概要
他の記事から参照するために、定理や公式をこの記事に書いておく。
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近接する2点と遠方の点との距離の差の極限
主張
$\bm{a},\bm{x}\in\realNumbers^3$とするとき次式が成り立つ。
\[ \lim_{\|\bm{x}\|\to\infty}\|\bm{x}-\bm{a}\| – \|\bm{x}\| = -\frac{\bm{x}}{\|\bm{x}\|}\cdot\bm{a} \]導出
$\bm{a}=\bm{0}$のときは明らかに成り立つ。以下では$\bm{a}\neq 0$とする。また、$\|\bm{x}\|$を十分大きくとり、$\|\bm{x}\|>\|\bm{a}\|/2$とする。$f(\bm{x}) := \|\bm{x}-\bm{a}\| – \|\bm{x}\|$とすると次式が成り立つ。
\[ \bigl(f(\bm{x}) + \|\bm{x}\|\bigr)^2 = \|\bm{x}-\bm{a}\|^2 \quad \therefore f(\bm{x})\bigl(f(\bm{x}) + 2\|\bm{x}\|\bigr) = \|\bm{a}\|^2 – 2\bm{x}\cdot\bm{a} \]$f(\bm{x})$が最小となるのは$\bm{a}$が、原点と$\bm{x}$を結ぶ線分の上にあるときで、その値は$-\|\bm{a}\|$である。$\|\bm{x}\|>\|\bm{a}\|/2$と仮定しているから次式を得る。
\[ f(\bm{x}) = \frac{\|\bm{a}\|^2 – 2\bm{x}\cdot\bm{a}}{f(\bm{x}) + 2\|\bm{x}\|} < \frac{\|\bm{a}\|^2 – 2\bm{x}\cdot\bm{a}}{2\|\bm{x}\| – \|\bm{a}\|} \to -\frac{\bm{x}}{\|\bm{x}\|}\cdot\bm{a} \quad \text{as} \quad \|\bm{x}\|\to\infty\]$\square$
