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 \(\displaystyle{f}{\left[{x}_{{i}},{x}_{{{i}+{1}}}\right]},\) where \(\displaystyle{f{{\left({x}\right)}}}={\sin{{x}}}\) and \(\displaystyle{x}_{{i}}={x}_{{{i}+{1}}}\) is set equal to \(\displaystyle{f}'{\left({x}_{{i}}\right)}.\)

\(\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\mathcal}\right\rbrace}{x}_{{i}}&{f}{\left[{x}_{{i}}\right]}&{f}'{\left[{x}_{{i}},{x}_{{{i}+{1}}}\right]}&&&&\backslash{h}{l}\in{e}{0.30}&{0.29552}\backslash&&{0.95534}\backslash&&&-{0.14200}\backslash&&&&-{1.0500}\backslash&&&&&{20.732}\backslash&&&&&&-{431.41}\backslash{0.30}&{0.29552}\backslash&&{0.95250}\backslash&&&-{0.16300}\backslash&&&&-{0.0134}\backslash&&&&&{0.83866}\backslash{0.32}&{0.31457}\backslash&&{0.94924}\backslash&&&-{0.16367}\backslash&&&&-{0.05533}\backslash{0.32}&{0.31457}\backslash&&{0.94433}\backslash&&&-{0.16533}\backslash{0.35}&{0.34290}\backslash&&{0.93937}\backslash{0.35}&{0.34290}\backslash{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\) It follows that Hermit interpolating polynomial is, \(\displaystyle{H}_{{5}}{\left({x}\right)}={0.29552}+{0.95532}{\left({x}-{0.30}\right)}-{0.142}{\left({x}-{0.30}\right)}^{{2}}{N}{S}{K}-{1.05}{\left({x}-{0.30}\right)}^{{2}}{\left({x}-{0.32}\right)}+{20.732}{\left({x}-{0.30}\right)}^{{2}}{\left({x}-{0.32}\right)}^{{2}}{N}{S}{K}-{431.41}{\left({x}-{0.30}\right)}^{{2}}{\left({x}-{0.32}\right)}^{{2}}{\left({x}-{0.35}\right)}\) Now, evaluating this polynomial at x = 0.34 using five-digit rounding. \(\displaystyle{H}_{{5}}{\left({0.34}\right)}={0.29552}+{0.95534}{\left({0.34}-{0.30}\right)}-{0.142}{\left({0.34}-{0.30}\right)}^{{2}}{N}{S}{K}-{1.05}{\left({0.34}-{0.30}\right)}^{{2}}{\left({0.34}-{0.32}\right)}+{20.732}{\left({0.34}-{0.30}\right)}^{{2}}{\left({0.34}-{0.32}\right)}^{{2}}{N}{S}{K}-{431.41}{\left({0.34}-{0.30}\right)}^{{2}}{\left({0.34}-{0.32}\right)}^{{2}}{\left({0.34}-{0.35}\right)}{N}{S}{K}{H}_{{5}}{\left({0.34}\right)}={0.33349}\)

\(\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\mathcal}\right\rbrace}{x}_{{i}}&{f}{\left[{x}_{{i}}\right]}&{f}'{\left[{x}_{{i}},{x}_{{{i}+{1}}}\right]}&&&&\backslash{h}{l}\in{e}{0.30}&{0.29552}\backslash&&{0.95534}\backslash&&&-{0.14200}\backslash&&&&-{1.0500}\backslash&&&&&{20.732}\backslash&&&&&&-{431.41}\backslash{0.30}&{0.29552}\backslash&&{0.95250}\backslash&&&-{0.16300}\backslash&&&&-{0.0134}\backslash&&&&&{0.83866}\backslash{0.32}&{0.31457}\backslash&&{0.94924}\backslash&&&-{0.16367}\backslash&&&&-{0.05533}\backslash{0.32}&{0.31457}\backslash&&{0.94433}\backslash&&&-{0.16533}\backslash{0.35}&{0.34290}\backslash&&{0.93937}\backslash{0.35}&{0.34290}\backslash{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}\) It follows that Hermit interpolating polynomial is, \(\displaystyle{H}_{{5}}{\left({x}\right)}={0.29552}+{0.95532}{\left({x}-{0.30}\right)}-{0.142}{\left({x}-{0.30}\right)}^{{2}}{N}{S}{K}-{1.05}{\left({x}-{0.30}\right)}^{{2}}{\left({x}-{0.32}\right)}+{20.732}{\left({x}-{0.30}\right)}^{{2}}{\left({x}-{0.32}\right)}^{{2}}{N}{S}{K}-{431.41}{\left({x}-{0.30}\right)}^{{2}}{\left({x}-{0.32}\right)}^{{2}}{\left({x}-{0.35}\right)}\) Now, evaluating this polynomial at x = 0.34 using five-digit rounding. \(\displaystyle{H}_{{5}}{\left({0.34}\right)}={0.29552}+{0.95534}{\left({0.34}-{0.30}\right)}-{0.142}{\left({0.34}-{0.30}\right)}^{{2}}{N}{S}{K}-{1.05}{\left({0.34}-{0.30}\right)}^{{2}}{\left({0.34}-{0.32}\right)}+{20.732}{\left({0.34}-{0.30}\right)}^{{2}}{\left({0.34}-{0.32}\right)}^{{2}}{N}{S}{K}-{431.41}{\left({0.34}-{0.30}\right)}^{{2}}{\left({0.34}-{0.32}\right)}^{{2}}{\left({0.34}-{0.35}\right)}{N}{S}{K}{H}_{{5}}{\left({0.34}\right)}={0.33349}\)