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Nothing's perfect. (Posted on 2016-12-24) Difficulty: 3 of 5

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(n-1). ; 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?

See The Solution Submitted by broll    
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Some Thoughts computer exploration Comment 1 of 1
To get continued-fraction approximations I used

DefDbl A-Z
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

4.01107584007607E-05
1 0      0
-24930 1     -4.01123144805455E-05
24931-1     -4.01107055473106E-05
-747929 30     -4.01107591763389E-05
1520789-61     -4.01107582971734E-05
-5310296 213     -4.01107584209995E-05
6831085-274     -4.01107583934324E-05
-18972466 761     -4.01107584011483E-05
120665881-4840     -4.01107584007115E-05
-139638347 5601     -4.01107584007708E-05

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 2016-12-26 08:56:34
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