mm-1589 === Subject: Help solving Equation I need help solving this equation. I have the answer but I dont know how to get it. (I got answer using mathematica) x^4+8x^3-368x^2+128x+256=0 Any help would be appreciated, === Subject: Re: Help solving Equation Distribution: aus > I need help solving this equation. I have the answer but I dont know > how to get it. (I got answer using mathematica) > x^4+8x^3-368x^2+128x+256=0 > Any help would be appreciated, Looking at the coefficients, there are obviously large factors of 2 involved. This suggests the substitution x = 4*u, leading to: 256*(u^4 + 2*u^3 - 3*u^2 + 2*u + 1) = 0 and we can remove the factor of 256 (which was the point of doing this substitution). So we now want to solve u^4 + 2*u^3 - 3*u^2 + 2*u + 1 = 0 and for each solution of u we get one for x by multiplying by 4. The new quartic has an obvious special form -- it is symmetric. By this I mean that the sequence of coefficients [1, 2, -3, 2, 1] is its own reverse. Such polynomials have a special name -- they are called reciprocal polynomials, because if t is a root then so is 1/t. There is a standard trick for such polynomials based on this symmetry; firstly, if the polynomial has odd degree, then -1 will be a root, so you can divide the polynomial by (u+1) to get a reciprocal polynomial of even degree. Secondly, once the polynomial has even degree, divide it by u^(degree/2) = u^2, giving this: u^2 + 2*u - 3 + 2/u + 1/u^2 = 0 (In this form, it may be clearer why the reciprocal of a root must also be a root.) Now group the terms with corresponding powers of u and 1/u: (u^2 + 1/u^2) + 2*(u + 1/u) - 3 = 0 We now make the substitutions v_i = u^i + 1/u^i: v_2 + 2*v_1 - 3 = 0 At this point, it is not clear that we have gained anything by these manipulations, but there is a point to this. :) First, for convenience let v = v_1, and note that v_0 = 2. Secondly, we have the following recurrence formula on the v_i values: v_0 = 2 v_1 = v v_(n+1) = u^(n+1) + 1/u^(n+1) = (u^n + 1/u^n)*(u + 1/u) - (u^(n-1) + 1/u^(n-1)) = v_n*v - v_(n-1) So each of the v_i is expressible in terms of v and 2. In particular (and this is what concerns us now) we have that v_2 = v^2 - 2. So the equation you want to solve is (v^2 - 2) + 2*v - 3 = 0 or v^2 + 2*v - 5 = 0 This is a simple quadratic equation, which I will assume you are able to solve. :) For each root v, you then need to solve for the possible corresponding values of u. Since v = u + 1/u, this is just solving u^2 - u*v + 1 = 0 for u given a particular value of v. Again, this is a quadratic equation. Once you have the possible values of u, multiply by 4 to get the original values of x. Geoff. ----------------------------------------------------------------------------\ - Geoff Bailey (Fred the Wonder Worm) | Programmer by trade -- ftww@maths.usyd.edu.au | Gameplayer by vocation. ----------------------------------------------------------------------------\ - === Subject: Re: Help solving Equation > I need help solving this equation. I have the answer but I dont know how > to get it. (I got answer using mathematica) > x^4+8x^3-368x^2+128x+256=0 > Any help would be appreciated, There are exakt solutions for all polynoms of the kind n a x + ... + a = 0 n 0 for n <= 4 (not so for n > 4 except of special cases). Use the algorithms! Ingo === Subject: Re: Help solving Equation no idea...without using a computer or calculator Someone let Andrew at andrew@spin.net.au loose on an internet connection and the result was: > I need help solving this equation. I have the answer but I dont know > how to get it. (I got answer using mathematica) > x^4+8x^3-368x^2+128x+256=0 > Any help would be appreciated, === Subject: Re: Help solving Equation same, Matlab or similar equation-solving program on the computer here would be the way to go I think. > no idea...without using a computer or calculator > Someone let Andrew at andrew@spin.net.au loose on an internet > connection and the result was: > I need help solving this equation. I have the answer but I dont know > how to get it. (I got answer using mathematica) > x^4+8x^3-368x^2+128x+256=0 > Any help would be appreciated, === Subject: Golf Tourney Paring Problem I have two teams (the Lets and the Nums) of six golfers (per team). Each player will play 5 rounds of golf . Each round will see two Lets (e.g. A,B) against two Nums (1,2). I am trying to plan the pairings to optimize who plays with and against whom. As there are 5 rounds, and each player has 5 team members, it is relatively easy to set the team pairings so that each player plays with a team mate only once. How can I set the matches so that each person plays against the others no more than twice, and at least once? http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- === Subject: ratio of surface area to volume This is a question about highschool level maths. I am studying surface are and volume and it has occured to me that there should be some rule regarding the ratio between surface area and volume of 3d shapes. Is there? Please point me to any good free online resources or answer in simple terms. Brenton impactbc@hotmail.com === Subject: Re: ratio of surface area to volume Distribution: aus > This is a question about highschool level maths. I am studying > surface area and volume and it has occured to me that there should > be some rule regarding the ratio between surface area and volume > of 3d shapes. Is there? Not really. You may be able to get a feel for it by considering things a dimension lower -- do you think there should be some rule regarding the ratio between the perimeter and area of 2d shapes? In particular, by considering rectangles with perimeter 1 you should be able to arrive at a large variety of possible areas. Geoff. ----------------------------------------------------------------------------\ - Geoff Bailey (Fred the Wonder Worm) | Programmer by trade -- ftww@maths.usyd.edu.au | Gameplayer by vocation. ----------------------------------------------------------------------------\ - === Subject: Re: ratio of surface area to volume Distribution: aus > This is a question about highschool level maths. I am studying > surface area and volume and it has occured to me that there should > be some rule regarding the ratio between surface area and volume > of 3d shapes. Is there? Consider a box with congruent square ends and congruent rectangular top, bottom and sides. For a fixed surface area the volume will be largest for a cube (all those rectangles are squares) and decreases as the end squares get smaller (and the length gets appropriately greater) only limited by zero when the end squares degenerate to points. === Subject: Re: ratio of surface area to volume Hi Assuming tha The reason I am asking this question is so that I can answer this question which I have created myself: can a 1 mm3 cube hold? (assume that amount of data cube can hold is directly proportional to the size of the cube) Is this any help? B. Fletcher > This is a question about highschool level maths. I am studying > surface area and volume and it has occured to me that there should > be some rule regarding the ratio between surface area and volume > of 3d shapes. Is there? > Consider a box with congruent square ends and congruent rectangular > top, bottom and sides. > For a fixed surface area the volume will be largest for a cube (all > those rectangles are squares) and decreases as the end squares get > smaller (and the length gets appropriately greater) only limited by > zero when the end squares degenerate to points. === Subject: Re: ratio of surface area to volume > Hi > Assuming tha > The reason I am asking this question is so that I can answer this question > which I have created myself: data > can a 1 mm3 cube hold? (assume that amount of data cube can hold is directly > proportional to the size of the cube) > Is this any help? > B. Fletcher By size do you mean surface area of the cube or volume of the cube? The method of recording data to the cube should make it clear which. For a cube, the surface area is proportional to the square , e^2 = e*e, of the edge length, e, and the volume is proportional to the cube (third power), e^3 = e*e*e, of the edge length.