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{SECT 0 {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 257 52 "High School Modul
es >  Geometry by Gregory A. Moore\n" }}{PARA 3 "" 0 "" {TEXT -1 4 "  \+
  " }{TEXT 256 16 "The Golden Ratio" }}{PARA 0 "" 0 "" {TEXT -1 112 "
\nAn exploration of the golden ratio. It is derived in several differe
nt ways, and applied to common rectangles.\n" }}{PARA 0 "" 0 "" {TEXT 
258 153 "[Directions : Execute the Code Resource section first. Althou
gh there will be no output immediately, these definitions are used lat
er in this worksheet.]" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 
{PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 260 7 "0. Code" }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 23 "restart; \nwith(plots): " }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 162 "cpk  := 'COLOR(RGB, .9, .4, .42)':
\ncsl  := 'COLOR(RGB, .5, .5, .85)':\ncsl2  := 'COLOR(RGB, .3, .3, .5)
':\n\ncbr  := 'color = COLOR(RGB, .6, .6, .4), filled = true':" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "scanft := 'scaling = constra
ined, axes = none, filled = true':" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 
262 35 "1. Solution of a Quadratic Equation" }}{PARA 0 "" 0 "" {TEXT 
-1 114 "\nWe begin our journey with this inauspicious quadratic equati
on, which has coefficients of 1 or -1, and solve it.\n" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "x^2 - x - 1 = 0;\nsolve(%,x);\n" }}
}{PARA 0 "" 0 "" {TEXT -1 231 "\nThere are two solutions, but the posi
tive root is called the golden ratio - for reasons which are not yet a
pparent. It is an irrational number prized by the ancient Greeks, and \+
supposedly used in the architecture of the Parthanon." }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 36 "g := (1 + sqrt(5) )/2;\nevalf(%, 40);" }}
}{PARA 0 "" 0 "" {TEXT -1 154 "\nAs we will see, this special number c
omes up in many places - although there are even more than we can go i
nto here. This number is sometimes denoted by " }{XPPEDIT 18 0 "phi" "
6#%$phiG" }{TEXT -1 109 ", after the Greek sculptor Phidias, however, \+
we will use g, for \"golden number\", here just for convenience.\n " }
}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 263 30 "2. Infinite Si
milar Rectangles" }}{PARA 0 "" 0 "" {TEXT -1 88 "\nImagine that we hav
e an ordinary looking rectangle of dimensions base x, and height 1.\n
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 168 "g := evalf( (1 + sqrt(5
) )/2):\nr1 := [[0,0],[g,0],[g,1],[0,1],[0,0]]:\n\ndisplay( plot( r1, \+
color = cpk, scanft ),\ntextplot( [g/2, -.1, `x`]), textplot( [-.1 , 1
/2, 1]));" }}}{PARA 0 "" 0 "" {TEXT -1 171 "\nThe side of height 1, is
 clearly smaller than the longer width of side x, so we could measure \+
off 1 unit on the base, and create a unit square, inside of this recta
ngle.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 273 "s1 := [[0,0],[1,
0],[1,1],[0,1],[0,0]]:\ndisplay( [ plot( s1, color = white),\n        \+
   plot( [s1,r1], color = [csl2,cpk], scanft),\n           textplot( [
g/2, -.1, `x`]), textplot( [-.1 , 1/2, 1]), \n           textplot( [1/
2 ,+.1, 1]) ], textplot( [(1+g)/2 ,+.05, `x-1`]) );" }}}{PARA 0 "" 0 "
" {TEXT -1 645 "\nThis procedure would work for any rectangle with bas
e wider than height. However, suppose we make an additional condition.
