复习笔记Day107：华中科技大学2023数学分析参考答案（上）

2023-02-23 23:38 作者:间宫_卓司 0人读过 | 我要投稿

之前在专栏里面说了，可能很长一段时间不会更新数分高代相关的内容了，但是复试的内容里面还有数分高代，所以我不得不花点时间复健一下数分高代，复健的方式大概就是写几套考研真题，然后再把课本和之前的笔记看一遍，然后我要侧重练习计算题，这次的高代就是吃了计算能力太差和基础不扎实的亏（我想了半天线性方程组要怎么解···）

1.求极限 $%5Cbegin%7Bequation%7D%0A%5Clim%20_%7Bx%20%5Crightarrow%200%7D%5Cleft(-%5Cfrac%7B%5Ccot%20x%7D%7Be%5E%7B-2%20x%7D%7D%2B%5Cfrac%7B1%7D%7Be%5E%7B-x%7D%20%5Csin%20%5E2%20x%7D-%5Cfrac%7B1%7D%7Bx%5E2%7D%5Cright)%0A%5Cend%7Bequation%7D$

这题可以通过直接通分来暴力求解，不过如果注意到极限 $%5Cunderset%7Bx%5Crightarrow%200%7D%7B%5Clim%7D%5Cleft(%20%5Cfrac%7B1%7D%7Bx%5E2%7D-%5Cfrac%7B1%7D%7B%5Csin%20%5E2x%7D%20%5Cright)%20$ 存在的话，可以减少计算量

$%5Cbegin%7Baligned%7D%0A%09%26-%5Cfrac%7B%5Ccot%20x%7D%7Be%5E%7B-2x%7D%7D%2B%5Cfrac%7B1%7D%7Be%5E%7B-x%7D%5Csin%20%5E2x%7D-%5Cfrac%7B1%7D%7Bx%5E2%7D%5C%5C%0A%09%26%3D%5Cfrac%7B-%5Ccos%20x%5Csin%20x%7D%7B%5Csin%20%5E2x%7De%5E%7B2x%7D%2B%5Cfrac%7B1%7D%7B%5Csin%20%5E2x%7De%5Ex-%5Cfrac%7B1%7D%7Bx%5E2%7D%5C%5C%0A%09%26%3D%5Cfrac%7B1%7D%7B%5Csin%20%5E2x%7D%5Cleft(%20-%5Cfrac%7B1%7D%7B2%7D%5Csin%202xe%5E%7B2x%7D%2Be%5Ex-1%20%5Cright)%20%2B%5Cleft(%20%5Cfrac%7B1%7D%7B%5Csin%20%5E2x%7D-%5Cfrac%7B1%7D%7Bx%5E2%7D%20%5Cright)%5C%5C%0A%5Cend%7Baligned%7D$

现在分别来计算最后一式中两项的极限，对于第一项，因为当 $x%5Crightarrow%200$ 时，有

$%5Csin%202xe%5E%7B2x%7D%3D%5Cleft(%202x%2Bo%5Cleft(%20x%5E2%20%5Cright)%20%5Cright)%20%5Cleft(%201%2B2x%2Bo%5Cleft(%20x%20%5Cright)%20%5Cright)%20%3D2x%2B4x%5E2%2Bo%5Cleft(%20x%5E2%20%5Cright)%20$

$e%5Ex-1%3Dx%2B%5Cfrac%7Bx%5E2%7D%7B2%7D%2Bo%5Cleft(%20x%5E2%20%5Cright)%20$

所以

$%5Cbegin%7Baligned%7D%0A%09%26-%5Cfrac%7B1%7D%7B2%7D%5Csin%202xe%5E%7B2x%7D%2Be%5Ex-1%5C%5C%0A%09%26%3D-%5Cfrac%7B1%7D%7B2%7D%5Cleft(%202x%2B4x%5E2%2Bo%5Cleft(%20x%5E2%20%5Cright)%20%5Cright)%20%2Bx%2B%5Cfrac%7Bx%5E2%7D%7B2%7D%2Bo%5Cleft(%20x%5E2%20%5Cright)%5C%5C%0A%09%26%3D-%5Cfrac%7B3%7D%7B2%7Dx%5E2%2Bo(x%5E2)%5C%5C%0A%5Cend%7Baligned%7D$

