Stirling's approximation: Difference between revisions
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James Stirling (1692-1770, Scotland) | James Stirling (1692-1770, Scotland) | ||
:<math>\left.\ln N!\right. = \ln 1 + \ln 2 + \ln 3 + ... + \ln N | :<math>\left.\ln N!\right. = \ln 1 + \ln 2 + \ln 3 + ... + \ln N = \sum_{k=1}^N \ln k</math> | ||
Revision as of 14:28, 28 March 2007
James Stirling (1692-1770, Scotland)
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left.\ln N!\right. = \ln 1 + \ln 2 + \ln 3 + ... + \ln N = \sum_{k=1}^N \ln k}
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ~\approx \int_1^N \ln x dx}
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ~= \left[ x \ln x - x \right]_1^N}
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ~= N \ln N -N +1}
Thus, for large N
