mersenneforum.org > Math The Golden Section.
 Register FAQ Search Today's Posts Mark Forums Read

2004-04-27, 12:06   #12
jinydu

Dec 2003
Hopefully Near M48

2·3·293 Posts

Quote:
 Originally Posted by 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.
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.

2004-04-27, 12:35   #13
R.D. Silverman

Nov 2003

26×113 Posts

Quote:
 Originally Posted by 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.
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".

 2004-04-27, 12:52 #14 jinydu     Dec 2003 Hopefully Near M48 110110111102 Posts 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
2004-04-27, 12:53   #15
xilman
Bamboozled!

"πΊππ·π·π­"
May 2003
Down not across

2·3·1,699 Posts

Quote:
 Originally Posted by 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".
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

2004-04-27, 13:35   #16
jinydu

Dec 2003
Hopefully Near M48

110110111102 Posts

Quote:
 Originally Posted by jinydu That way, it would be possible to generalize the technique to cases where I don't know what x equals to beforehand.
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))

2004-04-27, 13:38   #17
R.D. Silverman

Nov 2003

26·113 Posts

Quote:
 Originally Posted by 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
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.

2004-04-27, 13:46   #18
R.D. Silverman

Nov 2003

26·113 Posts

Quote:
 Originally Posted by 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))

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.

 2004-04-27, 13:55 #19 jinydu     Dec 2003 Hopefully Near M48 2×3×293 Posts 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.
2004-04-27, 13:57   #20
xilman
Bamboozled!

"πΊππ·π·π­"
May 2003
Down not across

2·3·1,699 Posts

Quote:
 Originally Posted by 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))
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

Last fiddled with by xilman on 2004-04-27 at 14:02 Reason: minor formatting change

2004-04-27, 14:42   #21
jinydu

Dec 2003
Hopefully Near M48

2×3×293 Posts

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:
 Originally Posted by 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?
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.

Last fiddled with by jinydu on 2004-04-27 at 14:50

2004-04-27, 15:02   #22
R.D. Silverman

Nov 2003

26·113 Posts

Quote:
 Originally Posted by 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.
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"

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.

 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

All times are UTC. The time now is 03:41.

Fri Nov 27 03:41:09 UTC 2020 up 78 days, 52 mins, 4 users, load averages: 2.19, 1.87, 1.63