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? :smile:

Mally.

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.

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...

Golden Intersection = 2*cos(pi/5)

The Golden Section
 

[QUOTE=Cyclamen Persicum]Golden Intersection = 2*cos(pi/5)[/QUOTE]

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. :smile:

Mally :coffee: .

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?

[QUOTE=jinydu]Anyway, here's the challenge. Prove that

cube root(10+sqrt(108)) + cube root(10-sqrt(108)) = 2[/QUOTE]

(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

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).

[QUOTE=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).[/QUOTE]
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. :smile:

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

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'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.

[QUOTE=Bob Silverman]
a^3 + b^3 = 20
ab = -2
x = 2
x^3 = (a+b)^3 = 8
I'm not sure what else you are looking for.[/QUOTE]

The problem is with the third line (x = 2). I wanted to simplify cbr(10 + sqrt(108)) + cbr(10 - sqrt(108)) without knowing a priori that the answer is 2.

That way, it would be possible to generalize the technique to cases where I don't know what x equals to beforehand.

[QUOTE=jinydu]The problem is with the third line (x = 2). I wanted to simplify cbr(10 + sqrt(108)) + cbr(10 - sqrt(108)) without knowing a priori that the answer is 2.

That way, it would be possible to generalize the technique to cases where I don't know what x equals to beforehand.[/QUOTE]

This just begs the question. When you say "simplify" what does this mean?
If you mean "can it be simplified to an integer and if so how?", the answer
is obvious. Even a very rough mental calculation shows that *if* the answer
is an integer it must be near cube root of 20, thus it must equal 2 or 3.
If you further look at the original equation you realize that if the root is an
integer it must be a divisor of 20 (the product of the roots is 20). Thus
we are led to x = 2. This is not guessing. Since the polynomial is monic,
we know that if there is a rational root, it will be an integer.

What else are you looking for? You have not defined what "simplify" means.
You must state what form you expect the final answer to take before you
can "simplify".

Ok, here is what I mean by simplify:

simple:integer---->rational number---->sqrt or cbrt of a rational number ----> cbrt(a+sqrt(b)), where a and b are rational numbers:complicated

Thus, cbrt(10 + sqrt(108)) + cbrt(10 - sqrt(108)) is complicated.

(1+sqrt(3)) + (1-sqrt(3)) is simpler

cbrt(8) is simpler still

2/1 is even simpler

2 is the simplest

[QUOTE=Bob Silverman]This just begs the question. When you say "simplify" what does this mean?
...
What else are you looking for? You have not defined what "simplify" means.
You must state what form you expect the final answer to take before you
can "simplify".[/QUOTE]
My guess is something like: express as an integer, if possible, as a rational if it's not possible as an integer, otherwise using the lowest n-th root of (integers if possible, rationals if not). There is presumably also some implied measure of simplicity based on the number of such terms, otherwise one could give any of a large number of weird and wonderful continued fractions.

Paul

[QUOTE=jinydu]
That way, it would be possible to generalize the technique to cases where I don't know what x equals to beforehand.[/QUOTE]

For example, simplify the following:

cbrt(-54+sqrt(2700))+cbrt(-54-sqrt(2700))

This time, I will not give out the answer a priori (although this particular case was also handpicked carefully). Also, the technique (maybe algorithm is an even better word) you use should also work for:

cbrt(10 + sqrt(108)) + cbrt(10 - sqrt(108))

[QUOTE=jinydu]Ok, here is what I mean by simplify:

simple:integer---->rational number---->sqrt or cbrt of a rational number ----> cbrt(a+sqrt(b)), where a and b are rational numbers:complicated

Thus, cbrt(10 + sqrt(108)) + cbrt(10 - sqrt(108)) is complicated.

(1+sqrt(3)) + (1-sqrt(3)) is simpler

cbrt(8) is simpler still

2/1 is even simpler

2 is the simplest[/QUOTE]

You have not given a rigorous, meaningful definition. You need to give
a rigorous metric for 'complexity of an expression'.

In any event, if you mean simplify into one of the forms:

integer
rational number
root of a rational number

You do it the way I outlined.

If you mean: simplify to the form (a + b sqrt(c)) it is clear that it can not
be done. The cubic extension Q(alpha) where alpha is a root of the original
equation has no such quadratic sub field.

The forms to which an expression alpha = cbr(a + sqrt(b)) + cbr(a - sqrt(c))
can be simplified are sharply limited. Any such form must lie within a
sub-field of Q(alpha).

To understand what forms are possible you need to understand some Galois
theory/algebraic number theory. Since the polynomial is monic its root
is an element of the maximal order of Q(alpha). In asking whether there
is a simpler form that is not in the ground field (Q in this case) you are
partially asking for separable extensions. See, for example Algorithmic
Algebraic Number Theory by Pohst & Zassenhaus, or H. Cohen's book.

My earlier example shows how to determine if the expression reduces
to an element in the ground field.

[QUOTE=jinydu]For example, simplify the following:

cbrt(-54+sqrt(2700))+cbrt(-54-sqrt(2700))

This time, I will not give out the answer a priori (although this particular case was also handpicked carefully). Also, the technique (maybe algorithm is an even better word) you use should also work for:

cbrt(10 + sqrt(108)) + cbrt(10 - sqrt(108))[/QUOTE]


I get cbr(-6). Once again if there is a simplified form it must lie
within the maximal order of one of the subfields of Q(alpha) [assuming
a monic polynomial]. Find a basis for the subfield. This gives an expression
for each element within its maximal order. Equate this expression to
your original number and solve for the coefficients.

