What methods can be used for finding a quadratic curve?

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Discussion Overview

The discussion revolves around methods for finding a quadratic curve that fits given data points, specifically focusing on extrapolating a quadratic curve between two known x values, along with their corresponding y values and derivatives. The scope includes mathematical reasoning and curve fitting techniques.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant expresses uncertainty about the methods available for extrapolating a quadratic curve and seeks guidance on where to start.
  • Another participant suggests setting up a system of linear equations based on the general form of a quadratic equation, providing an example with specific points.
  • A later reply indicates that the initial participant successfully implemented the suggested method.
  • Additional methods for curve fitting are proposed, including various types of functions such as linear, polynomial, exponential, logarithmic, and Gaussian functions.

Areas of Agreement / Disagreement

Participants do not reach a consensus on a single method for finding a quadratic curve, as multiple approaches and functions are discussed. The discussion remains open with various proposed methods.

Contextual Notes

Some methods mentioned may depend on the specific characteristics of the data points and the desired accuracy of the fit. The discussion does not resolve which method is most appropriate for the initial participant's specific case.

Bob Smith
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I'll be honest, I don't understand most of what's going on in this forum, so forgive me if this isn't the right place.

I'm trying to extrapolate a quadratic curve between two points (my x values). These are known, as are the corresponding y values and the y' (or dy/dx if you prefer) values.

Much as I like working things out on my own, I'm not sure where to start on this. Other than trial and error, what methods are there for working this out?

It's been a couple of years since I studied maths properly so I'm kinda rusty to say the least.

Cheers


Bob
 
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Set up a system of linear equations. In general a quadratic relation has the form

y = Ax^2 + Bx + C

Let's say I want the quadratic to go through three points, (2, 1), (3,1), and (0, 2). Just plug the three points into the equation:

1 = 4A + 2B + C
1 = 9A + 3B + C
2 = C

and solve. In your case it will be a little different because you will be using the derivative as well, but the principle is precisely the same.
 
Last edited:
Ahh excellent thanks, got it working now. :smile:
 
"Curve fitting". You can also try many other functions like:

y = A*x (Proportional)
y = m*x + b (Linear)
y = a*x^2 + b*x + c (Quadratic)
y = A + B*x + C*x^2 + D*x^3 + .. (Polynomial)
y = A*x^B (Power)
y = A/x (Inverse)
y = A*exp(-C*x) + B (Natural Exponent)
y = A*ln(B*x) (Natural Logarithm)
y = A*10^(B*x) (Base-10 Exponent)
y = A*sin(B*x+C)+D (Sine)
y = 1/(sqrt(2*3.14159)*S)*exp(-(x-M)^2/(2*S^2)) (Normalized Gaussian)
y = A*exp(-(x-B)^2/(C^2))+D (Gaussian)
 

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