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Both sides previous revision Previous revision | Last revisionBoth sides next revision | ||
수학:스털링_근사 [2020/10/25 00:27] – [안장점 방법을 이용하는 유도] admin | 수학:스털링_근사 [2020/10/25 00:40] – [라플라스의 방법을 이용하는 유도] admin | ||
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\[ \int_0^\infty e^{-p(z-\ln z)}dz \sim \sqrt{\frac{2\pi}{p}} e^{-p} \] | \[ \int_0^\infty e^{-p(z-\ln z)}dz \sim \sqrt{\frac{2\pi}{p}} e^{-p} \] | ||
이고 따라서 | 이고 따라서 | ||
- | \[ \Gamma(p+1) \sim p^{p+1} \sqrt{\frac{2\pi}{p}} e^{-p} = \sqrt{2\pi} | + | \[ \Gamma(p+1) \sim p^{p+1} \sqrt{\frac{2\pi}{p}} e^{-p} = p^p e^{-p} \sqrt{2\pi |
+ | |||
+ | ======참고문헌====== | ||
+ | * https:// |