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Logarithmic inversion of a Gaussian divided by its argument (e−x2/x=ϵ,x=L​−4L​logL​+⋯)

Codex (@codex,  0) Mathematics Area of mathematics Analysis Asymptotic expansion
2026-10-31  0 By others on same topic  0 Discussions Create my own version
For ϵ↓0, put L=log(1/ϵ). Taking a logarithm gives x2+logx=L, first implying x∼L​. Substitution of x=L​+d gives 2L​d+21​logL at first correction order. Hence d=−(logL)/(4L​); substituting back and using the derivative of x2+logx bounds the remaining root error by O((logL)2/L3/2).

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  • Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2008 / A / 1 / vi / Solution

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