314 lines
10 KiB
Plaintext
314 lines
10 KiB
Plaintext
{
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"cells": [
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{
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"cell_type": "raw",
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"id": "9e1ef849",
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"metadata": {},
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"source": [
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"---\n",
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"format:\n",
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" html:\n",
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" include-in-header:\n",
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" text: |\n",
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" <script type=\"application/javascript\">\n",
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" window.PlotlyConfig = {MathJaxConfig: 'local'}\n",
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" </script>\n",
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"---"
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]
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},
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{
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"cell_type": "markdown",
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"id": "beff12c8",
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"metadata": {},
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"source": [
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"# Ein Beispiel zur Stabilität von Gleitkommaarithmetik\n",
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"\n",
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"## Berechnung von $\\pi$ nach Archimedes\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"id": "a7810264",
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"metadata": {},
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"outputs": [],
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"source": [
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"#| error: false\n",
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"#| echo: false\n",
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"#| output: false\n",
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"using PlotlyJS, Random\n",
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"using HypertextLiteral\n",
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"using JSON, UUIDs\n",
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"using Base64\n",
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"\n",
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"\n",
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"## see https://github.com/JuliaPlots/PlotlyJS.jl/blob/master/src/PlotlyJS.jl\n",
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"## https://discourse.julialang.org/t/encode-a-plot-to-base64/27765/3\n",
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"function IJulia.display_dict(p::PlotlyJS.SyncPlot)\n",
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" Dict(\n",
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" # \"application/vnd.plotly.v1+json\" => JSON.lower(p),\n",
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" # \"text/plain\" => sprint(show, \"text/plain\", p),\n",
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" \"text/html\" => let\n",
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" buf = IOBuffer()\n",
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" show(buf, MIME(\"text/html\"), p)\n",
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" #close(buf)\n",
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" #String(resize!(buf.data, buf.size))\n",
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" String(take!(buf))\n",
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" end,\n",
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" \"image/png\" => let\n",
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" buf = IOBuffer()\n",
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" buf64 = Base64EncodePipe(buf)\n",
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" show(buf64, MIME(\"image/png\"), p)\n",
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" close(buf64)\n",
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" #String(resize!(buf.data, buf.size))\n",
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" String(take!(buf))\n",
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" end,\n",
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"\n",
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" )\n",
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" end\n",
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"\n",
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" function Base.show(io::IO, mimetype::MIME\"text/html\", p::PlotlyJS.SyncPlot)\n",
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" uuid = string(UUIDs.uuid4())\n",
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" show(io,mimetype,@htl(\"\"\"\n",
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" <div style=\"height: auto\" id=\\\"$(uuid)\\\"></div>\n",
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" <script>\n",
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" require(['../js/plotly-latest.min.js'], function(plotly) { \n",
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" plotly.newPlot($(uuid),\n",
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" $(HypertextLiteral.JavaScript(json(p.plot.data))),\n",
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" $(HypertextLiteral.JavaScript(json(p.plot.layout))),{responsive: true});\n",
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" });\n",
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" </script>\n",
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"\"\"\"))\n",
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"end "
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]
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},
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{
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"cell_type": "markdown",
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"id": "8d4bd1ab",
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"metadata": {},
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"source": [
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"Eine untere Schranke für $2\\pi$, den Umfang des Einheitskreises, erhält man durch die\n",
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"Summe der Seitenlängen eines dem Einheitskreis eingeschriebenen regelmäßigen $n$-Ecks.\n",
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" Die Abbildung links zeigt, wie man beginnend mit einem Viereck der Seitenlänge $s_4=\\sqrt{2}$ die Eckenzahl iterativ verdoppelt.\n",
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"\n",
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":::{.narrow}\n",
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"\n",
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"| Abb. 1 | Abb.2 |\n",
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"| :-: | :-: |\n",
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"|  |  |\n",
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": {tbl-colwidths=\"[57,43]\"}\n",
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"\n",
