Win Corduan
PAGE 3: PHI IN EQUATIONS
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Allow me to revisit what we’re doing here. We’re examining the nature of phi (1.61803…), the golden ratio constant, and for the moment we’re doing so as far away as we can get from the Fibonacci series and its exemplifications in art or nature. My point is to try to demonstrate its inherent elegance, which stems from its Creator just as much as the arrangement of flower petals and other “fibonaccied” items in nature.
Let’s go back to the question of two days ago: How contrived is the golden ratio? For example, take the golden triangle, which includes phi as an important ratio. Is it something that has been concocted just to show off phi, or is the golden triangle something that we can uncover and discover in other places. A partial response has been that it is included in the formation of a regular pentagon, and, thus, can be discovered without being guilty of fabricating an artificial instantiation just to smuggle in phi.
Phi is also found apart from geometry in number theory, which we can combine with a little algebra. Remember that phi is just one member of the infinity of all real numbers, which, as demonstrated by Georg Cantor, has turned out to be larger than some other infinities. However, it stands out from this uncountable crowd, right next to pi, e, i, √2, and a few others, due to its special properties or notoriety.
Phi and Some Nested Square Roots
Here are two ways of finding phi without geometry. We will take a look at a couple of somewhat unusual-looking formulas and turn them into equations, which will resolve into phi. Here is the key: We came up with the formula for phi by setting the length of the base of a golden triangle as 1, and one of its sides as x. We were able to manipulate those numbers into a quadratic equation
x2 -x -1 = 0
The method for finding phi tonight is going to consist of finding several equations that can also be rearranged into the same quadratic equations. I found the following two equations in Livio
Let’s start with a formula that consists of an endless fusion of square roots. Given:
By the way, this thing is a formula, not an equation. You can only have an equation if there are two (or more things) that are considered to be equal, as, e.g., in
It would be extremely tedious to work out a value for x from this equation if we were to calculate the square roots of square roots, etc. But there is an easier way to do so.
We can square both sides. Squaring x gives us x2, and squaring the formula to the right resolves the first square root into a 1.
Now it is apparent that the collection of square roots that follows after “1 +” is still identical to the original given formula since both extend to infinity,
and, what’s more, we have already designated it as x above.
Then, substituting x for that nest of square roots, we get
which we can reformulate to fit the pattern we had looked for as
This is, of course, the formula which has phi as one of its solutions. Just think: we have derived it from that unwieldy collection of square roots.
Phi and a Continuing Fraction
So, now you feel like the Lone Ranger and want to save the world from more confusing formulas. That’s great. I’m with you all the way. Let’s try this monstrosity, which in mathematical argot is counted among a large group of “continuing fractions.” Continuing fractions are characteristic of irrational numbers. We start with a 1 and add the fraction, which has 1 for its first numerator and is followed by a never-ending, always-repeating "1 + denominator/numerator of 1." Given:
Again, before we do anything else we need to turn this formula into an equation and label it with the variable x.
As you seek something wonderful in this equation, may I call your attention to the section underneath the topmost numerator of 1.
As in the previous equation, what you find there is actually identical to the originally given formula. There is no difference because both of them can be extended to infinity. And thus, we can safely apply the same letter variable, x to this new continuing fraction, which is also the old one. Remember now, that the area we have marked out is the denominator, and that the numerator of 1 still stands as before.
So, once we have substituted x for all of that clumsy denominator, we get a refreshingly simple equation.
Let’s get rid of the fraction 1/x by multiplying every term by x,
Then we have
and once more we can rearrange this equation into our favorite configuration:
Once again, we have turned a beast (the continuing fraction) into a beauty (phi). We have stumbled on yet another way of deriving phi without getting out our rulers or measuring tapes, let alone by bringing in the Fibonacci series.
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Just for Effect: A Quick Sidelight on PI
Again, because in this series I’m trying to stress the beauty of the numbers themselves, let me give you another example that strikes me as really astounding, and let ϕ take a breather. We all know about ϕ’s big brother pi (π) mostly for his part in the formula for the area of a circle:
A =πr2
But π also shows up in places where we would not expect to see him. If by this point you don't want to look at any more equations, please continue on with the next page.
Let us look at an interesting problem, known as the "Basel Problem" (Derbyshire 63) because it was in Basel, Switzerland, that it was posed. The problem, as originally stated by Jakob Bernoulli (1655-1705), went as follows.
Here is a convergent series, which means that it eventually approaches the limit of a specific number rather than diverging to infinity:6
As usual, we can come closer and closer to the number, but will never quite reach it. A precise enough approximation of the number is
1.64493406693 …
Are you getting tired of those three little dots ( … ), which indicate that you’ll never truly get to the end of calculating the formula’s value? Apparently Bernoulli was. In a publication on various related topics, he included this series and asked for input from other mathematicians about this question. Would anyone be able to state it in a form that did not lose itself in the forever of infinity, but in a “closed” formula that could be substituted for it?
Leonhard Euler (1707-1783), whose original home also happened to have been Basel, came up with a solution:
I don't think I'd be going out on a limb when I say that most of us are rather surprised to see pi popping up here.
Very quickly let's switch to a different topic, this time probability theory. Suppose you are asked to pick two numbers at random from a set of integers. What is the probability that they will not have any common divisors? E.g., 6 and 8 would share a divisor (2), but 4 and 9 would not. The probability of picking two numbers without a common divisor is:
This probability calculation turns out to be the reciprocal of Euler's solution to the Basel problem! Do you get goose bumps when you read about these things? I do, and when I really began to see these things, about fifteen years ago, my real infatuation with math took off. (We'll save that story for some other time.) The University of Illinois has set up an interactive webpage that illustrates this formula with some input from you.
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Note (6) Congratulations! You are looking at a special case of the famous Riemann zeta function.The Riemann Hypothesis actually is not really relevant to the point of this site. However, there are so many weird (and false) ideas circulating about it, often abetted by misrepresentations in films and television programs, that I thought I would take this opportunity to present a short summary of it. It has no mystical energy; it is not the key to unlock any spiritual power; and it does not represent the means of taking over all of the computers in the world.  The Basel series is one instantiation of this function, namely when s = 2. Its compact form can be transcribed as  You can read this collection of symbols as: designated as k, ranging from 1 to infinity (∞), raised to the power of s." Of course, neither Bernoulli nor Euler had any knowledge of the Riemann Hypothesis since Riemann did not come up with it until 1859. However, Euler did a lot of work that illuminates the issue and provided tools for its eventual proof—should that event ever come about. The Hypothesis only starts to raise its head when the function is applied in the complex plane ℂ, where numbers are expressed as a real number plus an imaginary one : (real number + (real coefficient x i) where i stands for √-1. Then the hypothesis stipulates that, when the function intersects with a "significant" 0, the real number a equals 1/2. To put it into its common form The non-trivial zeros of the zeta function have real part 1/2. If you're interested in understanding this problem further, I highly recommend John Derbyshire, Prime Obsession . Back to the text |
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