Question #111659
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
1
Expert's answer
2020-04-24T18:04:14-0400

We have h=xixi1=0.3h=x_i-x_{i-1}=0.3 and s=1/3,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=1n=1 is the number of the divided difference:


d1=f(1.3)f(1.0)hn=0.620080.765190.31=0.48371.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.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)f(1.5) with x0=1.6x_0 = 1.6 is


f(1.5)==0.45540+(13)(0.32)((0.54895)+(0.57861))++(13)2(0.3)2(0.04944)++12(13)((13)21)(0.3)3(0.06588+0.06807)++(13)2((13)21)(0.3)4(0.00183)==0.51182.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.



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