20040427, 12:06  #12  
Dec 2003
Hopefully Near M48
2·3·293 Posts 
Quote:
That way, it would be possible to generalize the technique to cases where I don't know what x equals to beforehand. 

20040427, 12:35  #13  
Nov 2003
2^{6}×113 Posts 
Quote:
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". 

20040427, 12:52  #14 
Dec 2003
Hopefully Near M48
11011011110_{2} 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)) + (1sqrt(3)) is simpler cbrt(8) is simpler still 2/1 is even simpler 2 is the simplest 
20040427, 12:53  #15  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
2·3·1,699 Posts 
Quote:
Paul 

20040427, 13:35  #16  
Dec 2003
Hopefully Near M48
11011011110_{2} Posts 
Quote:
cbrt(54+sqrt(2700))+cbrt(54sqrt(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)) 

20040427, 13:38  #17  
Nov 2003
2^{6}·113 Posts 
Quote:
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 subfield 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. 

20040427, 13:46  #18  
Nov 2003
2^{6}·113 Posts 
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. 

20040427, 13:55  #19 
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(54sqrt(2700)) a = cbrt(54+sqrt(2700)) b = cbrt(54sqrt(2700)) a^3 + b^3 = 108 a*b = cbrt((54+sqrt(2700))*(54sqrt(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. 
20040427, 13:57  #20  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
2·3·1,699 Posts 
Quote:
This time, a = c(54+s(2700)) and b = c(54s(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 20040427 at 14:02 Reason: minor formatting change 

20040427, 14:42  #21  
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(44))/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:
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 welldefined steps to arrive at the solution x = 2. Last fiddled with by jinydu on 20040427 at 14:50 

20040427, 15:02  #22  
Nov 2003
2^{6}·113 Posts 
Quote:
ground field (Q) or in some subfield 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 subfield, 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 subfields and their bases you will need to learn some algebraic number theory. 

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