Solve 2*C*sin(W)+P*cos(N*W)=P for W

  • Context: Undergrad 
  • Thread starter Thread starter SSGD
  • Start date Start date
  • Tags Tags
    Trig functions
Click For Summary

Discussion Overview

The discussion revolves around solving the equation 2*C*sin(W) + P*cos(N*W) = P for the variable W, where C and P are constants and N is an integer. Participants explore the possibility of finding solutions in closed form, approximations, and numerical methods, considering various values of N.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant notes that a closed-form solution for W is not generally possible for arbitrary N, but special cases like N=1 may allow for solutions.
  • Another participant suggests that N=2, N=3, and N=4 might have exact solutions, but larger values of N likely do not.
  • A participant proposes using complex exponentials to convert the equation into a polynomial form, indicating that general exact solutions exist only up to the 4th order.
  • There is a correction regarding the sign in the equation, but another participant argues that the constants C and P can be arbitrary, making the sign change irrelevant.
  • One participant inquires about approximating a solution under the assumption that W is positive and near zero, expressing a desire to find the smallest positive solution.
  • Another participant suggests using Taylor series for approximation and mentions the trade-off between the quality of the approximation and the complexity of the calculations involved.
  • A participant shares their experience with both Taylor series and numerical solutions, noting challenges with achieving high accuracy and the tendency of numerical methods to converge to the trivial solution of W=0.
  • There is an acknowledgment that finding a general analytic solution is unlikely, especially for larger values of N, which would lead to complex polynomial equations.

Areas of Agreement / Disagreement

Participants generally agree that a general closed-form solution for W is not feasible, particularly for larger values of N. However, there are differing opinions on the potential for exact solutions in specific cases and the effectiveness of various approximation methods.

Contextual Notes

The discussion highlights the complexity of the equation and the limitations of finding solutions, particularly for high values of N, which may lead to large polynomial equations. The participants express uncertainty regarding the best methods for approximation and the implications of their assumptions.

SSGD
Messages
49
Reaction score
4
Is there a way to solve for W in the below equation. There has to be multiple solution for W, but I am at a loss as to how to solve this.

2*C*sin(W)+P*cos(N*W)=P or 2*C/P*sin(W)+cos(N*W)=1

C and P are constants
N is an integer
 
Physics news on Phys.org
Not in a closed form for general N. There are special cases where a solution is possible, e.g. N=1.

I would expect N=2 and maybe N=3 and N=4 to have exact solutions as well, but not larger N. Careful: I didn't check those cases in detail.

You can write sine and cosine as sum of two complex exponentials, and convert your expression to a polynomial (with powers of eiW). Those polynomial equations have proper, general exact solutions only up to 4th order.
 
Made an error... change the sign before the P from + to -

2*C*sin(W)-P*cos(N*W)=P or 2*C/P*sin(W)-cos(N*W)=1
 
That doesn't matter. C and P can be arbitrary constants anyway.
 
Would there be a way to approximate a solution if we assume W is positive and near zero but always greater than zero.

Because I can't solve it for all the solutions. Would there be a way to find the smallest positive solution.
 
Taylor series might work well. You can always find solutions to arbitrary precision with numerical methods. It depends on your problem then. Do you prefer an approximation that is not so good, but can be written down as formula, or an approximation that is much better, but needs dedicated calculation in each case?
 
  • Like
Likes   Reactions: SSGD
Yeah I just did both. Taylor Series gives me a number, but I was hoping to get several decimals of accuracy. But, the number of terms I will need is going to be... I think... Large.

The numerical solution gives the right askers, but It always wants to go to the Trivial Solution "0". With the constraints it worked.

Thanks for the help on this. I didn't think there was a way to get an general analytic solution. I would need to solve it for N's in the range of 20 to 100. So we are talking about some really large polynomials.

Again that's for the help.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K