求函数f(x)=x^2ln(1+x)在x=0处的n阶导数f(n)(0)(n>=3)答案解析是把ln(1+x)进行泰勒展开代入原式得f(x)=x^2[x-x^2/2+...+ (-1)^(n-1) (x^(n-2))/(n-2)+0(x^(n-1))]=x^3-(x^4)/2+...+(-1)^(n-1) (x^n)/(n-2) +o(x^n)令f(n)(0)/n!=(-1)^(n-
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![求函数f(x)=x^2ln(1+x)在x=0处的n阶导数f(n)(0)(n>=3)答案解析是把ln(1+x)进行泰勒展开代入原式得f(x)=x^2[x-x^2/2+...+ (-1)^(n-1) (x^(n-2))/(n-2)+0(x^(n-1))]=x^3-(x^4)/2+...+(-1)^(n-1) (x^n)/(n-2) +o(x^n)令f(n)(0)/n!=(-1)^(n-](/uploads/image/z/3738048-24-8.jpg?t=%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%3Dx%5E2ln%281%2Bx%29%E5%9C%A8x%3D0%E5%A4%84%E7%9A%84n%E9%98%B6%E5%AF%BC%E6%95%B0f%28n%29%280%29%EF%BC%88n%3E%3D3%EF%BC%89%E7%AD%94%E6%A1%88%E8%A7%A3%E6%9E%90%E6%98%AF%E6%8A%8Aln%281%2Bx%29%E8%BF%9B%E8%A1%8C%E6%B3%B0%E5%8B%92%E5%B1%95%E5%BC%80%E4%BB%A3%E5%85%A5%E5%8E%9F%E5%BC%8F%E5%BE%97f%28x%29%3Dx%5E2%5Bx-x%5E2%2F2%2B...%2B+%28-1%29%5E%28n-1%29+%28x%5E%28n-2%29%29%2F%28n-2%29%2B0%28x%5E%28n-1%29%29%5D%3Dx%5E3-%28x%5E4%29%2F2%2B...%2B%28-1%29%5E%28n-1%29+%28x%5En%29%2F%28n-2%29+%2Bo%28x%5En%29%E4%BB%A4f%28n%29%280%29%2Fn%21%3D%28-1%29%5E%28n-)
求函数f(x)=x^2ln(1+x)在x=0处的n阶导数f(n)(0)(n>=3)答案解析是把ln(1+x)进行泰勒展开代入原式得f(x)=x^2[x-x^2/2+...+ (-1)^(n-1) (x^(n-2))/(n-2)+0(x^(n-1))]=x^3-(x^4)/2+...+(-1)^(n-1) (x^n)/(n-2) +o(x^n)令f(n)(0)/n!=(-1)^(n-
求函数f(x)=x^2ln(1+x)在x=0处的n阶导数f(n)(0)(n>=3)
答案解析是把ln(1+x)进行泰勒展开代入原式得f(x)=x^2[x-x^2/2+...+ (-1)^(n-1) (x^(n-2))/(n-2)+0(x^(n-1))]
=x^3-(x^4)/2+...+(-1)^(n-1) (x^n)/(n-2) +o(x^n)令f(n)(0)/n!=(-1)^(n-1) 1/(n-2),f(n)(0)=n!(-1)^(n-1) /(n-2)
在ln(1+x)泰勒展开中为什么要展开到第n-2项,而且(-1)的系数为什么不是n-3而是n-1
在ln(1+x)泰勒展开中展开到第n-2项是因为前面有x^2,要使
f(x)凑出x^n吗?
求函数f(x)=x^2ln(1+x)在x=0处的n阶导数f(n)(0)(n>=3)答案解析是把ln(1+x)进行泰勒展开代入原式得f(x)=x^2[x-x^2/2+...+ (-1)^(n-1) (x^(n-2))/(n-2)+0(x^(n-1))]=x^3-(x^4)/2+...+(-1)^(n-1) (x^n)/(n-2) +o(x^n)令f(n)(0)/n!=(-1)^(n-
你说的正确,求f(x)的n阶导数时需要知道泰勒展开的n次项的系数,因为前面有x^2,后面就展开到n-2次以凑出x^n.另外(-1)^(n-3)=(-1)^(n-1),两写法没什么不同.
这个题也可以用求高阶导数的牛顿莱布尼兹公式计算(即乘积uv的n阶导数公式计算).