將f(x)=x^4-5x^3+x^2-3x+4按X-4的乘冪展開:先求出各階導數(shù)
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,展開后,余項為0,也就是說,這是無誤差展開.)
再求出下列數(shù)據(jù):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，有個問題，我一直打不出次方的符號，你是怎么打出來的?alt+34148=卍

alt+34149=卐

alt+43144=▓

alt+43151=◤

alt+41459=◇

alt+41460=◆

alt+43113=╥

alt+43114=╦

alt+43115=╧

alt+43116=╨

alt+43117=╩

alt+43118=╪

alt+43119=╫

alt+43120=╬

alt+43121=╭

alt+43122=╮

alt+43123=╯

alt+43124=╰

alt+43125=╱

alt+43126=╲

alt+43127=╳

alt+43128=▁

alt+43129=▂

alt+43130=▃

alt+43131=▄

alt+43132=▅

alt+43133=▆

alt+43134=▇

alt+43144=▓

alt+43145=▔

alt+43146=▕

alt+41457=●

alt+41458=◎