Question

Consider the data of \( f(x) \) given in the table. \[ \begin{array}{|c|c|c|c|} \hline i & 0 & 1 & 2
\hline x_i & 1 & 2 & 3
\hline f(x_i) & 0 & 0.3010 & 0.4771
\hline \end{array} \] The value of \( f(1.5) \) estimated using second-order Newton's interpolation formula is \_\_\_\_\_\_\_\_ \textit{(rounded off to 2 decimal places).}

Hint:

Newton's interpolation formula is useful for estimating values within the given data range using divided differences.

Answer 1

Step 1: Given data points: \[ x_0 = 1, \quad x_1 = 2, \quad x_2 = 3 \] \[ f(x_0) = 0, \quad f(x_1) = 0.3010, \quad f(x_2) = 0.4771 \] Step 2: Compute divided differences: \[ f[x_0, x_1] = \frac{0.3010 - 0}{2 - 1} = 0.3010 \] \[ f[x_1, x_2] = \frac{0.4771 - 0.3010}{3 - 2} = 0.1761 \] \[ f[x_0, x_1, x_2] = \frac{0.1761 - 0.3010}{3 - 1} = -0.06245 \] Step 3: Newton's interpolation formula: \[ f(x) = f(x_0) + f[x_0, x_1](x - x_0) + f[x_0, x_1, x_2](x - x_0)(x - x_1) \] Substituting the values: \[ f(1.5) = 0 + (0.3010)(1.5 - 1) + (-0.06245)(1.5 - 1)(1.5 - 2) \] \[ = 0.3010(0.5) + (-0.06245)(0.5)(-0.5) \] \[ = 0.1505 + 0.0156 = 0.1661 \] Step 4: Final Answer (Rounded to 2 decimal places): \[ \approx 0.17 \] Conclusion: The estimated value of \( f(1.5) \) lies in the range \( 0.16 \) to \( 0.18 \).