从而

$%5Cunderset%7Bx%5Crightarrow%200%7D%7B%5Clim%7D%5Cfrac%7B1%7D%7B%5Csin%20%5E2x%7D%5Cleft(%20-%5Cfrac%7B1%7D%7B2%7D%5Csin%202xe%5E%7B2x%7D%2Be%5Ex-1%20%5Cright)%20%3D%5Cunderset%7Bx%5Crightarrow%200%7D%7B%5Clim%7D%5Cfrac%7B1%7D%7Bx%5E2%7D%5Cleft(%20-%5Cfrac%7B3%7D%7B2%7Dx%5E2%2Bo%5Cleft(%20x%5E2%20%5Cright)%20%5Cright)%20%3D-%5Cfrac%7B3%7D%7B2%7D$

对于第二项，有

$%5Cbegin%7Baligned%7D%0A%09%26%5Cunderset%7Bx%5Crightarrow%200%7D%7B%5Clim%7D%5Cleft(%20%5Cfrac%7B1%7D%7B%5Csin%20%5E2x%7D-%5Cfrac%7B1%7D%7Bx%5E2%7D%20%5Cright)%5C%5C%0A%09%26%3D%5Cunderset%7Bx%5Crightarrow%200%7D%7B%5Clim%7D%5Cfrac%7B%5Cleft(%20x%2B%5Csin%20x%20%5Cright)%20%5Cleft(%20x-%5Csin%20x%20%5Cright)%7D%7Bx%5E2%5Csin%20%5E2x%7D%5C%5C%0A%09%26%3D%5Cunderset%7Bx%5Crightarrow%200%7D%7B%5Clim%7D%5Cfrac%7B%5Cleft(%20x%2Bx%2Bo%5Cleft(%20x%20%5Cright)%20%5Cright)%20%5Cleft(%20x-%5Cleft(%20x-%5Cfrac%7Bx%5E3%7D%7B6%7D%2Bo%5Cleft(%20x%5E3%20%5Cright)%20%5Cright)%20%5Cright)%7D%7Bx%5E4%7D%5C%5C%0A%09%26%3D%5Cunderset%7Bx%5Crightarrow%200%7D%7B%5Clim%7D%5Cfrac%7B%5Cleft(%202x%2Bo%5Cleft(%20x%20%5Cright)%20%5Cright)%20%5Cleft(%20%5Cfrac%7Bx%5E3%7D%7B6%7D%2Bo%5Cleft(%20x%5E3%20%5Cright)%20%5Cright)%7D%7Bx%5E4%7D%3D%5Cfrac%7B1%7D%7B3%7D%5C%5C%0A%5Cend%7Baligned%7D$

所以 $%E5%8E%9F%E5%BC%8F%3D-%5Cfrac%7B3%7D%7B2%7D%2B%5Cfrac%7B1%7D%7B3%7D%3D-%5Cfrac%7B7%7D%7B6%7D$

2.计算 $%5Cbegin%7Bequation%7D%0A%5Cint_0%5E%7B%2B%5Cinfty%7D%20e%5E%7B-%5Cfrac%7B1%7D%7B4%20s%7D%7D%20s%5E%7B-%5Cfrac%7B3%7D%7B2%7D%7D%20e%5E%7B-s%7D%20%5Cmathrm%7B~d%7D%20s%0A%5Cend%7Bequation%7D$

