A Proof of Golden Section of Fibonacci Sequence

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Update on 2024/6/25 10:40 (UTF+8) : Add the Part Five and correct some words

There's a Chinese Version in This Blog's Part $$$7.4$$$ to $$$7.7$$$

Hello, I'm glad to show you one of the feasible proof methods for the following equation:

When n approaches positive infinity, then $$${f_{n-1}\over f_{n}}={{\sqrt{5}-1}\over 2}$$$ ，which one we usually call it golden section.

Although my own mathematical level is not very high, after studying an article about this conclusion, I found that the proof in that article is not very clear, but this conclusion is indeed very beautiful. So I will share this conclusion with everyone through this blog.

My English level is average, so if there are any errors in the article, please correct me!

Thanks for 5k_sync_closer's corrections

Following, we define $$$f_{i}$$$ as the $$$i_{th}$$$ number of a Fibonacci sequence.

Part One: The General Formula of Fibonacci Sequence

As we all know，we defines Fibonacci sequence using $$$f_{n}=f_{n-1}+f_{n-2}$$$

And we also know, if we have a sequence satisfy that $$$g_{i}=kg_{i-1}$$$, then its general formula is $$$g_{i}=k^{i-1}g_{1}$$$

So we transform the Fibonacci sequence into this form:

$$$f_{n}-\lambda f_{n-1}=\mu (f_{n-1}-\lambda f_{n-2})\tag{0}$$$

Substitute $$$f_{n}=f_{n-1}+f_{n-2}$$$ into this equation，we will get:

$$$f_{n-1}+f_{n-2}-\lambda f_{n-1}=\mu f_{n-1}-\mu\lambda f_{n-2}$$$ $$$(\mu +\lambda -1)f_{n-1}=(1+\mu\lambda)f_{n-2}$$$

Because $$$f_{n-1}\neq f_{n-2}$$$ , so：

$$$\begin{cases}\mu +\lambda -1=0\\1+\mu\lambda=0\end{cases}$$$

Solve the equation, we get the answer:

$$$\begin{cases} \lambda = \frac{1+\sqrt{5}}{2}\\ \mu = \frac{1-\sqrt{5}}{2} \end{cases}$$$

$$$\begin{cases} \lambda = \frac{1-\sqrt{5}}{2}\\ \mu = \frac{1+\sqrt{5}}{2} \end{cases}$$$

Substitute it into equation $$$(0)$$$

$$$f_n-\frac{1+\sqrt{5}}{2}f_{n-1}=(\frac{1-\sqrt{5}}{2})^{n-2}(f_2-\frac{1+\sqrt{5}}{2}f_1)\tag{1}$$$

$$$f_n-\frac{1-\sqrt{5}}{2}f_{n-1}=(\frac{1+\sqrt{5}}{2})^{n-2}(f_2-\frac{1-\sqrt{5}}{2}f_1)\tag{2}$$$

Based on the proof just now, Both of these equations hold relative to the original equation, So we try to eliminate $$$f_{n-1}$$$ ,then we finally get what we want:

$$$f_{n}=\frac{1}{\sqrt{5}}[(\frac{\sqrt{5}+1}{2})^{n}-(\frac{1-\sqrt{5}}{2})^{n}]$$$

Part Two: The Relationship of $$$f_{n}f_{n-2}$$$ and $$$f_{n-1}^{2}$$$

$$$f_{n}f_{n-2}-f_{n-1}^{2}=(-1)^{n-1}$$$

To simplify our proof, we define that $$$P=\frac{\sqrt{5}+1}{2},Q=\frac{1-\sqrt{5}}{2}$$$ ，then we have $$$\frac{1}{\sqrt{5}}(P^{n}-Q^{n})$$$

then we substitute the general formula to $$$f_{n}f_{n-2}-f_{n-1}^{2}$$$ :

$$$\frac{1}{\sqrt{5}\times\sqrt{5}}(P^{n}-Q^{n})(P^{n-2}-Q^{n-2})-\frac{1}{(\sqrt{5})^{2}}(P^{n-1}-Q^{n-1})^{2}$$$

