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parabolic equations - Quadratic equations
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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
gold_parabola.png [ 103.77 Kio | Consulté 23029 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 near-vertical 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 ... d-midrange ) 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 y-axis, 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://graph-plotter.cours-de-math.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)

667-256 = 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. :evil:

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://graph-plotter.cours-de-math.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 8-21-2011

This is the solution found by tsitrader:

Pièce jointe:
gold_parabola_8-21-2011.png
gold_parabola_8-21-2011.png [ 47.75 Kio | Consulté 23008 fois ]

Auteur:  Admin [ 25 Oct 2011, 22:39 ]
Sujet du message:  Re: parabolic equations - Quadratic equations

Image

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