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 Register FAQ Search Today's Posts Mark Forums Read  2004-04-23, 16:06 #1 mfgoode Bronze Medalist   Jan 2004 Mumbai,India 22·33·19 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.   2004-04-24, 06:00 #2 jinydu   Dec 2003 Hopefully Near M48 2×3×293 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.   2004-04-24, 06:21 #3 jinydu   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...   2004-04-24, 11:56 #4 Cyclamen Persicum   Mar 2003 34 Posts Golden Intersection = 2*cos(pi/5)   2004-04-24, 17:18   #5
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

1000000001002 Posts The Golden Section

Quote:
 Originally Posted by Cyclamen Persicum Golden Intersection = 2*cos(pi/5)
I am sorry, as I had a nagging doubt, that the way I printed my formula may cause a bit of confusion and this has indeed happened to Jinydu.
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 .   2004-04-26, 11:05 #6 jinydu   Dec 2003 Hopefully Near M48 6DE16 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(10-sqrt(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(10-sqrt(108)) = 2 without knowing a priori that 2 is a solution to the original cubic (and without using trial-and-error with the rational roots theorem). And in general: The cubic formula tends to give answers that look something like cube root (a+sqrt(b)) + (a-sqrt(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?   2004-04-27, 03:18   #7
wblipp

"William"
May 2003
New Haven

22·32·5·13 Posts Quote:
 Originally Posted by jinydu Anyway, here's the challenge. Prove that cube root(10+sqrt(108)) + cube root(10-sqrt(108)) = 2
(1+sqrt(3))^3 = 10+sqrt(108)
(1-sqrt(3))^3 = 10-sqrt(108)

So the expression is
1+sqrt(3)+1-sqrt(3) = 2

William   2004-04-27, 08:01 #8 jinydu   Dec 2003 Hopefully Near M48 110110111102 Posts Would it have been possible if you didn't know beforehand that the left-hand side equals 2? That is, if the question was: Simplify cube root(10+sqrt(108)) + cube root(10-sqrt(108)) as much as possible (if possible).   2004-04-27, 09:16   #9
xilman
Bamboozled!

"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across

5·7·172 Posts Quote:
 Originally Posted by jinydu Would it have been possible if you didn't know beforehand that the left-hand side equals 2? That is, if the question was: Simplify cube root(10+sqrt(108)) + cube root(10-sqrt(108)) as much as possible (if possible).
I'd do it this way, assuming that I was given only that information and
nothing else:

write x = a + b, where a = cbrt(10+sqrt(108)) and b = cbrt(10-sqrt(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 (100-108)
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 (x-2)(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   2004-04-27, 09:25 #10 jinydu   Dec 2003 Hopefully Near M48 175810 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 trial-and-error guessing of rational roots or fore-knowledge of the solution. Hint: Applying Cardano's method gives: x = cube root(10+sqrt(108)) + cube root(10-sqrt(108)), but this may or may not be the simplest possible way of expressing this solution.   2004-04-27, 11:55   #11
R.D. Silverman

Nov 2003

11100010000002 Posts Quote:
 Originally Posted by jinydu 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 trial-and-error guessing of rational roots or fore-knowledge of the solution. Hint: Applying Cardano's method gives: x = cube root(10+sqrt(108)) + cube root(10-sqrt(108)), but this may or may not be the simplest possible way of expressing this solution.

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

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