# So turns out the Fibanacci series isn't that special with respect to the Golden Ratio **URL:** \ **Category:** Miscellaneous and Personal Stuff I Must Share\ **Created:** [January 9, 2015, 6:14pm UTC](https://boards.straightdope.com/t/so-turns-out-the-fibanacci-series-isnt-that-special-with-respect-to-the-golden-ratio/709420 "2015-01-09T18:14:51Z")\ **Posts on this page:** 12\ **Page:** 1
**Author:** ![D18](https://avatars.discourse-cdn.com/v4/letter/d/c57346/32.png) [@D18](https://boards.straightdope.com/u/D18)\ **Post date:** [January 9, 2015, 6:14pm UTC](https://boards.straightdope.com/t/so-turns-out-the-fibanacci-series-isnt-that-special-with-respect-to-the-golden-ratio/709420/1 "2015-01-09T18:14:51Z")
I know that the math folks here will know this, but I found it interesting and thought others might like to learn it. Besides, I find when I post a rudimentary question about math on this board, I learn at least three knew things I hadn’t thought to ask about. So here goes. [So](http://boards.straightdope.com/sdmb/showthread.php?t=745662), I watched [this video](http://www.numberphile.com/videos/brady_numbers.html) from the Numberphile, and he shows that any quasi-Fibonacci series will predict the golden ratio. That is, randomly choose two starting numbers, make the third their sum, and then continue along. You get a good approximation of the golden ratio, and in fact, it will regress to it faster than Fibonacci will. MPSIMS (mathematical, pointless stuff I must share) ---
**Author:** ![friedo](https://avatars.discourse-cdn.com/v4/letter/f/8edcca/32.png) [@friedo](https://boards.straightdope.com/u/friedo)\ **Post date:** [January 9, 2015, 6:18pm UTC](https://boards.straightdope.com/t/so-turns-out-the-fibanacci-series-isnt-that-special-with-respect-to-the-golden-ratio/709420/2 "2015-01-09T18:18:12Z")
Yep. Any Fibonacci-type series will converge on the golden ratio if you iterate it enough. I wonder which one converges the fastest? ---
**Author:** ![billfish678](https://avatars.discourse-cdn.com/v4/letter/b/7bcc69/32.png) [@billfish678](https://boards.straightdope.com/u/billfish678)\ **Post date:** [January 9, 2015, 6:21pm UTC](https://boards.straightdope.com/t/so-turns-out-the-fibanacci-series-isnt-that-special-with-respect-to-the-golden-ratio/709420/3 "2015-01-09T18:21:21Z")
Two numbers that represent the golden ratio? 🙂 ---
**Author:** ![D18](https://avatars.discourse-cdn.com/v4/letter/d/c57346/32.png) [@D18](https://boards.straightdope.com/u/D18)\ **Post date:** [January 9, 2015, 6:24pm UTC](https://boards.straightdope.com/t/so-turns-out-the-fibanacci-series-isnt-that-special-with-respect-to-the-golden-ratio/709420/4 "2015-01-09T18:24:15Z")
Yes, but you don’t know that in advance. Is it the bigger the numbers that start the quasi-Fibonacci series, the faster? ---
**Author:** ![Hari\_Seldon](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hari_seldon/32/5173_2.png) [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)\ **Post date:** [January 10, 2015, 3:01am UTC](https://boards.straightdope.com/t/so-turns-out-the-fibanacci-series-isnt-that-special-with-respect-to-the-golden-ratio/709420/5 "2015-01-10T03:01:56Z")
Take any two numbers, say x and y and start a Fibonacci series starting with x and y and use the rule that each term is the sum of the two previous ones. It goes x, y, x+y, x+2y, 2x+3y, 3x+5y, 5x+8y,…, f\_{n-1}x+f\_{n}y,… where the f\_n are the standard Fibonacci numbers. That the ratio approaches the same golden ratio is now easy to prove. ---
**Author:** ![friedo](https://avatars.discourse-cdn.com/v4/letter/f/8edcca/32.png) [@friedo](https://boards.straightdope.com/u/friedo)\ **Post date:** [January 10, 2015, 3:24am UTC](https://boards.straightdope.com/t/so-turns-out-the-fibanacci-series-isnt-that-special-with-respect-to-the-golden-ratio/709420/6 "2015-01-10T03:24:11Z")
> [@Hari\_Seldon](#): > > Take any two numbers, say x and y and start a Fibonacci series starting with x and y and use the rule that each term is the sum of the two previous ones. It goes > x, y, x+y, x+2y, 2x+3y, 3x+5y, 5x+8y,…, f\_{n-1}x+f\_{n}y,… where the f\_n are the standard Fibonacci numbers. That the ratio approaches the same golden ratio is now easy to prove. I noticed this when, after learning about these sequences, I tried it with this series to see what I get: e, π, e+π , e+2π, 2e+3π, 3e+5π, 5e+8π…hey wait a minute! 🙂 ---
