将f(x)=x^4-5x^3+x^2-3x+4按X-4的乘幂展开:先求出各阶导数
f'(x)=4x^3-15x^2+2x-3.
f''(x)=12x^2-30x+2.
f'''(x)=24x-30
f''''(x)=24.
f'''''(x)=0(由此可知,展开后,余项为0,也就是说,这是无误差展开.)
再求出下列数据:f(4)=-56,f'(4)=21,f''(4)=74,f'''(4)=66,f''''(4)=24
于是f(x)=x^4-5x^3+x^2-3x+4
=-56+21(x-4)+(74/2!)(x-4)^2+(66/3!)(x-4)^3+(24/4!)(x-4)^4
=-56+21(x-4)+37(x-4)^2+11(x-4)^3+(x-4)^4大神，有个问题，我一直打不出次方的符号，你是怎么打出来的?alt+34148=卍

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