Golden powers

So, we know from a previous challenge that \varphi^2 = \varphi + 1. That’s a pretty interesting property, which is shared only by its cousin, \hat{\varphi}. I wonder whether other powers of \varphi have special properties too? Let’s see:

\varphi^3 = \varphi \cdot \varphi^2 = \varphi (\varphi + 1) = \varphi^2 + \varphi = 2\varphi + 1.

Interesting! What about \varphi^4?

\varphi^4 = \varphi \cdot \varphi^3 = \varphi (2\varphi + 1) = 2 \varphi^2 + \varphi = 3\varphi + 2.

And \varphi^5?

\varphi^5 = \varphi \cdot \varphi^4 = \varphi (3\varphi + 2) = 3 \varphi^2 + 2\varphi = 5\varphi + 3.

Curiouser and curiouser! Are you noticing a pattern? (Try working out \varphi^6 to see if it fits the pattern you suspect.)

Hopefully, you noticed that Fibonacci numbers are involved here. If this pattern holds, we would expect to find that \varphi^6 = 8\varphi + 5, \varphi^7 = 13\varphi + 8, and so on. To state it more generally, we suspect that

\varphi^n = F_n \varphi + F_{n-1}, when n \geq 2.

Is this true? It turns out, amazingly, that it is, and we can prove it using a technique called mathematical induction. The idea of mathematical induction is sort of like knocking over a chain of dominoes. First, we’ll show that the above statement is true for some particular value of n (the base case; for this statement, we’ll use n = 2). Then, we’ll show that if the statement is true for any given value of n, then it must be true for n + 1 as well (the inductive step). Do you see how it’s like knocking over dominoes? If the statement is true when n is 2, then it must be true for 3 as well, which means it must be true for 4, which means… and so on, for all positive integers!

Ready? First, recall that the Fibonacci numbers F_n are defined as follows:

\begin{array}{rcl} F_1 = F_2 & = & 1 \\ F_n &= & F_{n-1} + F_{n-2} \qquad (n > 2). \end{array}

Now for the base case: when n is 2, the statement says that \varphi^2 = F_2 \varphi + F_1; this is true since F_2 and F_1 are both equal to 1.

Finally, the inductive step. We’ll begin by supposing that \varphi^n = F_n \varphi + F_{n-1} for some n \geq 2, and see if this allows us to conclude that the statement is true for n + 1 as well. The method we used above to evaluate \varphi^3, \varphi^4 and so on suggests a method of attack:

\begin{array}{rcl} \varphi^{n+1} &=& \varphi \cdot \varphi^n \\ &=& \varphi \cdot (F_n \varphi + F_{n-1}) \\ &=& F_n \varphi^2 + F_{n-1} \varphi \\ &=& F_n (\varphi + 1) + F_{n-1} \varphi \\ &=& (F_n + F_{n-1}) \varphi + F_n \\ &=& F_{n+1} \varphi + F_n \end{array}

So, since we’ve proved that the statement is true when n is 2, and we’ve also proved that it must be true for n + 1 whenever it’s true for n, it must be true for all n \geq 2!

Next up: we’ll use this neat relationship between \varphi and the Fibonacci numbers to discover another one!

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6 Responses to Golden powers

  1. Jonathan says:

    Nice one! I didn’t know this.

  2. Pingback: Carnival of Math #18 « JD2718

  3. polymath says:

    Nice proof. The first equality you state (used for the cousin, not phi itself) is used prominently in the proof on my blog at http://polymathematics.typepad.com/polymath/trekking-into-the-desert.html

    It’s a great example of how phi shows up in the oddest places.

  4. George Bell says:

    The formula also works for negative powers (n) if you define F_{-n}=(-1)^{n+1}F_n. Rather amazing!

    I have used this formula on several recent papers on peg solitaire, for example

    http://arxiv.org/abs/math.CO/0612612

  5. Brent says:

    Indeed!

  6. Pingback: Explicit Fibonacci numbers | The Math Less Traveled

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