In my previous blog, I had concluded with a relation between the golden ratio and the Fibonacci sequence. It can be stated as below:
pn = Fn+1 p + Fn ---- (1)
Where p is the golden ratio
And the Fibonacci sequence starts off from 1, 1
i.e. 1, 1, 2, 3, 5, 8, 13, 21 etc.
Now I shall use this equality to prove something else about the relation between the Fibonacci sequence and the golden ratio, namely:
lim Fn+1
------- = p
n -> infinity Fn
Let us suppose that:
R = lim Fn+1
-------
n -> infinity Fn
By doing so, we implicitly assume that such a limiting value for the ratio of two successive Fibonacci terms exists.
R is the ratio of the greater of two successive Fibonacci numbers to the lesser of the two, as the numbers become infinitely large. Thus, R can also be expressed as:
R = lim Fn
------
n -> infinity Fn-1
Now, to start the proof, first divide both sides of (1) by Fn
pn Fn+1
--- = ------- p + 1
Fn Fn
pn Fn+1
--- - 1 = ------- p
Fn Fn
Fn+1 (pn / Fn) - 1
------ = -------------------
Fn p
[p(pn-1)/Fn] - 1
= -------------------
p
pn-1 = Fn p + Fn-1 so:
[p(Fn p + Fn-1)/Fn] - 1
= ------------------------------
p
p2 + p(Fn-1/Fn) – 1
= --------------------------
p
Fn-1 1
= p + ------- – --
Fn p
Fn-1 p2 - 1
= ------ + --------
Fn p
p2 = p + 1 so:
Fn-1 p + 1 - 1
= ------- + ----------
Fn p
Fn+1 Fn-1
------- = ------ + 1 ----- (2)*
Fn Fn
Now, as we discussed earlier,
R = lim Fn+1
-------
n -> infinity Fn
R = lim Fn
-------
n -> infinity Fn-1
Thus, as n tends to infinity, (2) becomes:
R = 1/R + 1
R = (1 + R)/R
R2 = 1 + R
(This is the same quadratic equation as the one for the golden ratio)
R2 – R – 1 = 0
Then R = [-b +- √ (b2 – 4ac)]/2a
R = [1 +- √ (1 + 4)]/2 = [1 +- √5]/2
Additionally, as all Fibonacci numbers are positive,
R = [1 + √5]/2 (which is equal to the golden ratio)
Therefore, the ratio of the greater to the smaller of two successive terms of the Fibonacci sequence, as we go on to bigger and bigger terms, tends to the Golden Ratio.
* The relation stated in (2) gives us:
Fn+1 Fn-1
------- = ------ + 1
Fn Fn
In other words, the ratio of two consecutive Fibonacci numbers (greater is to smaller) is equal to one added to the inverse of the ratio of the smaller number to the Fibonacci number immediately before it.
eg. The Fibonacci numbers 5, 8, 13
13/8 = 1/(8/5) + 1
13/8 = 5/8 + 1
13/8 = (5 + 8)/8
13/8 = 13/8
This relation can be derived in a much simpler way, by considering any sequence of three consecutive Fibonacci numbers.
If the first number is a and the second number is b, the third number must be a + b
i.e. Fn-1 = a
Fn = b
Fn+1 = a + b
Now the ratio of the third to the second is (a + b)/b
And the ratio of the second to the first is b/a
If we invert this second ratio and add 1 to it, we get:
1/(b/a) + 1
= a/b + 1
= (a + b)/b
This is the same as the ratio of the third Fibonacci number to the second Fibonacci number.
i.e. (a+b)/b = 1/(b/a) + 1
Thus we conclude:
Fn+1 Fn-1
------- = ------ + 1
Fn Fn
This would have been a much shorter method of deriving the proof (don’t ask why I didn’t just use this way).
Addendum:
Let Rn denote the ratio of the (n+1)th term to the nth term of the Fibonacci sequence:
1, 1, 2, 3, 5, 8, 13…
Then, from the equation in (2),
Rn = 1 + 1/Rn-1
Rn-1 = 1 + 1/Rn-2
Rn-2 = 1 + 1/Rn-3
Etc.
If we were to put them all together,
Rn = 1 + 1/Rn-1
= 1 + 1/(1 + 1/Rn-2)
= 1 + 1/(1 + 1/(1 + 1/Rn-3))
……..
= 1 + 1/(1 + 1/(1 + 1/(1 + 1/(….. + 1/R1)))))….)
Further, R1 = 1/1 = 1
So Rn = 1 + 1/(1 + 1/(1 + 1/(1 + 1/(….. + 1)))))….)
When n tends to infinity, Rn, as we found out, tends to the golden ratio.
Thus, as n tends to infinity, we get the infinite fraction for the golden ratio:
R = 1 + 1
---------------------------------
1 + 1
------------------------
1 + 1
----------------
1 + 1
-------
1 + …
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