Lennie Carroll

2021-01-19

Solve the given information Let $f\left(x\right)=\mathrm{sin}x.$ Use the following values and five-digit rounding arithmetic to construct the Hermite interpolating polynomial to approximate $f\left(0.34\right)=\mathrm{sin}0.34.$
$\begin{array}{ccc}x& f\left(x\right)& {f}^{\prime }\left(x\right)\\ 0.30& 0.29552& 0.95534\\ 0.32& 0.31457& 0.94924\\ 0.35& 0.34290& 0.93937\end{array}$

Bentley Leach

Consider the provided question, Given, $f\left(x\right)=\mathrm{sin}x$ We construct a divided- difference table, with each of the interpolation point included twice, that means, ${x}_{0}={x}_{1}=0.30,{x}_{2}={x}_{3}=0.32$ and ${x}_{4}={x}_{5}=0.35.$ each of the divided differences $f\left[{x}_{i},{x}_{i+1}\right],$ where $f\left(x\right)=\mathrm{sin}x$ and ${x}_{i}={x}_{i+1}$ is set equal to ${f}^{\prime }\left({x}_{i}\right).$
$\begin{array}{ccccccc}{x}_{i}& f\left[{x}_{i}\right]& {f}^{\prime }\left[{x}_{i},{x}_{i+1}\right]& & & & \\ 0.30& 0.29552\\ & & 0.95534\\ & & & -0.14200\\ & & & & -1.0500\\ & & & & & 20.732\\ & & & & & & -431.41\\ 0.30& 0.29552\\ & & 0.95250\\ & & & -0.16300\\ & & & & -0.0134\\ & & & & & 0.83866\\ 0.32& 0.31457\\ & & 0.94924\\ & & & -0.16367\\ & & & & -0.05533\\ 0.32& 0.31457\\ & & 0.94433\\ & & & -0.16533\\ 0.35& 0.34290\\ & & 0.93937\\ 0.35& 0.34290\end{array}$

It follows that Hermit interpolating polynomial is,

${H}_{5}\left(x\right)=0.29552+0.95532\left(x-0.30\right)-0.142\left(x-0.30{\right)}^{2}$

$-1.05\left(x-0.30{\right)}^{2}\left(x-0.32\right)+20.732\left(x-0.30{\right)}^{2}\left(x-0.32{\right)}^{2}$

$-431.41\left(x-0.30{\right)}^{2}\left(x-0.32{\right)}^{2}\left(x-0.35\right)$

Now, evaluating this polynomial at x = 0.34 using five-digit rounding. ${H}_{5}\left(0.34\right)=0.29552+0.95534\left(0.34-0.30\right)-0.142\left(0.34-0.30{\right)}^{2}$

$-1.05\left(0.34-0.30{\right)}^{2}\left(0.34-0.32\right)+20.732\left(0.34-0.30{\right)}^{2}\left(0.34-0.32{\right)}^{2}$

$-431.41\left(0.34-0.30{\right)}^{2}\left(0.34-0.32{\right)}^{2}\left(0.34-0.35\right)\phantom{\rule{0ex}{0ex}}{H}_{5}\left(0.34\right)=0.33349$

Jeffrey Jordon