Consider the function f (same as in the previous problem) defined on the interval [0, 4) as follows, F(x) = { 2/2 x. x € [0,2]. 2, x € [2, 4]Find the coefficients Cn of the eigenfunction expansion of function ff(x) = Σ[infinity], n=1 cnyn(x), where y... for n = 1,2,3,... are the unit eigenfunctions of the Regular Sturm-Liouville system - y^n = ꟾλy, y’(O) = 0, y(4) = 0Note: Label your eigenfunctions so the eigenfunction for the lowest eigenvalue corresponds ton = 1. Therefore, use 2n – 1 instead of 2n +1.C= ___

To find the coefficients Cn of the eigenfunction expansion of a function f(x), f(x) must be expanded with the eigenfunction yn(x). The expansion of f(x) with respect to the eigenfunction yn(x) is given by

f(x) = Σ[∞], n=1 cnyn(x)

To find the coefficient cn, we need to compute the dot product of f(x) and yn(x).

cn = (f,yn) = ∫[0,4]f(x)yn(x)dx

Since the eigenfunctions yn(x) are orthonormal, the scalar product is given by

cn = ∫[0,4]f(x)yn(x)dx = ∫[0,2]f(x)yn(x)dx + ∫[2,4]f(x)yn(x)dx

Since f(x) = 2/2 x for x in [0,2] and f(x) = 2 for x in [2,4], compute the coefficient cn as I can do it.

cn = ∫[0,2](2/2x)yn(x)dx + ∫[2,4](2)yn(x)dx

= ∫[0,2]xyn(x)dx + ∫[2,4]2yn(x)dx

= 1/2 ∫[0,2] (xyn(x) + 2yn(x)) dx

= 1/2 ∫[0,2] (x+2)yn(x) dx

Therefore, the coefficient Cn is given by

Cn = 1/2 ∫[0,2] (x+2)yn(x) dx

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