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Euler-Maclaurin 公式
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{ | |
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"# Euler Maclaurin 公式\n", | |
"このノートは、Dingle, Proc. Roy. Soc. A211, p.500 (1952)のAppendixの行間を埋めるノートである。" | |
] | |
}, | |
{ | |
"cell_type": "markdown", | |
"metadata": {}, | |
"source": [ | |
"## Diracの櫛関数\n", | |
"Diracの櫛関数を\n", | |
"$$\n", | |
" \\mathfrak{m}_T (x) = \\sum_{n=-\\infty}^{\\infty} \\delta(x-n T) \n", | |
"$$\n", | |
"と定義する。これは周期$T$の周期関数であり、$x$が$nT$と等しいところでデルタ関数ピークが立っている。\n", | |
"周期関数であるからFourier級数をつかって\n", | |
"$$\n", | |
"\\mathfrak{m}_T (x) = \\sum_{\\ell=-\\infty}^\\infty e^{2\\pi i\\ell x / T}\n", | |
"$$\n", | |
"と展開することができ、周期$T=1$とすれば\n", | |
"$$\n", | |
" \\mathfrak{m}_1 (x) = \\mathfrak{m} (x)=\n", | |
" \\sum_{n=-\\infty}^{\\infty} \\delta(x-n ) = \\sum_{\\ell=-\\infty}^\\infty e^{2\\pi i\\ell x }\n", | |
"$$\n", | |
"となる。\n" | |
] | |
}, | |
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"cell_type": "markdown", | |
"metadata": {}, | |
"source": [ | |
"## Poissonの和公式\n", | |
"Diracの櫛関数を使うと、Poissonの和公式を得ることができる。\n", | |
"$$\n", | |
"\\sum_{n=-\\infty}^{\\infty} f(n) = \\int_{-\\infty}^{\\infty} {dx} f(x) \\mathfrak{m} (x) = \\sum_{\\ell=-\\infty}^{\\infty} \\int_{-\\infty}^{\\infty} f(x) e^{2\\pi i \\ell x}dx\n", | |
"$$" | |
] | |
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"cell_type": "markdown", | |
"metadata": {}, | |
"source": [ | |
"## Euler-Maclaurin公式\n", | |
"Poissonの和公式より\n", | |
"$$\n", | |
"\\sum_{n=1}^{\\infty} f(n) + \\dfrac{1}{2} f(0) = \\sum_{\\ell=-\\infty}^{\\infty} \\int_{0}^{\\infty} f(x) e^{2\\pi i \\ell x}dx\n", | |
"$$\n", | |
"が成立する。\n", | |
"左辺の$f(0)$の項は、デルタ関数が積分の下限ちょうどでピークをもつことから$1/2$がついている。右辺で$ \\ell=0$の項を取り出すと、\n", | |
"$$\n", | |
"\\sum_{n=1}^{\\infty} f(n) + \\dfrac{1}{2} f(0) = \\int_{0}^{\\infty} f(x) dx+ \\sum_{\\ell=-\\infty}^{\\infty\\, \\prime} \\int_{0}^{\\infty} f(x) e^{2\\pi i \\ell x}dx\n", | |
"$$\n", | |
"となる。和のプライムは$\\ell=0$の項を取り除いていることを表している。\n" | |
] | |
}, | |
{ | |
"cell_type": "markdown", | |
"metadata": {}, | |
"source": [ | |
"$f(x)$が無限遠で十分にはやく$0$になることを仮定し、積分を逐次部分積分していくと、\n", | |
"$$\n", | |
"\\begin{align}\n", | |
" \\int_{0}^{\\infty} f(x) e^{2\\pi i \\ell x}dx & = \\left.\\dfrac{1}{2\\pi i \\ell}e^{2\\pi i \\ell x} f(x)\\right\\rvert_{0}^{\\infty} - \\int_{0}^{\\infty}\\dfrac{1}{2\\pi i \\ell}e^{2\\pi i \\ell x} f'(x) dx\\\\\n", | |