Ok, I'll try to apply that technique to the expression:

cbrt(-54+sqrt(2700))+cbrt(-54-sqrt(2700))

a = cbrt(-54+sqrt(2700))
b = cbrt(-54-sqrt(2700))
a^3 + b^3 = -108
a*b = cbrt((-54+sqrt(2700))*(-54-sqrt(2700))) = cbrt(216) = 6
x (the number we're trying to determine) = a + b

x^3 = (a+b)^3 = a^3 + 3(a^2)b + 3a(b^2) + b^3 = a^3 + b^3 + 3ab(a+b) = a^3 + b^3 + 3abx

x^3 = -108 + 18x

x^3 - 18x + 108 = 0 (which is exactly the cubic where I got my expression from). This is basically just the reverse of the depressed cubic formula.

[QUOTE=jinydu]For example, simplify the following:

cbrt(-54+sqrt(2700))+cbrt(-54-sqrt(2700))

This time, I will not give out the answer a priori (although this particular case was also handpicked carefully). Also, the technique (maybe algorithm is an even better word) you use should also work for:

cbrt(10 + sqrt(108)) + cbrt(10 - sqrt(108))[/QUOTE]

I'll use my previous approach, but a little more streamlined notation. Follow my earlier posting if you need more detail.

This time, a = c(-54+s(2700)) and b = c(-54-s(2700)), where x = a+b again.

Once more x^3 = (a+b)^3 = a^3 + b^3 + 3xab

x^3 = -108 + 3x * c(54^2 - 2700) = -108 +3x c(216) = -108 +18x.

Alternatively, x^3 -18x +108 = 0.

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.

The answer, of course, is -6.


Paul

I think there was somewhat of a misunderstanding. This is what I meant:

We all know that the solution to the quadratic equaion ax^2 + bx + c = 0 is x=(-b+-sqrt(b^2 - 4ac))/2a. Not only does the quadratic formula give you the root of the polynomial, it also allows you to simplify your answer to the fullest extent possible. That is, in "most" cases, the answer will look something like: x = p+sqrt(r). Now, if x happens to be a rational number, then taking the square root of r will yield a rational number. After just a few simple operations, you can express x in the form a/b, where a and b are integers. In the even simpler case where x is an integer, the numerator will clearly divide the denominator.

For example, with the quadratic formula applied to x^2 - 2x + 1 = 0, x = (-(-2)+-sqrt((2^2)-(4*1*1)))/(2*1). But this can be simplified:

x = (2+-sqrt(4-4))/2
x = (2+-sqrt(0))/2
x = (2+-0)/2
x = 2/2
x = 1

This is what I mean by simplifying it to the simplest form, rather than leaving the answer as x = x^2 - 2x + 1 = 0, x = (-(-2)+-sqrt((2^2)-(4*1*1)))/(2*1).

[QUOTE=jinydu]And in general: The cubic formula tends to give answers that look something like cube root (a+sqrt(b)) + cube root (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?[/QUOTE]

Basically, what I'm looking for is an analogous process for the cubic roots (which have already been calculated using the cubic formula). There are some special cases that are trivial to simplify. For example, if a=sqrt(b), then x=cbrt(a+sqrt(b))+c. But there are less trivial cases, such as when x is an integer. I would like to know how to simplify to each of the following forms (when they are possible, and some may be subsets of others):

a (integer)
p/q (rational number)
sqrt(p/q)+c
cbrt(p/q)+c

As an analogue to what I said earlier, I am not satisfied with saying that the solution to x^3 + 6x - 20 = 0 is

x = cbrt(10 + sqrt(108)) + cbrt(10 - sqrt(108)),

just as I would be unsatisfied with saying that the solution to x^2 - 2x + 1 = 0 is

x = (-(-2)+-sqrt((2^2)-(4*1*1)))/(2*1).

As in the quadratic example, I would like to go through a series of well-defined steps to arrive at the solution

x = 2.

[QUOTE=jinydu]I think there was somewhat of a misunderstanding. This is what I meant:

x = cbrt(10 + sqrt(108)) + cbrt(10 - sqrt(108)),

just as I would be unsatisfied with saying that the solution to x^2 - 2x + 1 = 0 is

x = (-(-2)+-sqrt((2^2)-(4*1*1)))/(2*1).

As in the quadratic example, I would like to go through a series of well-defined steps to arrive at the solution

x = 2.[/QUOTE]

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.

[QUOTE=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).[/QUOTE]
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.

[QUOTE=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[/QUOTE]
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

[QUOTE=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.[/QUOTE]

It seems I may be out of my league. I'm only a high school student...

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"...

Never mind...

cube root ((-25/432)+(5sqrt(15)/144)i) = (5/12) + (sqrt[15]/12)i

[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]


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]


:whistle: 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 :wink:

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) :smile:

Ans: (If) cube rt. of 72 - 32 sq.rt. 5 = x - sq.rt.y ( why not rt.x - rt.y ?) :question: --------( 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 :bounce: Q.E.D.

Mally :coffee:

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? :smile:

Mally.[/QUOTE]