**Author:** ![LSLGuy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lslguy/32/5813_2.png) [@LSLGuy](https://boards.straightdope.com/u/LSLGuy)\ **Post date:** [January 10, 2015, 1:25pm UTC](https://boards.straightdope.com/t/so-turns-out-the-fibanacci-series-isnt-that-special-with-respect-to-the-golden-ratio/709420/7 "2015-01-10T13:25:20Z")
> [@billfish678](#): > > Two numbers that represent the golden ratio? 🙂 [Golden ratio - Wikipedia](http://en.wikipedia.org/wiki/Golden_ratio) It’s a “magic” number equal to abut 1.6. ---
**Author:** ![JWT\_Kottekoe](https://avatars.discourse-cdn.com/v4/letter/j/3ec8ea/32.png) [@JWT\_Kottekoe](https://boards.straightdope.com/u/JWT_Kottekoe)\ **Post date:** [January 10, 2015, 2:16pm UTC](https://boards.straightdope.com/t/so-turns-out-the-fibanacci-series-isnt-that-special-with-respect-to-the-golden-ratio/709420/8 "2015-01-10T14:16:22Z")
It’s true unless you start with two numbers whose ratio is **exactly** the reciprocal of the golden ratio. In that case all subsequent numbers will be in that same ratio. Even the slightest amount of round off error will destroy this and the golden ratio will rapidly prevail. ---
**Author:** ![D18](https://avatars.discourse-cdn.com/v4/letter/d/c57346/32.png) [@D18](https://boards.straightdope.com/u/D18)\ **Post date:** [January 11, 2015, 3:48pm UTC](https://boards.straightdope.com/t/so-turns-out-the-fibanacci-series-isnt-that-special-with-respect-to-the-golden-ratio/709420/9 "2015-01-11T15:48:23Z")
> [@Hari\_Seldon](#): > > Take any two numbers, say x and y and start a Fibonacci series starting with x and y and use the rule that each term is the sum of the two previous ones. It goes > x, y, x+y, x+2y, 2x+3y, 3x+5y, 5x+8y,…, f\_{n-1}x+f\_{n}y,… where the f\_n are the standard Fibonacci numbers. That the ratio approaches the same golden ratio is now easy to prove. Huh. I wondered why it worked. Thank you for this. ---
**Author:** ![JWT\_Kottekoe](https://avatars.discourse-cdn.com/v4/letter/j/3ec8ea/32.png) [@JWT\_Kottekoe](https://boards.straightdope.com/u/JWT_Kottekoe)\ **Post date:** [January 12, 2015, 4:46am UTC](https://boards.straightdope.com/t/so-turns-out-the-fibanacci-series-isnt-that-special-with-respect-to-the-golden-ratio/709420/10 "2015-01-12T04:46:12Z")
> [@D18](#): > > Huh. I wondered why it worked. Thank you for this. It is easy to prove that, with any initial numbers, the series is the sum of an exponentially growing series and an exponentially decaying series, a\*(phi)^n and b\*(phi)^(-n), where phi is the golden ratio and n is the index of this term in the series. The coefficients a and b are determined by the starting numbers. Since phi is bigger than one, (phi)^n gets bigger exponentially, while (phi)^(-n) decays to zero exponentially. Thus, the a\*(phi)^n term always wins, with only one exception. If a is exactly zero, there is only the decaying term and the ratio of consecutive terms is always the 1/phi and not phi. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [January 12, 2015, 4:52am UTC](https://boards.straightdope.com/t/so-turns-out-the-fibanacci-series-isnt-that-special-with-respect-to-the-golden-ratio/709420/11 "2015-01-12T04:52:23Z")
> [@JWT\_Kottekoe](#): > > It’s true unless you start with two numbers whose ratio is **exactly** the reciprocal of the golden ratio. In that case all subsequent numbers will be in that same ratio. Even the slightest amount of round off error will destroy this and the golden ratio will rapidly prevail. It’s not the reciprocal of the “golden ratio” that works in the same way; it’s the _negated_ reciprocal of the “golden ratio” (equivalently, 1 - phi). > [@JWT\_Kottekoe](#): > > It is easy to prove that, with any initial numbers, the series is the sum of an exponentially growing series and an exponentially decaying series, a\*(phi)^n and b\*(phi)^(-n), where phi is the golden ratio and n is the index of this term in the series. In keeping with the above, the second term here should read “b \* (-phi)^(-n)”. [For those who don’t realize, these constants and the relation between them arise from solving the quadratic equation x[sup]2[/sup] = x + 1] ---
**Author:** ![JWT\_Kottekoe](https://avatars.discourse-cdn.com/v4/letter/j/3ec8ea/32.png) [@JWT\_Kottekoe](https://boards.straightdope.com/u/JWT_Kottekoe)\ **Post date:** [January 21, 2015, 6:35am UTC](https://boards.straightdope.com/t/so-turns-out-the-fibanacci-series-isnt-that-special-with-respect-to-the-golden-ratio/709420/12 "2015-01-21T06:35:27Z")
> [@Indistinguishable](#): > > It’s not the reciprocal of the “golden ratio” that works in the same way; it’s the _negated_ reciprocal of the “golden ratio” (equivalently, 1 - phi). Yes, I realized later that I made a sign error. Thanks for correcting it.