怎么求解泰勒公式?
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带佩亚诺余项的泰勒公式可以表示为:
f(x)=f(x0)+(x-x0) * f'(x0)/1! + (x-x0)^2 * f''(x0)/2! +… +(x-x0)^n * f^(n) (x0)/n! +o((x-x0)^n)
而x0→0时,
f(x)=f(0)+ x * f'(0)/1! + x^2 * f''(0)/2! +… +x^n * f^(n) (0)/n! +o(x^n)
显然当f(x)=arctanx时,
f(0)=0
f '(x)=1/(1+x^2),f ''(x)= -2x/(1+x^2)^2,
f '''(x)= -2/(1+x^2)^2 - 2x *(-2) * (2x)/(1+x^2)^3 = (6x^2-2)/(1+x^2)^3
所以当x0→0时,
f '(0)=1,f ''(0)=0,f '''(0)= -2
于是
arctanx=arctan0 + x * f'(0)/1! + x^2 * f''(0)/2! + x^3 * f''(0)/3! + o(x^3)
=0+ x +0*x^2/2 -2*x^3/6 +o(x^3)
= x - 1/3*x^3 + o(x^3)
f(x)=f(x0)+(x-x0) * f'(x0)/1! + (x-x0)^2 * f''(x0)/2! +… +(x-x0)^n * f^(n) (x0)/n! +o((x-x0)^n)
而x0→0时,
f(x)=f(0)+ x * f'(0)/1! + x^2 * f''(0)/2! +… +x^n * f^(n) (0)/n! +o(x^n)
显然当f(x)=arctanx时,
f(0)=0
f '(x)=1/(1+x^2),f ''(x)= -2x/(1+x^2)^2,
f '''(x)= -2/(1+x^2)^2 - 2x *(-2) * (2x)/(1+x^2)^3 = (6x^2-2)/(1+x^2)^3
所以当x0→0时,
f '(0)=1,f ''(0)=0,f '''(0)= -2
于是
arctanx=arctan0 + x * f'(0)/1! + x^2 * f''(0)/2! + x^3 * f''(0)/3! + o(x^3)
=0+ x +0*x^2/2 -2*x^3/6 +o(x^3)
= x - 1/3*x^3 + o(x^3)
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