octave3.ipynb

{
 "cells": [
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "A =\n",
      "\n",
      "   1   2   3\n",
      "   4   5   7\n",
      "   2   3   5\n",
      "\n",
      "b =\n",
      "\n",
      "   2100\n",
      "   5400\n",
      "   3400\n",
      "\n",
      "ans = -2\n",
      "x =\n",
      "\n",
      "   200.00\n",
      "   500.00\n",
      "   300.00\n",
      "\n",
      "ans =\n",
      "\n",
      "   2100.0\n",
      "   5400.0\n",
      "   3400.0\n",
      "\n"
     ]
    }
   ],
   "source": [
    "# Inhomogén lineáris egyenletrendszer egyértelmű megoldása Ax = b\n",
    "#piaci vásárlás: alma 200, körte 500, citrom 300\n",
    "A =[1,2,3;4,5,7;2,3,5]\n",
    "b =[2100;5400;3400]\n",
    "det(A)\n",
    "x=inv(A)*b\n",
    "A*x #ellenőrzés"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "x =\n",
      "\n",
      "   200.00\n",
      "   500.00\n",
      "   300.00\n",
      "\n"
     ]
    }
   ],
   "source": [
    "#Túlhatározott lineáris egyenletrendszer megoldása\n",
    "\n",
    "A =[1,2,3;4,5,7;2,3,5;6,8,9;2,2,1;1,2,1]; #túlhatározott\n",
    "b =[2100;5400;3400;7900;1700;1500];\n",
    "x=inv(A'*A)*A'*b"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "bz =\n",
      "\n",
      "   2120\n",
      "   5420\n",
      "   3420\n",
      "   7920\n",
      "   1720\n",
      "   1520\n",
      "\n",
      "A =\n",
      "\n",
      "   1   2   3   1\n",
      "   4   5   7   1\n",
      "   2   3   5   1\n",
      "   6   8   9   1\n",
      "   2   2   1   1\n",
      "   1   2   1   1\n",
      "\n",
      "x =\n",
      "\n",
      "   200.000\n",
      "   500.000\n",
      "   300.000\n",
      "    20.000\n",
      "\n"
     ]
    }
   ],
   "source": [
    "#Konstans tag szerepeltetése az e.h. mátrixban \n",
    "\n",
    "bz=b+20\n",
    "A(:,4)=1 \n",
    "x=inv(A'*A)*A'*bz"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Inhomogén lineáris egyenletrendszer egyértelmű megoldása - Cramer szabály\n",
    "\n",
    "$Ax=b$<br>\n",
    "$Ac_{i}$ mátrix: az $A$ együttható mátrix $i$-edik oszlopát helyettesítjük a $b$ eredményvektorral:\n",
    "$\\scriptsize Ac_{i}=\\begin{bmatrix}\n",
    "a_{11} & \\ldots&b_{1}&\\ldots& a_{1n}\\\\\n",
    "a_{21} & \\ldots&b_{2}&\\ldots& a_{2n}\\\\\n",
    "\\vdots & \\vdots&\\vdots&\\vdots& \\vdots\\\\\n",
    "a_{n1} & \\ldots&b_{n}&\\ldots& a_{nn}\n",
    "\\end{bmatrix}$<br>\n",
    "$x_{i}=\\frac{\\mathrm{det}(Ac_{i})}{\\mathrm{det}(A)}$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "A =\n",
      "\n",
      "   1   2   3   1\n",
      "   4   5   7   1\n",
      "   2   3   5   1\n",
      "   6   8   9   1\n",
      "\n",
      "b =\n",
      "\n",
      "   2120\n",
      "   5420\n",
      "   3420\n",
      "   7920\n",
      "\n",
      "ans =  200.00\n",
      "ans =  20.000\n"
     ]
    }
   ],
   "source": [
    "A = resize(A,4,4)\n",
    "b = resize(bz,4,1)\n",
    "\n",
    "Ac1=A;\n",
    "Ac1(:,1)=b;\n",
    "det(Ac1)/det(A)\n",
    "\n",
    "Ac4=A;\n",
    "Ac4(:,4)=b;\n",
    "det(Ac4)/det(A)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Valószínűségi változó függvényének eloszlása <br>\n",
