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You are not logged in. #1 20130620 21:14:04
Define the intersection points of polynomialsI am facing the following problem. I know that from these 2 unknown points For each of these polynomials I know one point Is it possible to find the intersection points (i.e. the 2 unknown points) of the aforementioned polynomials? Last edited by Herc11 (20130620 21:22:12) #2 20130620 21:33:57
Re: Define the intersection points of polynomialsHi Herc11; 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. #3 20130620 21:39:35
Re: Define the intersection points of polynomialshi Herc11 You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #4 20130620 21:45:05
Re: Define the intersection points of polynomialsHmmm.Thats the problem. I know the leading coefficient and one point from each of the polynomials. #5 20130620 21:49:29
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. #6 20130620 22:03:03
Re: Define the intersection points of polynomialsOk, That's right. #7 20130620 22:10:56
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. #9 20130620 22:31:14
Re: Define the intersection points of polynomialsLet me give you a concrete example here are two polynomials with the the leading coefficient known. Also each has one point known. Can you determine point A and B? 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. #10 20130620 22:40:48
Re: Define the intersection points of polynomialsOk...What I am thinking is: Last edited by Herc11 (20130620 22:41:59) #11 20130620 22:43:56
Re: Define the intersection points of polynomials
That is what I am saying. You need more relationships between the variables. We call these equations. You have to make sure they are unique equations and not just the same as the others. 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. #14 20130620 22:50:34
Re: Define the intersection points of polynomialsYou want A, B in my drawing. Therefore you (x0,y0) and (x1,y1) that is 4 variables. To have a chance at getting those 4 as a number I need 4 equations. Then to get the 2 polynomials I need 4 more equations. 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. #16 20130620 22:58:56
Re: Define the intersection points of polynomialsOnly if you knew the equations of those polynomials. If you did not you would be just adding more variables! 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. #17 20130620 22:58:58
Re: Define the intersection points of polynomialsI have a feeling that the intersections can be determined. You can set up three equations in x0,y0,x1,y1,a12, and a13. Then each three new equations add just 2 new variables, ai2 and ai3. So, with 12 equations and 12 variables, there is aa chance that the system will be determined. 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 #18 20130620 22:59:51
Re: Define the intersection points of polynomialsWith 8 and 8 it could be determined. He has more variables than relationships that is the problem. 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. #19 20130620 23:06:40
Re: Define the intersection points of polynomialsa2=2= (7+ (y0 ( (y1y0) / (x1x0))*(1x1)))/ ( (1x1)*(1x0) #20 20130620 23:09:10
Re: Define the intersection points of polynomialsWhere is (2,6) and (8,2) coming from? 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. #21 20130620 23:12:06
Re: Define the intersection points of polynomials
Consider that there are two more polynomials passing from the intersection points illustrated in your figure and I know their leading coefficient and one point form each of them i.e. 2,6 8,2. #22 20130620 23:14:32
Re: Define the intersection points of polynomialsThat will add more variables because you will not know two terms from each of them. You are adding 2 equations and 4 more unknowns. See post #4. 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. #24 20130620 23:18:52
Re: Define the intersection points of polynomialsYou will have the x^2 coefficient but not the x and the constant term for both. You will need to know what the equation is for each. 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. 