Truncate - Maple Help

OrthogonalSeries

 Truncate
 truncate a series

 Calling Sequence Truncate(S, k) Truncate(S1, [k1,.., kn])

Parameters

 S, S1 - orthogonal series k, k1, .., kn - non-negative integers

Description

 • The Truncate(S, k) calling sequence truncates S after the kth coefficient.  S must have dimension 1.
 • The Truncate(S1, [k1,.., kn]) calling sequence truncates S1 by removing ith index coefficients after the kith index. S must have dimension $n$.

Examples

 > $\mathrm{with}\left(\mathrm{OrthogonalSeries}\right):$
 > $S≔\mathrm{Create}\left(\left\{\left[1,2\right],u\left(n\right),n=4..8\right\},\mathrm{GegenbauerC}\left(n,1,x\right)\right)$
 ${S}{≔}{\mathrm{GegenbauerC}}{}\left({0}{,}{1}{,}{x}\right){+}{2}{}{\mathrm{GegenbauerC}}{}\left({1}{,}{1}{,}{x}\right){+}\left({\sum }_{{n}{=}{4}}^{{8}}{}{u}{}\left({n}\right){}{\mathrm{GegenbauerC}}{}\left({n}{,}{1}{,}{x}\right)\right)$ (1)
 > $\mathrm{Truncate}\left(S,3\right);$$\mathrm{Truncate}\left(S,6\right);$$\mathrm{Truncate}\left(S,10\right)$
 ${\mathrm{GegenbauerC}}{}\left({0}{,}{1}{,}{x}\right){+}{2}{}{\mathrm{GegenbauerC}}{}\left({1}{,}{1}{,}{x}\right)$
 ${\mathrm{GegenbauerC}}{}\left({0}{,}{1}{,}{x}\right){+}{2}{}{\mathrm{GegenbauerC}}{}\left({1}{,}{1}{,}{x}\right){+}\left({\sum }_{{n}{=}{4}}^{{6}}{}{u}{}\left({n}\right){}{\mathrm{GegenbauerC}}{}\left({n}{,}{1}{,}{x}\right)\right)$
 ${\mathrm{GegenbauerC}}{}\left({0}{,}{1}{,}{x}\right){+}{2}{}{\mathrm{GegenbauerC}}{}\left({1}{,}{1}{,}{x}\right){+}\left({\sum }_{{n}{=}{4}}^{{8}}{}{u}{}\left({n}\right){}{\mathrm{GegenbauerC}}{}\left({n}{,}{1}{,}{x}\right)\right)$ (2)
 > $S≔\mathrm{Create}\left(\frac{n+1}{m+1},\mathrm{GegenbauerC}\left(n,1,x\right),\mathrm{LaguerreL}\left(m,1,y\right)\right)$
 ${S}{≔}{\sum }_{{m}{=}{0}}^{{\mathrm{\infty }}}{}\left({\sum }_{{n}{=}{0}}^{{\mathrm{\infty }}}{}\frac{\left({n}{+}{1}\right){}{\mathrm{GegenbauerC}}{}\left({n}{,}{1}{,}{x}\right)}{{m}{+}{1}}\right){}{\mathrm{LaguerreL}}{}\left({m}{,}{1}{,}{y}\right)$ (3)
 > $\mathrm{Truncate}\left(S,\left[2,2\right]\right)$
 ${\mathrm{GegenbauerC}}{}\left({0}{,}{1}{,}{x}\right){}{\mathrm{LaguerreL}}{}\left({0}{,}{1}{,}{y}\right){+}\frac{{\mathrm{GegenbauerC}}{}\left({0}{,}{1}{,}{x}\right){}{\mathrm{LaguerreL}}{}\left({1}{,}{1}{,}{y}\right)}{{2}}{+}\frac{{\mathrm{GegenbauerC}}{}\left({0}{,}{1}{,}{x}\right){}{\mathrm{LaguerreL}}{}\left({2}{,}{1}{,}{y}\right)}{{3}}{+}{2}{}{\mathrm{GegenbauerC}}{}\left({1}{,}{1}{,}{x}\right){}{\mathrm{LaguerreL}}{}\left({0}{,}{1}{,}{y}\right){+}{\mathrm{GegenbauerC}}{}\left({1}{,}{1}{,}{x}\right){}{\mathrm{LaguerreL}}{}\left({1}{,}{1}{,}{y}\right){+}\frac{{2}{}{\mathrm{GegenbauerC}}{}\left({1}{,}{1}{,}{x}\right){}{\mathrm{LaguerreL}}{}\left({2}{,}{1}{,}{y}\right)}{{3}}{+}{3}{}{\mathrm{GegenbauerC}}{}\left({2}{,}{1}{,}{x}\right){}{\mathrm{LaguerreL}}{}\left({0}{,}{1}{,}{y}\right){+}\frac{{3}{}{\mathrm{GegenbauerC}}{}\left({2}{,}{1}{,}{x}\right){}{\mathrm{LaguerreL}}{}\left({1}{,}{1}{,}{y}\right)}{{2}}{+}{\mathrm{GegenbauerC}}{}\left({2}{,}{1}{,}{x}\right){}{\mathrm{LaguerreL}}{}\left({2}{,}{1}{,}{y}\right)$ (4)