 The red rectangle you see on the right is similar to the original rec
tangle - much like similar trianagles. That is, the ratio of its width
 to height is equal to the ratio of the width to height of the origina
l rectangle. This is not true in general. However, it's true for just \+
the right number x. \n\nLet's find x - by writing this equation of rat
ios and solving. Look carefully at the diagram above, and find the wid
th and height of two rectangles.\n\n                                  \+
                                    " }{TEXT 267 13 "Smaller Side " }
{TEXT -1 10 "          " }{TEXT 268 11 "Longer Side" }{TEXT -1 193 "\n
    \n                        Original Rectangle                 1    \+
                           x\n\n                        Smaller Rectan
gle                  x  - 1                         1\n\n" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "1/x = (x-1)/1;" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 20 "lhs(%)*x = rhs(%)*x;" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 40 "sort(simplify(  rhs(%) - lhs(%)),x) = 0;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "solve(%);\nevalf(%, 15);" }}
}{PARA 0 "" 0 "" {TEXT -1 563 "\nWell there it is. The positive soluti
on is the golden ratio. (Obviously, the negative solution must be thro
wn out, because we can make negative measurement of sides.) The golden
 ratio, or golden number, is the length of the side which makes this p
roportionality true - and it's the only number that does.\n\n\nWe can \+
carry this procedure a little further - in fact, quite a bit further. \+
Since the smaller rectangle is a smaller copy of the original, we can \+
do the same thing to it - square the shorter side, and cut off that sq
uare to leave a yet smaller triangle.\n" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 430 "g2 := g - 1;\ns2 := [[1,g2],[g,g2],[g,1],[g,g2],[1,g
2]]:\ndisplay([plot( \{s1,s2\}, color = white),\n         plot( [s2,s1
,r1], color = [csl2,csl,cpk], scanft ),\n         textplot( [g/2, -.1,
 '`g`']), textplot( [-.1 , 1/2, 1]), \n         textplot( [1/2 ,+.1, 1
]),  \n         textplot( [(1+g)/2 ,+.05, `g-1`]),textplot( [g -.05, g
2/2, `g-1`]),\n         textplot( [1+.08,(1+g2)/2 ,`2-g`]),textplot([(
1+g)/2 ,g2 +.05, `g-1`]) \n        ]);" }}}{PARA 0 "" 0 "" {TEXT -1 
205 "\nNow this third rectangle is similar to the second one, and thus
 to the original one too! Lets now square its base. We can continue th
is process ... infinitely! Each rectangle is similar to the original. \+
\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 557 "g3 := 1 - g2;\ns3 := \+
[[1,g2],[1 + g3,g2],[1 + g3,1],[1,1],[1,g2]]:\n\ndisplay([ plot( \{s1,
s2,s3\}, color = white),\n          plot( [s3,s2,s1,r1], color = [csl,
csl,csl,cpk], scanft ),\n           textplot( [g/2, -.1, '`g`']), text
plot( [-.1 , 1/2, 1]), \n           textplot( [1/2 ,+.1, 1]),  \n     \+
      textplot( [(1+g)/2 ,+.05, `g-1`]),textplot( [g -.05, g2/2, `g-1`
]),\n           textplot( [1+.08,(1+g2)/2 ,`2-g`]),textplot([(1+g)/2 ,
1 +.05, `g-1`]),\n           textplot( [1+g3+.08,(1+g2)/2 ,`2-g`]),\n \+
          textplot( [1+g3/2 ,g2 +.05, `2-g`]) \n        ]);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 197 "g4 := g2 - g3;\ns4 := [[1+g
3,g2],[g,g2],[g,g2+g4],[1+g3,g2+g4],[1+g3,g2]]:\ndisplay( [plot( \{s1,
s2,s3,s4\}, color = white),\n          plot( [s4,s3,s2,s1,r1], color =
 [csl,csl,csl,csl,cpk], scanft )]);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 215 "g5 := g3 - g4;\ns5 := [[1+g3,g2+g4],[1+g3+g5,g2+g4],
[1+g3+g5,1],[1+g3,1],[1+g3,g2+g4]]:\ndisplay( [plot( \{s1,s2,s3,s4,s5
\}, color = white),\n          plot( [s5,s4,s3,s2,s1,r1],color=[csl,cs
l,csl,csl,csl,cpk],scanft)]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 253 "g6 := g4 - g5;\ns6 := [[1+g3+g5,g2+g4],[g,g2+g4],[g,g2+g4+g6],[
1+g3+g5,g2+g4+g6],[1+g3+g5,g2+g4]]:\ndisplay( [plot( \{s1,s2,s3,s4,s5,
s6\}, color = white),\n          plot( [s6,s5,s4,s3,s2,s1,r1],color=[c
sl,csl,csl,csl,csl,csl,cpk],\n                scanft)]);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 272 "g7 := g5 - g6;\ns7 := [[1+g3+g5,1-
g7],[1+g3+g5+g7,1-g7],[1+g3+g5+g7,1],\n       [1+g3+g5,1],[1+g3+g5,1-g
7]]:\ndisplay( [plot( \{s1,s2,s3,s4,s5,s6,s7\}, color = white),\n     \+
     plot( [s7,s6,s5,s4,s3,s2,s1,r1],color=[csl,csl,csl,csl,csl,csl,cs
l,cpk],\n                scanft)]);" }}}{PARA 0 "" 0 "" {TEXT -1 287 "
 As you can see, there is no end to this process. It works because of \+
the similarity of the first two rectangles is continued. \n\nAlso, not
ice that the entire original rectangle is gradually being covered with
 a series of squares. The lengths of these the sides of these squares \+
are ... \n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "g||1:= 1: for \+
k from 1 to 7 do g||k; od;" }}}{PARA 0 "" 0 "" {TEXT -1 249 "\nIf we a
dd up the areas of these rectangles, which are the sums of the squares
 of these numbers, we get a number very close to the golden number, g.
