关于泰勒公式的问题 求(√(1+x))*cosx的带皮亚诺余项的三阶麦克劳林公式.(√(1+x))*cosx=[1+(1/2)x-(1/8)x^2+(1/16)x^3+o(x^3)]*[1-(1/2)x^2+o(x^3)]=1+(1/2)x-(1/8)x^2+(1/16)x^3-(1/2)x^2-(1/4)x^3+o(x^3)=1+(1/2)x-(5/8)x^2-(3/
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![关于泰勒公式的问题 求(√(1+x))*cosx的带皮亚诺余项的三阶麦克劳林公式.(√(1+x))*cosx=[1+(1/2)x-(1/8)x^2+(1/16)x^3+o(x^3)]*[1-(1/2)x^2+o(x^3)]=1+(1/2)x-(1/8)x^2+(1/16)x^3-(1/2)x^2-(1/4)x^3+o(x^3)=1+(1/2)x-(5/8)x^2-(3/](/uploads/image/z/5247207-63-7.jpg?t=%E5%85%B3%E4%BA%8E%E6%B3%B0%E5%8B%92%E5%85%AC%E5%BC%8F%E7%9A%84%E9%97%AE%E9%A2%98+%E6%B1%82%28%E2%88%9A%281%2Bx%29%29%2Acosx%E7%9A%84%E5%B8%A6%E7%9A%AE%E4%BA%9A%E8%AF%BA%E4%BD%99%E9%A1%B9%E7%9A%84%E4%B8%89%E9%98%B6%E9%BA%A6%E5%85%8B%E5%8A%B3%E6%9E%97%E5%85%AC%E5%BC%8F.%28%E2%88%9A%281%2Bx%29%29%2Acosx%3D%EF%BC%BB1%2B%281%2F2%29x-%281%2F8%29x%5E2%2B%281%2F16%29x%5E3%2Bo%28x%5E3%29%EF%BC%BD%2A%EF%BC%BB1-%281%2F2%29x%5E2%2Bo%28x%5E3%29%EF%BC%BD%3D1%2B%281%2F2%29x-%281%2F8%29x%5E2%2B%281%2F16%29x%5E3-%281%2F2%29x%5E2-%281%2F4%29x%5E3%2Bo%28x%5E3%29%3D1%2B%281%2F2%29x-%285%2F8%29x%5E2-%283%2F)
关于泰勒公式的问题 求(√(1+x))*cosx的带皮亚诺余项的三阶麦克劳林公式.(√(1+x))*cosx=[1+(1/2)x-(1/8)x^2+(1/16)x^3+o(x^3)]*[1-(1/2)x^2+o(x^3)]=1+(1/2)x-(1/8)x^2+(1/16)x^3-(1/2)x^2-(1/4)x^3+o(x^3)=1+(1/2)x-(5/8)x^2-(3/
关于泰勒公式的问题 求(√(1+x))*cosx的带皮亚诺余项的三阶麦克劳林公式.
(√(1+x))*cosx=[1+(1/2)x-(1/8)x^2+(1/16)x^3+o(x^3)]*[1-(1/2)x^2+o(x^3)]
=1+(1/2)x-(1/8)x^2+(1/16)x^3-(1/2)x^2-(1/4)x^3+o(x^3)
=1+(1/2)x-(5/8)x^2-(3/16)x^3+o(x^3)
我的问题在于,如何从第一步得到第一个等号的结论?以及从第一个等号得到第二个等号 的结论?本人基础薄弱,
关于泰勒公式的问题 求(√(1+x))*cosx的带皮亚诺余项的三阶麦克劳林公式.(√(1+x))*cosx=[1+(1/2)x-(1/8)x^2+(1/16)x^3+o(x^3)]*[1-(1/2)x^2+o(x^3)]=1+(1/2)x-(1/8)x^2+(1/16)x^3-(1/2)x^2-(1/4)x^3+o(x^3)=1+(1/2)x-(5/8)x^2-(3/
首先要搞清楚(1+x)^α和cosx的泰勒展开式
(1+x)^α=1+αx+α(α-1)/2!*x^2+...+α(α-1)...(α-n+1)/n!*x^n+o(x^n)
令α=1/2,取前4项,即得(1+x)^(1/2)=√(1+x)=1+(1/2)x-(1/8)x^2+(1/16)x^3+o(x^3)
cosx=1-1/2!*x^2+1/4!*x^4-...+(-1)^n/(2n)!*x^(2n)+o[x^(2n)]
取前2项,即得cosx=1-(1/2)x^2+o(x^3)
第一个等号到第二个等号,按照多项式的乘法进行即可,计算时3次以上的项一律放入o(x^3)