An interesting expression.
I saw this problem posed in Professor Tom Cleaver's column
"Brainbusters" in Louisville's Courier-Journal newspaper.
The solution is 'mine,' although I imagine it's pretty much
the same as his or anyone else's. I don't remember where I
first heard about fibonacci sequence and golden ratio.
Resolve the expression
__________________
/ _____________
/ / ________
\/1 + \/1 + \/1 + ....
Where the .... means that you keep taking the square root 'forever,'
or, infinitely.
Set the variable y equal to the big mess, as follows:
__________________
/ _____________
y = / / ________
\/1 + \/1 + \/1 + ....
Now, square both sides, giving
2 _____________
y = / ________
1 + \/1 + \/1 + ....
Subtracting 1 from both sides gives,
_____________
2 / ________
y - 1 = \/1 + \/1 + ....
We note that the big mess on the right looks a lot like
our 'original' big mess that we had set equal to y. If
we look closer, we'll note that it's exactly equal to the
original big mess because no matter how many square root
symbols we actually show, the '...' indicates that we are
taking the square root forever and so we can substitute y
into the equation giving,
That is,
__________________
/ _____________ _____________
y = / / ________ = / ________ , so...
\/1 + \/1 + \/1 + .... \/1 + \/1 + ....
2
y - 1 = y , or
2
y - y - 1 = 0
This is a quadratic equation with solutions
___________
1 +/- \/1 - 4(1)(-1)
y = -------------------- (where +/- denotes plus or minus)
2(1)
_ _
1 + \/5 1 - \/5
y1 = ------- and y2 = -------
2 2
y2 is an extraneous root, so y1 is our answer.
_
\/5 + 1
y = -------
2
--------------------------------------------------
A curious note, here, is the following.
[Prepared to be bored to tears about the relationship between
this result and the golden ratio! And then about the relationship
between the golden ratio and the Fibbonacci sequence!]
There is this thing called 'the golden ratio.' In ancient times they
believed that rectangles with this ratio were the ones that were most
pleasing to the eye. It is defined as the ratio of adjacent sides of
a rectangle such that the width divided by the length is equal to the
length divided by the sum of the length and width. Let's clarify this
verbiage with a formula. The sides of a rectangle form a golden ratio
when
w l i.e. width length
- = ----- ====> ------ = --------------
l l + w length length + width
We can multiply through by the factor l(l + w) to give:
2
w(l + w) = l
2 2
w + lw - l = 0
Now, set l = 1 and solve for w. We can do this because we're only
interested in the ratio w/l, and not in particular values of w or l.
2
w + w - 1 = 0
Hey, wow. This is vaguely similar to the equation from our other problem!
____________
-1 +/- \/1 - 4(1)(-1)
w = ---------------------
2(1)
_
-1 +/- \/5
w = ---------- (Since lengths and widths shouldn't be negative,
2 let's ignore the negative root as being extraneous.)
So we have,
_ _
-1 + \/5 \/5 - 1
w = -------- = -------
2 2
And so, it just seems kinda interesting that y = w + 1.
-------------------------------------------------------------
The Fibbonacci sequence is defined like this:
f(0) = 1
f(1) = 1
f(n) = f(n-1) + f(n-2), for n > 1.
(I know it's kinda funny to start out the first element of an 'array'
with a 'subscript' of zero, but math guys do that a lot. I'm not sure
why. There's probably some interesting bit of history associated with
it.)
So the first few elements are:
f(0) = 1
f(1) = 1
f(2) = f(1) + f(0) = 1 + 1 = 2
f(3) = f(2) + f(1) = 2 + 1 = 3
f(4) = f(3) + f(2) = 3 + 2 = 5
f(5) = f(4) + f(3) = 5 + 3 = 8 and so on.
In any case, a number of people have found this sequence to be very
interesting. No doubt these people have had too much free time on
their hands, and an insufficient number of real life problems to keep
them busy.
In any case, we're going to take the ratio of two adjacent elements
of the sequence and look at their ratio as we 'slide' on out the
sequence. That is we're going to look at the ratio of adjacent elements
in the limit.
f(n-1)
r = lim ------
n->oo f(n)
Now, it turns out that f(n) is defined in terms of
previous elements in the sequence, so we have
f(n) = f(n-1) + f(n-2).
So we can rewrite r as
f(n-1)
r = lim ---------------
n->oo f(n-1) + f(n-2)
Now we can divide both the numerator and denominator by f(n-1) to give:
f(n-1)/f(n-1) 1
r = lim ----------------------------- = lim -----------------
n->oo f(n-1)/f(n-1) + f(n-2)/f(n-1) n->oo 1 + f(n-2)/f(n-1)
However, f(n-2)/f(n-1), IF IT CONVERGES in the limit (i.e. as n
'approaches' infinity), is the same as f(n-1)/f(n)! And so we have
f(n-1) f(n-2)
lim ------ = lim ------ !!!! Wow! And so, we can now write,
n->oo f(n) n->oo f(n-1)
1 1
r = lim ---------------- = lim --------------
n->oo 1 + f(n-2)/f(n-1) n->oo 1 + f(n-1)/f(n)
1
r = -----
1 + r
2
r + r - 1 = 0,
which is the identical equation we had for w when we calculated the
golden ratio. This means that the ratio of successive elements of the
fibonacci sequence do converge in the limit, and that they converge to
the golden ratio.
Whoa. Be still my heart.
(
geocities.com/athens)