    "\n",
    "$\\displaystyle f_{y}(Y) = f_{x}(x(Y))\\frac{\\mathrm{d}x}{\\mathrm{d}y}=f_{x}\\circ x(Y)\\frac{\\mathrm{d}x}{\\mathrm{d}y}$<br>\n",
    "ellenőrzéshez: sűrűségfüggvény tulajdonsága volt: $\\int\\limits_{\\infty}^{\\infty}f(t)\\mathrm{d}t=1$\n",
    "\n",
    "a) $f_{x}=\\lambda\\mathrm{e}^{-\\lambda x}$, $\\lambda=1$ paraméterű exponenciális eloszlású sűrűségfv.<br> \n",
    "$\\frac{\\mathrm{d}x}{\\mathrm{d}y}$   (x=x(y) deriválása y szerint)<br>\n",
    "$y = \\sqrt{x}\\rightarrow$ $x=y^{2}\\rightarrow\\frac{\\mathrm{d}x}{\\mathrm{d}y}=2y$ <br>\n",
    "$f_{y} = 2y\\;\\mathrm{e}^{-y^{2}}$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "intx =  0.94848\n",
      "inty =  0.99833\n"
     ]
    },
    {
     "data": {
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2UZCfzVar5XleEATdbrdUKnU6ncVikR6fz+cbX2VXqVSSTzswXnQHQCZTV8XtWRAkpZTv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\nYII=\n",
      "text/plain": [
       "<IPython.core.display.Image object>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "# x és y kölcsönösen egyértelmű függvények adott szakaszon: egymás inverzei adott szakaszon\n",
    "# ha az egyik eloszlása ismert (fx) akkor a másik eloszlása számítható (fy) \n",
    "\n",
    "clear\n",
    "for i=1:60 x(i)=0.1*i; endfor; \n",
    "#for i=1:60 fx(i)=exp(-x(i)); endfor;\n",
    "fx = exp(-x);\n",
    "\n",
    "for i=1:60 y(i)=0.1*i; endfor; \n",
    "#for i=1:60 fy(i)=2*y(i)*exp(-y(i)*y(i)); endfor; \n",
    "fy = 2*y .* exp(-y .^ 2);\n",
    "\n",
    "intx=sum(fx)*0.1 #görbe alatti terület ellenőrzése -téglányösszeg\n",
    "inty=sum(fy)*0.1 #görbe alatti terület ellenőrzése \n",
    "\n",
    "subplot (2, 1, 1)\n",
    "plot (x,fx)\n",
    "subplot (2, 1, 2)\n",
    "plot (y,fy)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "b) $x$ standard normális eloszlású: $f_{x}=\\frac{1}{\\sqrt{2\\pi}}\\mathrm{e}^{-\\frac{x^{2}}{2}}$<br> \n",
    "$y = x^{2}\\rightarrow$ $x=\\sqrt{y}\\rightarrow\\frac{\\mathrm{d}x}{\\mathrm{d}y}=\\frac{1}{2\\sqrt{y}}$ <br>\n",
    "\n",
    "$\\displaystyle f_{y}=f_{x}\\circ x(Y)\\frac{\\mathrm{d}x}{\\mathrm{d}y}=\\frac{1}{\\sqrt{2\\pi}}\\mathrm{e}^{-\\frac{y}{2}}\\;\\frac{1}{2\\sqrt{y}}$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "ans =  0.46011\n",
      "ans =  0.92021\n",
      "ans =  0.74313\n",
      "ans =  0.94313\n"
     ]
    },
    {
     "data": {
      "image/png": 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iQV86tjbi2+zXBwAwl/R7SACAK0EgAQBEIJA+w8PfAJAVAgkAIAKBBAAQgUACAIhAIK3j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\nVgR9KQAAAABJRU5ErkJggg==\n",
      "text/plain": [
       "<IPython.core.display.Image object>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "clear\n",
    "\n",
    "for i=1:100 x(i)=0.2*(i); endfor; #0.2...20\n",
    "for i=1:100 fx(i)=1/sqrt(2*pi)*exp(-x(i)*x(i)/2); endfor;\n",
    "#fx = 1/sqrt(2*pi)*exp(-x .^ 2 / 2);\n",
    "\n",
    "for i=1:100 y(i)=0.2*(i); endfor; #0.2...20\n",
    "for i=1:100 fy(i)=1/sqrt(2*pi)*exp(-y(i)/2)*(1/2/sqrt(y(i))); endfor;\n",
    "#fy = 1/sqrt(2*pi)*exp(-y / 2) .* (1./(2*sqrt(y)));\n",
