Question 11.2.3: Expand the function f(x) = x, 0 ≤ x ≤ 1 (35) in terms of the......

In Example 2, where r(x) = 1, we found the normalized eigenfunctions to be

\phi_{n}(x)=k_{n}\sin({\sqrt{\lambda_{n}}} x), (36)

where k_{n} is given by equation (28)

k_{n}=\left(\frac{2}{1+\cos^{2}\sqrt{\lambda_{n}}}\right)^{1/2}. (28)

and \lambda_{n} satisfies equation (26).

\sin{\sqrt{\lambda_{n}}}+{\sqrt{\lambda_{n}}}\cos{\sqrt{\lambda_{n}}}=0 (26)

To find the expansion for f in terms of the eigenfunctions \phi_{n}, we write

f(x)=\sum_{n=1}^{\infty}c_{n}\phi_{n}(x), (37)

where the coefficients are given by equation (34).

c_{m}=\int_{0}^{1}r(x)\,f(x)\phi_{m}(x)d x=(f,r\phi_{m}),\quad m=1,2,\,\dots\,. (34)

Thus

c_{n}=\int_{0}^{1}r(x)f(x)\phi_{n}(x)d x=k_{n}\int_{0}^{1}x\sin\biggl(\sqrt{\lambda_{n}}x\biggr)d x.

Integrating by parts, we obtain

c_{n}=k_{n}\left(\frac{\sin\sqrt{\lambda_{n}}}{\lambda_{n}}-\frac{\cos\sqrt{\lambda_{n}}}{\sqrt{\lambda_{n}}}\right)=k_{n}\,\frac{2\sin\sqrt{\lambda_{n}}}{\lambda_{n}},

where we have used equation (26) in the last step. Upon substituting for k_{n} from equation (28), we obtain

c_{n}=\frac{2\sqrt{2}\sin\sqrt{\lambda_{n}}}{\lambda_{n}\Big(1+\cos^{2}\sqrt{\lambda_{n}}\Big)^{1/2}}. (38)

Thus

f(x)=4\sum_{n=1}^{\infty}{\frac{\sin({\sqrt{\lambda_{n}})}}{\lambda_{n}{\Big(}1+\cos^{2}{\sqrt{\lambda_{n}}}\Big)}}\sin({\sqrt{\lambda_{n}}}x). (39)

In Example 1 of Section 10.4, we found the Fourier sine series for f(x) = x (with L = 1) to be

f(x)={\frac{2}{\pi}}\sum_{n=1}^{\infty}{\frac{(-1)^{n+1}}{n}}\sin(n\pi x).

While equation (39) is a series of sine functions, it is not a Fourier sine series. This means its convergence cannot be ascertained from the Fourier Convergence Theorem (Theorem 10.3.1). But from Theorem 11.2.4, we conclude that series (39) converges pointwise to f(x) = x for 0 ≤ x ≤ 1.