Fitting a Cosine Function to Inflection & Known Derivative Point

Click For Summary
SUMMARY

This discussion focuses on fitting a cosine function to specific points, particularly an inflection point and a known derivative point. The cosine function is defined as y(x)=A.cos(B.x)+C, with its derivative y'(x)=B.A.sin(B.x). Given the conditions y(0)=Y0, y(X1)=Y1, and y'(X1)=DY1, the analytical solution for the parameters A, B, and C can be derived. The discussion concludes that while the solution may not be unique, using the inflection point allows for multiple values of B, which can subsequently determine A.

PREREQUISITES
  • Understanding of trigonometric functions, specifically cosine functions.
  • Knowledge of calculus, particularly derivatives and inflection points.
  • Familiarity with analytical methods for solving equations.
  • Basic algebra skills for manipulating equations and solving for variables.
NEXT STEPS
  • Research methods for solving nonlinear equations in calculus.
  • Explore the concept of inflection points in greater detail.
  • Study the application of trigonometric identities in solving equations.
  • Learn about numerical methods for finding roots of equations when analytical solutions are complex.
USEFUL FOR

Mathematicians, physics students, and engineers who are involved in modeling periodic phenomena or require fitting functions to data points.

adamharrybrow
Messages
1
Reaction score
0
I am trying to fit a cosine function to two points knowing that the first is an inflection point (e.g. a trough) and also knowing the gradient at the second. I have a gut feeling this has a unique solution it just needs the right identities and massaging but as of yet I haven't found the way:

Consider a cosine function:

y(x)=A.cos(B.x)+C

and derivative:

y'(x)=B.A.sin(B.x)

and given:

y(0)=Y0
y(X1)=Y1
y'(X1)=DY1

where X1, Y0,Y1 and DY1 are known constants

find the analytical solution for A,B and C

My boundless gratitude to anyone who can solve this.

Adam
 
Physics news on Phys.org


In general, I would expect that your solution is not unique.

Using the inflection point, you know where y''(x1)=0, this will give you several possible values for B. In addition, y(x1)=C fixes C.
For each value of B, you can then calculate the gradient at your second point and solve that for A.
 

Similar threads

Replies
17
Views
3K
Replies
3
Views
2K
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
7
Views
2K
  • · Replies 14 ·
Replies
14
Views
4K
Replies
5
Views
8K
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K