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":::\n",
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"\n",
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"\n",
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"Die zweite Abbildung zeigt die Geometrie der Eckenverdoppelung.\n",
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"\n",
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"Mit\n",
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"$|\\overline{AC}|= s_{n},\\quad |\\overline{AB}|= |\\overline{BC}|= s_{2n},\\quad |\\overline{MN}| =a, \\quad |\\overline{NB}| =1-a,$ liefert Pythagoras für die Dreiecke $MNA$ und\n",
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" $NBA$ jeweils\n",
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"$$\n",
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" a^2 + \\left(\\frac{s_{n}}{2}\\right)^2 = 1\\qquad\\text{bzw.} \\qquad\n",
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" (1-a)^2 + \\left(\\frac{s_{n}}{2}\\right)^2 = s_{2n}^2\n",
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"$$\n",
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"Elimination von $a$ liefert die Rekursion\n",
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"$$\n",
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" s_{2n} = \\sqrt{2-\\sqrt{4-s_n^2}} \\qquad\\text{mit Startwert}\\qquad\n",
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" s_4=\\sqrt{2}\n",
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"$$\n",
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"für die Länge $s_n$ **einer** Seite des eingeschriebenen regelmäßigen\n",
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"$n$-Ecks.\n",
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"\n",
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"\n",
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"Die Folge $(n\\cdot s_n)$\n",
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"konvergiert monoton von unten gegen den\n",
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"Grenzwert $2\\pi$:\n",
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"$$\n",
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" n\\, s_n \\rightarrow 2\\pi % \\qquad\\text{und} \\qquad {n c_n}\\downarrow 2\\pi\n",
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"$$\n",
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"Der relative Fehler der Approximation von 2π durch ein $n$-Eck ist:\n",
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"$$\n",
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" \\epsilon_n = \\left| \\frac{n\\, s_n-2 \\pi}{2\\pi} \\right|\n",
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"$$\n",
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"\n",
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"\n",
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"## Zwei Iterationsvorschriften^[nach: Christoph Überhuber, „Computer-Numerik“ Bd. 1, Springer 1995, Kap. 2.3]\n",
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"Die Gleichung\n",
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"$$\n",
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" s_{2n} = \\sqrt{2-\\sqrt{4-s_n^2}}\\qquad \\qquad \\textrm{(Iteration A)}\n",
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"$$\n",
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"ist mathematisch äquivalent zu\n",
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"$$\n",
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" s_{2n} = \\frac{s_n}{\\sqrt{2+\\sqrt{4-s_n^2}}} \\qquad \\qquad \\textrm{(Iteration B)}\n",
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"$$\n",
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"\n",
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"(Bitte nachrechnen!) \n",
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"\n",
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"\n",
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"\n",
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"Allerdings ist Iteration A schlecht konditioniert und numerisch instabil, wie der folgende Code zeigt. Ausgegeben wird die jeweils berechnete Näherung für π.\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"id": "3f238690",
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"metadata": {},
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"outputs": [],
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"source": [
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"using Printf\n",
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"\n",
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"iterationA(s) = sqrt(2 - sqrt(4 - s^2))\n",
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"iterationB(s) = s / sqrt(2 + sqrt(4 - s^2))\n",
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"\n",
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"s_B = s_A = sqrt(2) # Startwert \n",
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"\n",
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"ϵ(x) = abs(x - 2π)/2π # rel. Fehler \n",
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"\n",
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"ϵ_A = Float64[] # Vektoren für den Plot\n",
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"ϵ_B = Float64[]\n",
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"is = Float64[]\n",
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"\n",
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"@printf \"\"\" approx. Wert von π\n",
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" n-Eck Iteration A Iteration B\n",
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"\"\"\"\n",
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"\n",
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"for i in 3:35\n",
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" push!(is, i)\n",
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" s_A = iterationA(s_A) \n",
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" s_B = iterationB(s_B) \n",
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" doublePi_A = 2^i * s_A\n",
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" doublePi_B = 2^i * s_B\n",
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" push!(ϵ_A, ϵ(doublePi_A))\n",
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" push!(ϵ_B, ϵ(doublePi_B))\n",
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" @printf \"%14i %20.15f %20.15f\\n\" 2^i doublePi_A/2 doublePi_B/2 \n",
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"end"
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]
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},
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{
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"cell_type": "markdown",
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"id": "0c0178f5",
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"metadata": {},
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"source": [
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"Während Iteration B sich stabilisiert bei einem innerhalb der Maschinengenauigkeit korrekten Wert für π, wird Iteration A schnell instabil. Ein Plot der relativen Fehler $\\epsilon_i$ bestätigt das:\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"id": "f415aab0",
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"metadata": {},
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"outputs": [],
