20040423, 16:06  #1 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
The Golden Section.
The Golden section has been known to the ancient Greeks since antiquity.
It had been worked out geometrically as 1.618033989.... and called phi. Phi was used in building the Great Pyramid of Giza about 3070 B.C. They referred to it as the 'sacred ratio' In the Fibonacci series the ratio of successive terms Fn+1/Fn tends to phi as the series progresses. In dividing a line x+y In parts x,y such that (x+y)/x=x/y Then x/y= (1+sqr.rt.5)/2=1.618033989..... There is a novel way that the ratio can be expressed trigonometrically using the well known constants 'e' and 'i'=sqr.rt (1) The Golden ratio can be shown as 2*cos(log ((i^2))/5*i)) Can anyone show that this is equivalent to phi the golden ratio? Mally. 
20040424, 06:00  #2 
Dec 2003
Hopefully Near M48
1758_{10} Posts 
I don't really know how to take the log of complex numbers, but I'll start off like this:
phi = 2 cos (log ((i^2))/5*i) phi = 2 cos (log (1)/5i) phi = 2 cos (log (i/5)) phi = 2 cos (log i  log 5) I would suggest applying the compound angle formula for cosine at this point. 
20040424, 06:21  #3 
Dec 2003
Hopefully Near M48
2×3×293 Posts 
Ok, it seems that ln i = (pi*i)/2, so continuing on:
phi = 2 cos ([pi*i]/2  log 5) phi = 2 * ([cos(pi*i)/2]*[cos(log 5)] + [sin(pi*i)/2]*[sin(log 5)]) Its getting too complicated for this text box... 
20040424, 11:56  #4 
Mar 2003
3^{4} Posts 
Golden Intersection = 2*cos(pi/5)

20040424, 17:18  #5  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
The Golden Section
Quote:
I should have clarified that in the expression the numerator is Log (i^2) only and the denominator of this is 5*i i.e. 5*i is not under log with i^2. Euler discovered the remarkable formula e^(i*pi) = 1. Substitute this in log(i^2) and then divide by 5*i and you will get 2*cos (pi/5) as Cyclamen Periscum has done correctly. To complete the derivation 2* cos pi/5 ==2*cos36 and cos 36 =(1+sq rt.5)/4 so 2*cos 36 =(1+sq.rt.5)/2 =1.6180339... =Phi the golden section as given in my thread. or use a calculator. Mally . 

20040426, 11:05  #6 
Dec 2003
Hopefully Near M48
1758_{10} Posts 
Ok, here's another derivation question:
If you're trying to solve a quadratic equation that (lucky you!) has an integer solution, the expression you obtain using the quadratic formula (after straightforward simplifications) will be an obvious integer. But this doesn't seem to be the case with cubics. Suppose I have the cubic equation: x^3 + 6x  20 = 0 (this particular equation was chosen on purpose for relative simplicity). Now, if I didn't know how to solve cubics, I might try plugging in a few small integer values of x. In this case, I would be lucky, because x = 2 is a solution, and I can easily use polynomial division to find the other two. But suppose that I instead try to solve the equation using the cubic formula (i.e. Cardano's method). After working through the entire process, I would find that one root of the equation is: x = cube root(10+sqrt(108)) + cube root(10sqrt(108)). My calculator easily verifies that the much more complicated expression above is indeed equal to 2. But my calculator just uses numerical approximations for expressions like sqrt(108). Therefore, it couldn't really be considered a mathematical proof. Anyway, here's the challenge. Prove that cube root(10+sqrt(108)) + cube root(10sqrt(108)) = 2 without knowing a priori that 2 is a solution to the original cubic (and without using trialanderror with the rational roots theorem). And in general: The cubic formula tends to give answers that look something like cube root (a+sqrt(b)) + (asqrt(b)) + c. In the particular cases where the solution has a simpler form (such as an integer, rational number or expression with only a single radical), how can I get to this simpler form? 
20040427, 03:18  #7  
"William"
May 2003
New Haven
2^{3}×5×59 Posts 
Quote:
(1sqrt(3))^3 = 10sqrt(108) So the expression is 1+sqrt(3)+1sqrt(3) = 2 William 

20040427, 08:01  #8 
Dec 2003
Hopefully Near M48
1758_{10} Posts 
Would it have been possible if you didn't know beforehand that the lefthand side equals 2?
That is, if the question was: Simplify cube root(10+sqrt(108)) + cube root(10sqrt(108)) as much as possible (if possible). 
20040427, 09:16  #9  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
10100100011111_{2} Posts 
Quote:
nothing else: write x = a + b, where a = cbrt(10+sqrt(108)) and b = cbrt(10sqrt(108). Then cube both sides, to get x^3 = (a^3 + b^3) +3ab(a+b) = 20 + 3abx. Now x^3 = 20+3x * cbrt((10+sqrt(108)) * (10  sqrt(108))) or x^3 = 20 +3x * cbrt (100108) or x^3= 20 6x where I took the integer cube root of 8. This now gives me a polynomial with integer coefficients. It is, of course, the original equation. It's possible to solve this one by inspection to find the root x=2. Indeed, the equation is (x2)(x^2 + 2x +10). Before you all jump on me for cheating, please remember that the problem as proposed specified that the only information to be assumed known was the sum of the two cube roots. Deducing a reducible cubic from that information is entirely allowable in my view. Paul 

20040427, 09:25  #10 
Dec 2003
Hopefully Near M48
2·3·293 Posts 
Ok, I'll try to frame my question more clearly this time.
Goal: Find a solution of x^3 + 6x  20 = 0 and express it in the simplest possible form. Condition: Not allowed to use trialanderror guessing of rational roots or foreknowledge of the solution. Hint: Applying Cardano's method gives: x = cube root(10+sqrt(108)) + cube root(10sqrt(108)), but this may or may not be the simplest possible way of expressing this solution. 
20040427, 11:55  #11  
Nov 2003
2^{2}×5×373 Posts 
Quote:
I'm not sure I understand the difficulty. It is easy to show that 2 = x from x = cbr(10 + sqrt(108)) + cbr(10  sqrt(108)) = a + b We have a^3 + b^3 = 20 ab = 2 x = 2 x^3 = (a+b)^3 = 8 But (a+b)^3 = a^3 + b^3 + 3abx = 20  6x = 20  6*2 = 8 = 2^3 8 = 8 QED I'm not sure what else you are looking for. 

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