In the figure below, expansion $\frac{de{l}^{2}T}{del{x}^2}=-1$ when $x=0\mathrm{.}5m$ and $x=0\mathrm{.}1m$ and when $T_1=0,T_6=1\mathrm{.}0,$ and obtain it by Thomas method.

$($Thomas method$)$

$x_n={}_n$

$x_i={}_i-\frac{e_i{x}_{i+1}}{{}_i}(i=n-1,n-2,,$dots,$1)$

${}_1=d_1$

${}_1=\frac{b_1}{{}_1}$

${}_i=d_i-\frac{c_i{e}_{i-1}}{{}_{i-1}}(i=2,3,$dots,$n)$

${}_i=\frac{b_i-c_i{}_{i-1}}{{}_i}(i=2,3,$dots,$n)$