The bisection of the Fibonacci series, Sloane A001906 {1, 3, 8, 21, 55, 144,...}, naturally produces approximations to phi^2, by the division of the nth term by its predecessor: a(n)/a(n1). ; e.g 55/21, 144/55, etc.
WolframAlpha also lists these fractions as convergents to 5pi/6.
In fact, there will always be a small shortfall between the two: (phi)^2  5pi/6 is not zero.
For sufficiently large n, how is the shortfall best approximated, in terms of a rational fraction, say 1/x?
To get continuedfraction approximations I used
DefDbl AZ
Dim crlf$
Private Sub Form_Load()
Form1.Visible = True
Text1.Text = ""
crlf = Chr$(13) + Chr$(10)
pi = Atn(1) * 4
phi = (1 + Sqr(5)) / 2
x = phi * phi  5 * pi / 6: y = 1
a = 1: b = 0
c = 0: d = 1
Text1.Text = Text1.Text & x & crlf
For i = 1 To 10
q = Int(x / y)
z = x  q * y
x = y: y = z
newa = c: newb = d
c = a  q * c: d = b  q * d
a = newa: b = newb
Text1.Text = Text1.Text & c & Str(d) & " " & Str(d / c) & crlf
Next
Text1.Text = Text1.Text & crlf & " done"
End Sub
The output was
4.01107584007607E05
1 0 0
24930 1 4.01123144805455E05
249311 4.01107055473106E05
747929 30 4.01107591763389E05
152078961 4.01107582971734E05
5310296 213 4.01107584209995E05
6831085274 4.01107583934324E05
18972466 761 4.01107584011483E05
1206658814840 4.01107584007115E05
139638347 5601 4.01107584007708E05
Indicating the difference itself is given as approximately 0.0000401107584007607 and successive approximations are
1/24930
1/24931
30/747929
61/1520789
213/5310296
...
5601/139638347
...

Posted by Charlie
on 20161226 08:56:34 