    "\n",
    "sum(fx)*0.2 # fél gauss-t integráltunk, szükséges 2-vel szorozni\n",
    "sum(fx)*0.2*2\n",
    "\n",
    "# emiatt fy téglányösszeget is szorozzuk 2-vel \n",
    "sum(fy)*0.2*2 # miért kapunk 0.74313-t? \n",
    "sum(fy)*0.2*2+0.2 # hozzáadunk 0.2 (téglányösszegből adódó pontatlanság)\n",
    "\n",
    "subplot (2, 1, 1)\n",
    "plot (x,fx)\n",
    "subplot (2, 1, 2)\n",
    "plot (y,fy)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Gauss-féle hibaterjedési szabály \n",
    "$y=f(x_{1},\\ldots,x_{n})$ sokváltozós fv.<br>\n",
    "$(x_{1},\\ldots,x_{n})\\rightarrow(\\xi_{1},\\ldots,\\xi_{n})$ \\[nem ismert $\\boldsymbol{x}$ változók becslése $\\boldsymbol{\\xi}$ valószínűségi változókkal\\]<br>\n",
    "$(\\xi_{1},\\ldots,\\xi_{n})$ mindegyikéhez számolható: <br>\n",
    ">$(E_{1},\\ldots,E_{n})$ becsült várható érték,<br>\n",
    ">$(s_{1},\\ldots,s_{n})$ becsült szórás,<br>\n",
    ">$(s_{1}^{2},\\ldots,s_{n}^{2})$ becsült variancia, <br>\n",
    ">$(s_{11}^{2},s_{12}^{2}\\ldots,s_{1n}^{2},s_{21}^{2}\\ldots,s_{nn}^{2})$ becsült kovariancia<br>\n",
    "\n",
    "Kérdés: $s_{y}^{2}$ ($y$ varianciája)?\n",
    "\n",
    "Válasz: $s_{y}^{2}=\\sum\\limits_{i=1}^{n}\\left(\\frac{\\partial y}{\\partial x_{i}}\\right)^{2}s_{i}^{2}+2\\sum\\limits_{i<j}\\sum\\limits_{j=1}^{n}\\left(\\frac{\\partial y}{\\partial x_{i}}\\right)\\left(\\frac{\\partial y}{\\partial x_{j}}\\right)s_{ij}^{2}=\\sum\\limits_{i=1}^{n}\\sum\\limits_{j=1}^{n}\\left(\\frac{\\partial y}{\\partial x_{i}}\\right)\\left(\\frac{\\partial y}{\\partial x_{j}}\\right)s_{ij}^{2}$\n",
    "\n",
    "\n",
    "Mi található a képletben:<br>\n",
    "parciális deriváltak szorzatából álló mátrix ($PD$), kihasználva, hogy szimmetrikus megadható tömörebb alakban:\n",
    "\n",
    "$PD=\\begin{bmatrix}\n",
    "\\left(\\frac{\\partial y}{\\partial x_{1}}\\right)\\left(\\frac{\\partial y}{\\partial x_{1}}\\right)&\\left(\\frac{\\partial y}{\\partial x_{1}}\\right)\\left(\\frac{\\partial y}{\\partial x_{2}}\\right)&\\left(\\frac{\\partial y}{\\partial x_{1}}\\right)\\left(\\frac{\\partial y}{\\partial x_{3}}\\right)\\\\\n",
    "\\left(\\frac{\\partial y}{\\partial x_{2}}\\right)\\left(\\frac{\\partial y}{\\partial x_{1}}\\right)&\\left(\\frac{\\partial y}{\\partial x_{2}}\\right)\n",
    "\\left(\\frac{\\partial y}{\\partial x_{2}}\\right)&\\left(\\frac{\\partial y}{\\partial x_{2}}\\right)\\left(\\frac{\\partial y}{\\partial x_{3}}\\right)\\\\\n",
    "\\left(\\frac{\\partial y}{\\partial x_{3}}\\right)\\left(\\frac{\\partial y}{\\partial x_{1}}\\right)&\\left(\\frac{\\partial y}{\\partial x_{3}}\\right)\n",
    "\\left(\\frac{\\partial y}{\\partial x_{2}}\\right)&\\left(\\frac{\\partial y}{\\partial x_{3}}\\right)\\left(\\frac{\\partial y}{\\partial x_{3}}\\right)\n",
    "\\end{bmatrix}=\\begin{bmatrix}\n",
    "\\left(\\frac{\\partial y}{\\partial x_{1}}\\right) \\\\ \n",
    "\\left(\\frac{\\partial y}{\\partial x_{2}}\\right)  \\\\\n",
    "\\left(\\frac{\\partial y}{\\partial x_{3}}\\right) \\end{bmatrix}\\begin{bmatrix}\n",
