davids@utcsstat.UUCP (David Scollnik) (07/03/85)
Undoubtedly, the great majority of you are familiar with Pascal's
triangle, that is, this triangle which continues off
to infinity ...
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
note, every entry in the triangle is the sum of
the two entries directly above.
Now, consider this, which I shall refer to as Pascal's Inverse triangle...
1
1 1
1 1/2 1
1 2/3 2/3 1
1 3/5 3/4 3/5 1
this triangle also continues on for infinity.
Each entry is the INVERSE of the sum of the two entries
directly above, for example
1/2 = 1/(1 + 1)
and 2/3 = 1/(1/2 + 1)
and 3/4 = 1/(2/3 + 2/3) .
Of course, the outermost layer will always be one,
and it can be shown (for example with the
use of continued fractions) that the next
layer inward will converge to (sqrt(5)-1)/2.
This is apparent since we have, as the number of levels in
the triangle goes off to infinity, that
1
X = -----------------
1 + 1
-----------
1 + 1
---------
1 + etc.
and this is the same as
1
X = ---------------
1 + X
X * X + X - 1 = 0
and this equations one positive root is
(sqrt(5)-1)/2
Has anyone come across this triangle before, and if so
have you anything of interest to share concerning it ?skinner@saber.UUCP (Robert Skinner) (07/08/85)
> Now, consider this, which I shall refer to as Pascal's Inverse triangle... > > 1 > 1 1 > 1 1/2 1 > 1 2/3 2/3 1 > 1 3/5 3/4 3/5 1 > > this triangle also continues on for infinity. > > Each entry is the INVERSE of the sum of the two entries > directly above, for example > > 1/2 = 1/(1 + 1) > > and 2/3 = 1/(1/2 + 1) > > .... > > > 1 > X = ----------------- > 1 + 1 > ----------- > 1 + 1 > --------- > 1 + etc. > > and this is the same as > > > 1 > X = --------------- > 1 + X > > X * X + X - 1 = 0 > > and this equations one positive root is > > (sqrt(5)-1)/2 > > Has anyone come across this triangle before, and if so > have you anything of interest to share concerning it ? *** REPLACE THIS LINE WITH YOUR MESSAGE *** I've never seen the triangle you describe here, but the root you show is also the Golden Ratio. (Or the inverse: (sqrt(5)-1)/2=.618=1/1.618= 1/(sqrt(5)+1)/2). The way I first heard it stated, the ancient Greeks thought that the most pleasing ratio for a (Golden) rectangle were such that if a square were chopped off one end, it would yield another Golden rectangle. In other words, x/1=1/(x-1) ==> x=1.618. The Golden Ratio also pops up in Fibonacci numbers, the ratio of fn to fn-1 converges to 1.618, regardless of the values of f0 and f1. Yet another odd place: the five pointed star. This is hard for me to describe, but suppose you draw a five pointed star like you did in grade school. Assuming you draw a perfect star, the ratio of the length from the body to the tip of one point, to the length of the rest of the same line segment is .618=1/1.618. It's nice to know the Golden Ratio lives in other places as well. Anybody know of other places in math or geometry where the Golden Ratio occurs? ------------------------------------------------------------------------------ Name: Robert Skinner Mail: Saber Technology, 2381 Bering Drive, San Jose, California 95131 AT&T: (408) 945-0518, or 945-9600 (mesg. only) UUCP: ...{decvax,ucbvax}!decwrl!saber!skinner ...{amd,ihnp4,ittvax}!saber!skinner
wjhe@hlexa.UUCP (Bill Hery) (07/08/85)
> > > Undoubtedly, the great majority of you are familiar with Pascal's > triangle, that is, this triangle which continues off > to infinity ... > >...... > Now, consider this, which I shall refer to as Pascal's Inverse triangle... > > > 1 > 1 1 > 1 1/2 1 > 1 2/3 2/3 1 > 1 3/5 3/4 3/5 1 > > > Each entry is the INVERSE of the sum of the two entries > directly above, for example >........ > Of course, the outermost layer will always be one, > and it can be shown (for example with the > use of continued fractions) that the next > layer inward will converge to (sqrt(5)-1)/2. >......... > Has anyone come across this triangle before, and if so > have you anything of interest to share concerning it ? The terms in the 'next layer' (if the fractions are not reduced ) can be shown to be the inverse ratio of the successive terms of the Fibonacci series (a(0)=1, a(1)=1, a(n)=a(n-1)+a(n-2) for n>1); the limit of this ratio is well known (and easily shown to be) (sqrt(5)-1)/2. The Fibonacci series (actually a sequence in modern mathematical parlance) comes up in many places in nature (e. g., the number of petals in succesive layers in a flower), and has been studied by mathematicians for many hundreds of years. There is a society (called the Fibonacci Society) which is devoted to it's study, and publishes a scholarly journal (the Fibonacci Quarterly) which publishes articles on the theory and applications of Fibonacci Series. I suggest consulting that journal for possible leads. Another source would be texts on finite difference equations, of which the general form for an element of your triangle is an example. This area has been somewhat dormant for a while, but Dave Jaegerman of ATT Bell Labs (Holmdel NJ) is about to release a new book on the topic. Bill Hery ATT Bell Labs
perlman@wanginst.UUCP (Gary Perlman) (07/10/85)
For more information about the Golden Ratio, Pascal's Triangle, Pentagrams, and Fibonacci, see: H E Huntley (1970) The Devine Proportion (A Study in Mathematical Beauty) Dover Publications 180 Varick Street New York, NY 10014 -- Gary Perlman Wang Institute Tyngsboro, MA 01879 (617) 649-9731 UUCP: decvax!wanginst!perlman CSNET: perlman@wanginst
leeper@mtgzz.UUCP (m.r.leeper) (07/11/85)
Skinner points out that the Golden Ratio pops up in Fibonocci Numbers. That is WHY is is involved with the inverse triangle. It is just the ratio of the nth Fibonocci th the (n+1)st that shows up in that column of the triangle. Mark Leeper ...ihnp4!mtgzz!leeper
franka@mmintl.UUCP (Frank Adams) (07/12/85)
The center terms of the inverse triangle approach 1/sqrt(2). This can be seen by noting that, assuming that a limit x is approached, x = 1/(x + x). Algebra gives us the result stated. This isn't a formal proof, but it can be extended to one. Generally, each column converges to a distinct limit. If a column converges to a, the next column will converge to b = 1/(a + b), which when solved gives b = (-a + sqrt(a^2 + 4)) / 2. The first case is a = 1 and b = (-1 + sqrt(5))/2; generally we get a nested sequence of square roots. These results are quite straightforward. Off hand, I see no approach to getting any deeper results. This seems to be any interesting problem.