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First- and second order recurrence equations, homogenous and nonhomogenous

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$y_2 = 2y_{2-1}-y_{2-2}=2y_1-y_0 = 2 \\cdot 1 - 0 = 2$

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$y_3 = 2y_{3-1}-y_{3-2}=2y_2-y_1 = 2 \\cdot 2 - 1 = 3$

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$y_4 = 2y_{4-1}-y_{4-2}=2y_3-y_2 = 2 \\cdot 3 - 2 = 4$

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Det er snublende nært å mene at $y_5=5, y_6=6,...$ - og dermed $y_n=n$.

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En rekursjonslikning er gitt som   $y_n = 2y_{n-1}-y_{n-2},\\;\\;n\\geq2,\\;\\;y_0=0,\\;\\;y_1=1$

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Beregn de 3 neste $y$-verdiene,

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$y_2$ = [[0]] ,   $y_3$ = [[1]] ,   $y_4$ = [[2]]

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Prøv om du kan gjette funksjonsuttrykket    $y_n$ = [[3]]

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Sett inn for  $n$  i  $y_n = 2y_{n-1}-y_{n-2}$

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$y_2 = 2y_{2-1}-y_{2-2}=2y_1-y_0 = 2 \\cdot 1 - 0 = 2$

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$y_3 = 2y_{3-1}-y_{3-2}=2y_2-y_1 = ...$

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Recurrence equations - rekursjon

", "licence": "Creative Commons Attribution 4.0 International"}, "type": "question", "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Recurrence - rekursjon 5", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Tore Gaupseth", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/28/"}], "functions": {}, "ungrouped_variables": [], "tags": [], "preamble": {"css": "", "js": ""}, "advice": "
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  1. Ordne likninga, yn - 6 yn-1 + 9 yn-2 = 2n , n≥0 , y0=1 , y1=2
    2. \n
  2. Sett opp karakteristisk likning, λ2 - 6 λ + 9 = 0
    med røtter λ1 = λ2 = 3 .
    Generell homogen løsning, yn(h) = A 3n + B n 3n .
    3. \n
  3. Tipper partikulær løsning yn(p) = K⋅2n
    som settes inn i OL, K⋅2n - 6 K⋅2n-1 + 9 K ⋅ 2n-2 = 2n
    dividerer med 2n-2 og forkorter, K⋅22 - 6 K⋅21 + 9 K⋅20 = 22 med løsning K = 4
    og partikulær løsning yn(p) = 4⋅2n = 2n+2
    og foreløpig løsning, yn = yn(h) + yn(p) = A 3n + B n 3n + 2n+2
    4. \n
  4. Setter inn for startverdier,
    y0 = 1 = A 30 + B⋅0⋅30 + 20+2 = A + 4
    y1 = 2 = A 31 + B⋅1⋅31 + 21+2 = 3A + 3B + 8
    som gir A=-3 og B=1 og løsningen yn = -3n+1 + n 3n + 2n+2
    5. \n
", "rulesets": {}, "parts": [{"stepsPenalty": 0, "prompt": "

Løs rekursjonslikninga   $y_n - 6y_{n-1} +9y_{n-2}= 2^n, \\;\\;n\\geq2,\\;\\;y_0=1,\\;\\;y_1=2$

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yn - 6 yn-1 + 9 yn-2 = 2n , n≥0 , y0=1 , y1=2

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$y_n$ = [[0]]

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Ordnet likning,
        yn - a1 yn-1 - a2 yn-1 = f(n) + def.område + startverdier
Det inhomogene leddet f(n) er gjerne et polynom i n (c nr...), en eksponent i n, (c αn...) eller en kombinasjon (c nr αn...), som tidligere vist for 1. orden.
Løsningen er generelt yn = yn(h) + yn(p),
der yn(h) er løsningen av den homogene likninga yn - a1 yn-1 - a2 yn-1 = 0
og yn(p) er en partikulær løsning av yn - a1 yn-1 - a2 yn-1 = f(n) .
Den partikulære løsningen vil være av samme type som det inhomogene leddet.
Dersom yn(p) inngår i yn(h) multipliseres yn(p) med n, n2, n3, ...
Likninga har en karakteristisk likning λ2 - a1 λ - a2 = 0 med 3 mulige løsningstyper:
(1) to ulike, reelle røtter, λ1 ≠ λ2:
        yn = A λ1n + B λ2n
(2) sammenfallende røtter, λ1 = λ2 = λ:
        yn = A λn + B n λn
(3) komplekskonjugerte røtter, λ1,2 = r e±φi
        yn = rn(C cos(nφ) + D sin(nφ))

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Recurrence equations - rekursjon

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    \n
  1. Karakteristisk likning, λ2 - 3λ +2 = 0 med røtter λ1=1 og λ2=2 .
    2. \n
  2. Generell løsning, yn = A ⋅1n + B ⋅ 2n .
    3. \n
  3. Finner A og B ut fra startbetingelser, y1=0=A+B⋅21 og y2=2=A+B⋅22
    som gir A=-2 og B=1 og løsningen yn = 2n - 2 .
    4. \n
", "rulesets": {}, "parts": [{"stepsPenalty": 0, "prompt": "

Løs rekursjonslikninga   $y_n - 3y_{n-1} +2y_{n-2}= 0, \\;\\;n\\geq3,\\;\\;y_1=0,\\;\\;y_2=2$

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$y_n$ = [[0]]

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Ordnet likning,
        yn - a1 yn-1 - a2 yn-1 = 0 + def.område + startverdier
Likninga har en karakteristisk likning λ2 - a1 λ - a2 = 0 med 3 mulige løsningstyper:
(1) to ulike, reelle røtter, λ1 ≠ λ2:
        yn = A λ1n + B λ2n
(2) sammenfallende røtter, λ1 = λ2 = λ:
        yn = A λn + B n λn
(3) komplekskonjugerte røtter, λ1,2 = r e±φi
        yn = rn(C cos(nφ) + D sin(nφ))

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Recurrence equations - rekursjon

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Løs rekursjonslikninga   $y_n - 3y_{n-1} = -4n, \\;\\;n\\geq1,\\;\\;y_0=2$

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$y_n$ = [[0]]

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    \n
  1. Ordne likninga:  $y_n - ay_{n-1}=f(n)$ + def.område + startverdi
    2. \n
  2. Generell løsning på homogen liking er: $y_n^{(h)}= C a^n$.
    3. \n
  3. Tipp generell partikulær løsning $y_n^{(p)}$ ut fra homogent ledd og skjemaet ovenfor.
    Finn de ukjente konstantene i generell, partikulær løsning ved innsetting av $y_n^{(p)}$ i opprinnelig likning.
    Sett verdiene inn i $y_n^{(p)}$.
    4. \n
  4. Løsningen er $y_n=y_n^{(h)}+ y_n^{(p)} = Ca^n+ y_n^{(p)}$ , der den ukjente konstanten C kan bestemmes av startverdien.
    5. \n
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Recurrence equations - rekursjon

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Løs rekursjonslikninga   $y_n - 3y_{n-1} = 0, \\;\\;n\\geq1,\\;\\;y_0=2$

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$y_n$ = [[0]]

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Lag karakteristisk likning,  $\\lambda + A = 0$

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Recurrence equations - rekursjon

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