$$$\frac{1}{5}[P^{2n-2}+Q^{2n-2}-P^{n}Q^{n-2}-P^{n-2}Q^{n}-(P^{2n-2}+Q^{2n-2}-2P^{n-1}Q^{n-1})]$$$

$$$\frac{1}{5}[-P^{n}Q^{n-2}-P^{n-2}Q^{n}+2P^{n-1}Q^{n-1}]$$$

$$$-\frac{1}{5}[P^{n-1}Q^{n-1}(\frac{P}{Q}+\frac{Q}{P}-2)]$$$

$$$-\frac{1}{5}[(PQ)^{n-1}(\frac{P}{Q}+\frac{Q}{P}-2)]$$$

Remember that we defined $$$P=\frac{\sqrt{5}+1}{2},Q=\frac{1-\sqrt{5}}{2}$$$ , so $$$PQ=-1,\frac{P}{Q}=-\frac{3+\sqrt{5}}{2},\frac{Q}{P}=-\frac{3-sqrt{5}}{2},\frac{P}{Q}+\frac{Q}{P}-2=-5$$$ , $$$-5\times -\frac{1}{5}=1$$$ , then we get:

$$$f_{n}f_{n-2}-f_{n-1}^{2}=(-1)^{n-1}$$$

Part Three: The Final Proof I

According to Part Two, $$$f_{n}f_{n-2}-f_{n-1}^{2}=(-1)^{n-1}$$$, Transfer the term to this equation.

$$$f_{n}f_{n-2}=(-1)^{n-1}+f_{n-1}^{2}$$$

Then we divide both sides of the equation by $$$f_{n-1}f_{n-2}$$$ simultaneously.

$$$\frac{f_{n}}{f_{n-1}}=\frac{(-1)^{n-1}}{f_{n-1}f_{n-2}}+\frac{f_{n-1}}{f_{n-2}}$$$

In our definition, $$$n$$$ is approaching positive infinity, namely $$$n\rightarrow +\infty$$$ , $$$\frac{(-1)^{n-1}}{f_{n-1}f_{n-2}}\rightarrow 0$$$ , this item has a negligible impact on our answer, so we will omit it.

$$$\frac{f_{n}}{f_{n-1}}=\frac{f_{n-1}}{f_{n-2}}$$$

Part Four: The Final Proof II

According to Part Three, we define k that $$$\frac{f_{n}}{f_{n-1}}=\frac{f_{n-1}}{f_{n-2}}=k$$$ , notice that $$$f_{i}=f_{n-1}+f_{n-2}$$$

$$$\frac{f_{n-1}+f_{n-2}}{f_{n-1}}=\frac{f_{n-1}}{f_{n-2}}=k$$$

$$$1+\frac{f_{n-2}}{f_{n-1}}=\frac{f_{n-1}}{f_{n-2}}=k$$$

$$$f_{n-1}f_{n-2}+f_{n-2}^{2}=f_{n-1}^{2}$$$

We define that $$$f_{n-1}=kf_{n-2}$$$

$$$kf_{n-2}^{2}+f_{n-2}^{2}=k^{2}f_{n-2}^{2}$$$

divide both sides of the equation by

$$$f_{n-2}^{2}$$$

simultaneously.

$$$k+1=k^{2}$$$

Solve this equation, we finally get $$$k=\frac{1+\sqrt{5}}{2}$$$

Part Five: Promotion

Notice that we completely did not use a very important property of the Fibonacci sequence: $$$f_{1}=f_{2}=1$$$

Actually, for every sequence satisfy that $$$f_{n}=f_{n-1}+f_{n-2}$$$ , not only the Fibonacci sequence , the above conclusions are all valid. Just because we are familiar with the Fibonacci sequence in our daily lives, we use it as an example to prove it.

By the way, have you ever tried that $$$\frac{f_{1}}{f_{2}}=1,\frac{f_{2}}{f_{3}}=0.5,\frac{f_{3}}{f_{4}}=0.6667,\frac{f_{4}}{f_{5}}=0.6,\frac{f_{5}}{f_{6}}=0.625$$$ . We can observe that for adjacent $$$n$$$ , one is always greater than $$$0.618$$$ and the other is less than $$$0.618$$$ . This indicates another pattern we have discovered.

That's all I have to say. Thank you for reading!