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Mar 25, 2016 at 6:26 vote accept jlperla
Mar 25, 2016 at 6:26 comment added jlperla I was able to build a basis matrix, as described, given the standard recurrence formula $\int T_n(x) dx = \frac{1}{2}\left( \frac{T_{n+1}(x)}{n+1} - \frac{T_{n-1}(x)}{n-1} \right)$ with special cases for $n=0,1$ and with the affine transformation from my domain. Thanks for the suggestion.
Mar 20, 2016 at 14:06 comment added jlperla Thank you so much. So are you saying that I am overthinking this, and all I need to do is find a new basis matrix above for the integra like my $B$ and $B'$l, say $B^{int}$, where $B^{int}[i,j] \equiv \int_0^{z_i}T_j(z) dz$? Then my cumulative integrals is just $\vec{F} = B^{int} \cdot d$?
Mar 19, 2016 at 8:30 history answered GertVdE CC BY-SA 3.0