 If we continue infinitely, we will get exactly it, since the area of \+
the original triangle is g too." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 58 "'g1^2 + g2^2 + g3^2 + g4^2 + g5^2 + g6^2 + g7^2': '%' = %;" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "g;" }}}}{SECT 0 {PARA 4 "" 
0 "" {TEXT -1 1 " " }{TEXT 264 32 "3. The Most Beautiful Proportion" }
}{PARA 0 "" 0 "" {TEXT -1 684 "\nIt is said that the golden ratio is t
he most pleasing rectangular ratio to the eye. In various kinds of art
 and architecture, it is supposedly the most natural look. It was call
ed the \"divine proportion\" by Leonardo Da Vinci, and some say that t
he face of the Mona Lisa fits into a golden rectangle. Also, it report
edly, is the ratio of the height to the width of the Greek Parthenon, \+
a famous landmark of Athens built in 447 b.c., and  called \"the world
's most perfect poem in stone\" by one French author.\n\nLets compare \+
it to some common rectangles we see in every day life. Here are some c
ommon rectangles, followed by their ratio, and the percent comparison \+
to the golden number.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "g
 := evalf( (1 + sqrt(5))/2, 15);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 48 "`5 x 8 photograph`;  evalf(8/5,8); evalf(%/g,3);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "`8 x 10 photograph`;  evalf(
10/8,8); evalf(%/g,3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "`
5 x 7 photograph`;  evalf(7/5,8); evalf(%/g,3);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 49 "`11 x 17 artwork`;  evalf(17/11,8); evalf(%/g,
3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "`640 x 480 web page`
;  evalf(640/480,8); evalf(%/g,3);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 64 "`1024 x 768 Computer monitor`;  evalf(1024/768,8); ev
alf(%/g,3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "`8.5 x 11 pi
ece of paper`;  evalf(11/8.5,8); evalf(%/g,3);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 61 "`2.25 x 1.5 business card`;  evalf(2.25/1.5,8)
; evalf(%/g,3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "`94 x 50
 Official Basketball Court`;  evalf(94/50,8); evalf(%/g,3);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "`64 x 100 Smaller Soccer/Foo
tball Field `;  evalf(100/64,8); evalf(%/g,3);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 73 "`75 x 110 Larger Soccer/Football Field `;  eva
lf(100/64,8); evalf(%/g,3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
74 "`64 x 105 Smaller Soccer/Football Field `;  evalf(105/64,8); evalf
(%/g,3);" }}}{PARA 0 "" 0 "" {TEXT -1 244 "\n\nAnother application of \+
the golden ratio is the Egyptian pyramids. The Great pyramid at Giza r
ises at an angle of 51 degrees, 54 minutes. What is the measure of a g
olden right triangle (formed by drawing the diagonal of the golden rec
tangle).\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "evalf( arctan(
g)*180/Pi);\nfloor(%), `deg`, round((%-floor(%))*60), `min`;" }}}
{PARA 0 "" 0 "" {TEXT -1 169 "Well this is not particularly close. How
ever, if you create a triangle where the hypotenuse is g, and the base
 is 1, then you get the angle...which is very close to this" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "evalf( arcsec(g)*180/Pi);\nfloor(%)
, `deg`, round((%-floor(%))*60), `min`;" }}}{PARA 0 "" 0 "" {TEXT -1 
129 "\nThis means that the length of the face of the pyramid is about \+
g times the base. Did the ancient Egyptians do this on purpose?\n " }}
}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 265 27 "4. Fibonacci Nu
mbers & Gold" }}{PARA 0 "" 0 "" {TEXT -1 333 "\nIf we turn our attenti
on away from geometry and toward a simple recursive sequence of number
s, the Fibonnaci numbers, we'll find a totally unexpected connection. \+
The Fibonnaci numbers are defined by two seeds, the first two elements
 are 0 and 1. Then you get the next, and subsequent members of the lis
t by adding the previous two.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 21 "F||0 := 0; F||1 := 1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 54 "for k from 2 to 20 do F||k := F||(k-1) + F||(k-2); od;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "plot( F||(floor(x)), x = 0..
