划线部分求解释
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Tylor expansion
f(x) = e^x =>f(0) =1
f'(x)= e^x =>f'(0)/1! = 1
f^(n)(x)= e^x => f^(n)(0)/n! = 1/n!
f(x) = f(0) + [f'(0)/1!]x + [f''(0)/2!]x^2+...
=> e^x = 1+x/1! +x^2/2! +....+ x^n/n! +...
(e^x -1)/x = 1 + x/2! +...+x^(n-1)/n! +...
f(x) = e^x =>f(0) =1
f'(x)= e^x =>f'(0)/1! = 1
f^(n)(x)= e^x => f^(n)(0)/n! = 1/n!
f(x) = f(0) + [f'(0)/1!]x + [f''(0)/2!]x^2+...
=> e^x = 1+x/1! +x^2/2! +....+ x^n/n! +...
(e^x -1)/x = 1 + x/2! +...+x^(n-1)/n! +...
追问
第二步那个怎么化的，-1怎么没了
追答
e^x = 1+x/1! +x^2/2! +....+ x^n/n! +...
-1+e^x =x/1! +x^2/2! +....+ x^n/n! +...
(-1+e^x)/x =1 + x/2! +x^2/3!+...+x^(n-1)/n!+....
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