Constraining the parameter space of a function

In summary, the function ##f(x) = - \frac{1}{2}a^{2}x^{2} + \frac{1}{4}bx^{4} + d^{4}\cos(x/c)## has regions in its parameter space where there are oscillations. These regions are found when ##c \sim a##, regardless of the values of ##b## and ##d##. The derivative of the function, ##f'(x) = - ax + bx^{3} - \frac{d^{4}}{c}\sin(x/c)##, can help identify these regions by finding the solutions to the equation ##bcx^{3} - acx = d^{4}\sin(x
  • #1
spaghetti3451
1,344
34
Consider the function

$$f(x) = - \frac{1}{2}a^{2}x^{2} + \frac{1}{4}bx^{4} + d^{4}\cos(x/c),$$

where ##a##, ##b##, ##c## and ##d## are arbitrary parameters.

For some region(s) of the parameter space, there are oscillations in the function. My goal is to identify these regions of the parameter space.

Here are three examples:

faaa55451b.png
faa9ea7e73.png
faaacd391c.png
fac54bca94.png
Does this mean that there are bumps in the function only for ##c \sim a## (and any arbitrary value of ##b## and ##d##)?
 
Mathematics news on Phys.org
  • #2
Did you calculate the derivative?
What does it tell you?
 
  • #3
I find that

$$f'(x) = - ax + bx^{3} - \frac{d^{4}}{c}\sin(x/c).$$

Therefore, the minima are given by the solutions to the equation

$$ bcx^{3} - acx = d^{4}\sin(x/c).$$

Finding a closed-form solution to this equation is a big ask, I suppose. But the question is really about the number of solutions to this equation, and not the solutions themselves.
 
  • #4
There is no closed-form solution, but you know that the right side is limited by ##\pm d^4##.
 
  • #5
mfb said:
There is no closed-form solution, but you know that the right side is limited by ##\pm d^4##.

So, ##bcx^{3} - acx = 0## is a cubic equation with solutions at ##x = 0, \pm \sqrt{\frac{a}{b}}##.

And, ##d^{4}\sin(x/c)## is a sinuosidal function with amplitude ##d^{4}## and period ##2\pi c##.

How can I set up constraints now?
 
  • #6
You can consider what happens for small d or large c, for example. In which region can it be relevant for the full equation?
How often does the sine term change its sign in this region?
 

Related to Constraining the parameter space of a function

1. What does it mean to constrain the parameter space of a function?

Constraining the parameter space of a function involves limiting the possible values of the parameters that can be used in a mathematical function. This is done in order to narrow down the range of possible solutions and make the function more useful and relevant in a specific context.

2. Why is constraining the parameter space important?

Constraining the parameter space is important because it helps eliminate irrelevant solutions and allows for a more precise and accurate analysis of the function. It also helps in making the function more applicable to real-world situations and can improve its predictive power.

3. How do you determine the appropriate constraints for a function's parameter space?

The appropriate constraints for a function's parameter space can be determined through a combination of mathematical analysis and practical considerations. This may involve considering the physical limitations of the system, the expected range of values for the parameters, and any known relationships between the parameters.

4. Can constraining the parameter space affect the accuracy of a function?

Yes, constraining the parameter space can affect the accuracy of a function. By limiting the range of possible values for the parameters, the function may not be able to accurately represent all scenarios. However, in most cases, constraining the parameter space can actually improve the accuracy of a function by eliminating unrealistic or irrelevant solutions.

5. Are there any drawbacks to constraining the parameter space of a function?

One potential drawback of constraining the parameter space of a function is that it may limit the flexibility of the function and make it less applicable to a wider range of situations. Additionally, if the constraints are not carefully chosen, it could lead to inaccurate or biased results. It is important to carefully consider the purpose and scope of the function when constraining its parameter space.

Similar threads

Replies
2
Views
1K
Replies
2
Views
1K
Replies
3
Views
1K
Replies
6
Views
3K
Replies
4
Views
966
Replies
7
Views
959
Replies
2
Views
973
Replies
6
Views
1K
Replies
7
Views
1K
Replies
9
Views
2K
Back
Top