Use the term 1/(12p)in (11.5) to show that the error in Stirling’s formula (11.1) is < 10%for p > 1; < 1%for p > 10; < 0.1%for p > 100; < 0.01%for p > 1000.

Short Answer

Expert verified

If p > 1000 the error will be lesser than 0.01%.

Step by step solution

01

Definition of the logarithmic function

The Sterling formula is defined by n!~nne-n2πn.

02

Take the ratio of the first omitted term to the last omitted term

Truncate an infinite series to a finite number of terms.

Take the ratio of the first and last truncated term.

ppe-p2πp12pppe-p2πp=112p

03

Make comparisons in different domains and analyze

If p > 1, the ratio gives 110>112. The error becomes smaller than 0.1 and hence it is 10%.

Similarly, if p > 10 the error will be lesser than 1%.

If p > 100 the error will be lesser than 0.1%.

Similarly, if p > 1000 the error will be lesser than 0.01%.

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