就是换元,令 u = x²,
e^u = 1+u + u²/2!+ u³/3!+ .+ u^n /n!+ o(x^n)
再代入 u = x²,这个是利用间接法把函数展成Maclaurin公式.
更简单的,1/(1-x) = 1+ x + x² + x³ + ,+ x^n + o(x^n)
1/(1-x²) = 1 + x² + x(²)²+ .+ x^(2n) + o(x^2n)
若令f(x) = e^(x²),f(0) = 1,
f'(x) = 2x * e^(x²),f'(0) = 0,
f''(x) = (2+4x²) * e^(x²),f''(0) = 2
f'''(x) = (12x + 8x³) * e^(x²),f'''(0) = 0
f''''(0) = 12 .