{ "cells": [ { "cell_type": "markdown", "metadata": { "slideshow": { "slide_type": "slide" } }, "source": [ "### ústřední téma\n", "**jak co nejpřesněji určit polohu maxima - a s jakou nejistotou**" ] }, { "cell_type": "markdown", "metadata": { "nbpresent": { "id": "1c321046-112d-48c4-88a8-889e5b254db2" }, "slideshow": { "slide_type": "subslide" } }, "source": [ "podle [teorému implicitní funkce](https://en.wikipedia.org/wiki/Implicit_function_theorem)\n", "lze nejistotu určení maxima $\\mu$ funkce $a + bx + cx^2 = q$ stanovit z diferenciálu první derivace\n", "\n", "$$b + 2cx = 0$$\n", "$$\\mathrm{d} b+ 2x \\mathrm{d} c + 2c \\mathrm{d} x = 0$$\n", "\n", "$$\\frac{\\partial x}{\\partial b}= -\\frac{1}{2c}$$\n", "$$\\frac{\\partial x}{\\partial c}= -\\frac{x}c = \\frac{b}{2c^2}$$\n", "s použitím explicitního řešení $x=-b/2c$. Totéž dostaneme přímým derivováním explicitní funkce.\n", "\n", "Pro vyšší řád $$\\frac{\\partial^2 x}{\\partial c \\partial b}= \\frac{1}{2c^2}$$." ] }, { "cell_type": "markdown", "metadata": { "nbpresent": { "id": "5646f4d7-fa8b-4b7e-9b63-942fa945040b" } }, "source": [ "Tayloruv rozvoj do 2. řádu\n", "\n", "$$x=x_0 + \\frac{\\partial x}{\\partial a} \\Delta a + \\frac{\\partial x}{\\partial b} \\Delta b + \\frac{1}2 \\left (\\frac{\\partial^2 x}{\\partial a^2} \\Delta a^2 + \\frac{\\partial^2 x}{\\partial b^2} \\Delta b^2 + 2\\frac{\\partial^2 x}{\\partial a\\partial b} \\Delta a \\Delta b \\right) + \\dots$$" ] }, { "cell_type": "markdown", "metadata": { "nbpresent": { "id": "2b3e924c-5c8c-49d3-a6dc-8ab0ebfd48c6" } }, "source": [ "$$E(x)=x_0 + \\frac{1}2 \\left (\\frac{\\partial^2 x}{\\partial a^2} E(\\Delta a^2) + \\frac{\\partial^2 x}{\\partial b^2} E(\\Delta b^2) + 2\\frac{\\partial^2 x}{\\partial a\\partial b} E(\\Delta a \\Delta b) \\right) + \\dots$$\n", "\n", "pokud $E(\\Delta a)=E(\\Delta b)=0$, ozn. kovarianci $E(\\Delta a \\Delta b)=\\sigma_{ab}$ a $E(\\Delta a^2)=\\sigma_a^2$, $E(\\Delta b^2)=\\sigma_b^2$" ] }, { "cell_type": "markdown", "metadata": { "nbpresent": { "id": "c37380c0-350f-4f3f-a32e-9d5211156381" } }, "source": [ "$$E(x^2)=x_0^2 + x_0 \\left (\\frac{\\partial^2 x}{\\partial a^2} \\sigma^2_{a} + \\frac{\\partial^2 x}{\\partial b^2} \\sigma^2_{b} + 2\\frac{\\partial^2 x}{\\partial a\\partial b} \\sigma_{ab} \\right) + \\left(\\frac{\\partial x}{\\partial a}\\right)^2 \\sigma^2_{a} + \\left(\\frac{\\partial x}{\\partial b}\\right)^2 \\sigma^2_{b} + 2\\frac{\\partial x}{\\partial a} \\frac{\\partial x}{\\partial b} \\sigma_{ab} + \\dots$$" ] }, { "cell_type": "markdown", "metadata": { "nbpresent": { "id": "74c91cf9-e330-4035-b13c-67b75f24658a" } }, "source": [ "nakonec (opět do 2. řádu v $\\Delta a, \\Delta b$) se velké závorky odečtou\n", "\n", "$$\\sigma^2_x = E(x^2)-E(x)^2 = \\left(\\frac{\\partial x}{\\partial a}\\right)^2 \\sigma^2_{a} + \\left(\\frac{\\partial x}{\\partial b}\\right)^2 \\sigma^2_{b} + 2\\frac{\\partial x}{\\partial a} \\frac{\\partial x}{\\partial b} \\sigma_{ab} + \\dots$$" ] }, { "cell_type": "markdown", "metadata": { "nbpresent": { "id": "2e46f19c-b9e5-46d0-9b8c-ee98a04830e7" } }, "source": [ "Konkrétně pro fit paraboly\n", "\n", "$$ \\sigma_x^2=\\frac{1}{4c^2} \\sigma_b^2 + \\frac{b^2}{4c^4} \\sigma_c^2 + \\frac{b}{2c^3} \\sigma_{bc}$$" ] }, { "cell_type": "markdown", "metadata": { "nbpresent": { "id": "5d33656c-f38a-4c2d-9ad6-9dafac67a7c0" } }, "source": [ "### vyšší polynom\n", "\n", "Často pík nelze dobře proložit parabolou - vykazuje jistou míru asymetrie.