这题还是比较哈人的，一开始我觉得它和90.2有点像，但是好像又不太一样，经过一番尝试后，我发现通过换元法可能可以把夹在两个 $e$ 中间的 $s$ 消去

$%5Cbegin%7Baligned%7D%0A%09%26%5Cint_0%5E%7B%2B%5Cinfty%7D%7Be%5E%7B-%5Cfrac%7B1%7D%7B4s%7D%7Ds%5E%7B-%5Cfrac%7B3%7D%7B2%7D%7De%5E%7B-s%7D%5Cxlongequal%7Bu%5E%7B%5Calpha%7D%3Ds%7D%7D%5Cint_0%5E%7B%2B%5Cinfty%7D%7Be%5E%7B-%5Cfrac%7B1%7D%7B4u%5E%7B%5Calpha%7D%7D%7Du%5E%7B-%5Cfrac%7B3%7D%7B2%7D%5Calpha%7De%5E%7B-u%5E%7B%5Calpha%7D%7D%5Calpha%20u%5E%7B%5Calpha%20-1%7D%5Cmathrm%7Bd%7Du%7D%5C%5C%0A%09%26%3D%5Calpha%20%5Cint_0%5E%7B%2B%5Cinfty%7D%7Be%5E%7B-%5Cfrac%7B1%7D%7B4u%5E%7B%5Calpha%7D%7D%7Du%5E%7B-%5Cfrac%7B1%7D%7B2%7D%5Calpha%20-1%7De%5E%7B-u%5E%7B%5Calpha%7D%7D%5Cmathrm%7Bd%7Du%7D%5C%5C%0A%5Cend%7Baligned%7D$

取 $%5Calpha%3D-2$ （注意上下限要互换，上面那个式子只在 $%5Calpha%3E0$ 的时候是对的）

可得 $%5Cint_0%5E%7B%2B%5Cinfty%7D%7Be%5E%7B-%5Cfrac%7B1%7D%7B4s%7D%7Ds%5E%7B-%5Cfrac%7B3%7D%7B2%7D%7De%5E%7B-s%7D%3D%7D2%5Cint_0%5E%7B%2B%5Cinfty%7D%7Be%5E%7B-%5Cfrac%7Bu%5E2%7D%7B4%7D-%5Cfrac%7B1%7D%7Bu%5E2%7D%7D%5Cmathrm%7Bd%7Du%7D$

进而

$%5Cint_0%5E%7B%2B%5Cinfty%7D%7Be%5E%7B-%5Cfrac%7Bu%5E2%7D%7B4%7D-%5Cfrac%7B1%7D%7Bu%5E2%7D%7D%5Cmathrm%7Bd%7Du%7D%5Cxlongequal%7By%3D2c%7D2%5Cint_0%5E%7B%2B%5Cinfty%7D%7Be%5E%7B-y%5E2-%5Cfrac%7B1%7D%7B4y%5E2%7D%7D%5Cmathrm%7Bd%7Dy%7D$

所以依90.2的结论， $%5Ctext%7B%E5%8E%9F%E5%BC%8F%7D%3D2%5Ccdot%202%5Ccdot%20%5Cleft(%20%5Cfrac%7B1%7D%7B2%7D%5Csqrt%7B%5Cpi%7De%5E%7B-2%5Ccdot%20%5Cfrac%7B1%7D%7B2%7D%7D%20%5Cright)%20%3D2%5Csqrt%7B%5Cpi%7De%5E%7B-1%7D$

3.判断积分 $%5Cbegin%7Bequation%7D%0A%5Cint_0%5E%7B2%20%5Cpi%7D%20e%5E%7B-x%5E2%7D%20%5Ccos%20x%20%5Cmathrm%7B~d%7D%20x%0A%5Cend%7Bequation%7D$ 的正负，并证明你的结论

