Question

Consider the following data
x f(x)
1.0 0.7651977
1.3 0.6200860
1.9 0.2818186
2.2 0.1103623
Use Stirling's formula to approximate f(1.5) with x0= 1.6

Answer 1

Answer on Question #75522 – Math – Quantitative Methods

Question

Consider the following data

x f(x)

1.0 0.7651977

1.3 0.6200860

1.6 0.4554022

1.9 0.2818186

2.2 0.1103623

Use Stirling's formula to approximate $f(1.5)$ with $x_0 = 1.6$

Solution

First divided differences

$$\frac{0.6200860 - 0.7651977}{1.3 - 1.0} = -0.4837057$$ $$\frac{0.4554022 - 0.6200860}{1.6 - 1.3} = -0.5489460$$ $$\frac{0.2818186 - 0.4554022}{1.9 - 1.6} = -0.5786120$$ $$\frac{0.1103623 - 0.2818186}{2.2 - 1.9} = -0.571521$$

Second divided differences

$$\frac{-0.5489460 - (-0.4837057)}{1.6 - 1.0} = -0.1087338$$ $$\frac{-0.5786120 - (-0.5489460)}{1.9 - 1.3} = -0.0494433$$ $$\frac{-0.0494433 - (-0.5786120)}{2.2 - 1.6} = 0.0118183$$

Third divided differences

$$\frac{-0.0494433 - (-0.1087338)}{1.9 - 1.0} = 0.0658783$$ $$\frac{0.0118183 - (-0.0494433)}{2.2 - 1.3} = 0.0680684$$

Fourth divided differences

$$\frac {0 . 0 6 8 0 6 8 4 - (0 . 0 6 5 8 7 8 3)}{2 . 2 - 1 . 0} = 0. 0 0 1 8 2 5 1$$

To apply Stirling's formula we use the underlined entries in the difference

The Stirling's formula, with $h = 0.3$ , $x_0 = 1.6$ and $s = -\frac{1}{3}$ , becomes

$$\begin{array}{l} f (1. 5) = 0. 4 5 5 4 0 2 2 + \left(- \frac {1}{3}\right) \left(\frac {0 . 3}{2}\right) \left((- 0. 5 4 8 9 4 6 0) + (- 0. 5 7 8 6 1 2 0)\right) + \left(- \frac {1}{3}\right) ^ {2} (0. 3) ^ {2} (- 0. 0 4 9 4 4 3 3) \\ + \frac {1}{2} \left(- \frac {1}{3}\right) \left(\left(- \frac {1}{3}\right) ^ {2} - 1\right) (0. 3) ^ {3} (0. 0 6 5 8 7 8 3 + 0. 0 6 8 0 6 8 4) \\ + \left(- \frac {1}{3}\right) ^ {2} \left(\left(- \frac {1}{3}\right) ^ {2} - 1\right) (0. 3) ^ {4} (0. 0 0 1 8 2 5 1) = 0. 5 1 1 8 2 0 0 \\ \end{array}$$

Answer: 0.5118200.

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