\n", "\n", "Uvažujeme-li fit 4. řádu s podmínkou pro maximum\n", "$$b + 2cx + 3 ex^2 + 4 fx^3 = 0$$\n", "a diferenciálem \n", "$$\\d b+ 2x \\d c + 2c \\d x + 3 x^2 \\d e + 6 e x \\d x + 4 x^3 \\d f + 12 fx^2 \\d x= \\d b+ 2x \\d c + 3 x^2 \\d e + 4 x^3 \\d f + (2 c+ 6 e x + 12 fx^2) \\d x= 0$$\n" ] }, { "cell_type": "markdown", "metadata": { "nbpresent": { "id": "3a7ca843-fd06-45b9-8304-9fae993baeec" } }, "source": [ "$$\\frac{\\partial x}{\\partial b}= -\\frac{1}{m}$$\n", "$$\\frac{\\partial x}{\\partial c}= -\\frac{2x}{m}$$\n", "$$\\frac{\\partial x}{\\partial e}= -\\frac{3x^2}{m}$$\n", "$$\\frac{\\partial x}{\\partial f}= -\\frac{4x^3}{m}$$\n", "\n", "kde $m=2 c+ 6 e x + 12 fx^2$, $x$ je třeba vyřešit (např. numericky)." ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": false, "nbpresent": { "id": "47597766-84c1-4568-b77d-4340df57f788" } }, "outputs": [ { "data": { "text/plain": [ "[]" ] }, "execution_count": 1, "metadata": {}, "output_type": "execute_result" }, { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "#modelova funkce\n", "%matplotlib inline\n", "from matplotlib import pyplot as pl\n", "import numpy as np\n", "x=np.r_[:3:0.1]\n", "mat=np.array([x**i for i in range(5)]) #polynom 4. stupne\n", "p0=np.r_[1,7.2,-4.,0.4,0.05]\n", "y0=p0.dot(mat)\n", "y=y0+np.random.normal(size=x.shape[0])*0.1\n", "pl.plot(x,y)" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "collapsed": false, "nbpresent": { "id": "450950b8-757a-4d6f-adfd-190ff2e66cd4" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "chi2: 1.10415500218\n" ] }, { "data": { "text/plain": [ "array([ 1.09156457, 6.70893116, -3.34866818, 0.07813442, 0.10271255])" ] }, "execution_count": 2, "metadata": {}, "output_type": "execute_result" } ], "source": [ "#nevazena linearni regrese polynomem\n", "cov=np.linalg.inv(mat.dot(mat.T))\n", "sol=cov.dot(mat.dot(y))\n", "print(\"chi2:\",((sol.dot(mat)-y)**2).sum()*100/(len(x)-5))\n", "sol" ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "collapsed": false, "nbpresent": { "id": "47ab1853-758f-4707-a606-8ee8ae272b54" } }, "outputs": [ { "data": { "text/plain": [ "1.137216337552478" ] }, "execution_count": 3, "metadata": {}, "output_type": "execute_result" } ], "source": [ "allsol=np.roots((sol*np.r_[:5])[::-1])\n", "sel=(allsol>0.5)*(allsol<2)\n", "xm=allsol[sel][0] #vybereme relevantni koren v intervalu (0.5,2)\n", "xm" ] }, { "cell_type": "code", "execution_count": 4, "metadata": { "collapsed": false, "nbpresent": { "id": "19747e90-6225-4530-9a23-f27c67f539d8" } }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "maxim. hodnota 4.84636210855\n" ] }, { "data": { "text/plain": [ "1.3081475451951126" ] }, "execution_count": 4, "metadata": {}, "output_type": "execute_result" } ], "source": [ "print(\"maxim. hodnota\",y.max())\n", ".37/(np.sqrt(2)*0.2)" ] }, { "cell_type": "code", "execution_count": 4, "metadata": { "collapsed": false, "nbpresent": { "id": "c1a0a5e4-9b0d-47d4-b28f-e678431d074d" } }, "outputs": [ { "data": { "text/plain": [ "array([ 0. , 0.21932186, 0.49963058, 0.85364515, 1.29644237])" ] }, "execution_count": 4, "metadata": {}, "output_type": "execute_result" } ], "source": [ "# spocteme parcialni derivace\n", "m=np.polyval([12*sol[4],6*sol[3],2*sol[2]],xm)\n", "ders=np.r_[0,-1/m,-2*xm/m,-3*xm**2/m,-4*xm**3/m]\n", "ders" ] }, { "cell_type": "code", "execution_count": 5, "metadata": { "collapsed": false, "nbpresent": { "id": "aa425d2c-55bd-4f11-883d-40d8c373b351" } }, "outputs": [ { "data": { "text/plain": [ "0.015280005546266531" ] }, "execution_count": 5, "metadata": {}, "output_type": "execute_result" } ], "source": [ "np.sqrt(ders.dot(cov.dot(ders)))*0.1" ] }, { "cell_type": "code", "execution_count": 6, "metadata": { "collapsed": false, "nbpresent": { "id": "3932226a-bb3b-44b8-8edf-c98c8bbe7512" } }, "outputs": [ { "data": { "text/plain": [ "array([ 0.07571108, 0.37485728, 0.53889655, 0.28150165, 0.04813487])" ] }, "execution_count": 6, "metadata": {}, "output_type": "execute_result" } ], "source": [ "errs=np.sqrt(cov.diagonal())*0.1\n", "errs" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "**numerický experiment** - 1000 náhodných sad parametrů podle hodnot `sol` a příslušných nejistot" ] }, { "cell_type": "code", "execution_count": 8, "metadata": { "collapsed": false, "nbpresent": { "id": "5e2b866d-1f67-490f-a8f8-406f1d611497" } }, "outputs": [ { "data": { "text/plain": [ "0.91911573343335662" ] }, "execution_count": 8, "metadata": {}, "output_type": "execute_result" }, { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "ids=np.array([sol[i]+errs[i]*np.random.normal(size=1000) for i in range(5)]).T\n", "allmax=np.array([np.polyval(k[::-1],xm) for k in ids])\n", "pl.hist(allmax,20)\n", "allmax.std()" ] }, { "cell_type": "code", "execution_count": 14, "metadata": { "collapsed": false, "nbpresent": { "id": "1275dd98-2030-4b83-b00a-490ef4ce8d09" } }, "outputs": [], "source": [ "#poloha maxima urcena z derivace \n", "allpos=[np.roots((k*np.r_[:5])[::-1])[2] for k in ids]" ] }, { "cell_type": "code", "execution_count": 15, "metadata": { "collapsed": false, "nbpresent": { "id": "d1c651be-aeee-48ec-95f3-5dd07b959ef1" } }, "outputs": [ { "data": { "text/plain": [ "1.1901803626699805" ] }, "execution_count": 15, "metadata": {}, "output_type": "execute_result" } ], "source": [ "np.array(allpos).real.mean()" ] }, { "cell_type": "code", "execution_count": 16, "metadata": { "collapsed": false, "nbpresent": { "id": "3eafa0bf-b3a0-416b-8fa1-ec459078ab77" } }, "outputs": [ { "data": { "text/plain": [ "0.31102091875588916" ] }, "execution_count": 16, "metadata": {}, "output_type": "execute_result" } ], "source": [ "np.array(allpos).real.std()" ] }, { "cell_type": "code", "execution_count": 18, "metadata": { "collapsed": false, "nbpresent": { "id": "fa64dbd5-3e37-4492-a2d4-0b2c17d9b19b" } }, "outputs": [ { "data": { "image/png": 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"outputs": [ { "data": { "text/plain": [ "(array([-9.31374046, 3.1218292 , 1.12744772]),\n", " array([-0.00314229, 0.20909618, -0.42477534, 0.25725955, -0.04767332]))" ] }, "execution_count": 86, "metadata": {}, "output_type": "execute_result" } ], "source": [ "k=rvals[:,2]\n", "np.roots(((k+sol)*np.r_[:5])[:0:-1]),k" ] }, { "cell_type": "code", "execution_count": 95, "metadata": { "collapsed": false, "nbpresent": { "id": "b3f9413e-46d7-4ba9-bb05-9422bca74105" } }, "outputs": [ { "data": { "text/plain": [ "(1.135317759198921, 0.01510632918891267)" ] }, "execution_count": 95, "metadata": {}, "output_type": "execute_result" } ], "source": [ "almax=np.array([np.roots(((sol+k)*np.r_[:5])[:0:-1])[2] for k in rvals.T[:2000]])\n", "almax.mean(),almax.std()" ] }, { "cell_type": "code", "execution_count": 96, "metadata": { "collapsed": false, "nbpresent": { "id": "694b70fd-84db-4b0c-bac7-7a24615382a7" } }, "outputs": [ { "data": { "image/png": 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