    "\\left(\\frac{\\partial y}{\\partial x_{1}}\\right)& \\left(\\frac{\\partial y}{\\partial x_{2}}\\right)&\\left(\\frac{\\partial y}{\\partial x_{3}}\\right)\n",
    "\\end{bmatrix}=pd\\cdot pd^{\\mathsf{T}}$\n",
    "\n",
    "változókat a várható értéküknél vesszük figyelembe<br>\n",
    "továbbá elemenként szorozzuk a $PD$-t a kovariancia mátrixszal: $CD = PD.*CV$ <br>\n",
    "végül két egymást követő összegzés (oszlopok és elemek szerint) adja a becsült varianciát sum(sum($CD$))<br>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [],
   "source": [
    "D=randn(10,3); # D mátrix oszlopaiban az x1, x2, x3 változók értékei\n",
    "D(:,3)=D(:,1)+2*D(:,2)+0.1*randn(); #3. oszlop az első kettő lineáris kombinációja + zaj"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### a) lineáris feladat: $y=x_{1}+2x_{2}-3x_{3}$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "parciális deriváltak<br>\n",
    "$\\frac{\\partial y}{\\partial x_{1}}=1\\quad \\frac{\\partial y}{\\partial x_{2}}=2\\quad\\frac{\\partial y}{\\partial x_{3}}=-3$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "PD =\n",
      "\n",
      "   1   2  -3\n",
      "   2   4  -6\n",
      "  -3  -6   9\n",
      "\n"
     ]
    }
   ],
   "source": [
    "pd=[1;2;-3];\n",
    "PD=pd*pd'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$\\small CV=\\tiny\\begin{bmatrix}\n",
    " s_{1,1}^{2}=s_{1}^{2}&s_{1,2}^{2}  &s_{1,3}^{2} \\\\\n",
    " s_{2,1}^{2}&s_{2,2}^{2}=s_{2}^{2}  &s_{2,3}^{2} \\\\\n",
    " s_{3,1}^{2}&s_{3,2}^{2}  &s_{3,3}^{2}=s_{3}^{2} \n",
    "\\end{bmatrix}$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "CV =\n",
      "\n",
      "   1.030015   0.056880   1.143774\n",
      "   0.056880   1.026307   2.109494\n",
      "   1.143774   2.109494   5.362763\n",
      "\n"
     ]
    }
   ],
   "source": [
    "CV=cov(D)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "CD =\n",
      "\n",
      "    1.03001    0.11376   -3.43132\n",
      "    0.11376    4.10523  -12.65697\n",
      "   -3.43132  -12.65697   48.26486\n",
      "\n"
     ]
    }
   ],
   "source": [
    "CD=CV.*PD"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "ans =\n",
      "\n",
      "   -2.2875   -8.4380   32.1766\n",
      "\n",
      "ans =  21.451\n",
      "ellenőrzés:\n",
      "ans =  21.451\n"
     ]
    }
   ],
   "source": [
    "sum(CD)\n",
    "sum(sum(CD)) #lineáris esetben a gauss hibaterjedés pontos, nemlineáris esetben csak egy becslést ad\n",
    "\n",
    "disp(\"ellenőrzés:\")\n",
    "y=D(:,1)+2*D(:,2)-3*D(:,3);\n",
    "#size(y)\n",
    "var(y)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**b) nemlineáris példa:** $y=x_{1}^{2}+\\cos(x_{2})x_{3}$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "parciális deriváltak<br>\n",
    "$\\frac{\\partial y}{\\partial x_{1}}=2x_{1}\\quad \\frac{\\partial y}{\\partial x_{2}}=-\\sin(x_{2})x_{3}\\quad\\frac{\\partial y}{\\partial x_{3}}=\\cos(x_{2})$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "e =\n",
      "\n",
      "   0.4795458   0.0096168   0.5343070\n",
      "\n",
      "pd =\n",
      "\n",
      "   0.9590916\n",
      "  -0.0051383\n",
      "   0.9999538\n",
      "\n"
     ]
