In expansion of a function in half range Fourier cosine series, only a_n and a₀ appear. As function is considered to be an even function, b_n term becomes a null value as integral becomes zero. So, only b_n doesn?t appear in parseval?s relation of half range Fourier cosine series.

\(a_0 = \frac{1}{?} (\int_0^? x dx)(\int_?^{2?} (2?-x) dx) \)
\( = \frac{1}{?} [?^2+4?^2-4 \frac{?^2}{2}-2?^2+\frac{?^2}{2}] \)
\( = ?.\)
\(a_n = \frac{1}{?} (\int_0^? (xcos(nx)dx) )(\int_?^{2?}((2?-x)cos(nx))dx) \)
\( = \frac{-4}{?n^2} \) when n is odd and 0 when n is even
\( b_n = 0 \)
now, substituting x=0 in given function and Fourier series expansion, we get,
\(0 = \frac{?}{2}-\frac{4}{?} (\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} +….) \)
Therefore, \(\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} +………= \frac{?^2}{8} \).

The Parseval?s formula for half range Fourier cosine series is \(\int_0^l(f(x))^2 dx=\frac{l}{2} [\frac{a_0^2}{2}+?_{n=1}^?(a_n^2)] \)
The Parseval?s formula for half range Fourier sine series is \(\int_0^l(f(x))^2 dx=\frac{l}{2} ?_{n=1}^?(b_n^2) .\)

b_n = \(\frac{2}{3} \int_0^3 x^{2} sin\left(n\pi\frac{x}{3}\right)dx \)
= \(\frac{2}{3} \left(-27\frac{cos(n?)}{n?}\right) \)
= \( -18\frac{cos(n?)}{n?}.\)

Since function is odd function, odd times even function (cos(nx)) is odd function. So in given interval, coefficient is zero.

Using parseval?s relation for half range Fourier sine series,
\(\int_{-l}^l(f(x))^2 dx=l[\frac{a_0^2}{2}+?_{n=1}^?(a_n^2+b_n^2 ) ] \)
L.H.S. = \(\int_0^\pi(1)^2 dx = \pi \)
b_n = \(\frac{2}{\pi} \int_0^?1.sin(nx)dx \)
= \(\frac{2}{\pi n} (1-(-1)^n ) \)
Using parseval?s formula,
\(\pi = \frac{\pi}{2} * \frac{4}{\pi^2} * 2 (\frac{1}{1^2} +\frac{1}{3^2} +\frac{1}{5^2} +\frac{1}{7^2} +….) \)
Therefore,
\(\frac{1}{1^2} +\frac{1}{3^2} +\frac{1}{5^2} +\frac{1}{7^2} +….= \frac{\pi^2}{8}.\)

Since to find b_n we have sin(nx) and function we have is also odd function, product of odd function and anor odd function yields even function, only coefficient which exists is b_n.

In half range Fourier series expansion, nature of function in half of its period is only known. So when we find half range cosine series, re are only cosine terms which imply that function is even function. f(x) = f(-x).

When re is a point of discontinuity, value of function at that point is found by taking average of limit of function in left hand side of discontinuous point and right hand side of discontinuous point. Hence value of function at that point of discontinuity is \(\frac{1}{2} [f(x+0) + f(x-0)] \).

\(f(x) = \sqrt{\frac{1-cosx}{2}} = sin(\frac{x}{2}) \)
\(a_0 = \frac{1}{?} \int_0^{2?} sin(\frac{x}{2})dx \)
\( = \frac{1}{?} (-cos(\frac{x}{2}))^2 \)
\( = \frac{(-2)}{ ? (-2)} \)
\( = \frac{4}{?} \)