Solve the given information Let f(x)=sin⁡x. Use the following values and five-digit rounding arithmetic to...

Consider the provided question, Given, ${\displaystyle f\left(x\right)=\sin x}$ We construct a divided- difference table, with each of the interpolation point included twice, that means, ${\displaystyle x_0=x_1=0.30,x_2=x_3=0.32}$ and ${\displaystyle x_4=x_5=0.35.}$ each of the divided differences ${\displaystyle f[x_i,x_{i+1}],}$ where ${\displaystyle f\left(x\right)=\sin x}$ and ${\displaystyle x_i=x_{i+1}}$ is set equal to ${\displaystyle f^{\prime }\left(x_i\right).}$
$\begin{array}{ccccccc}{x}_i& f[x_i]& f^{\prime }[x_i,x_{i+1}]& & & & \\ 0.30& 0.29552\\ & & 0.95534\\ & & & -0.14200\\ & & & & -1.0500\\ & & & & & 20.732\\ & & & & & & -431.41\\ 0.30& 0.29552\\ & & 0.95250\\ & & & -0.16300\\ & & & & -0.0134\\ & & & & & 0.83866\\ 0.32& 0.31457\\ & & 0.94924\\ & & & -0.16367\\ & & & & -0.05533\\ 0.32& 0.31457\\ & & 0.94433\\ & & & -0.16533\\ 0.35& 0.34290\\ & & 0.93937\\ 0.35& 0.34290\end{array}$

It follows that Hermit interpolating polynomial is,

$H_5(x)=0.29552+0.95532(x-0.30)-0.142(x-0.30{)}^2$

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

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

Now, evaluating this polynomial at x = 0.34 using five-digit rounding. $H_5(0.34)=0.29552+0.95534(0.34-0.30)-0.142(0.34-0.30{)}^2$

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

$$\begin{gathered} -431.41(0.34-0.30{)}^2(0.34-0.32{)}^2(0.34-0.35) \\ {H}_5(0.34)=0.33349 \end{gathered}$$

Answer is given below (on video)