# Find the coefficient of an equation?

Posted 1 year ago
1407 Views
|
4 Replies
|
2 Total Likes
|
 I have an equation which involves a lot of trigonometric functions. I want to find the coefficients of function Y (also include trigonometric functions) and without Y. I have tried Coefficient command but could not get the desired result because this command is only for the polynomial. So, which command I need to use to find the results. Anyone can help me it will be very appreciable. Attachments:
Answer
4 Replies
Sort By:
Posted 1 year ago
 What is wrong with the results you found? They seem correct to me. Here is a variation: expr = 1/( 8 Bi (1 + k)^2 S^2 (Bi + Bi k - 4 k S^2)) (-2 Bi (1 + k) (-(1 + k) (Bi - 4 S^2) Cosh[2 S (-1 + Y)] + Cosh[2 S] (Bi + Bi k - 4 k S^2 - 4 S^2 Cos[(Sqrt[Bi] Sqrt[-1 - k] (-1 + Y))/Sqrt[k]] Sec[( Sqrt[Bi] Sqrt[-1 - k])/Sqrt[k]])) Subscript[A, 2] + 1/(Bi + Bi k - k S^2) 8 Sqrt[ Bi] (Bi + Bi k - 4 k S^2) (Sqrt[Bi] (1 + k) S (Bi + Bi k - k S^2) Y + Sqrt[-1 - k] Sqrt[k] S^3 Sec[(Sqrt[Bi] Sqrt[-1 - k])/Sqrt[k]] Sin[( Sqrt[Bi] Sqrt[-1 - k] Y)/Sqrt[k]] + Sqrt[Bi] (1 + k) ((-Bi (1 + k) + k S^2 + S^2 Cos[(Sqrt[Bi] Sqrt[-1 - k] (-1 + Y))/Sqrt[k]] Sec[( Sqrt[Bi] Sqrt[-1 - k])/Sqrt[k]]) Sinh[ S] + (1 + k) (Bi - S^2) Sinh[S - S Y])) Subscript[A, 3]); vars = {Y, Cosh[2 S (-1 + Y)], Cos[(Sqrt[Bi] Sqrt[-1 - k] (-1 + Y))/Sqrt[k]], Sin[(Sqrt[Bi] Sqrt[-1 - k] Y)/Sqrt[k]], Sinh[S - S Y]}; coeff = Coefficient[expr, vars, 1] // Simplify withoutY = expr - coeff.vars // Simplify coeff.vars + withoutY - expr // Simplify Collect[expr, vars, Simplify] 
Answer
Posted 1 year ago
 Thanks, for your help. I got my results. When I solved for 1, it gives me the coefficients of all the functions of Y but without Y it does not give me the same result as yours. I directly used the coefficient command and find for 0 (without Y), it gives me the very complicated answers involving all the functions of Y. See the attached file. Attachments:
Answer
Posted 1 year ago
 You are using Coefficient with a list as second argument. This usage does not seem to be documented. I don't understand what you are trying to do. However, you can turn your expression into a polynomial with a replacement rule: expr = 1/( 8 Bi (1 + k)^2 S^2 (Bi + Bi k - 4 k S^2)) (-2 Bi (1 + k) (-(1 + k) (Bi - 4 S^2) Cosh[2 S (-1 + Y)] + Cosh[2 S] (Bi + Bi k - 4 k S^2 - 4 S^2 Cos[(Sqrt[Bi] Sqrt[-1 - k] (-1 + Y))/Sqrt[k]] Sec[( Sqrt[Bi] Sqrt[-1 - k])/Sqrt[k]])) Subscript[A, 2] + 1/(Bi + Bi k - k S^2) 8 Sqrt[ Bi] (Bi + Bi k - 4 k S^2) (Sqrt[Bi] (1 + k) S (Bi + Bi k - k S^2) Y + Sqrt[-1 - k] Sqrt[k] S^3 Sec[(Sqrt[Bi] Sqrt[-1 - k])/Sqrt[k]] Sin[( Sqrt[Bi] Sqrt[-1 - k] Y)/Sqrt[k]] + Sqrt[Bi] (1 + k) ((-Bi (1 + k) + k S^2 + S^2 Cos[(Sqrt[Bi] Sqrt[-1 - k] (-1 + Y))/Sqrt[k]] Sec[( Sqrt[Bi] Sqrt[-1 - k])/Sqrt[k]]) Sinh[ S] + (1 + k) (Bi - S^2) Sinh[S - S Y])) Subscript[A, 3]); vars = {Y, Cosh[2 S (-1 + Y)], Cos[(Sqrt[Bi] Sqrt[-1 - k] (-1 + Y))/Sqrt[k]], Sin[(Sqrt[Bi] Sqrt[-1 - k] Y)/Sqrt[k]], Sinh[S - S Y]}; Collect[expr /. Thread[vars -> {a, b, c, d, e}], {a, b, c, d, e}] 
Answer
Posted 1 year ago
 Thank you for your help. I understand I was solving without converting into a polynomial.
Answer
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Reply Preview
Attachments