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You are not logged in. #76 20130621 02:08:11
Re: Define the intersection points of polynomialsWhy is that? 2 points? Last edited by Herc11 (20130621 02:13:49) #77 20130621 02:12:35
Re: Define the intersection points of polynomialsBecause we need the number of equations equal to the number of unknowns. If we have only one point, the equality of their numbers is unreachable. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #79 20130621 03:45:21
Re: Define the intersection points of polynomialsThat is true. And I have good news  4 will be enough! The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #81 20130621 03:54:34
Re: Define the intersection points of polynomialsWell, that is when the number of unknown variables and equations will be the same. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #83 20130621 04:09:20
Re: Define the intersection points of polynomialsNo, there are more. For each cubic you introduce into the system, you get 3 new variables. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #85 20130621 04:21:02
Re: Define the intersection points of polynomialsHm, you are right. It seems that for nth degree polynomials, you will need 2n quadratics. It isn't large for very, very small n, but in terms of system solving, 2n(n+1) variables can be a lot to handle, even for some smaller n, let alone larger ones. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #86 20130621 04:26:34
Re: Define the intersection points of polynomialsIn Newton Interpolation. Last edited by Herc11 (20130621 04:27:21) #87 20130621 04:29:31
Re: Define the intersection points of polynomialsNot quadratics, sorry. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #88 20130621 04:37:50
Re: Define the intersection points of polynomialsOk. You think that I m right? Last edited by Herc11 (20130621 04:56:36) #89 20130621 05:03:22
Re: Define the intersection points of polynomialsI'm on the phone now, so I am not able to post the code at the moment, but it's nothing spectacular anyway. I just used the builtin Solve function and entered the system of equations into it. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #90 20130621 05:08:22
Re: Define the intersection points of polynomialsOk no problem! Whenever you can, I am not familiar with Mathematica, thats why I am asking... Last edited by Herc11 (20130621 06:30:41) #91 20130621 10:08:04
Re: Define the intersection points of polynomialsActually, there are 24 equations and unknowns, it's just that we can set them up from 6 cubics. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #92 20130621 15:23:30
Re: Define the intersection points of polynomials24??? Can you explain please? #93 20130621 15:29:18
Re: Define the intersection points of polynomialsHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #94 20130621 15:36:56
Re: Define the intersection points of polynomialsHi bobbym, #95 20130621 15:46:17
Re: Define the intersection points of polynomialsIf you want to latex them go here In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #97 20130621 15:58:05
Re: Define the intersection points of polynomialsJust type what you need into the box they provide. Or use the pull down menus. It will spit out latex underneath. Copy what is in the box and put it in here between the math tags. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #100 20130621 16:18:57
Re: Define the intersection points of polynomialsHi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. 