st\u0105d<\/a>. Jest z nim zwi\u0105zana liczba<\/p>\n\\[\\varphi=\\dfrac{\\sqrt{5}+1}{2}\\approx 1,61803\\]<\/p>\n
znana bardziej jako z\u0142ota liczba lub z\u0142ota proporcja. Dok\u0142adniej, okazuje si\u0119, \u017ce ci\u0105g iloraz\u00f3w \\[a_n=\\dfrac{F_{n+1}}{F_n}\\] ma granic\u0119, kt\u00f3r\u0105 jest w\u0142a\u015bnie \\(\\varphi\\).<\/p>\n
Aby to pokaza\u0107, udowodnimy najpierw tzw. to\u017csamo\u015b\u0107 Cassiniego.<\/em><\/p>\n\\(F_{n+1}F_{n-1}-F_n^2=(-1)^2 \\textrm{ dla } n>1.\\)<\/p>\n
\nDow\u00f3d przeprowadzimy indukcyjnie. Dla \\(n=2\\) to\u017csamo\u015b\u0107 Cassiniego jest spe\u0142niona. Za\u0142\u00f3\u017cmy teraz, \u017ce jest ona spe\u0142niona dla pewnego \\(k>1\\). Poka\u017cemy, \u017ce z tego wynika jej prawdziwo\u015b\u0107 dla \\(k+1\\). Istotnie,<\/p>\n
\\[ \\begin{array}{lll} F_{k+2}F_k-F_{k+1}^2 & = & (F_{k+1}+F_k)\\cdot F_k-(F_k+F_{k-1})^2 =\\\\ & = & F_kF_{k+1}+F_k^2-F_k^2-2F_kF_{k-1}-F^2_{k-1} = \\\\ & = & F_kF_{k+1}-2F_kF_{k-1}-F_{k-1}^2 = \\\\ & = & F_k(F_k+F_{k-1})- 2F_kF_{k-1}-F_{k-1}^2 =\\\\ & = &F_k^2+F_kF_{k-1}- 2F_kF_{k-1}-F_{k-1}^2 =\\\\ & = & F_k^2-F_kF_{k-1}-F_{k-1}^2= F_k^2-F_{k-1}(F_k+F_{k-1}) = \\\\ & = & F_k^2-F_{k+1}F_{k-1} = – (F_{k+1}F_{k-1}-F_k^2)=-(-1)^k=(-1)^{k+1}. \\end{array} \\]<\/p>\n
Zatem to\u017csamo\u015b\u0107 Cassiniego jest prawdziwa.<\/p>\n
Teraz zauwa\u017cmy, \u017ce dla \\(n>1\\) mamy<\/p>\n
\\[|a_n-a_{n-1}|=\\left|\\dfrac{F_{n+1}}{F_n}-\\dfrac{F_n}{F_{n-1}}\\right|= \\left|\\dfrac{F_{n+1}F_{n-1}-F_n^2}{F_nF_{n-1}}\\right|=\\dfrac{1}{F_nF_{n-1}}.\\]<\/p>\n
Poniewa\u017c<\/p>\n
\\[\\sum\\limits_{n=2}^\\infty\\dfrac{1}{F_nF_{n-1}}\\leq\\sum\\limits_{n=2}^\\infty\\dfrac{1}{n(n-1)},\\]<\/p>\n
to szereg<\/p>\n
\\[\\sum\\limits_{n=2}^\\infty\\dfrac{1}{F_nF_{n-1}}\\]<\/p>\n
jest zbie\u017cny, a wi\u0119c ci\u0105g \\(a_n\\) r\u00f3wnie\u017c jest zbie\u017cny. Oznaczmy przez \\(g\\) jego granic\u0119. Poniewa\u017c \\[g=\\lim\\limits_{n\\to\\infty}a_n=\\lim\\limits_{n\\to\\infty}a_{n+1}\\] oraz (co \u0142atwo sprawdzi\u0107) \\[a_{n+1}=1+\\dfrac{1}{a_n},\\] to<\/p>\n
\\[g=1+\\dfrac{1}{g}.\\]<\/p>\n
Prowadzi to do r\u00f3wnania \\(g^2=g+1\\), kt\u00f3rego dodatnim pierwiastkiem jest \\(\\varphi\\).<\/p>\n","protected":false},"excerpt":{"rendered":"
W tym wpisie poka\u017cemy, \u017ce ci\u0105g iloraz\u00f3w s\u0105siednich wyraz\u00f3w ci\u0105gu Fibonacciego ma granic\u0119 r\u00f3wn\u0105 s\u0142ynnej z\u0142otej proporcji.<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7,5],"tags":[8,11,9],"yoast_head":"\n
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