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The golden ratio

  1. Jun 11, 2008 #1

    madmike159

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    Can someone please explain the golden ratio to me. I looked on wikipedia (http://en.wikipedia.org/wiki/Golden_ratio) for an explaination but I couldn't make sense of it. How does (a + b) / a = a/b. What does a+b is to segment a, as a is to the shorter segment b.
     
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  3. Jun 11, 2008 #2

    symbolipoint

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    madmike159, you might become more comfortable if you focus on how the ratio value relates to the fibonnaci sequence; and then examine the other parts of the article in Wikipedia. Also study the relevant sections from a high school Geometry book.
     
  4. Jun 11, 2008 #3

    matt grime

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    There's nothing really to explain.

    The golden ratio a/b is defined by the rule that a/b = (a+b)/a. If we set b=1, then a=(a+1)/a, and we see that it is the larger of the two roots of x^2-x-1.
     
  5. Jun 11, 2008 #4
    Sure, the golden ratio is such that (a+b)/a = a/b. Now let's work this equation. First let's multiply both sides by a to get (a+b)=a^2/b now let's multiply both sides by b to get b(a+b) = ab + b^2 = a^2 now we rearrange to get b^2 + ab - a^2 = 0 or a^2 - ab - b^2 and this is simply a quadratic formula so we want to find out for what values of a this is valid, thus using the quadratic formula we get a = (b +/- sqrt(b^2+4b^2))/2 which equals b*(1 +/- sqrt(5))/2 but since 1 - sqrt(5) is negative (and there's no such thing as a negative length) the only solution of interest is a = b*(1 + sqrt(5))/2 and by a quick rearrange we get a/b = (1+sqrt(5))/2 the pseudo-mystical golden ratio
     
  6. Jun 11, 2008 #5
    Remember that a and b both represents segments. You can think of a+b as the length of a plus the length of b. If you picture a and b connected at an endpoint than a+b becomes one large line-segment that can be split into two line-segments a and b. Then a and b are in a sense sub line-segments.

    Continuing with that idea, then you can re-write this: " a+b is to segment a, as a is to the shorter segment b."
    As
    The total lineament is to the longer sub line-segment, as the longer sub line-segment is to the the shorter sub line-segment.

    I hope that made sense.
     
  7. Jun 11, 2008 #6

    madmike159

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    Thanks I understand now. I was mostly confused how they got from (a+b)/a to a/b.
     
  8. Jun 11, 2008 #7
    I used to think he golden ratio was a lame constant whose entire importance is related to the fact that it shows up in some DEs. However, it actually is a pretty interesting number, and heres why - the golden ratio, phi, can be defined as the number whose reciprocal is equal to 1 minus itself.

    phi-1 = 1/phi.

    Now, thats all fine and good, but something interesting happens when you take the continued fraction expansion:
    phi = 1+1/phi = 1+1/(1+1/phi) = 1+1/(1+1/(1+1/phi)) = 1+1/(1+1/(1+1/(1+1/(...))))

    It is the only number whose continued fraction expression is 1 1 1 1 1 1 1 1....

    This can be used to show that phi's rational approximation converges as slowly as possible, meaning that in some sense that phi is "the most irrational number". Now that is interesting!
    http://www.ams.org/featurecolumn/archive/irrational1.html
     
  9. Jun 12, 2008 #8
    The golden ratio can be found by solving the following problem :

    Find two numbers that have a difference of 1, and when multiplied together equal 1.

    The numbers are 1.61803.... and .61803.... ; the golden ratio being former.

    Another way to find the golden ratio is to use the Fibinachi sequance, where every term is the sum of the previous two terms, starting with 1,1

    1,1,2,3,5,8,13,21,34,55.......

    The golden ration can be approximated by the ratio of any two succesive numbers in this sequance. The approximation gets better the larger numbers you use.

    ie. 55/34 = 1.617....
     
  10. Jun 12, 2008 #9

    HallsofIvy

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    The "golden rectangle" (it is said that the side view of the Parthenon in Athens is a golden rectangle and that DaVinci's painting "The Last Supper" is in the proportions of a golden rectangle. More generally it is claimed that the golden rectangle has the "most elegant" proportion of any rectangle. That is, of course, not a mathematical claim.) is a rectangle, with width w and height h, such that, if you mark off distance h from one corner of the width and draw a perpendicular to make a new rectangle; that is, construct a new rectangle having width w-h and height w, the ratio of "height to width" is still the same: you have constructed a new golden rectangle.

    The "height to width" ratio of the first rectangle is h/w, the "height to width" ratio of the second is w/(w-h). If those are the same h/w= w/(w-h). Taking a= w-h and b= h, then a+ b= w-h+ h= w so h/w= (a+b)/b and w/(x-h)= b/a: the proportion h/w= w/(w-h) is (a+b)/b= b/a.

    From h/w= w/(w-h) we can multiply both sides by w(w-h) and get h(w-h)= w2 or hw- h2= w2. Dividing both sides by h2, (w/h)- 1= (w/h)2. (w/h)2- (w/h)+ 1= 0. Using the quadratic formula to solve that equation gives phi as the positive solution for the ratio w/h.
     
  11. Jun 12, 2008 #10

    madmike159

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    Wow that is confusing. I'm going to have another look in the moarning =D. Thanks for the help every one. I think I know enough for what I need to do, but any more places where the GR appears would be nice to know (wiki article said something about the golden ratio in trees and humans =S).
     
  12. Jun 13, 2008 #11
    It's actually very simple. It's a rectangle which you try to make square by cutting out the excess material. Try to picture:
    • a square of dimensions a x a;
    • a rectangle which is (a+b) x a (so that the square fits in);
    • the remainder of the rectangle, after you cut out the square, which is a smaller rectangle of dimensions a x b.
    If the big rectangle is proportional to the smaller one, the ratio of their sides, (a+b) / a = a / b gives you the golden ratio.

    Try this:
    http://golden-rectangle.lcpdesign.com/construct.htm
     
    Last edited: Jun 13, 2008
  13. Jun 13, 2008 #12

    madmike159

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    Yea it is really easy. Best not to do maths late at night though. Thanks for the help everyone.
     
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