这是一份exeter埃克塞特大学MTH1002作业代写的成功案例
Now we determine the curves for which $L^{2}=4 \pi A$. To do this we find conditions for which $A$ is equal to the upper bound we obtained for it above. First note that
$$
\sum_{n=1}^{\infty} n\left(a_{n}^{2}+b_{n}^{2}+c_{n}^{2}+d_{n}^{2}\right)=\sum_{n=1}^{\infty} n^{2}\left(a_{n}^{2}+b_{n}^{2}+c_{n}^{2}+d_{n}^{2}\right)
$$
implies that all the coefficients except $a_{0}, c_{0}, a_{1}, b_{1}, c_{1}$ and $d_{1}$ are zero. The constraint,
$$
\pi \sum_{n=1}^{\infty} n\left(a_{n} d_{n}-b_{n} c_{n}\right)=\frac{\pi}{2} \sum_{n=1}^{\infty} n\left(a_{n}^{2}+b_{n}^{2}+c_{n}^{2}+d_{n}^{2}\right)
$$
then becomes
$$
a_{1} d_{1}-b_{1} c_{1}=a_{1}^{2}+b_{1}^{2}+c_{1}^{2}+d_{1}^{2}
$$
MTH1002/MTH1002-JD COURSE NOTES ：
The Fourier sine series has the form
$$
x(1-x)=\sum_{n=1}^{\infty} a_{n} \sin (n \pi x)
$$
The norm of the eigenfunctions is
$$
\int_{0}^{1} \sin ^{2}(n \pi x) d x=\frac{1}{2}
$$
The coefficients in the expansion are
$$
\begin{aligned}
a_{n} &=2 \int_{0}^{1} x(1-x) \sin (n \pi x) \mathrm{d} x \
&=\frac{2}{\pi^{3} n^{3}}(2-2 \cos (n \pi)-n \pi \sin (n \pi)) \
&=\frac{4}{\pi^{3} n^{3}}\left(1-(-1)^{n}\right)
\end{aligned}
$$