    }
   ],
   "source": [
    "e = mean(D) #várható értékek\n",
    "# ha változó marad a parciális deriváltakban, akkor azt a változó várható értékével helyettesítjük\n",
    "\n",
    "pd = [2*e(1);-sin(e(2))*e(3);cos(e(2))]\n",
    "PD = pd*pd';"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "ans =  8.4814\n",
      "ellenőrzés:\n",
      "ans =  5.2732\n"
     ]
    }
   ],
   "source": [
    "CV = cov(D);\n",
    "CD = CV.*PD;\n",
    "sum(sum(CD))\n",
    "\n",
    "disp(\"ellenőrzés:\")\n",
    "y = D(:,1).*D(:,1)+cos(D(:,2)).*D(:,3);\n",
    "var(y) \n",
    "#nem lineáris esetben a gauss hibaterjedés nem pontos, csak egy becslést ad"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "c) $y=x_{1}^{2}+\\cos(x_{2})+x_{3}$<br>\n",
    "parciális deriváltak:<br>\n",
    "$\\frac{\\partial y}{\\partial x_{1}}=2x_{1}\\quad \\frac{\\partial y}{\\partial x_{2}}=-\\sin(x_{2})\\quad\\frac{\\partial y}{\\partial x_{3}}=1$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "ans =  8.4627\n",
      "ellenőrzés:\n",
      "ans =  11.651\n"
     ]
    }
   ],
   "source": [
    "e = mean(D); #várható értékek\n",
    "pd = [2*e(1);-sin(e(2));1];\n",
    "PD = pd*pd';\n",
    "CV = cov(D);\n",
    "CD = CV.*PD;\n",
    "sum(sum(CD))\n",
    "\n",
    "disp(\"ellenőrzés:\")\n",
    "y = D(:,1).*D(:,1)+cos(D(:,2))+D(:,3);\n",
    "var(y)\n",
    "#nem lineáris esetben a gauss hibaterjedés nem pontos, csak egy becslést ad"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Integrálás \n",
    "\n",
    "határozatlan integrál: $\\displaystyle \\int f(x)\\;\\mathrm{d}x=F(x)+C$<br>$F(x)$: az $f(x)$ függvény primitív függvénye (határozatlan integrálja)\n",
    "\n",
    "$\\displaystyle f(x)=\\frac{\\mathrm{d}F(x)}{\\mathrm{d}x}=F^{\\prime}(x)$\n",
    "\n",
    "határozott integrál (Newton-Leibniz tétel): $\\int_{a}^{b}f(x)\\mathrm{d}x=F(b)-F(a)\\quad$<br>\n",
    "\n",
    "numerikus integrálás: pl.:Trapézszabály: $\\displaystyle \\int_{a}^{b}f(x)\\mathrm{d}x\\approx\\frac{\\Delta x}{2}\\left(f(x_0)+2f(x_1)+\\ldots+2f(x_{N-1}) +f(x_N)\\right)  ;\\quad$ $\\displaystyle \\Delta x=\\frac{b-a}{N}$<br>\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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3qrOOY19CNMCsMuHo1laUpGYdPb34ag4AYxBYFnJqieRUZ0FgoRnWr2sqTVLrj14aAgv4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\nwSYUAIxBYAEwBoEFwBgEFgBjEFgAjPEfLt65UpPqJMUAAAAASUVORK5CYII=\n",
      "text/plain": [
       "<IPython.core.display.Image object>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "u=rand(1,20);\n",
    "for i=1:20 v(i)=i*0.1; endfor;\n",
    "\n",
    "figure(1, 'position',[0,0,400,150]);\n",
    "plot(v,u,\"linewidth\",3)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "ans =  0.89925\n",
      "ans =  1.00000\n"
     ]
    }
   ],
   "source": [
    "#számoljuk ki a görbe alatti területet (integráljunk)\n",
    "trapz(v,u) #integrálás trapézszabállyal\n",
    "\n",
    "quad('cos',0,pi/2) #integrálás gauss kvadratúrával, egzakt\n",
    "#quadnak az integrálandó függvényt adjuk meg (integrandus) és az integrálási határokat"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$\\int x^{n}\\mathrm{d}x=\\frac{x^{n+1}}{n+1}+c, \\quad n\\in \\mathbb{R}, n\\neq-1$<br>\n",
