Let $x(t) = \cosh(t)$ and the basis $B = \{1,t,t^2, ...\}$ I know $c_k = \frac{\langle x,e_k \rangle}{\lVert e_k \rVert ^2} $ so I attempted to find the coefficients:
$c_0 = \frac{\langle x,e_0 \rangle}{\lVert e_0 \rVert ^2} = \frac{\langle \cosh(t), 1 \rangle}{\lVert 1 \rVert ^2} = \frac{\int_0^1\cosh(t)dt}{1} = sinh(1)$
$c_1 = \frac{\langle x,e_1 \rangle}{\lVert e_1 \rVert ^2} = \frac{\langle \cosh(t), t \rangle}{\lVert t \rVert ^2} = \frac{\int_0^1\cosh(t)tdt}{\lVert t \rVert ^2} = \frac{1-e^{-1}}{\lVert t \rVert ^2}$ What is $\lVert t \rVert ^2$? Is it just $t^2$ or do I leave it as $\lVert t \rVert ^2?$
$c_2 = \frac{\langle x,e_2 \rangle}{\lVert e_2 \rVert ^2} = \frac{\langle \cosh(t), t^2 \rangle}{\lVert t^2 \rVert ^2} = \frac{\int_0^1\cosh(t)t^2dt}{\lVert t^2 \rVert ^2} = \frac{e^2-5}{2e\lVert t^2 \rVert ^2}$ Again, what is ${\lVert t^2 \rVert ^2}$?
From here I just plug these coefficients into $\sum c_ke_k$ to get the terms. Is what I have done here correct?
One general approach would be to use Gram Schmidt orthogonalisation to the basis $t \mapsto t^k$. In this case, this would generate the shifted Legendre polynomials $P_k$ (look for shifted in http://mathworld.wolfram.com/LegendrePolynomial.html) and then one would find the Fourier coefficients with respect to this Schauder basis using the formula for $c_k$ above.
However, in this example, there is a simple approach using the fact that the Fourier coefficients are unique.
Since $\cosh t = 1 + {t^2 \over 2!} + {t^4 \over 4!}+ \cdots$, we can read off the coefficients as $c_0 = 1, c_1 = 0, c_2={1 \over 2!}$, etc.