Last night I was reading in The Golden Section: Nature's Greatest Secretabout the golden section. Imagine a line segment (C), split into 2 unequal parts (A & B), such that A/B = B/C. Next assume that B=1. So how do we determine what A and C are? Follow below:
I have Gambit Scheme (programming language) on my cellphone and not only can it do calculations but it can send them to my e-mail!
So here are the solutions and code:
Gambit v4.6.1
(/ (+ -1 (expt 5 .5)) 2)
.6180339887498949
(/ (- -1 (expt 5 .5)) 2)
-1.618033988749895
The main solution for A is .6180339887498949, and because B = 1, then C = 1.6180339887498949
Many thanks to Scott Olsen author of The Golden Secret, Marc Feeley (main author of Gambit Scheme), Keith Flower (for porting of Gambit to the Android OS), Google (for helping me to remember/find the solution to a quadratic equation and providing picture-capable Gmail), Microsoft (for making Word handle equations) and my math teachers for a stroll down memory lane and a most interesting adventure for today.
Steve
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On 2011-10-27, at 9:46 AM, Steve Graham wrote:
So here are the solutions and code:
Gambit v4.6.1
(/ (+ -1 (expt 5 .5)) 2)
.6180339887498949
(/ (- -1 (expt 5 .5)) 2)
-1.618033988749895
The main solution for A is .6180339887498949, and because B = 1, then C = 1.6180339887498949
Many thanks to Scott Olsen author of The Golden Secret, Marc Feeley (main author of Gambit Scheme), Keith Flower (for porting of Gambit to the Android OS), Google (for helping me to remember/find the solution to a quadratic equation and providing picture-capable Gmail), Microsoft (for making Word handle equations) and my math teachers for a stroll down memory lane and a most interesting adventure for today.
Steve
No problem!
By the way, one of the nice properties of the fibonacci sequence is its relation to the golden ratio (the limit n->oo of fib(n)/fib(n-1) is the golden ratio, or its inverse, I can never remember). Here is a demonstration which you can try on your iPhone:
(define (fib n)
(let loop ((i n) (a 1) (b 1)) (if (= i 0) a (loop (- i 1) b (+ a b)))))
(fib 20)
10946
(exact->inexact (/ (fib 1000) (fib 999)))
1.618033988749895
(exact->inexact (/ (fib 999) (fib 1000)))
.6180339887498949
(fib 1000)
70330367711422815821835254877183549770181269836358732742604905087154537118196933579742249494562611733487750449241765991088186363265450223647106012053374121273867339111198139373125598767690091902245245323403501
Marc
Thanks for the memory nudge, Marc. I think the author mentioned such, too. Steve
________________________________ From: Marc Feeley feeley@iro.umontreal.ca To: Steve Graham jsgrahamus@yahoo.com Cc: Gambit List Gambit-list@iro.umontreal.ca Sent: Thursday, October 27, 2011 11:16 AM Subject: Re: [gambit-list] Golden Section and Gambit Scheme
On 2011-10-27, at 9:46 AM, Steve Graham wrote:
So here are the solutions and code:
Gambit v4.6.1
(/ (+ -1 (expt 5 .5)) 2)
.6180339887498949
(/ (- -1 (expt 5 .5)) 2)
-1.618033988749895
The main solution for A is .6180339887498949, and because B = 1, then C = 1.6180339887498949
Many thanks to Scott Olsen author of The Golden Secret, Marc Feeley (main author of Gambit Scheme), Keith Flower (for porting of Gambit to the Android OS), Google (for helping me to remember/find the solution to a quadratic equation and providing picture-capable Gmail), Microsoft (for making Word handle equations) and my math teachers for a stroll down memory lane and a most interesting adventure for today.
Steve
No problem!
By the way, one of the nice properties of the fibonacci sequence is its relation to the golden ratio (the limit n->oo of fib(n)/fib(n-1) is the golden ratio, or its inverse, I can never remember). Here is a demonstration which you can try on your iPhone:
(define (fib n)
(let loop ((i n) (a 1) (b 1)) (if (= i 0) a (loop (- i 1) b (+ a b)))))
(fib 20)
10946
(exact->inexact (/ (fib 1000) (fib 999)))
1.618033988749895
(exact->inexact (/ (fib 999) (fib 1000)))
.6180339887498949
(fib 1000)
70330367711422815821835254877183549770181269836358732742604905087154537118196933579742249494562611733487750449241765991088186363265450223647106012053374121273867339111198139373125598767690091902245245323403501
Marc
On 2011-10-27, at 10:48 AM, Steve Graham wrote:
Thanks for the memory nudge, Marc. I think the author mentioned such, too. Steve
From: Marc Feeley feeley@iro.umontreal.ca To: Steve Graham jsgrahamus@yahoo.com Cc: Gambit List Gambit-list@iro.umontreal.ca Sent: Thursday, October 27, 2011 11:16 AM Subject: Re: [gambit-list] Golden Section and Gambit Scheme
On 2011-10-27, at 9:46 AM, Steve Graham wrote:
So here are the solutions and code:
Gambit v4.6.1
(/ (+ -1 (expt 5 .5)) 2)
.6180339887498949
(/ (- -1 (expt 5 .5)) 2)
-1.618033988749895
The main solution for A is .6180339887498949, and because B = 1, then C = 1.6180339887498949
Many thanks to Scott Olsen author of The Golden Secret, Marc Feeley (main author of Gambit Scheme), Keith Flower (for porting of Gambit to the Android OS), Google (for helping me to remember/find the solution to a quadratic equation and providing picture-capable Gmail), Microsoft (for making Word handle equations) and my math teachers for a stroll down memory lane and a most interesting adventure for today.