" & = -\\dfrac{1}{2\\pi i \\ell} f(0) +\\dfrac{1}{(2\\pi i \\ell)^{2}} f'(0) + \\int_{0}^{\\infty}\\dfrac{1}{(2\\pi i \\ell)^{2}}e^{2\\pi i \\ell x} f''(x) dx\\\\ & = \\sum_{r=1}^{\\infty} \\dfrac{(-1)^{r}}{(2\\pi i \\ell)^{r}} f^{(r-1)}(0)\n", | |
"\\end{align}\n", | |
"$$\n", | |
"となる。\n" | |
] | |
}, | |
{ | |
"cell_type": "markdown", | |
"metadata": {}, | |
"source": [ | |
"さらに$\\ell$に関する和の項はRiemannのゼータ関数 $\\zeta (r)$ をつかって\n", | |
"$$\n", | |
"\\begin{align}\n", | |
"\\sum_{\\ell=-\\infty}^{\\infty\\, \\prime} \\int_{0}^{\\infty} f(x) e^{2\\pi i \\ell x}dx & = \n", | |
"\\sum_{\\ell=-\\infty}^{\\infty\\, \\prime}\\sum_{r=1}^{\\infty} \\dfrac{(-1)^{r}}{(2\\pi i \\ell)^{r}} f^{(r-1)}(0)\\\\\n", | |
"& = \n", | |
"\\sum_{r=1}^{\\infty}f^{(r-1)}(0) \\dfrac{(-1)^{r}}{(2\\pi i )^{r}} \\sum_{\\ell=-\\infty}^{\\infty\\, \\prime} \\dfrac{1}{ \\ell^{r}} \\\\\n", | |
"& = \\sum_{r=1}^{\\infty}f^{(r-1)}(0) \\dfrac{(-1)^{r}}{(2\\pi i )^{r}} ( 1 + (-1)^{r} ) \\zeta(r)\n", | |
"\\end{align}\n", | |
"$$\n", | |
"と表す事ができる。これをみると$r$が奇数の項は落ち、偶数の項だけ残ることがわかる。\n" | |
] | |
}, | |
{ | |
"cell_type": "markdown", | |
"metadata": {}, | |
"source": [ | |
"\n", | |
"偶数引き数のRiemannゼータ関数はBernoulli数で表す事ができる。\n", | |
"[参考 Wikipedia](https://en.wikipedia.org/wiki/Bernoulli_number)\n", | |
"$$\n", | |
"B_{2\\nu} = \\dfrac{(-1)^{\\nu+1} 2 (2\\nu)! }{(2\\pi)^{2\\nu}} \\zeta(2\\nu)\n", | |
"$$\n", | |
"よって\n", | |
"$$\n", | |
"\\sum_{\\ell=-\\infty}^{\\infty\\, \\prime} \\int_{0}^{\\infty} f(x) e^{2\\pi i \\ell x}dx = - \\sum_{\\nu=1}^{\\infty}\n", | |
" \\dfrac{ B_{2\\nu}}{ (2\\nu)! } f^{(2\\nu-1)}(0)\n", | |
"$$\n", | |
"となる。\n", | |
"初めの数項を展開すると\n", | |
"$$\n", | |
"\\begin{align}\n", | |
"\\sum_{\\ell=-\\infty}^{\\infty\\, \\prime} \\int_{0}^{\\infty} f(x) e^{2\\pi i \\ell x}dx & = - \\dfrac{ B_{2}}{ 2! } f^{(1)}(0) - \\dfrac{ B_{4}}{ 4! } f^{(3)}0) - \\dfrac{ B_{6}}{ 6! } f^{(5)}(0) -\\dots\\\\\n", | |
"& = - \\dfrac{ 1}{6 } \\dfrac{ 1}{ 2! } f^{(1)}(0) + \\dfrac{ 1}{ 30 } \\dfrac{ 1}{ 4! } f^{(3)}(0) - \\dfrac{ 1}{42 } \\dfrac{ 1}{ 6! } f^{(5)}(0) - \\dots\n", | |
"\\end{align}\n", | |
"$$\n", | |
"となる。\n", | |
"\n" | |
] | |
}, | |
{ | |
"cell_type": "markdown", | |
"metadata": {}, | |
"source": [ | |
"すべて合わせると\n", | |
"$$\n", | |
"\\begin{align}\n", | |
"\\sum_{n=1}^{\\infty} f(n) & = \\int_{0}^{\\infty} f(x) dx - \\dfrac{1}{2} f(0)- \\dfrac{ 1}{6 } \\dfrac{ 1}{ 2! } f^{(1)}(0) + \\dfrac{ 1}{ 30 } \\dfrac{ 1}{ 4! } f^{(3)}(0) - \\dfrac{ 1}{42 } \\dfrac{ 1}{ 6! } f^{(5)}(0) - \\dots\n", | |
"\\end{align}\n", | |
"$$\n", | |
"または\n", | |
"$$\n", | |
"\\begin{align}\n", | |