这题我一开始想了很长时间，用积分第二中值定理搞了半天都没有搞出来，后来我把被积函数的图像画了出来

虽然我知道 $e%5E%7B-x%5E2%7D$ 收敛的很快，但是没想到这么快，所以我觉得实际上证明这个结论并不需要很精确的估计，随便估计一下就够了

所以现在先来估计 $%5Cint_0%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%7Be%5E%7B-x%5E2%7D%7D%5Ccos%20x%5Cmathrm%7Bd%7Dx$

$%5Cint_0%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%7Be%5E%7B-x%5E2%7D%5Ccos%20x%5Cmathrm%7Bd%7Dx%7D%5Cge%20%5Cint_0%5E%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%7D%7Be%5E%7B-x%5E2%7D%5Ccos%20x%5Cmathrm%7Bd%7Dx%7D%5Cge%20%5Cfrac%7B%5Cpi%7D%7B4%7D%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7De%5E%7B-%5Cfrac%7B%5Cpi%20%5E2%7D%7B4%5E2%7D%7D%3D%5Cfrac%7B%5Cpi%20%5Csqrt%7B2%7D%7D%7B8%7De%5E%7B-%5Cfrac%7B%5Cpi%20%5E2%7D%7B16%7D%7D$

再来估计 $%5Cint_%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%5E%7B%5Cfrac%7B3%5Cpi%7D%7B2%7D%7D%7Be%5E%7B-x%5E2%7D%5Ccos%20x%5Cmathrm%7Bd%7Dx%7D$

$%5Cint_%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%5E%7B%5Cfrac%7B3%5Cpi%7D%7B2%7D%7D%7Be%5E%7B-x%5E2%7D%5Ccos%20x%5Cmathrm%7Bd%7Dx%7D%5Cge%20%5Cint_%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%5E%7B%5Cfrac%7B3%5Cpi%7D%7B2%7D%7D%7Be%5E%7B-x%7D%5Ccos%20x%5Cmathrm%7Bd%7Dx%7D$ ，

$%5Cleft(%20e%5E%7B-x%7D%5Ccos%20x%20%5Cright)%20'%3D-e%5E%7B-x%7D%5Csin%20x-e%5E%7B-x%7D%5Ccos%20x$

所以 $e%5E%7B-x%7D%5Ccos%20x$ 在 $%5Cleft%5B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%2C%5Cfrac%7B3%7D%7B2%7D%5Cpi%20%5Cright%5D%20$ 上有最小值 $-%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7De%5E%7B-%5Cfrac%7B5%5Cpi%7D%7B4%7D%7D$ ，所以

$%5Cint_%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%5E%7B%5Cfrac%7B3%5Cpi%7D%7B2%7D%7D%7Be%5E%7B-x%5E2%7D%5Ccos%20x%5Cmathrm%7Bd%7Dx%7D%5Cge%20-%5Cfrac%7B%5Csqrt%7B2%7D%5Cpi%7D%7B2%7De%5E%7B-%5Cfrac%7B5%5Cpi%7D%7B4%7D%7D$

做比可得

$%5Cfrac%7B%5Cfrac%7B%5Cpi%20%5Csqrt%7B2%7D%7D%7B8%7De%5E%7B-%5Cfrac%7B%5Cpi%20%5E2%7D%7B16%7D%7D%7D%7B%5Cfrac%7B%5Csqrt%7B2%7D%5Cpi%7D%7B2%7De%5E%7B-%5Cfrac%7B5%5Cpi%7D%7B4%7D%7D%7D%3D%5Cfrac%7Be%5E%7B%5Cfrac%7B5%7D%7B4%7D%5Cpi%20-%5Cfrac%7B%5Cpi%20%5E2%7D%7B16%7D%7D%7D%7B4%7D%5Cge%20%5Cfrac%7Be%5E%7B%5Cfrac%7B15%7D%7B4%7D-1%7D%7D%7B4%7D%3D%5Cfrac%7Be%5E%7B%5Cfrac%7B11%7D%7B4%7D%7D%7D%7B4%7D%5Cge%20%5Cfrac%7B2%5E2%7D%7B4%7D%5Cge%201$

