数学 - JavaScript - 線形代数学 - R^nにおけるベクトル – 複素数(共役複素数、加法、乗法、絶対値の積、内積、規則の証明)

ラング線形代数学(上)
原著

学習環境

• Surface 3 (4G LTE)、Surface 3 タイプ カバー、Surface ペン(端末)
• Windows 10 Pro (OS)
• 数式入力ソフト(TeX, MathML): MathType
• MathML対応ブラウザ: Firefox、Safari
• MathML非対応ブラウザ(Internet Explorer, Google Chrome...)用JavaScript Library: MathJax

ラング線形代数学(上)(S.ラング (著)、芹沢 正三 (翻訳)、ちくま学芸文庫)の1章(R^nにおけるベクトル)、6(複素数)、練習問題1、2、3.を取り組んでみる。

4. 
   $\begin{array}{l}\alpha =a+bi\\ \beta =c+di\\ \\ \overline{\alpha \beta }\\ =\overline{ac-bd+\left(ad+bc\right)i}\\ =ac-bd-\left(ad+bc\right)i\\ \\ \overline{\alpha }\overline{\beta }\\ =\left(a-bi\right)\left(c-di\right)\\ =ac-bd-\left(ad+bc\right)i\\ \\ \overline{\alpha \beta }=\overline{\alpha }\overline{\beta }\\ \\ \overline{\alpha +\beta }\\ =\overline{a+c+\left(b+d\right)i}\\ =a+c-\left(b+d\right)i\\ \\ \overline{\alpha }\overline{\beta }\\ =\left(a-bi\right)+\left(c-di\right)\\ =a+c-\left(b+d\right)i\end{array}$
5. 
   $\begin{array}{l}\alpha =a+bi\\ \beta =c+di\\ \\ \left|\alpha \beta \right|\\ =\left|ac-bd+\left(ad+bc\right)i\right|\\ =\sqrt{{\left(ac-bd\right)}^2+{\left(ad+bc\right)}^2}\\ =\sqrt{a^2{c}^2+a^2{d}^2+b^2{c}^2+b^2{d}^2}\\ \\ \left|\alpha \right|\left|\beta \right|\\ =\sqrt{a^2+b^2}\sqrt{c^2+d^2}\\ =\sqrt{a^2{c}^2+a^2{d}^2+b^2{c}^2+b^2{d}^2}\end{array}$
6. 1. 
      $\begin{array}{l}{\alpha }_k=a_k+b_{k}i\\ {\beta }_k=c_k+d_{k}i\\ \\ \langle B,A\rangle \\ ={\displaystyle \sum _{k=1}^n\left(a_k+b_{k}i\right)\overline{\left(c_k+d_{k}i\right)}}\\ ={\displaystyle \sum _{k=1}^n\left(a_k+b_{k}i\right)\left(c_k-d_{k}i\right)}\\ ={\displaystyle \sum _{k=1}^n\left(a_k{c}_k+b_k{d}_k+\left(b_k{c}_k-a_k{d}_k\right)i\right)}\\ \\ \overline{\langle B,A\rangle }\\ =\overline{{\displaystyle \sum _{k=1}^n\left(c_k+d_{k}i\right)\overline{\left(a_k+b_{k}i\right)}}}\\ ={\displaystyle \sum _{k=1}^n\overline{\left(c_k+d_{k}i\right)}\overline{\left(a_k-b_{k}i\right)}}\\ ={\displaystyle \sum _{k=1}^n\left(c_k-d_{k}i\right)\left(a_k+b_{k}i\right)}\\ ={\displaystyle \sum _{k=1}^n\left(a_k{c}_k+b_k{d}_k+\left(b_k{c}_k-a_k{d}_k\right)i\right)}\end{array}$
   2. 
      $\begin{array}{l}{\gamma }_k=e_k+f_{k}i\\ \\ \langle A,B+C\rangle \\ ={\displaystyle \sum _{k=1}^n\left(a_k+b_{k}i\right)\overline{\left(\left(c_k+e_k\right)+\left(d_k+f_k\right)i\right)}}\\ ={\displaystyle \sum _{k=1}^n\left(a_k+b_{k}i\right)\left(\left(c_k+e_k\right)-\left(d_k+f_k\right)i\right)}\\ ={\displaystyle \sum _{k=1}^n\left(a_k{c}_k+a_k{e}_k+b_k{d}_k+b_k{f}_k+\left(b_k{c}_k+b_k{e}_k-a_k{d}_k-a_k{f}_k\right)i\right)}\\ \\ \langle A,B\rangle +\langle A,C\rangle \\ ={\displaystyle \sum _{k=1}^n\left(\left(a_k+b_{k}i\right)\left(c_k-d_{k}i\right)+\left(a_k+b_{k}i\right)\left(e_k-f_{k}i\right)\right)}\\ ={\displaystyle \sum _{k=1}^n\left(\left(a_k+b_{k}i\right)\left(c_k-d_{k}i\right)+\left(a_k+b_{k}i\right)\left(e_k-f_{k}i\right)\right)}\\ ={\displaystyle \sum _{k=1}^n\left(a_k{c}_k+a_k{e}_k+b_k{d}_k+b_k{f}_k+\left(b_k{c}_k+b_k{e}_k-a_k{d}_k-a_k{f}_k\right)i\right)}\\ \\ \langle A,B+C\rangle =\langle A,B\rangle +\langle A,C\rangle \end{array}$
   3. 
      $\begin{array}{l}\langle \alpha A,B\rangle \\ ={\displaystyle \sum _{k=1}^n\left(\left(a+bi\right)\left(a_k+b_{k}i\right)\left(c_k-d_{k}i\right)\right)}\\ =\left(a+bi\right){\displaystyle \sum _{k=1}^n\left(a_k+b_{k}i\right)\left(c_k-d_{k}i\right)}\\ =\alpha \langle A,B\rangle \\ \\ \langle A,\alpha \beta \rangle \\ ={\displaystyle \sum _{k=1}^n\left(\left(a_k+b_{k}i\right)\left(a-bi\right)\left(c_k-d_{k}i\right)\right)}\\ =\left(a-bi\right){\displaystyle \sum _{k=1}^n\left(a_k+b_{k}i\right)\left(c_k-d_{k}i\right)}\\ =\overline{\alpha }\langle A,\beta \rangle \end{array}$
   4. 
      $\begin{array}{l}\langle A,A\rangle \\ ={\displaystyle \sum _{k=1}^n{\alpha }_k\overline{{\alpha }_k}}\\ ={\displaystyle \sum _{k=1}^n{\left|a_k\right|}^2}\\ \\ \langle O,O\rangle =0\\ A\ne O\Rightarrow \langle A,A\rangle \ne 0\end{array}$
7. 
   

