# Maths

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 Sujet du message: parabolic equations - Quadratic equationsPublié: 20 Aoû 2011, 18:33

Inscrit le: 20 Aoû 2011, 18:30
Messages: 3
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,

Haut

 Sujet du message: How far gold's path strays from it's mathematically derived Publié: 20 Aoû 2011, 18:38

Inscrit le: 20 Aoû 2011, 18:30
Messages: 3
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 24721 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.

Haut

 Sujet du message: Re: parabolic equations - Quadratic equationsPublié: 20 Aoû 2011, 18:54

Inscrit le: 19 Mar 2009, 20:46
Messages: 146
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.

Haut

 Sujet du message: Re: parabolic equations - Quadratic equationsPublié: 20 Aoû 2011, 20:15

Inscrit le: 20 Aoû 2011, 18:30
Messages: 3
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.

a should not equal b

but I do not know what I did wrong.

Haut

 Sujet du message: Re: parabolic equations - Quadratic equationsPublié: 21 Aoû 2011, 09:14

Inscrit le: 19 Mar 2009, 20:46
Messages: 146
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.

Haut

 Sujet du message: gold parabola 8-21-2011Publié: 22 Aoû 2011, 13:51

Inscrit le: 19 Mar 2009, 20:46
Messages: 146
This is the solution found by tsitrader:

Pièce jointe:

gold_parabola_8-21-2011.png [ 47.75 Kio | Consulté 24700 fois ]

Haut

 Sujet du message: Re: parabolic equations - Quadratic equationsPublié: 25 Oct 2011, 22:39

Inscrit le: 19 Mar 2009, 20:46
Messages: 146

Cours de l'or entre 1960 et 2011

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