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parabolic equations  Quadratic equations http://forum.coursdemath.eu/viewtopic.php?f=20&t=524 
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Auteur:  tsitrader [ 20 Aoû 2011, 18:33 ] 
Sujet du message:  parabolic equations  Quadratic equations 
Hi  by chance I found your website and email address on the Internet when trying to find some information of parabolic equations. I am interested in deriving the parabolic equation of gold's price movement from January 2001 forward, but do not have a clue where to begin. Obviously we have 10 years of (x,y) points that are to be used to figure the equation. But how? Suggestions where to begin? Much appreciation, 
Auteur:  tsitrader [ 20 Aoû 2011, 18:38 ] 
Sujet du message:  How far gold's path strays from it's mathematically derived 
I have attached a chart to this email that details the (x,y) coordinates to be considered. Pièce jointe: gold_parabola.png [ 103.77 Kio  Consulté 24695 fois ] What I intend to do, ultimately, is study how far gold's path strays from it's mathematically derived parabola  both above and below this imaginary line. I also hope to gain some clue as to the timeframe that the parabola will essentially begin to achieve nearvertical direction. Will that be in 2 years, 5 years, 8 years, or what? Also, I have found a single link (http://www.traddr.com/profiles/blogs/sh ... dmidrange ) to work on this question done by another person. He uses a different starting date than I purpose on the chart attached. But I include it for your consideration. My idea is to imagine that the parabola is shaped equidistant from the yaxis, using the first day of the parabola as essentially (0,0) or, in this case (0,256). I am very excited to see what you share with me. I know I already said that, but, well, it's true! Thank you for your contribution to my understanding. No doubt I will share it widely with others. 
Auteur:  Admin [ 20 Aoû 2011, 18:54 ] 
Sujet du message:  Re: parabolic equations  Quadratic equations 
Hi, The general equation of a parabola is also called a quadratic equation : y = ax² + bx + c if you know at least the coordinates of 3 points on the curve, you can find its equation. So your job now is to find the value of the parameters a, b and c. In your case you have 7 known points, so you just have to pick three points and find manually the parabola equation. Let's pick for instance the following 3 points : (0 ; 256) (1000 ; 667) (1800 ; 1414) We can now write the following equations based on y = ax² + bx + c : 256 = a*0² + b*0 + c 667 = a*1000² + b*1000 + c 1414 = a*1800² + b*1800 + c You must now solve that system of 3 equations and find the three unknown parameters a, b and c. For c it's very easy, we immediately see that c = 256. Can you now find a, b and show me your work and result please ? Once you have found your equation, draw it with http://graphplotter.coursdemath.eu/ and check if you've got what you wanted. Kind regards. 
Auteur:  tsitrader [ 20 Aoû 2011, 20:15 ] 
Sujet du message:  Re: parabolic equations  Quadratic equations 
256 = a*0² + b*0 + c 667 = a*1000² + b*1000 + c 1414 = a*1800² + b*1800 + c c=256 *********************************** 667  256 = 1000000a + 1000b 408 = 1000(1000a + b) 667256 = 411 !! 408/1000 = 1000a + b .408  b = 1000a a = (.408  b) / 1000 b = 1000a + .408 *********************************** 1414 = 3240000a + 1800b +256 1158 = 1800(1800a + b) 1158/1800 = 1800a + b 0.6433  b = 1000a 0.6433  b = 1800a a = (.6433  b) / 1000 b = 1000a + .6433 *********************************** a = (.408  b) / 1000 = (.6433  b) / 1000 a = .408  b = .6433  b a = .408  .6433 = b + b a = (.2353) ************************************ b = 1000a + .408 = 1000a + .6433 b = .408  .6433 = 1000a + 1000a b = (.2353) ************************************ No way this is correct. a should not equal b but I do not know what I did wrong. 
Auteur:  Admin [ 21 Aoû 2011, 09:14 ] 
Sujet du message:  Re: parabolic equations  Quadratic equations 
Hi, I have put in red what was wrong in your calculation. And in green the right answer. I have sorted this out and I got this : c = 256 a = 0,0002904 b = 1,1661 And I get the following graph : http://graphplotter.coursdemath.eu/graph.php?a0=2&a1=0.0002904%2Ax%5E2%2B1%2C1661%2Ax%2B256&a2=&a3=&a4=3&a5=3&a6=7&a7=1&a8=1&a9=1&b0=500&b1=500&b2=0&b3=3000&b4=256&b5=3000&b6=10&b7=10&b8=5&b9=5&c0=3&c1=0&c2=1&c3=1&c4=1&c5=1&c6=1&c7=0&c8=0&c9=0&d0=1&d1=20&d2=20&d3=0&d4=&d5=&d6=&d7=&d8=&d9=&e0=&e1=&e2=&e3=&e4=14&e5=14&e6=13&e7=12&e8=0&e9=0&f0=0&f1=1&f2=1&f3=0&f4=0&f5=&f6=&f7=&f8=&f9=&g0=&g1=1&g2=&g3=0&g4=0&g5=0&zalt= I bet this is not exactly what you expected ! right ? Let us think about what happened. I am going to find other equations to force the parabola to be concave up ("smiling") and to pass by all the points we want it to go through. Give me a few days to think about that. Hint: to have a smiling parabola (concave up) we need a>0 or the second derivative of the parabola to be positive. Another problem is that the following point (1000 ; 667) is not on the parabola I drew. We might have made an error in our calculation. Let's check that too ! I'll come back to you later. 
Auteur:  Admin [ 22 Aoû 2011, 13:51 ] 
Sujet du message:  gold parabola 8212011 
This is the solution found by tsitrader: Pièce jointe: gold_parabola_8212011.png [ 47.75 Kio  Consulté 24674 fois ] 
Auteur:  Admin [ 25 Oct 2011, 22:39 ] 
Sujet du message:  Re: parabolic equations  Quadratic equations 
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