Golden Ratio is considered to be one of the greatest beauties in mathematics. Two numbers \(a\) and \(b\) are said to be in Golden Ratio if \[a>b>0,\quad and\quad \frac { a }{ b } =\frac { a+b }{ a } \] If we consider this ratio to be equal to some \(\varphi \) then we have \[\varphi =\frac { a }{ b } =\frac { a+b }{ a } =1+\frac { b }{ a } =1+\frac { 1 }{ \varphi } \] Solving in quadratic we get two values of \(\varphi \), viz. \(\frac { 1+\sqrt { 5 } }{ 2 } \) and \(\frac { 1-\sqrt { 5 } }{ 2 } \) one of which (the second one) turns out to be negative (extraneous) which we eliminate. So the first one is taken to be the golden ratio (which is obviously a constant value). It is considered that objects with their features in golden ratio are aesthetically more pleasant. A woman's face is in general more beautiful than a man's face since different features of a woman's face are nearly in the golden ratio.

Now let us come to Fibonacci sequence. The Fibonacci sequence \({ \left( { F }_{ n } \right) }_{ n\ge 1 }\) is a natural sequence of the following form:\[{ F }_{ 1 }=1,\quad { F }_{ 2 }=1,\quad { F }_{ n-1 }+{ F }_{ n }={ F }_{ n+1 }\] The sequence written in form of a list, is \(1,1,2,3,5,8,13,21,34,..\).

The two concepts: The Golden Ratio and The Fibonacci Sequence, which seem to have completely different origins, have an interesting relationship, which was first observed by Kepler. He observed that the golden ratio is the limit of the ratios of successive terms of the Fibonacci sequence or any Fibonacci-like sequence (by Fibonacci-like sequence, I mean sequences with the recursion relation same as that of the Fibonacci Sequence, but the seed values different). In terms of limit:\[\underset { n\rightarrow \infty }{ lim } \left( \frac { { F }_{ n+1 } }{ { F }_{ n } } \right) =\varphi \] We shall now prove this fact. Let \[{ R }_{ n }=\frac { { F }_{ n+1 } }{ { F }_{ n } } ,\forall n\in N\]Then we have \(\forall n\in N\) and \(n\ge 2\), \[{ F }_{ n+1 }={ F }_{ n }+{ F }_{ n-1 }\\ \] and \[{ R }_{ n }=1+\frac { 1 }{ { R }_{ n-1 } } >1\]We shall show that this ratio sequence goes to the Golden Ratio \(\varphi \) given by: \[\varphi =1+\frac { 1 }{ \varphi } \]We see that: \[\left| { R }_{ n }-\varphi \right| =\left| \left( 1+\frac { 1 }{ { R }_{ n-1 } } \right) -\left( 1+\frac { 1 }{ \varphi } \right) \right| \\ =\left| \frac { 1 }{ { R }_{ n-1 } } -\frac { 1 }{ \varphi } \right| \\ =\left| \frac { \varphi -{ R }_{ n-1 } }{ \varphi { R }_{ n-1 } } \right| \\ \le \left( \frac { 1 }{ \varphi } \right) \left| \varphi -{ R }_{ n-1 } \right|\\ \le { \left( \frac { 1 }{ \varphi } \right) }^{ n-2 }\left| { R }_{ 2 }-\varphi \right| \] Which clearly shows that\[\left( { R }_{ n } \right) \longrightarrow \varphi \] (since \(\left| { R }_{ 2 }-\varphi \right| \) is a finite positive real whose value depends on the seed values)

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## Comments

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TopNewestWow,very helpful!Thank you!

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You are welcome! Did you like it?

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Yes!

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I like Fibonacci very much.It is really The beauty of Mathematics.

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Very nice knowledge.. Loved it...The Magic of Maths!!!

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https://brilliant.org/problems/wow-12/?group=w3HWB8GobVLl&ref_id=1095702

i posted a problem about the same thing

my solution was almost the same as your proof of it (:

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I have seen your proof. Your idea is essentially the same. Only some of your steps are erroneous.

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why?can you elaborate please?

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There is one more interesting thing I found yesterday. The Ratio of the diagonal and the side of a regular Pentagon is exactly equal to the golden ratio.

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Ok then I will write a note on it..

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Didn't you find it extremely interesting? This is the beauty of Mathematics.

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Nice

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Thanks..don't you think whatever is written above is a reconciliation of two apparently different mathematical ideas?..

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Nice work ! I read this in the book Da Vinci Code by Dan Brown.

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That is one book that I want to read but haven't read yet..thank you for your compliments..:-)

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Have you read any other book by Dan Brown ? If not then try them ,they are awesome .

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