[VA] SRC #007 - 2020 April 1st Ramblings

[Valentin Albillo]

 
Hi, all:

I see that Paul Dale and Juan14 recently addressed my Rambling 5, which was left unanswered so far, so thanks to both of you for addressing it and consequently I'm giving here my original solution and comments (all code is for the HP-71B + Math ROM):

Quote:5) Finding some simple way to express the Nth Fibonacci number using them, perhaps as simple as a few lines of RPN/RPL code or a short single-line user-defined function, used like this: FNF(1) = 1, FNF(2) = 1, FNF(3) = 2, ..., FNF(10) = 55, ...

My original solution consists of the following code, which simply initializes two numeric variables (one real, one complex) to hold some constants for efficiency's sake, and then implements a single-line user-defined function FNF(N), which takes N (whether integer or not) and uses a single instance of the Hyperbolic Sine (SINH) function (as required) of a complex argument to directly return the N-th Fibonacci number, F_N :

      1  DESTROY ALL @ COMPLEX Z @ X=2/SQR(5) @ Z=(LN((1+SQR(5))/2),PI/2)
      2  DEF FNF(N)=REPT(X*SINH(N*Z)/(0,1)^N)

Unlike Juan14's solution, mine doesn't distinguish between odd N and even N, a single SINH suffices for every real N. Also notice that (0,1) is the value i (the imaginary unit) and (1+√5)/2 is φ, the Golden Ratio. Let's check it:

      >RUN      (just once, to initialize the two needed constants)

and now we'll use FNF directly from the command line to compute and display the first 16 Fibonacci numbers F₁ .. F₁₆

      >FOR N=1 TO 16 @ DISP USING "2D,2X,3D.8D";N;FNF(N) @ NEXT N

            1      1.00000000
            2      1.00000000
            3      2.00000000
            4      3.00000000
            5      5.00000000
            6      8.00000000
            7     13.00000000
                  ...
           15    610.00000000
           16    987.00000000

which are fully correct. The argument N isn't constrained to be an integer so we can compute F_20.20 like this:

      >FNF(20.20)

            7448.4401

and we can check the basic recurrence F_N + F_N+1 = F_N+2 for N = 20.20 like this:

      >FNF(20.20)+FNF(21.20),FNF(22.20)

            19500.2694   19500.2694

Just for fun, let's compute F_Pi and F_e :

      >STD @ FNF(PI),FNF(EXP(1))

            (F_Pi =) 2.11702705791   (F_e =) 1.7308152698

(notice that when rounded to 4 digits we coincidentally have F_e = 1.731 while √3 = 1.732 and thus we have F_e ~ √3 )

Last, and also for fun, let's compute N such that F_N is a given value, i.e: inverse Fibonacci. Add this 3^rd program line:

      3  INPUT F @ FNROOT(1,100,FNF(FVAR)-F) @ GOTO 3

and let's try some values:

      >FIX 8
      >RUN
           ? 55
                     10.00000000    (F₁₀ = 55)
           ? 6765
                     20.00000000    (F₂₀ = 6765)
           ? PI
                     4.09041655     (F_4.09041655 = Pi)
           ? 2020
                     17.48828958    (F_17.48828958 = 2020)

Regards.
V.