Q6Advanced Engineering Mathematics

When tasked with interpolating a value that lies firmly in the middle region of a tabulated dataset (rather than at the beginning or the extreme end), Newton's Forward and Backward formulas become less accurate due to truncation of higher-order differences. Central difference formulas, such as Stirling's formula (which is the mathematical arithmetic mean of Gauss's Forward and Backward formulas), provide significantly superior accuracy by symmetrically utilizing data points radiating outward from the center.

1. Structuring the Central Difference Table

We are provided with data at uniform intervals $x=\{20,25,30,35,40,45\}$ (step size $h=5$). We seek the value at $x=32$. We logically designate the closest tabulated $x$-value as the central origin, $x_0=30$. The corresponding index parameter is $p=\frac{x-x_0}{h}=\frac{32-30}{5}=0.4$.

We systematically construct the central difference table, meticulously labeling the entries relative to the chosen $x_0$ origin.

• $y_{-2}(20)=14.035,y_{-1}(25)=13.674,y_0(30)=13.257,y_1(35)=12.734,y_2(40)=12.089$
• First Differences: $\Delta y_{-2}=-0.361,\Delta y_{-1}=-0.417,\Delta y_0=-0.523,\Delta y_1=-0.645$
• Second Differences: ${\Delta }^2{y}_{-2}=-0.056,{\Delta }^2{y}_{-1}=-0.106,{\Delta }^2{y}_0=-0.122$
• Third Differences: ${\Delta }^3{y}_{-2}=-0.050,{\Delta }^3{y}_{-1}=-0.016$
• Fourth Difference: ${\Delta }^4{y}_{-2}=0.034$

2. Applying Stirling's Formula

Stirling's mathematical interpolation formula utilizes the central differences located symmetrically horizontally across the $y_0$ row of the table:

We carefully substitute the precisely extracted values from our table and the parameter $p=0.4$ into the formula:

The rigorous mathematical calculation confirms that the highly accurate interpolated value at $x=32$ is approximately $13.062$.