    "$\\int_{0}^{1}x^{2}\\mathrm{d}x=\\left[\\frac{x^{3}}{3}\\right]_{0}^{1}=\\frac{1}{3}-0=0.3333$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "ans =  0.33333\n"
     ]
    }
   ],
   "source": [
    "function y = f(x) y = x*x; endfunction;\n",
    "quad('f',0,1) #vs Newton-Leibniz tétel"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Polinomok<br>\n",
    "polinom: $f(x)=\\sum\\limits_{i=0}^{n}a_{i}x^{i}=a_{0}x^{0}+a_{1}x^{1}+a_{2}x^{2}+\\ldots+a_{n}x^{n}=a_{0}+a_{1}x+a_{2}x^{2}+\\ldots+a_{n}x^{n}$, $n$-ed fokú polinom, $a_{0}$ konstans tag<br>\n",
    "\n",
    "konvolúció: szorzok két polinomot, jelölés csillaggal<br>\n",
    "$f*g=\\sum\\limits_{k=0}^{m+n}(\\sum\\limits_{j=0}^{k}a_{k-j}b_{j})x^{k}$, octave elvégzi<br>\n",
    "\n",
    "\n",
    "polinom megadása octaveban: koefficiens vektorral, fokszám szerinti csökkenő sorrendben, utolsó a konstans tag<br>\n",
    "\n",
    "$c=2x^{4}+3x^{3}+4x^{2}+5x+1\\quad\\quad\\quad$ c = \\[2,3,4,5,1\\];<br> \n",
    "$a=x^{3}+2x^{2}+3x+4\\quad\\quad\\quad\\quad\\quad\\quad$    a = \\[1,2,3,4\\];<br> \n",
    "$a*c=2x^{7}+7x^{6}+16x^{5}+30x^{4}+35x^{3}+33x^{2}+23x+4\\quad$ conv(a,c)\n",
    "\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "ans =\n",
      "\n",
      "    2    7   16   30   35   33   23    4\n",
      "\n"
     ]
    }
   ],
   "source": [
    "c=[2,3,4,5,1] ;     \n",
    "a=[1,2,3,4]; \n",
    "conv(a,c) #konvolúció - két polinom szorzata"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "ans =\n",
      "\n",
      "   3   4   3\n",
      "\n",
      "ans =\n",
      "\n",
      "    14    42    80   120   105    66    23\n",
      "\n",
      "k =\n",
      "\n",
      "    14    42    80   120   105    66    23\n",
      "\n",
      "q =\n",
      "\n",
      "   -0.50000   -2.00000   -5.00000  -10.00000   -8.75000   -7.00000   -4.25000\n",
      "\n",
      "r =\n",
      "\n",
      " Columns 1 through 7:\n",
      "\n",
      "    1.00000    3.00000    6.25000   11.00000   12.50000   11.50000    8.25000\n",
      "\n",
      " Columns 8 and 9:\n",
      "\n",
      "    2.50000    0.25000\n",
      "\n"
     ]
    }
   ],
   "source": [
    "polyder(a)           # polinom deriváltja\n",
    "\n",
    "polyder(conv(a,c))   # két polinom konvolúciójának deriváltja\n",
    "#vagy\n",
    "[k]=polyder(a,c)\n",
    "\n",
    "[q,r]=polyder(a,c)   # a/c hányados (racionális törtfüggvény) deriváltja, q: derivált számláló, r: derivált nevező"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 20,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "r =\n",
      "\n",
      "  -1.65063 + 0.00000i\n",
      "  -0.17469 + 1.54687i\n",
      "  -0.17469 - 1.54687i\n",
      "\n",
      "ans =\n",
      "\n",
      "  -8.8818e-16 + 0.0000e+00i\n",
      "  -4.4409e-15 + 6.6613e-16i\n",
      "  -4.4409e-15 - 6.6613e-16i\n",
      "\n"
     ]
    }
   ],
   "source": [
    "# polinom gyökei\n",
    "a=[1,2,3,4]; \n",
    "r = roots(a) # valós gyökök, komplex gyökök, polinomnak annyi gyöke van ahányad rendű\n",
