Answer to Question #111659 in Quantitative Methods for Anju Jayachandran

Question

Consider the following data
x 1.0 1.3 1.6 1.9 2.2
f(x) 0.76519 0.62008 0.45540 0.28181 0.11036
Use stirling’s formula to approximate f(1.5) with x0 = 1.6

Answer 1

We have $h=x_i-x_{i-1}=0.3$ and $s=-1/3,$ therefore, make a table with first to fourth divided differences. For example, calculate the first value in the First divided differences column (where $n=1$ is the number of the divided difference:

$$d_1=\frac{f(1.3)-f(1.0)}{h\cdot n}=\frac{0.62008 -0.76519 }{0.3\cdot1}=-0.48371.$$

Consequently, for the second, third and fourth divided difference $n=2,3,4.$

The table will look like this:

The table is adapted from Burden & Faires (2004)

We will use the underlined numbers in the Stirling's formula. The Stirling’s formula to approximate $f(1.5)$ with $x_0 = 1.6$ is

$$f(1.5)=\\=0.45540+\bigg(-\frac{1}{3}\bigg)\bigg(\frac{0.3}{2}\bigg)((-0.54895)+(-0.57861))+\\ +\bigg(-\frac{1}{3}\bigg)^2(0.3)^2(-0.04944)+\\ +\frac{1}{2}\bigg(-\frac{1}{3}\bigg)\Bigg(\bigg(-\frac{1}{3}\bigg)^2-1\Bigg)(0.3)^3(0.06588+0.06807)+\\ +\bigg(-\frac{1}{3}\bigg)^2\Bigg(\bigg(-\frac{1}{3}\bigg)^2-1\Bigg)(0.3)^4(0.00183)=\\ =0.51182.$$

References

R. Burden, J. Faires (2004). Numerical Analysis. 8 edition.