Question

e的sinx次方泰勒展开

Answer 1

f(x) = e^(sinx) =>f(0) =1
f'(x) = cosx. e^(sinx) => f'(0)/1! = 1
f''(x) = [(cosx)^2-sinx ].e^(sinx) => f''(0)/2! = 1/2
e^(sinx)
=1+x +(1/2)x^2+....

Answer 2

e^sinx = 1 + sinx + sin^2x/2! + sin^3x/3! + ... + sin^nx/n!, {n->oo}
= 1 + [x - x^3/3! + x^5/5! - ... + x^(2n+1)/(2n+1)!] + [x - x^3/3! + x^5/5! - ... + x^(2n+1)/(2n+1)!]^2/2! + [x - x^3/3! + x^5/5! - ... + x^(2n+1)/(2n+1)!]^3/3! + ...
= 1 + x + x^2/2 - x^4/8 + O(x^5)