所以

$%5Cbegin%7Bequation%7D%0A%5Cint_0%5E%7B2%20%5Cpi%7D%20e%5E%7B-x%5E2%7D%20%5Ccos%20x%20%5Cmathrm%7B~d%7D%20x%5Cge%5Cint_0%5E%7B%5Cfrac%7B3%5Cpi%7D%7B2%7D%7D%7Be%5E%7B-x%5E2%7D%5Ccos%20x%5Cmathrm%7Bd%7Dx%7D%5Cge%200%0A%5Cend%7Bequation%7D$

4.设 $%5Cbegin%7Bequation%7D%0AI%5Cleft(x_0%2C%20x_1%5Cright)%3D%5Ciint_%7B%5CSigma%7D%20%5Cfrac%7Be%5E%7B-x%5E%5Calpha%7D%7D%7B%5Csqrt%7By%5E2%2Bz%5E2%7D%7D%20%5Cmathrm%7B~d%7D%20y%20%5Cmathrm%7B~d%7D%20z%0A%5Cend%7Bequation%7D$ ，其中 $%5CSigma$ 为抛物面 $x%3Dy%5E2%2Bz%5E2$ 与平面 $x%3Dx_0%2Cx%3Dx_1$ 所围立体表面的内侧， $%5Calpha%3E0$ ， $x_1%3Ex_0%3E0$ ，求极限

$%5Cunderset%7B%5Cbegin%7Barray%7D%7Bc%7D%0A%09x_0%5Crightarrow%200%5C%5C%0A%09x_1%5Crightarrow%20%2B%5Cinfty%5C%5C%0A%5Cend%7Barray%7D%7D%7B%5Clim%7DI%5Cleft(%20x_0%2Cx_1%20%5Cright)%20$

做换元 $y%3Dr%5Ccos%20%5Ctheta%20%2Cz%3Dr%5Csin%20%5Ctheta%20%2Cx%3Dr%5E2$ ，可得

$I%5Cleft(%20x_0%2Cx_1%20%5Cright)%20%3D%5Cint_0%5E%7B2%5Cpi%7D%7B%5Cmathrm%7Bd%7D%5Ctheta%20%5Cint_%7B%5Csqrt%7Bx_0%7D%7D%5E%7B%5Csqrt%7Bx_1%7D%7D%7Be%5E%7B-r%5E%7B2%5Calpha%7D%7D%5Cmathrm%7Bd%7Dr%7D%7D$

那么

$%5Cunderset%7B%5Cbegin%7Barray%7D%7Bc%7D%0A%09x_0%5Crightarrow%200%5C%5C%0A%09x_1%5Crightarrow%20%2B%5Cinfty%5C%5C%0A%5Cend%7Barray%7D%7D%7B%5Clim%7DI%5Cleft(%20x_0%2Cx_1%20%5Cright)%20%3D2%5Cpi%20%5Cint_0%5E%7B%2B%5Cinfty%7D%7Be%5E%7B-r%5E%7B2%5Calpha%7D%7D%5Cmathrm%7Bd%7Dr%7D%5Cxlongequal%7Br%5E%7B2%5Calpha%7D%3Dx%7D2%5Cpi%20%5Cint_0%5E%7B%2B%5Cinfty%7D%7B%5Cfrac%7B1%7D%7B2%5Calpha%7Dx%5E%7B%5Cfrac%7B1%7D%7B2%5Calpha%7D-1%7De%5E%7B-x%7D%5Cmathrm%7Bd%7Dx%7D%3D%5Cfrac%7B%5Cpi%7D%7B%5Calpha%7D%5CGamma%20%5Cleft(%20%5Cfrac%7B1%7D%7B2%5Calpha%7D%20%5Cright)%20$