コード(Emacs)

HTML5

<pre id="output0"></pre>
<button id="run0">run</button>
<button id="clear0">clear</button>

<script src="sample4.js"></script>

JavaScript

let pre0 = document.querySelector('#output0'),
    btn0 = document.querySelector('#run0'),
    btn1 = document.querySelector('#clear0'),
    p = (x) => pre0.textContent += x + '\n',
    range = (start, end, step=1) => {
        let result = [];

        for (let i = start; i < end; i += step) {
            result.push(i);
        }
        return result;
    };

let Complex = (x, y) => {
    let that = {},
        toString = () => `${x} + ${y}i`,
        real = () => x,
        imag = () => y,
        add = (z) => Complex(x + z.real(), y + z.imag()),
        mul = (z) => Complex(x * z.real() - y * z.imag(),
                             x * z.imag() + y * z.real()),
        conjugate = () => Complex(x, -y),
        inv = () => {
            let den = Math.pow(x, 2) + Math.pow(y, 2);
            return Complex(x / den, -y / den);
        },
        mag = () => Math.sqrt(Math.pow(x, 2) + Math.pow(y, 2)),
        isEqual = (z) => x === z.real() && y === z.imag();

    that.toString = toString;
    that.real = real;
    that.imag = imag;
    that.add = add;
    that.mul = mul;
    that.conjugate = conjugate;
    that.inv = inv;
    that.mag = mag;
    that.isEqual = isEqual;
    
    return that;
};

let innerProd = (A, B) => {
    return A.reduce((prev, next, i) => prev.add(next.mul(B[i].conjugate())),
                    Complex(0, 0));
};
let output = () => {
    let a = Complex(Math.floor(Math.random() * 100),
                    Math.floor(Math.random() * 100)),
        b = Complex(Math.floor(Math.random() * 100),
                    Math.floor(Math.random() * 100)),
        c = Complex(Math.floor(Math.random() * 100),
                    Math.floor(Math.random() * 100));
    p('4.');
    let l = a.mul(b).conjugate(),
        r = a.conjugate().mul(b.conjugate());

    p(l.isEqual(r));
    p(l);
    p(r);

    l = a.add(b).conjugate();
    r = a.conjugate().add(b.conjugate());
    p(l.isEqual(r));
    p(l);
    p(r);

    p('5.');
    l = a.mul(b).mag();
    r = a.mag() * b.mag();
    p(l === r);
    p(l);
    p(r);

    p('6.');
    let A = [],
        B = [],
        C = [],
        O = [],
        n = Math.floor(Math.random() * 100);

    for (let k = 0; k < n; k += 1) {
        A.push(Complex(Math.floor(Math.random() * 100),
                       Math.floor(Math.random() * 100)));
        B.push(Complex(Math.floor(Math.random() * 100),
                       Math.floor(Math.random() * 100)));
        C.push(Complex(Math.floor(Math.random() * 100),
                       Math.floor(Math.random() * 100)));
        O.push(Complex(0, 0));
    }
    p('6.');
    p('SP1');
    l = innerProd(A, B);
    r = innerProd(B, A).conjugate();
    p(l.isEqual(r));
    p(l);
    p(r);
    p('SP2');
    l = innerProd(A, B.map((z, i) => z.add(C[i])));
    r = innerProd(A, B).add(innerProd(A, C));
    p(l.isEqual(r));
    p(l);
    p(r);
    p('SP3');
    l = innerProd(A.map((z) => a.mul(z)), B);
    r = a.mul(innerProd(A, B));
    p(l.isEqual(r));
    p(l);
    p(r);
    l = innerProd(A, B.map((z) => a.mul(z)));
    r = a.conjugate().mul(innerProd(A, B));
    p(l.isEqual(r));
    p(l);
    p(r);
    p('SP4');
    p(innerProd(O, O));
    p(innerProd(A, A));
};

btn0.onclick = output;
btn1.onclick = () => pre0.textContent = '';

output();

4.
true
455 + -3575i
455 + -3575i
true
74 + -101i
74 + -101i
5.
false
3603.838231663569
3603.8382316635693
6.
6.
SP1
true
124226 + 5967i
124226 + 5967i
SP2
true
258059 + 33951i
258059 + 33951i
SP3
true
4288240 + 1451105i
4288240 + 1451105i
true
4407580 + -1033415i
4407580 + -1033415i
SP4
0 + 0i
168583 + 0i