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"The successive ratios of the terms, i.e. 1/1, 2/1, 3/2, 5/3 ... tend to a number called the Golden Ratio by the Greeks." Fun fact: take any two numbers (e.g.

"The successive ratios of the terms, i.e. 1/1, 2/1, 3/2, 5/3 ... tend to a number called the Golden Ratio by the Greeks." Fun fact: take any two numbers (e.g. chosen randomly), and use them as the seeds for a Fibonacci-like sequence by summing the last two terms to generate the next term. The ratio of any two consecutive terms in that series will tend towards the golden ratio.

cscheid 6y ago
If you read this and you’re curious, there’s a proof that is fairly easy to follow if you understand eigenvectors. Write the operation that takes the two last elements of the sequence and produces the following two, notice it’s linear, then analyze the eigenvalues of the associated matrix and relate the original operation to the power method.
- redvenom 6y ago
  You actually don't need eigenvectors for this proof. Sketch: Write x = lim a_n/a_{n-1} where a_n is the nth term of Fibonacci. Replace a_n with a_{n-1} + a_{n-2}, which will give you a quadratic equation in x. Solving this quadratic equation gives you the golden ratio.
  - cscheid 6y ago
    I believe that gives you the result starting from Fibonacci, but not that any starting pair of numbers eventually lands at the golden ratio.
    - redvenom 6y ago
      No, it works for any pair of numbers, as long as they are not both zero. I am only using the recursive relation of the Fibonacci sequence, not the starting terms.
cleansingfire 6y ago
This seems like a pretty basic calculus problem. Constants become irrelevant at the limit, so that the relationship (ratio) is all that's left.
- dan-robertson 6y ago
  Well it isn’t really calculus as there’s no differentiation (I guess you could consider the last step where you take a limit to be calculus), but it’s a bit like differential equations. You can write down the recurrence relation: a_(n+2) = a_(n+1) + a_n Observe that there is a linear solution space (I.e. if you add solutions point wise or multiply each value by the same scalar, you get solutions), and the values a_0 and a_1 are sufficient to determine the sequence. Now guess that a_n = k^n is a solution: k^2 = k + 1 (k - 1/2)^2 = 5/4 k = (1 +/- sqrt(5))/2 k = φ or -1/φ, where φ is the golden ratio Due to linearity, there are a family of solutions a_n = Rφ^n + S(-1/φ)^n for any values of R and S. Because this family provides a solution for any choices of a_0 and a_1, it contains all the solutions. Because |1/φ|<1, we find that asymptotically a_n ~ Rφ^n as n grows. Therefore the ratio of terms tends to φ in the limit.
  - cleansingfire 6y ago
    You're right, of course. Thanks for the explanation. I was thinking particularly of the comment about the behavior with arbitrary initial numbers. Start with 1, 5000, and it still converges quickly to the golden ratio, as the parent comment mentioned. I enjoy watching the initial constants disappear.

cscheid 6y ago

If you read this and you’re curious, there’s a proof that is fairly easy to follow if you understand eigenvectors. Write the operation that takes the two last elements of the sequence and produces the following two, notice it’s linear, then analyze the eigenvalues of the associated matrix and relate the original operation to the power method.

redvenom 6y ago
You actually don't need eigenvectors for this proof. Sketch: Write x = lim a_n/a_{n-1} where a_n is the nth term of Fibonacci. Replace a_n with a_{n-1} + a_{n-2}, which will give you a quadratic equation in x. Solving this quadratic equation gives you the golden ratio.
- cscheid 6y ago
  I believe that gives you the result starting from Fibonacci, but not that any starting pair of numbers eventually lands at the golden ratio.
  - redvenom 6y ago
    No, it works for any pair of numbers, as long as they are not both zero. I am only using the recursive relation of the Fibonacci sequence, not the starting terms.

cleansingfire 6y ago

This seems like a pretty basic calculus problem. Constants become irrelevant at the limit, so that the relationship (ratio) is all that's left.

dan-robertson 6y ago
Well it isn’t really calculus as there’s no differentiation (I guess you could consider the last step where you take a limit to be calculus), but it’s a bit like differential equations. You can write down the recurrence relation: a_(n+2) = a_(n+1) + a_n Observe that there is a linear solution space (I.e. if you add solutions point wise or multiply each value by the same scalar, you get solutions), and the values a_0 and a_1 are sufficient to determine the sequence. Now guess that a_n = k^n is a solution: k^2 = k + 1 (k - 1/2)^2 = 5/4 k = (1 +/- sqrt(5))/2 k = φ or -1/φ, where φ is the golden ratio Due to linearity, there are a family of solutions a_n = Rφ^n + S(-1/φ)^n for any values of R and S. Because this family provides a solution for any choices of a_0 and a_1, it contains all the solutions. Because |1/φ|<1, we find that asymptotically a_n ~ Rφ^n as n grows. Therefore the ratio of terms tends to φ in the limit.
- cleansingfire 6y ago
  You're right, of course. Thanks for the explanation. I was thinking particularly of the comment about the behavior with arbitrary initial numbers. Start with 1, 5000, and it still converges quickly to the golden ratio, as the parent comment mentioned. I enjoy watching the initial constants disappear.