12.9, thickness = 2, title=`Fibonnacci Plot`, cbr);" }}}{PARA 0 "" 0 "
" {TEXT -1 208 "\nWhat if we take the ratio of consecutive Fibonacci n
umbers? That is take one number and divide by the previous one, then t
he next number, and divide by its previous number. What will happen to
 these ratios?\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "for k fr
om 2 to 20 do; \nF||k/F||(k-1) = evalf(F||k/F||(k-1), 15); od;" }}}
{PARA 0 "" 0 "" {TEXT -1 68 "\nThese numbers look quite similar to the
 golden number. Could it be?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
35 "(1 + sqrt(5))/2;\ng := evalf(%, 15);" }}}{PARA 0 "" 0 "" {TEXT -1 
185 "\nLets see how close these numbers get to the golden ratio. If we
 divide each of these by g, and multiply by 100, we will get the perce
nt that these numbers agree with the golden number." }}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 74 "for k from 2 to 20 do; \n   print(evalf( 10
0* (F||k/F||(k-1))/g, 15)); \nod:" }}}{PARA 0 "" 0 "" {TEXT -1 232 "\n
\nIts really interesting because these Fibonacci ratios get a little b
igger then a little smaller than g, but by the 20th, they agree to wit
hin 1 ten-thousandeth of a percent! If we were to go further it might \+
be even more accurate." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 114 "f
or k from 2 to 50 do; \n       F||k := F||(k-1) + F||(k-2):\n       pr
int(evalf( 100* (F||k/F||(k-1))/g, 15)); \nod:" }}}{PARA 0 "" 0 "" 
{TEXT -1 193 "\nThats pretty close to 100%! We will actually prove tha
t these numbers converge to the golden ration in the next section.\n\n
By the way, here is F50/F49, just to show you how big the numbers get.
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "F||40/F||39;" }}}{PARA 
0 "" 0 "" {TEXT -1 3 "\n\n " }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " \+
" }{TEXT 266 29 "5. Continued Fractions & Gold" }{TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 110 "\nHere is another place where the golden
 ratio comes up. Let's recall what g and its decimal approximation is.
\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "g := (1 + sqrt(5))/2;
\ng := evalf(%, 25);" }}}{PARA 0 "" 0 "" {TEXT -1 102 "\nNow we form a
 series of complex fractions with all 1, which are ever more complicat
ed :\n    1,       " }{XPPEDIT 18 0 "1 + 1/1" "6#,&\"\"\"F$*&F$F$F$!\"
\"F$" }{TEXT -1 8 ",       " }{XPPEDIT 18 0 "1+1/(1+1/1);" "6#,&\"\"\"
F$*&F$F$,&F$F$*&F$F$F$!\"\"F$F(F$" }{TEXT -1 7 ",      " }{XPPEDIT 18 
0 "1+1/(1+1/(1+1/1));" "6#,&\"\"\"F$*&F$F$,&F$F$*&F$F$,&F$F$*&F$F$F$!
\"\"F$F*F$F*F$" }{TEXT -1 10 ",         " }{XPPEDIT 18 0 "1+1/(1+1/(1+
1/(1+1/1)));" "6#,&\"\"\"F$*&F$F$,&F$F$*&F$F$,&F$F$*&F$F$,&F$F$*&F$F$F
$!\"\"F$F,F$F,F$F,F$" }{TEXT -1 7 ",      " }{XPPEDIT 18 0 "1+1/(1+1/(
1+1/(1+1/(1+1/1))));" "6#,&\"\"\"F$*&F$F$,&F$F$*&F$F$,&F$F$*&F$F$,&F$F
$*&F$F$,&F$F$*&F$F$F$!\"\"F$F.F$F.F$F.F$F.F$" }{TEXT -1 9 ",     ..." 