Steve
No problem!
By the way, one of the nice properties of the fibonacci sequence is its relation to the golden ratio (the limit n->oo of fib(n)/fib(n-1) is the golden ratio, or its inverse, I can never remember). Here is a demonstration which you can try on your iPhone:
(define (fib n)
(let loop ((i n) (a 1) (b 1)) (if (= i 0) a (loop (- i 1) b (+ a b)))))
(fib 20)
10946
(exact->inexact (/ (fib 1000) (fib 999)))
1.618033988749895
(exact->inexact (/ (fib 999) (fib 1000)))
.6180339887498949
(fib 1000)
70330367711422815821835254877183549770181269836358732742604905087154537118196933579742249494562611733487750449241765991088186363265450223647106012053374121273867339111198139373125598767690091902245245323403501
Marc
The neat thing is that with Scheme you are not limited to the precision of floating point numbers. For example, here is how to compute the 400 first decimals of the golden ratio:
(round (* (expt 10 400) (/ (fib 1000) (fib 999))))
16180339887498948482045868343656381177203091798057628621354486227052604628189024497072072041893911374847540880753868917521266338622235369317931800607667263544333890865959395829056383226613199282902678806752087668925017116962070322210432162695486262963136144381497587012203408058879544547492461856953648644492410443207713449470495658467885098743394422125448770664780915884607499887124007652170575179788
Marc
That is simply amazing. Thanks, Marc. Steve
________________________________ From: Marc Feeley feeley@iro.umontreal.ca To: Steve Graham jsgrahamus@yahoo.com Cc: Gambit List Gambit-list@iro.umontreal.ca Sent: Thursday, October 27, 2011 6:09 PM Subject: Re: [gambit-list] Golden Section and Gambit Scheme
On 2011-10-27, at 10:48 AM, Steve Graham wrote:
Thanks for the memory nudge, Marc. I think the author mentioned such, too. Steve
From: Marc Feeley feeley@iro.umontreal.ca To: Steve Graham jsgrahamus@yahoo.com Cc: Gambit List Gambit-list@iro.umontreal.ca Sent: Thursday, October 27, 2011 11:16 AM Subject: Re: [gambit-list] Golden Section and Gambit Scheme
On 2011-10-27, at 9:46 AM, Steve Graham wrote:
So here are the solutions and code:
Gambit v4.6.1
(/ (+ -1 (expt 5 .5)) 2)
.6180339887498949
(/ (- -1 (expt 5 .5)) 2)
-1.618033988749895
The main solution for A is .6180339887498949, and because B = 1, then C = 1.6180339887498949
Many thanks to Scott Olsen author of The Golden Secret, Marc Feeley (main author of Gambit Scheme), Keith Flower (for porting of Gambit to the Android OS), Google (for helping me to remember/find the solution to a quadratic equation and providing picture-capable Gmail), Microsoft (for making Word handle equations) and my math teachers for a stroll down memory lane and a most interesting adventure for today.
Steve
No problem!
By the way, one of the nice properties of the fibonacci sequence is its relation to the golden ratio (the limit n->oo of fib(n)/fib(n-1) is the golden ratio, or its inverse, I can never remember). Here is a demonstration which you can try on your iPhone:
(define (fib n)
(let loop ((i n) (a 1) (b 1)) (if (= i 0) a (loop (- i 1) b (+ a b)))))
(fib 20)
10946
(exact->inexact (/ (fib 1000) (fib 999)))
1.618033988749895
(exact->inexact (/ (fib 999) (fib 1000)))
.6180339887498949
(fib 1000)
70330367711422815821835254877183549770181269836358732742604905087154537118196933579742249494562611733487750449241765991088186363265450223647106012053374121273867339111198139373125598767690091902245245323403501
Marc
The neat thing is that with Scheme you are not limited to the precision of floating point numbers. For example, here is how to compute the 400 first decimals of the golden ratio:
(round (* (expt 10 400) (/ (fib 1000) (fib 999))))
16180339887498948482045868343656381177203091798057628621354486227052604628189024497072072041893911374847540880753868917521266338622235369317931800607667263544333890865959395829056383226613199282902678806752087668925017116962070322210432162695486262963136144381497587012203408058879544547492461856953648644492410443207713449470495658467885098743394422125448770664780915884607499887124007652170575179788
Marc