Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 17

With the standard unweighted inner product on real square-integrable functions, orthogonality on $[-L,L]$ meansMultiplying this inner product by a common nonzero normalising constant does not change orthogonality.

The Fourier coefficients follow by multiplying the expansion by $1$, by the appropriate cosine, or by the appropriate sine, then using orthogonality. In the convention with constant term $a_0/2$,The factor $1/L$ comes from the squared norms of the nonconstant sine and cosine functions: the integral of each square over $[-L,L]$ is $L$. The constant function has squared norm $2L$.

For $L=1$, the even function $h(x)=x^2$ has $b_n=0$, because $x^2\sin (n\pi x)$ is an odd function. Its constant coefficient is $a_0={\int }_{-1}^1{x}^{2}dx=2/3$. The remaining Fourier coefficients areTwo applications of integration by parts give the antiderivative, with $k=n\pi$,Its endpoint difference is $2(-1{)}^n/(n\pi {)}^2$, so $a_n=4(-1{)}^n/(n^2{\pi }^2)$. ThereforeThe periodic extension is continuous and piecewise smooth, so pointwise convergence of a piecewise smooth Fourier series gives equality, including at the endpoints.

Termwise differentiation of a Fourier series, applied to the continuous periodic extension in part (c), gives for $-1<x<1$Hence the required Fourier sine series isThe endpoint deserves care: at $x=1$ every term is zero, so the Fourier series sums to $0$, the average of the periodic one-sided limits $1$ and $-1$. It therefore does not reproduce the specified endpoint value $g(1)=1$. Changing a value at one point does not change any Fourier coefficient, so this is still the Fourier series of the stated function on $(-1,1]$.

Evaluate the Fourier sine series of part (d) at the interior point $x=1/2$. Only odd indices remain, with alternating signs:Thus $\boxed{A=4}$.

For the second identity, apply Parseval's identity to $h(x)=x^2$. From part (c), $a_0=2/3$, $a_n=4(-1{)}^n/(n^2{\pi }^2)$ and $b_n=0$, soThe left side is $2/5$, giving $\frac{16}{{\pi }^4}\sum n^{-4}=2/5-2/9=8/45$. ConsequentlyThe first series is conditionally convergent, while the second converges absolutely by the P-series criterion.