    "polyval(a,r) # ellenőrzés, polinom helyettesítési értéke (Horner-módszer) a kapott gyököknél zérus"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 21,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "image/png": 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      "text/plain": [
       "<IPython.core.display.Image object>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "#polinomillesztés, legkisebb négyzetek módszerével\n",
    "for i=1:20 v(i)=i*0.1; endfor; # 0.1...2.0\n",
    "for i=1:20 u(i)=v(i)+cos(v(i)*0.3); endfor;\n",
    "\n",
    "figure(1, 'position',[0,0,400,150]);\n",
    "plot(v,u,\"linewidth\",3)\n",
    "ylim([0,3])"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 22,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "p =\n",
      "\n",
      "  -6.2529e-06   3.5244e-04  -1.7165e-05  -4.4990e-02   1.0000e+00   1.0000e+00\n",
      "\n",
      "pint =\n",
      "\n",
      "  -0.00000   0.00007  -0.00000  -0.01500   0.50000   1.00000   0.00000\n",
      "\n"
     ]
    }
   ],
   "source": [
    "p = polyfit(v,u,5)  # illesztés ötödfokú polinommal\n",
    "#p = polyfit(v,u,4) # illesztés negyedfokú polinommal\n",
    "pint = polyint(p)   # illesztett polinom határozatlan integrálja"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 23,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "ans =  3.8821\n",
      "ans =  3.7770\n",
      "ans =  3.8821\n"
     ]
    }
   ],
   "source": [
    "polyval(pint,2)-polyval(pint,0) # Newton-Leibniz: pint határozatlan integrál helyettesítési értékét (polyval) számoljuk\n",
    "                                # a megadott integrálási határok között (0,2) a határozott integrál értékének kiszámítására\n",
    "#VS\n",
    "trapz(v,u)                      # trapézszabály\n",
    "#VS\n",
    "function u=fv(v) u=v+cos(v*0.3); endfunction;\n",
    "quad('fv',0,2)                  # gauss-kvadratúra"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Interpoláció"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 24,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "ans =  2.00068941632700\n",
      "ans =  2.00068941632700\n",
      "ans =  2.00079635831301\n",
      "ans =  2.00079637690771\n",
      "ans =  2.00079635991402\n"
     ]
    }
   ],
   "source": [
    "format long\n",
    "# interp1 1D interpoláció\n",
    "interp1(v,u,1.05) # u(v=1.05)?, u értéke v = 1.05-nél\n",
    "interp1(v,u,1.05,'linear') # default (legkisebb négyzetek módszere)\n",
    "interp1(v,u,1.05,'cubic')  # köbös interpoláció\n",
    "interp1(v,u,1.05,'spline') # köbös spline\n",
    "\n",
    "polyval(p,1.05) # pontosabbnak kell lennie bármelyiknél\n",
    "format"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
 ],
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   "display_name": "Octave",
   "language": "octave",
   "name": "octave"
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   "help_links": [
    {
     "text": "GNU Octave",
     "url": "https://www.gnu.org/software/octave/support.html"
    },
    {
     "text": "Octave Kernel",
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