"\\sum_{n=0}^{\\infty} f(n) & = \\int_{0}^{\\infty} f(x) dx + \\dfrac{1}{2} f(0)- \\dfrac{ 1}{6 } \\dfrac{ 1}{ 2! } f^{(1)}(0) + \\dfrac{ 1}{ 30 } \\dfrac{ 1}{ 4! } f^{(3)}(0) - \\dfrac{ 1}{42 } \\dfrac{ 1}{ 6! } f^{(5)}(0) - \\dots\n", | |
"\\end{align}\n", | |
"$$\n", | |
"となりEuler-Maclaurin公式を得る。\n" | |
] | |
}, | |
{ | |
"cell_type": "markdown", | |
"metadata": {}, | |
"source": [ | |
"## 半整数公式\n", | |
"統計力学への応用(Landau反磁性)では、半整数での和を積分で表さなければならない。" | |
] | |
}, | |
{ | |
"cell_type": "markdown", | |
"metadata": {}, | |
"source": [ | |
"Euler-Maclaurin公式を示すときに出発点としたPoissonの和公式で$f(n)= g(n+1/2)$とすると\n", | |
"$$\n", | |
"\\sum_{n=1}^{\\infty} g(n+1/2) + \\dfrac{1}{2} g(1/2) = \\sum_{\\ell=-\\infty}^{\\infty} \\int_{0}^{\\infty} g(x+1/2) e^{2\\pi i \\ell x}dx\n", | |
"$$\n", | |
"和を$0$からに取り直し、あらためて$g$を$f$と書くと、\n", | |
"$$\n", | |
"\\sum_{n=0}^{\\infty} f(n+1/2) - \\dfrac{1}{2} f(1/2) = \\sum_{\\ell=-\\infty}^{\\infty} \\int_{0}^{\\infty} f(x+1/2) e^{2\\pi i \\ell x}dx\n", | |
"$$\n", | |
"この右辺の積分を\n", | |
"$$\n", | |
" \\sum_{\\ell=-\\infty}^{\\infty}\\int_{0}^{\\infty} f(x+1/2) e^{2\\pi i \\ell x}dx =- \\sum_{\\ell=-\\infty}^{\\infty}\\int_{-1/2}^{0} f(x+1/2) e^{2\\pi i \\ell x}dx + \\sum_{\\ell=-\\infty}^{\\infty}\\int_{-1/2}^{\\infty} f(x+1/2) e^{2\\pi i \\ell x}dx \n", | |
"$$\n", | |
"と分解する。\n" | |
] | |
}, | |
{ | |
"cell_type": "markdown", | |
"metadata": {}, | |
"source": [ | |
"ここでDiracの櫛関数のFourier級数の表現を思い出して、この分解の第一項は、\n", | |
"$$\n", | |
" \\sum_{\\ell=-\\infty}^{\\infty}\\int_{-1/2}^{0} f(x+1/2) e^{2\\pi i \\ell x}dx = \n", | |
"\\int_{-1/2}^{0} f(x+1/2) \\sum_{n=-\\infty}^{\\infty} \\delta (x-n) dx = \\dfrac{1}{2} f(\\frac{1}{2})\n", | |
"$$\n", | |
"となる。\n", | |
"よって\n", | |
"$$\n", | |
"\\sum_{n=0}^{\\infty} f(n+1/2) = \\sum_{\\ell=-\\infty}^{\\infty} \\int_{-1/2}^{\\infty} f(x+1/2) e^{2\\pi i \\ell x}dx\n", | |
"$$\n" | |
] | |
}, | |
{ | |
"cell_type": "markdown", | |
"metadata": {}, | |
"source": [ | |
"積分で$x+1/2\\to x$と変数変換すると\n", | |
"$$\n", | |
"\\sum_{n=0}^{\\infty} f(n+1/2) = \\sum_{\\ell=-\\infty}^{\\infty} e^{-\\pi i \\ell } \\int_{0}^{\\infty} f(x) e^{2\\pi i \\ell x}dx\n", | |
"= \\sum_{\\ell=-\\infty}^{\\infty} (-1)^{ \\ell } \\int_{0}^{\\infty} f(x) e^{2\\pi i \\ell x}dx\n", | |
"$$\n", | |
"となる。\n", | |
"$\\ell=0$の項を取りだし、\n", | |
"$$\n", | |
"\\sum_{n=0}^{\\infty} f(n+1/2) \n", | |
"= \n", | |
" \\int_{0}^{\\infty} f(x) dx\n", | |
"+\\sum_{\\ell=-\\infty}^{\\infty\\prime} (-1)^{ \\ell } \\int_{0}^{\\infty} f(x) e^{2\\pi i \\ell x}dx\n", | |
"$$\n", | |
"最後の項で、Euler-Maclaurinの時と同様に積分を逐次部分積分すると、\n", | |
"$$\n", | |
"\\sum_{\\ell=-\\infty}^{\\infty\\prime} (-1)^{ \\ell } \\int_{0}^{\\infty} f(x) e^{2\\pi i \\ell x}dx \n", | |
