{1}{{{a}_{2}}+{{b}_{2}}} & \cdots & \frac{1}{{{a}_{2}}+{{b}_{n-1}}} & 1 \\\vdots & \vdots & \ddots & \vdots & \vdots \\\frac{1}{{{a}_{n-1}}+{{b}_{1}}} & \frac{1}{{{a}_{n-1}}+{{b}_{2}}} & \cdots & \frac{1}{{{a}_{n-1}}+{{b}_{n-1}}} & 1 \\0 & 0 & \cdots & 0 & 1 \\\end{matrix} \right|}$$

$${{{\Delta }_{n}}=\frac{\prod\limits_{i=1}^{n-1}{\left( {{a}_{n}}-{{a}_{i}} \right)\left( {{b}_{n}}-{{b}_{i}} \right)}}{\left( {{a}_{n}}+{{b}_{n}} \right)\prod\limits_{i=1}^{n-1}{\left( {{a}_{n}}+{{b}_{i}} \right)\left( {{b}_{n}}+{{a}_{i}} \right)}}\left| \begin{matrix}{} & {} & \cdots & {} & 1 \\{} & {{\Delta }_{n-1}} & \cdots & {} & 1 \\\vdots & \vdots & \ddots & \vdots & \vdots \\{} & {} & \cdots & {} & 1 \\0 & 0 & \cdots & 0&1\\\end{matrix} \right|}$$

これは最初に述べた求めたい結果である。