请问sin²x如何展开为幂级数？
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f(x)=(sinx)^2 =>f(0) =0
f'(x)= sin2x =>f'(0)/1! = 0
f''(x)= 2cos2x =>f''(0)/2! = 1
f'''(x)= -4sin2x =>f'''(0)/3! = 0
f''''(x)=-8cos2x =>f''''(0)/4! = -2^3/4!
f^(5)(x)=16sin2x =>f^(5)(0)/5! = 0
f^(6)(x)= 32cos2x =>f^(6)(0)/5! = 2^5/6!
(sinx)^2
=x^2 +(-2^3/4!)x^4 + (2^5/6!)x^6 +.....+(-1)^(n-1) .[2^(2n-1)/(2n)! ]. x^(2n)+...
f'(x)= sin2x =>f'(0)/1! = 0
f''(x)= 2cos2x =>f''(0)/2! = 1
f'''(x)= -4sin2x =>f'''(0)/3! = 0
f''''(x)=-8cos2x =>f''''(0)/4! = -2^3/4!
f^(5)(x)=16sin2x =>f^(5)(0)/5! = 0
f^(6)(x)= 32cos2x =>f^(6)(0)/5! = 2^5/6!
(sinx)^2
=x^2 +(-2^3/4!)x^4 + (2^5/6!)x^6 +.....+(-1)^(n-1) .[2^(2n-1)/(2n)! ]. x^(2n)+...
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简单计算一下即可，答案如图所示
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