" = \\sum_{r=1}^{\\infty} f^{(r-1)}(0) \\dfrac{(-1)^{r}}{(2\\pi i )^{r}} \\left( -1 + (-1)^{r+1}\n", | |
"\\right) \\eta(r)\n", | |
"$$\n", | |
"となる。ここで$\\eta(r)$はDirichletのエータ関数である。\n" | |
] | |
}, | |
{ | |
"cell_type": "markdown", | |
"metadata": {}, | |
"source": [ | |
"この場合も$r$の奇数べきはおちて\n", | |
"$$\n", | |
"\\sum_{\\ell=-\\infty}^{\\infty\\prime} (-1)^{ \\ell } \\int_{0}^{\\infty} f(x) e^{2\\pi i \\ell x}dx \n", | |
"= - \\sum_{\\nu=1}^{\\infty} f^{(2\\nu-1)}(0) \\dfrac{2(-1)^{\\nu}}{(2\\pi )^{2\\nu}} \\eta(2\\nu)\n", | |
"$$\n", | |
"となる。\n", | |
"\n", | |
"Dirichletのエータ関数はRiemannのゼータ関数と\n", | |
"$$\n", | |
"\\begin{align}\n", | |
"\\zeta(s) - \\eta(s) & = \\sum_{n=1}^{\\infty} \\dfrac{1}{n^{s}} - \\sum_{n=1}^{\\infty} \\dfrac{(-1)^{n+1}}{n^{s}} \n", | |
"= \\sum_{n=1}^{\\infty} \\dfrac{1-(-1)^{n+1}}{n^{s}} \\\\\n", | |
"& = \\sum_{n=1}^{\\infty} \\dfrac{2}{(2n)^{s}} = 2^{1-s} \\zeta (s)\n", | |
"\\end{align}\n", | |
"$$\n", | |
"という関係があり、ゼータ関数で表すことができる。\n" | |
] | |
}, | |
{ | |
"cell_type": "markdown", | |
"metadata": {}, | |
"source": [ | |
"よって\n", | |
"$$\n", | |
"\\sum_{\\ell=-\\infty}^{\\infty\\prime} (-1)^{ \\ell } \\int_{0}^{\\infty} f(x) e^{2\\pi i \\ell x}dx \n", | |
"= \\sum_{\\nu=1}^{\\infty} (1-2^{1-2\\nu}) \n", | |
"\\dfrac{B_{2\\nu}} { (2\\nu)! } f^{(2\\nu-1)}(0) \n", | |
"$$\n", | |
"を得る。\n", | |
"最初の数項を書くと\n", | |
"$$\n", | |
"\\begin{align}\n", | |
"\\sum_{\\ell=-\\infty}^{\\infty\\prime} (-1)^{ \\ell } \\int_{0}^{\\infty} f(x) e^{2\\pi i \\ell x}dx \n", | |
"&= \\dfrac{1}{2} \\dfrac{1}{6} \\dfrac{1} { 2! } f^{(1)}(0)- \\dfrac{7}{8} \\dfrac{1}{30} \\dfrac{1} { 4! }f^{(3)}(0) + \\dfrac{31}{32} \\dfrac{1}{42} \\dfrac{1} { 6! } f^{(5)}(0) \n", | |
" + \\dots \n", | |
"\\end{align}\n", | |
"$$\n", | |
"よって\n", | |
"$$\n", | |
"\\sum_{n=0}^{\\infty} f(n+1/2) \n", | |
"= \\int_{0}^{\\infty} f(x) dx+ \\dfrac{1}{2} \\dfrac{1}{6} \\dfrac{1} { 2! } f^{(1)}(0)- \\dfrac{7}{8} \\dfrac{1}{30} \\dfrac{1} { 4! }f^{(3)}(0) + \\dfrac{31}{32} \\dfrac{1}{42} \\dfrac{1} { 6! } f^{(5)}(0) + \\dots \n", | |
"$$\n", | |
"となる。\n" | |
] | |
}, | |
{ | |
"cell_type": "markdown", | |
"metadata": {}, | |
"source": [ | |
"## 物理を学ぶ人向け問題" | |
] | |
}, | |
{ | |
"cell_type": "markdown", | |
"metadata": {}, | |
"source": [ | |
"半整数公式を量子力学にしたがう一次元の調和振動子の分配関数\n", | |
"$$\n", | |
"Z=\\sum_{n=0}^\\infty e^{\\beta \\hbar \\omega(n+1/2) }\n", | |
"$$\n", | |
"に応用せよ。エネルギーの期待値を計算し厳密解と比較することで、この場合の展開の物理的意味を考察せよ。(結果をグラフにして比較すると分かり易い。)" | |
] | |
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