SLIDE 1
THE GOLDEN RATIO AND THE FIBONACCI NUMBERS
Common Measures
3 feet 2 feet 1 foot 3 2 Ratio = Common measure = 1 foot 2 1 Ratio =
THE GOLDEN RATIO AND THE FIBONACCI NUMBERS Common Measures 1 foot - - PowerPoint PPT Presentation
THE GOLDEN RATIO AND THE FIBONACCI NUMBERS Common Measures 1 foot - - PowerPoint PPT Presentation
THE GOLDEN RATIO AND THE FIBONACCI NUMBERS Common Measures 1 foot 2 feet 3 feet 3 2 Ratio = Ratio = 2 1 Common measure = 1 foot 25 inches 11 inches 36 inches 36 25 Ratio = Ratio = 25 11 Common measure = 1 inch a b a + b
SLIDE 2
SLIDE 3
36 inches 25 inches 11 inches 36 25 Ratio = Common measure = 1 inch 25 11 Ratio =
SLIDE 4
a b
Incommensurable!
π + π π = π π a + b (no fraction of a foot can be used to measure this distance)
SLIDE 5
οͺ
1.618033988749894β¦
SLIDE 6
The origins of οͺ are shrouded in the mists of time
SLIDE 7
The Golden Ratio: οͺ
- In modern times is denoted by the symbol phi: οͺ
- Known to Euclid (300 B.C.) as a result of solving:
- A number of painters and architects have used
the golden ratio in their work
- The length of a diagonal of a regular pentagram,
whose sides have unit length, is οͺ
- Occurs in nature β represents a growth pattern
π¦2 β π¦ β 1 = 0
SLIDE 8
Legend and Speculation
- Was known to the ancient Egyptians.
- Was used to form the dimensions of the
Great Pyramids of Egypt.
- Was applied to the design of the Parthenon.
- Was used in the design of Notre Dame in
Paris.
- Was used in the construction of the Taj
Mahal.
SLIDE 9
The Parthenon
SLIDE 10
Ratio of a Rectangle
a b
π π Ratio =
SLIDE 11
Another Ratio
a a b
Ratio = π + π /π These two rectangles have a divine proportion if: π + π π = π π
SLIDE 12
The Algebra
π + π π = π π π π + π = π2 π2 β ππ β π2 = 0 Letting π = 1 gives us: π2 β π β 1 = 0 Whose only positive solution is οͺ
SLIDE 13
Golden Ratio
- Numerically the golden ratio is:
- This comes from solving π¦2 β π¦ β 1 = 0 using
the quadratic formula:
- All rectangle pairs that are in divine proportion
to each other will have this ratio. οͺ =
1+β5 2
= 1.61803 β¦
βπ Β± π2 β 4ππ 2π
SLIDE 14
a
Ratio = π + π /π If these two rectangles have a divine proportion then:
π+π π = π π = οͺ
a b
SLIDE 15
Fibonacci Numbers
SLIDE 16
The Original Problem
- Stated by Fibonacci (whose original name
was Leonardo of Pisa) in the year 1202
- Gives a recursive rule for computing the total
number of rabbit pairs under βidealβ reproductive circumstances.
SLIDE 17
Problem Statement
- Start with a rabbit pool containing one pair of newly born
rabbits (one male and one female)
- A newly born rabbit takes one month to reach reproductive
maturity
- The gestation period of a reproductively mature female
rabbit is one month
- A female rabbit will always give birth to two rabbits β one
male and one female
οΊ This newly born pair is added to the rabbit pool
- Question: How big is the rabbit pool after
οΊ 12 months? οΊ n months?
SLIDE 18
Young Pair Mature Pair Mature Pair Mature Pair Young Pair Mature Pair Mature Pair Mature Pair Month 1 Young Pair Mature Pair Mature Pair Mature Pair Young Pair Mature Pair Mature Pair Mature Pair Month 2 Month 3 Month 4
SLIDE 19
Fibonacci Sequence
- Starting from 1
- Starting from 0
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, β¦ 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, β¦
SLIDE 20
The pedal count of many flowers are Fibonacci numbers (this is a known growing pattern)
In Flowers
SLIDE 21
,2,3,5,8,13,21,34,55,89,β¦
white calla lily
1
SLIDE 22
1, ,3,5,8,13,21,34,55,89,β¦ 2
Euphorbia
SLIDE 23
1,2, ,5,8,13,21,34,55,89,β¦ 3
Trillium
SLIDE 24
1,2,3, ,8,13,21,34,55,89,β¦
Buttercup
5
SLIDE 25
1,2,3,5, ,13,21,34,55,89,β¦
Bloodroot
8
SLIDE 26
1,2,3,5,8, ,21,34,55,89,β¦ 13
Black eyed Susan
SLIDE 27
1,2,3,5,8,13, ,34,55,89,β¦
Shasta Daisy
21
SLIDE 28
1,2,3,5,8,13,21, ,55,89,β¦ 34
Field Daisies
SLIDE 29
1,2,3,5,8,13,21,34, , ,β¦ 55 89
Michelmas Daisies
SLIDE 30
lim
πββ
πππ π + 1 πππ π = οͺ
SLIDE 31
fib(n+1) fib(n)
Approaching the Golden Ratio
SLIDE 32
1, 1, 2, 3, 5, 8, 13, 21, 34, β¦
οͺ
1 1 2 1 3 2 5 3 8 5 13 8 ο