5.(1)证明：方程 $%5Cbegin%7Bequation%7D%0A(x%2B1)%5E%7Bx%2B1%7D%3De%20x%5Ex%0A%5Cend%7Bequation%7D$ 有唯一正根

$(x%2B1)%5E%7Bx%2B1%7D%3Dex%5Ex%5CLeftrightarrow%20e%5E%7B%5Cleft(%20x%2B1%20%5Cright)%20%5Cln%20%5Cleft(%20x%2B1%20%5Cright)%7D%3De%5E%7Bx%5Cln%20x%2B1%7D%5CLeftrightarrow%20e%5E%7B%5Cleft(%20x%2B1%20%5Cright)%20%5Cln%20%5Cleft(%20x%2B1%20%5Cright)%20-x%5Cln%20x-1%7D%3D1$

所以只要证明 $f%5Cleft(%20x%20%5Cright)%20%3D%5Cleft(%20x%2B1%20%5Cright)%20%5Cln%20%5Cleft(%20x%2B1%20%5Cright)%20-x%5Cln%20x-1$ 有唯一正根就好了，而

$f'%5Cleft(%20x%20%5Cright)%20%3D%5Cln%20%5Cleft(%20%5Cfrac%7Bx%2B1%7D%7Bx%7D%20%5Cright)%20%3E0$

所以 $f(x)$ 是单调递增的，进而 $f(0%5E%2B)%3D-1%3C0$ ， $f(1)%3D%5Cln4-1%3E0$ ，所以有唯一正根

(2)设 $p(x)%3Dx-x%5E2$ ， $f(x)$ 是二阶连续可微函数，证明对任意非负整数 $k$ ，成立

$%5Cbegin%7Bequation%7D%0A%5Cint_k%5E%7Bk%2B1%7D%20f(x)%20%5Cmathrm%7Bd%7D%20x%3D%5Cfrac%7Bf(k%2B1)%2Bf(k)%7D%7B2%7D-%5Cfrac%7B1%7D%7B2%7D%20%5Cint_k%5E%7Bk%2B1%7D%20f%5E%7B%5Cprime%20%5Cprime%7D(x)%20p(x-%5Bx%5D)%20%5Cmathrm%7Bd%7D%20x%0A%5Cend%7Bequation%7D$

这题考察的是欧拉-麦克劳林公式，谢惠民上面就有，一开始看到取整函数的时候可能会被吓到，但是只需要注意到

$%5Cint_k%5E%7Bk%2B1%7D%7Bf''%7D(x)p(x-%5Bx%5D)%5Cmathrm%7Bd%7Dx%3D%5Cint_k%5E%7Bk%2B1%7D%7Bf''%7D(x)p(x-k)%5Cmathrm%7Bd%7Dx$

也就没那么吓人了

$%5Cbegin%7Baligned%7D%0A%09%26%5Cint_k%5E%7Bk%2B1%7D%7Bf%5Cleft(%20x%20%5Cright)%20%5Cmathrm%7Bd%7Dx%7D%3D%5Cint_k%5E%7Bk%2B1%7D%7Bf%5Cleft(%20x%20%5Cright)%20%5Cmathrm%7Bd%7D%5Cleft(%20x-k-%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%7D%5C%5C%0A%09%26%3D%5Cfrac%7Bf%5Cleft(%20k%2B1%20%5Cright)%20%2Bf%5Cleft(%20k%20%5Cright)%7D%7B2%7D-%5Cint_k%5E%7Bk%2B1%7D%7B%5Cleft(%20x-k-%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20f'%5Cleft(%20x%20%5Cright)%20%5Cmathrm%7Bd%7Dx%7D%5C%5C%0A%5Cend%7Baligned%7D$

