Number Theory Number Theory Math Forum

 February 22nd, 2017, 04:32 PM #11 Newbie   Joined: Aug 2015 From: china Posts: 6 Thanks: 4 In fact, it is a recursive problem. Known $\displaystyle f(0)=1,f(1)=1,f(n)=f(n-1)+f(n-2)$ One obtains $\displaystyle f(n)=(\frac{\sqrt{5}+1}{2})^n+(\frac{\sqrt{5}-1}{2})^n$ Similarly Known $\displaystyle f(0)=2,f(1)=6,f(n)=3f(n-1)-f(n-2)$ One obtains $\displaystyle f(n)=(\frac{3+\sqrt{5}}{2})^n+(\frac{3-\sqrt{5}}{2})^n=(\frac{\sqrt{5}+1}{2})^{2n}+(\frac {\sqrt{5}-1}{2})^{2n}$ Thanks from topsquark and JeffM1 Tags show Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post fahad nasir Trigonometry 2 October 2nd, 2016 08:51 AM jiasyuen Calculus 12 January 25th, 2015 01:47 AM notnaeem Real Analysis 4 August 16th, 2010 12:32 PM naserellid Algebra 2 August 15th, 2010 02:20 AM 450081592 Calculus 2 January 25th, 2010 09:38 AM

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