Re: 問題~Maclaurin's series

[yupisces (liqueur)]

※ 引述《puppy110330 (慵懶的葉子)》之銘言：
: 求        -1
:   f(x)=sin  x  的Maclaurin's series
: 煩請各路高手幫忙~謝謝~^^~


        -1
f(x)=sin  x   -1<x<1


             1/2
f'(x)=(1-x^2)      (二項氏定理)

=1+1/2(x^2)+3/8(x^4)+5/16(x^6)+...

一個一個積分
            -1
= >  f(x)=sin  x=  x+1/6(x^3)+3/40(x^5)+5/112(x^7)+...    -1<x<1



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