HP 12C Fibonacci Sequence

08162017, 05:16 AM
Post: #1




HP 12C Fibonacci Sequence
Here is the HP 12C program to find Fibonacci Sequence accurate up to 40
Anyone have a better solution? Gamo 

08162017, 09:06 AM
Post: #2




RE: HP 12C Fibonacci Sequence
(08162017 05:16 AM)Gamo Wrote: Here is the HP 12C program to find Fibonacci Sequence accurate up to 40 Hi, for interest see posts #2 & #3 in this thread in the Software Library: http://www.hpmuseum.org/forum/thread835...i#pid73315 Best regards 

08162017, 04:38 PM
(This post was last modified: 08162017 05:28 PM by Dieter.)
Post: #3




RE: HP 12C Fibonacci Sequence
(08162017 05:16 AM)Gamo Wrote: Here is the HP 12C program to find Fibonacci Sequence accurate up to 40 Your program implements an interesting formula that calculates the Fibonacci numbers directly instead of the usual approach with adding the two previous ones. I did not know this before, but it seems to be the MoivreBinet formula: \[F_n = \frac{1}{\sqrt{5}} \left [ \left ( \frac{1+\sqrt{5}}{2} \right )^n  \left ( \frac{1\sqrt{5}}{2} \right )^n \right ] \] A straightforward implementation does not require any registers and could be done like this: Code: 01 ENTER (08162017 05:16 AM)Gamo Wrote: Anyone have a better solution? Since for n≥0 the second term is always less than 1/2 it doesn't have to be calculated. Simply round \(\frac{1}{\sqrt{5}} \left (\frac{1+\sqrt{5}}{2} \right )^n\) to the nearest integer. That's what the following program does. Code: 01 ENTER On the 10digit 12C this also works correctly up to n=40. Above that the results are a bit larger than expected because the 10digit value of \(\Phi\) is slightly high while \(\sqrt{5}\) in the denominator is slightly low. That's also why the rounding function in the above code does not add 0,5 but 0,3. BTW, using code boxes like this instead of attaching a PDF (which is only available after login) may be the better solution. Dieter 

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