再次使用第一行的技巧，可得

$%5Cbegin%7Baligned%7D%0A%09%26%5Cint_k%5E%7Bk%2B1%7D%7B%5Cleft(%20x-k-%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20f'%5Cleft(%20x%20%5Cright)%20%5Cmathrm%7Bd%7Dx%7D%3D%5Cint_k%5E%7Bk%2B1%7D%7Bf'%5Cleft(%20x%20%5Cright)%20%5Cmathrm%7Bd%7D%5Cfrac%7B1%7D%7B2%7D%5Cleft(%20x-k-%5Cfrac%7B1%7D%7B2%7D%20%5Cright)%20%5E2%7D%5C%5C%0A%09%26%3D%5Cint_k%5E%7Bk%2B1%7D%7Bf'%5Cleft(%20x%20%5Cright)%20%5Cmathrm%7Bd%7D%5Cfrac%7B1%7D%7B2%7D%5Cleft(%20x-k%20%5Cright)%20%5Cleft(%20x-k-1%20%5Cright)%7D%5C%5C%0A%09%26%3D%5Cint_k%5E%7Bk%2B1%7D%7Bf''%5Cleft(%20x%20%5Cright)%20%5Cfrac%7B1%7D%7B2%7D%5Cleft(%20x-k%20%5Cright)%20%5Cleft(%20x-k-1%20%5Cright)%20%5Cmathrm%7Bd%7Dx%7D%5C%5C%0A%09%26%3D%5Cfrac%7B1%7D%7B2%7D%5Cint_k%5E%7Bk%2B1%7D%7Bf''%5Cleft(%20x%20%5Cright)%20p%5Cleft(%20x-k%20%5Cright)%20%5Cmathrm%7Bd%7Dx%7D%5C%5C%0A%5Cend%7Baligned%7D$

这就证明了结论

(3)计算极限 $%5Cbegin%7Bequation%7D%0A%5Clim%20_%7Bn%20%5Crightarrow%20%5Cinfty%7D%5Cleft(%5Cbeta%2B%5Cfrac%7B1%7D%7Bn%7D%5Cright)%5Cleft(%5Cbeta%2B%5Cfrac%7B2%7D%7Bn%7D%5Cright)%20%5Ccdots%5Cleft(%5Cbeta%2B%5Cfrac%7Bn%7D%7Bn%7D%5Cright)%0A%5Cend%7Bequation%7D$

没有想法，甚至怀疑是题目记错了，凑不出定积分的形式

6.(1)对任意的 $x%3E0%2Cy%5Cin%5Cmathbb%7BR%7D$ ，证明： $%5Cbegin%7Bequation%7D%0Ax%20y%20%5Cleq%20x%20%5Cln%20x-x%2Be%5Ey%0A%5Cend%7Bequation%7D$

这题好像是陈纪修上面的例题

记 $g%5Cleft(%20y%20%5Cright)%20%3Dxy-x%5Cln%20x%2Bx-e%5Ey$ ，则 $g'%5Cleft(%20y%20%5Cright)%20%3Dx-e%5Ey$ ，故 $g(y)$ 在 $y%3D%5Cln%20x$ 处取最大值，即 $g%5Cleft(%20y%20%5Cright)%20%5Cle%20g%5Cleft(%20%5Cln%20x%20%5Cright)%20%3D0$

(2)设 $%5Calpha%2C%5Cbeta$ 是任意非零实数，对正整数 $n$ ，证明：

$%5Cbegin%7Bequation%7D%0A%5Csum_%7Bk%3D0%7D%5En%5Cleft(%5Cbegin%7Barray%7D%7Bc%7D%0A%5Calpha%20%5C%5C%0Ak%0A%5Cend%7Barray%7D%5Cright)%5Cleft(%5Cbegin%7Barray%7D%7Bc%7D%0A%5Cbeta%20%5C%5C%0An-k%0A%5Cend%7Barray%7D%5Cright)%3D%5Cleft(%5Cbegin%7Barray%7D%7Bc%7D%0A%5Calpha%2B%5Cbeta%20%5C%5C%0An%0A%5Cend%7Barray%7D%5Cright)%20%0A%5Cend%7Bequation%7D$

