Yet another Fibonacci mini-challenge (HP-42S/Free42)
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12-07-2018, 03:41 PM
(This post was last modified: 12-07-2018 03:50 PM by Gerson W. Barbosa.)
Post: #15
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RE: Yet another Fibonacci mini-challenge (HP-42S/Free42)
(12-07-2018 10:42 AM)Thomas Klemm Wrote:(12-05-2018 04:14 PM)Gerson W. Barbosa Wrote: … if you notice that asinh(1/2) = ln(phi) then you can easily do it in ten steps. By spoiling the fun I meant my somewhat premature disclosing of the inverse hyperbolic trick. Yes, I remember that old thread. Thanks for bringing it back to our attention! Meanwhile, I've tried an HP-15C version: 001 { 42 21 11 } f LBL A 002 { 48 } . 003 { 5 } 5 004 { 43 22 23 } g HYP-¹ SIN 005 { 20 } × 006 { 12 } e^x 007 { 5 } 5 008 { 11 } √x̅ 009 { 10 } ÷ 010 { 48 } . 011 { 5 } 5 012 { 40 } + 013 { 43 44 } g INT 014 { 43 32 } g RTN It seems this will work for n = 0 to 41, unlike the standard formula which fails for n = 40. While testing it, I noticed that \(\varphi^{39}\approx \frac{\pi ^{17}}{2}\). A little tweaking, making use of the apparent \(\sqrt{2}\) factors present in both sides of the expression, gives \(\varphi \approx \sqrt[39]{\frac{\pi }{2}\left ( \pi^{16}+19\sqrt{2} \right )}\) Cheers, Gerson. |
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