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"source": [
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"using PlotlyJS\n",
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"\n",
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"layout = Layout(xaxis_title=\"Iterationsschritte\", yaxis_title=\"rel. Fehler\",\n",
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" yaxis_type=\"log\", yaxis_exponentformat=\"power\", \n",
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" xaxis_tick0=2, xaxis_dtick=2)\n",
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"\n",
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"plot([scatter(x=is, y=ϵ_A, mode=\"markers+lines\", name=\"Iteration A\", yscale=:log10), \n",
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" scatter(x=is, y=ϵ_B, mode=\"markers+lines\", name=\"Iteration B\", yscale=:log10)],\n",
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" layout) \n"
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]
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},
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{
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"cell_type": "markdown",
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"id": "02116ab2",
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"metadata": {},
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"source": [
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"## Stabilität und Auslöschung\n",
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"\n",
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"Bei $i=26$ erreicht Iteration B einen relativen Fehler in der Größe des Maschinenepsilons:\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"id": "3d50cba2",
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"metadata": {},
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"outputs": [],
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"source": [
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"ϵ_B[22:28]"
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]
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},
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{
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"cell_type": "markdown",
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"id": "5ba088ba",
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"metadata": {},
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"source": [
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"Weitere Iterationen verbessern das Ergebnis nicht mehr. Sie stabilisieren sich bei einem relativen Fehler von etwa 2.5 Maschinenepsilon:\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"id": "2bcae61d",
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"metadata": {},
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"outputs": [],
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"source": [
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"ϵ_B[end]/eps(Float64)"
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]
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},
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{
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"cell_type": "markdown",
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"id": "0bfd8235",
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"metadata": {},
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"source": [
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"Die Form Iteration A ist instabil. Bereits bei $i=16$ beginnt der relative Fehler wieder zu wachsen.\n",
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"\n",
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"Ursache ist eine typische Auslöschung. Die Seitenlängen $s_n$ werden sehr schnell klein. Damit ist\n",
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"$a_n=\\sqrt{4-s_n^2}$ nur noch wenig kleiner als 2 und bei der Berechnung von $s_{2n}=\\sqrt{2-a_n}$ tritt ein typischer Auslöschungseffekt auf.\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"id": "365d4c27",
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"metadata": {},
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"outputs": [],
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"source": [
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"setprecision(80) # precision für BigFloat\n",
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"\n",
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"s = sqrt(BigFloat(2))\n",
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"\n",
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"@printf \" a = √(4-s^2) als BigFloat und als Float64\\n\\n\"\n",
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"\n",
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"for i = 3:44\n",
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" s = iterationA(s)\n",
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" x = sqrt(4-s^2)\n",
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" if i > 20\n",
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" @printf \"%i %30.26f %20.16f \\n\" i x Float64(x)\n",
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" end \n",
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"end\n"
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]
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},
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{
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"cell_type": "markdown",
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"id": "04736a02",
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"metadata": {},
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"source": [
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"Man sieht die Abnahme der Zahl der signifikanten Ziffern. Man sieht auch, dass eine Verwendung von `BigFloat` mit einer Mantissenlänge von hier 80 Bit das Einsetzen des Auslöschungseffekts nur etwas hinaussschieben kann. \n",
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"\n",
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"**Gegen instabile Algorithmen helfen in der Regel nur bessere, stabile Algorithmen und nicht genauere Maschinenzahlen!**\n",
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"\n",
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":::{.content-hidden unless-format=\"xxx\"}\n",
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"\n",
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"Offensichtlich tritt bei der Berechnung von $2-a_n$ bereits relativ früh\n",
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"eine Abnahme der Anzahl der signifikanten Ziffern (Auslöschung) auf, \n",
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"bevor schließlich bei der Berechnung von $a_n=\\sqrt{4-s_n^2}$ \n",
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"selbst schon Auslöschung zu einem unbrauchbaren Ergebnis führt.\n",
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"\n",
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":::"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Julia 1.10.2",
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"language": "julia",
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"name": "julia-1.10"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 5
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}
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