其中

$%5Cleft(%5Cbegin%7Barray%7D%7Bc%7D%0A%5Calpha%20%5C%5C%0Ak%0A%5Cend%7Barray%7D%5Cright)%3D%5Cfrac%7B%5Calpha(%5Calpha-1)%20%5Ccdots(%5Calpha-k%2B1)%7D%7Bk%20!%7D%2C%5Cleft(%5Cbegin%7Barray%7D%7Bl%7D%0A%5Calpha%20%5C%5C%0A0%0A%5Cend%7Barray%7D%5Cright)%3D1$

这题是史济怀上面的例题，一方面，在收敛范围内，有

$%5Cleft(%201%2Bx%20%5Cright)%20%5E%7B%5Calpha%20%2B%5Cbeta%7D%3D%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%7B%5Cleft(%20%5Cbegin%7Barray%7D%7Bc%7D%0A%09%5Calpha%20%2B%5Cbeta%5C%5C%0A%09n%5C%5C%0A%5Cend%7Barray%7D%20%5Cright)%7Dx%5En$

另一方面，有

$%5Cbegin%7Baligned%7D%0A%09%26%5Cleft(%201%2Bx%20%5Cright)%20%5E%7B%5Calpha%20%2B%5Cbeta%7D%3D%5Cleft(%201%2Bx%20%5Cright)%20%5E%7B%5Calpha%7D%5Cleft(%201%2Bx%20%5Cright)%20%5E%7B%5Cbeta%7D%5C%5C%0A%09%26%3D%5Cleft(%20%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%7B%5Cleft(%20%5Cbegin%7Barray%7D%7Bc%7D%0A%09%5Calpha%5C%5C%0A%09n%5C%5C%0A%5Cend%7Barray%7D%20%5Cright)%20x%5En%7D%20%5Cright)%20%5Cleft(%20%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%7B%5Cleft(%20%5Cbegin%7Barray%7D%7Bc%7D%0A%09%5Cbeta%5C%5C%0A%09n%5C%5C%0A%5Cend%7Barray%7D%20%5Cright)%20x%5En%7D%20%5Cright)%5C%5C%0A%09%26%3D%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%7B%5Cleft(%20%5Csum_%7Bn%3D0%7D%5En%7B%5Cleft(%20%5Cbegin%7Barray%7D%7Bc%7D%0A%09%5Calpha%5C%5C%0A%09k%5C%5C%0A%5Cend%7Barray%7D%20%5Cright)%20%5Cleft(%20%5Cbegin%7Barray%7D%7Bc%7D%0A%09%5Cbeta%5C%5C%0A%09n-k%5C%5C%0A%5Cend%7Barray%7D%20%5Cright)%7D%20%5Cright)%7D%5C%5C%0A%5Cend%7Baligned%7D$

这说明了 $%5Cbegin%7Bequation%7D%0A%5Csum_%7Bk%3D0%7D%5En%5Cleft(%5Cbegin%7Barray%7D%7Bc%7D%0A%5Calpha%20%5C%5C%0Ak%0A%5Cend%7Barray%7D%5Cright)%5Cleft(%5Cbegin%7Barray%7D%7Bc%7D%0A%5Cbeta%20%5C%5C%0An-k%0A%5Cend%7Barray%7D%5Cright)%3D%5Cleft(%5Cbegin%7Barray%7D%7Bc%7D%0A%5Calpha%2B%5Cbeta%20%5C%5C%0An%0A%5Cend%7Barray%7D%5Cright)%20%0A%5Cend%7Bequation%7D$

因为b站的专栏一个只能放100张图片，而一个公式算一张图片，所以剩下的题目只能放到下去了

标签：华中科技大学考研数学分析

复习笔记Day107：华中科技大学2023数学分析参考答案（上）

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