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "x||1 := 1;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "for k from 2 to 15 do\nx||k := 1 + \+
1/( 1 + x||(k-1)); od;" }}}{PARA 0 "" 0 "" {TEXT -1 92 "\nWe may recog
nize these numbers as the Fibonnaci numbers....which tend to the golde
n number!" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "for k from 1 to
 15 do x||k = evalf(x||k); od;" }}}{PARA 0 "" 0 "" {TEXT -1 275 "\nLet
s see if we can prove that these numbers tend to golden number. Notice
 that an infinite complex fraction of this type has a certain self sim
ilarity - because of the infinite nature of this list of embedded radi
cals. One less than infinity is still infinity.\n\n            " }
{XPPEDIT 18 0 "x = 1+1/(1+1/(1+1/(1+1/(1+1/`...`))));" "6#/%\"xG,&\"\"
\"F&*&F&F&,&F&F&*&F&F&,&F&F&*&F&F&,&F&F&*&F&F&,&F&F&*&F&F&%$...G!\"\"F
&F1F&F1F&F1F&F1F&" }{TEXT -1 57 "\n\nThus we have these kinds of relat
ionships.\n            " }{XPPEDIT 18 0 "x = 1+1/x;" "6#/%\"xG,&\"\"\"
F&*&F&F&F$!\"\"F&" }{TEXT -1 139 "\n\n\nIf we solve this equation, we'
ll find out what the value would be of an expression of this sort with
 an infinite number of 1's included." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "x:='x':\nx = 1 + 1/x;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "lhs(%)*x = simplify(rhs(%)*x) ;" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 16 "x^2 - x - 1 = 0;" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 9 "solve(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
12 "evalf(%,25);" }}}{PARA 0 "" 0 "" {TEXT -1 29 "\nThese numbers look
 familiar!" }}{PARA 0 "" 0 "" {TEXT -1 2 "\n " }}}{SECT 0 {PARA 4 "" 
0 "" {TEXT -1 1 " " }{TEXT 261 27 "6. Infinite Radicals & Gold" }}
{PARA 0 "" 0 "" {TEXT -1 110 "\nHere is another place where the golden
 ratio comes up. Let's recall what g and its decimal approximation is.
\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "g := (1 + sqrt(5))/2;
\ng := evalf(%, 25);" }}}{PARA 0 "" 0 "" {TEXT -1 55 "\nNow we form a \+
series of radicals formed all with 1's.\n" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 16 "x||1 := sqrt(1);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 54 "for k from 2 to 15 do\nx||k := sqrt( 1 + x||(k-1)); o
d;" }}}{PARA 0 "" 0 "" {TEXT -1 47 "\nIf we evaluate these numerically
, we get this." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "for k from
 1 to 15 do x||k = evalf(x||k); od;" }}}{PARA 0 "" 0 "" {TEXT -1 226 "
\nHmmm. Curious. Those numbers look suspiciously similar to the golden
 ratio. Let's get some idea of how close those numbers are to the gold
en number. If we divide each by g, and then multiply by 100 to express
 as a percent...\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "for k \+
from 1 to 15 do 100*evalf(x||k / g)  ; od;" }}}{PARA 0 "" 0 "" {TEXT 
-1 338 "\nIt appears that these numbers get REALLY close to being the \+
same as g. Can we prove it? Notice that an infinite radical of this ty
pe has a certain self similarity - because of the infinite nature of t
his list of embedded radicals. One less than infinity is still infinit
y. From this observation, we can get these equations ...\n \n\n       \+
" }{XPPEDIT 18 0 "x = sqrt(1 + sqrt( 1 sqrt(1 + sqrt( 1 + sqrt(1 + `..
.` )))))" "6#/%\"xG-%%sqrtG6#,&\"\"\"F)-F&6#*&F)F)-F&6#,&F)F)-F&6#,&F)
F)-F&6#,&F)F)%$...GF)F)F)F)F)" }{TEXT -1 11 "\n\n\n        " }
{XPPEDIT 18 0 "x = sqrt(1 +x)\n" "6#/%\"xG-%%sqrtG6#,&\"\"\"F)F$F)" }
{TEXT -1 170 "\n\nIf we take the first example as an equation, and sol
ve it, we'll find out what the value would be of an expression of this
 sort with an infinite number of terms inside." }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 17 "x = sqrt( 1 + x);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "lhs(%)^2 = rhs(%)^2;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 16 "x^2 - x - 1 = 0;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 9 "solve(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
12 "evalf(%,25);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{PARA 0 "" 0 "" {TEXT -1 31 "\nThese numbers look familiar!\n " }}}
{PARA 0 "" 0 "" {TEXT 259 36 "\n         \251 2002 Waterloo Maple Inc \+
" }}}{MARK "0 1" 27 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 
1 2 33 1 1 }