Equation Fitting: Seeking Insight to Meet Requirements

  • Context: Graduate 
  • Thread starter Thread starter CRGreathouse
  • Start date Start date
  • Tags Tags
    Fitting
Click For Summary
SUMMARY

The discussion centers on fitting equations to data with specific requirements, including that the first derivative f'(x) must be positive for all x > 0, the second derivative f''(x) must be negative for all x > 0, and the function must approach a bounded limit as x approaches infinity. A proposed equation that meets these criteria is f(x) = k(1 - e^{-ax}), where k > 0 and a > 0, which ensures the necessary properties of increasing cost and decreasing average cost. The user expresses gratitude for the suggestion and plans to further explore this equation.

PREREQUISITES
  • Understanding of calculus, specifically derivatives and limits
  • Familiarity with exponential functions and their properties
  • Knowledge of equation fitting techniques
  • Basic proficiency in mathematical modeling
NEXT STEPS
  • Explore the properties of exponential functions in mathematical modeling
  • Research techniques for fitting equations to data, including least-squares fitting
  • Investigate other models that exhibit similar characteristics, such as logistic growth models
  • Learn about stability analysis in mathematical equations
USEFUL FOR

Mathematicians, data analysts, and anyone involved in mathematical modeling or equation fitting who seeks to understand how to create equations that meet specific criteria for data representation.

CRGreathouse
Science Advisor
Homework Helper
Messages
2,832
Reaction score
0
I'm almost too embarrassed to post, but I thought someone might have insight that could help me here. I'm trying to fit an equation to data, but I'm just not sure what sort of equation I should use. I have some requirements on the form of the equation, and then I have points (yet to be determined, actually).

[tex]f'(x)>0\,\,\forall x>0[/tex] (cost is increasing)
[tex]f''(x)<0\,\,\forall x>0[/tex] (essentially, average cost is decreasing)
[tex]\lim_{x\rightarrow\infty} f(x)=k[/tex] (price is bounded above in this fashion)

It would make the most sense if the origin was included, but this can be one of the data points.

So before I even think about least-squares vs. minimal points on the curve, I wanted to consider different forms that could make sense. The first that comes to mind was the elementary

[tex]y=a-\frac{ac}{x+c}[/tex]

but this has some instabilities and oddities. I guess I'm just looking for thoughts on widely-used models that fit my criteria or could be modified to fit them. In the perfect case I'd have a smooth equation that was nearly linear for small x and essentially hyperbolic for large x.

Any suggestions would be welcomed.
 
Mathematics news on Phys.org
CRGreathouse said:
I'm almost too embarrassed to post, but I thought someone might have insight that could help me here. I'm trying to fit an equation to data, but I'm just not sure what sort of equation I should use. I have some requirements on the form of the equation, and then I have points (yet to be determined, actually).

[tex]f'(x)>0\,\,\forall x>0[/tex] (cost is increasing)
[tex]f''(x)<0\,\,\forall x>0[/tex] (essentially, average cost is decreasing)
[tex]\lim_{x\rightarrow\infty} f(x)=k[/tex] (price is bounded above in this fashion)

It would make the most sense if the origin was included, but this can be one of the data points.

So before I even think about least-squares vs. minimal points on the curve, I wanted to consider different forms that could make sense. The first that comes to mind was the elementary

[tex]y=a-\frac{ac}{x+c}[/tex]

but this has some instabilities and oddities. I guess I'm just looking for thoughts on widely-used models that fit my criteria or could be modified to fit them. In the perfect case I'd have a smooth equation that was nearly linear for small x and essentially hyperbolic for large x.

Any suggestions would be welcomed.
One equation that fits your requirements is:
[tex]f(x) = k(1 - e^{-ax})[/tex] k > 0 a > 0
Then
[tex]f´(x) = ak e^{-ax} > 0[/tex]
[tex]f"(x) = -a^2k e^{-ax} < 0[/tex]
 
SGT said:
One equation that fits your requirements is:
[tex]f(x) = k(1 - e^{-ax})[/tex] k > 0 a > 0
Then
[tex]f´(x) = ak e^{-ax} > 0[/tex]
[tex]f"(x) = -a^2k e^{-ax} < 0[/tex]

Thanks. You know, I was just so out of it I didn't think to use this. :blushing:

I'm going to play with this for a while and make it work. You gave me just what I needed to get my brain working again.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 19 ·
Replies
19
Views
7K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
6K