 mersenneforum.org > Math The Golden Section.
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wblipp

"William"
May 2003
New Haven

23×103 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).
You could always look to see if the cube root simplifies in the same way it did here. First reduce it as 10+6*sqrt(3) and ask if there are integers a and b so that
(a+b*sqrt(3))^3 = 10+6*sqrt(3)

the left expands as a^3 + 3a^2b*sqrt(3) + 3ab^2*3+b^3*sqrt(3), so we need

a^3+9*a*b^2 = 10
3a^2b+3b^3=6

a*(a^2+9*b^2)=10
b*(a^2+b^2)=2

The second equation b is a factor of 2, so it's +/-1 or +/-2. Two can't work because the secod term is then 4 or more, so the the only solutions are b=1 a=+/-1. Moving to the first equation, a=1 and b=1 is the only solution, and it then allows you to simplify the cube root.

Applying to -54+sqrt(2700), you get
a(a^2+9b^2)=-54
b(a^2+b^2)=10

a is negative from the first equation, b is positive from the second equation.

From the second equation, b divides 10 and b^3<10, so b=1 or 2. Both cases have solutions: a=-3 or a=-1. Checking back to the first equation, only a=-3 b=1 works, so the simplification comes from
(-3+sqrt(3))^3 = -54+sqrt(2700)

Note that this works because there were integer solutions to a and b - rational solutions would also simplify the answer but would not have been found. Also the trial and error solution is similar to but less effective than trying all the rational possibilities in the original equation.   2004-04-27, 16:32   #24
xilman
Bamboozled!

"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across

97×113 Posts Quote:
 Originally Posted by xilman If this cubic has a rational solution, it must be an integer (because it is monic) and it must be a divisor of 108, which is 2*2*2*3*3*3. There are very few of these and we don't need to search them all because c(-54+s(2700)) + c(-54 -s(2700)) is a root. The first of these is about c(-54+52) and the second about c(-54 -52). The first term is about -1, the second about c(-106). We know that 4^3 =64 and 5^3 = 125, so we are looking for a root fairly close to -6. The only candidates from the factorization of 108 are -4, -6 and -8 with the numerical estimate strongly suggesting -6. Paul
Doh!

I really must stop trying to factor small integers in my head. 108=2*2*3*3*3.

Doesn't change my argument but the correct factorization removes x=-8
from candidates to be considered.

Paul   2004-04-28, 01:05   #25
jinydu

Dec 2003
Hopefully Near M48

2×3×293 Posts Quote:
 Originally Posted by Bob Silverman I gave the steps. If the expression can be simplified it must lie in the ground field (Q) or in some sub-field of the full splitting field of your polynomial. If it is in the ground field then a quick, rough, numerical approximation to the root will tell you the answer. If it is in a subfield you must first find a basis for that sub-field, express the elements of the subfield as a linear combination of basis elements, equate to your root and solve for the coefficients. The coefficients will be integers and can be found by a number of methods. Look up the "relation finding" algorithm of Ferguson & Forcade. What more do you want? If you want the details of how to compute the sub-fields and their bases you will need to learn some algebraic number theory.
It seems I may be out of my league. I'm only a high school student...   2004-05-03, 09:51 #26 jinydu   Dec 2003 Hopefully Near M48 2×3×293 Posts Sorry if I'm Being Annoying, But Looks like I've run into another problem like this: cube root (10+sqrt(108)) = 1 + sqrt(3) cube root ((-25/432)+(5sqrt(15)/144)i) = ? This time, I'm not working with "nice integers"...   2004-05-03, 12:25 #27 jinydu   Dec 2003 Hopefully Near M48 2×3×293 Posts Never mind... cube root ((-25/432)+(5sqrt(15)/144)i) = (5/12) + (sqrt/12)i   2004-05-04, 17:21   #28
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

1000000001002 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.

QUOTE=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.[/QUOTE] I will solve the questions in 2 parts.
1) Further simplication of the cube rt. obtained from Cardano's method.

2) Solve the eq. x^5 +6x -20 = 0

This will be done in two posts one for each point.

1) Take the first term in brackets without the cube rt. viz. (10 +sq.rt.108)

Now 10+ sq.rt. 108 = 10 + sq.rt (2*2*3*3*3) = 10 + 6sq.rt.3

This can be further be simplified to 1 + sq.rt 3 after taking the cube rt. of

10 + 6 sq rt 3.

The second term similarly simplifies to 1 - sq.rt. 3.

Now adding both terms we get (1 + sq.rt. 3) + (1 -sq.rt 3).

This gives 2 which is the rational root Please try this out after you have gone thru the general method as given in the foll. worked example. I leave your problem as homework to do.

Ques: Find the cube rt. of 72 - 32 sq.rt.5 (and simplify) Ans: (If) cube rt. of 72 - 32 sq.rt. 5 = x - sq.rt.y ( why not rt.x - rt.y ?) --------( 1 )
Note: I am dispensing of the symbol sq.rt. and calling it just rt.

Then , cube rt of 72 + rt.32 = x + rt.y (Theorem) -------( 2)

By multiplication of (1 ) and (2 )we get

cube rt. (5184 - (1024 * 5 ) = x^2 - y

cube rt. 64 = 4 = x^2 - y Hence y = x^2 - 4 -------(3)

Now by removing cube rt. 72 - 32 rt. 5 = ( x + rt.y ) ^ 3
= ( x^3 + 3xy ) + irrational
terms ----------(4)

Equating rational terms on both sides we get 72 = x^3 + 3xy

From (3) and ( 4) 72 = x^3 +3x ( x^2 - 4 ))

Therefore 72 = x^3 + 3x^3 - 12x

72 = 4x^3 -12x
Hence dividing by 4 18 = x^3 - 3x

Therefore x^3 - 3x - 18 = 0

This is a cubic equation in itself which can be solved by Cardano's formula.

To cut short here allow yourself to solve by trial which is permissible as in factoring.

We find that x = 3 solves the eqn.

Hence from (3) y = 5.

Therefore the cube root of 72 - 32 rt 5 = 3 - rt 5 or

72 = ( 3 - rt. 5) ^ 3

You can check it out by actual multiplying Q.E.D.

Mally    2004-05-31, 10:40 #29 devarajkandadai   May 2004 4748 Posts g.s. Not thought about it yet.regards Devaraj 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.[/QUOTE]   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post kar_bon Riesel Prime Data Collecting (k*2^n-1) 6 2010-11-25 13:39 edron1011 Software 5 2008-10-31 00:17 mfgoode Puzzles 1 2007-01-31 16:26 Citrix Forum Feedback 1 2006-05-03 09:29 eepiccolo Teams 1 2003-05-13 11:52

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