Yet another Fibonacci mini-challenge (HP-42S/Free42)

[Gerson W. Barbosa]

(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.
… Hopefully I haven't spoiled the fun :-)

No worries, you just made me remember an older thread where the same trick is used to solve \(x(x+1)=2n\) (a.k.a. Inverse Little Gauss).
Similarly, we can use \(x (x-1) = 1 \) to calculate the golden ratio.

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.