复习笔记Day107：华中科技大学2023数学分析参考答案（下）

2023-02-23 23:57 作者:间宫_卓司 0人读过 | 我要投稿

（续上）

7.设 $f%3A%5Cleft%5B%200%2C1%20%5Cright%5D%20%5Crightarrow%20%5Cleft(%200%2C%2B%5Cinfty%20%5Cright)%20$ 为连续函数，常数 $a%5Cge1$ ，证明

$%5Cunderset%7Bn%5Crightarrow%20%5Cinfty%7D%7B%5Clim%7D%5Csqrt%5Bn%5D%7B%5Cint_0%5E1%7B%5Cleft(%20a%2Bx%5En%20%5Cright)%20%5Enf%5Cleft(%20x%20%5Cright)%20%5Cmathrm%7Bd%7Dx%7D%7D%3Da%2B1$

这个感觉就是把典中典的题目魔改了一下，一方面，设 $f%5Cleft(%20x%20%5Cright)%20%3CM%2CM%3E0$ ，则

$%5Cint_0%5E1%7B%5Cleft(%20a%2Bx%5En%20%5Cright)%20%5Enf%5Cleft(%20x%20%5Cright)%20%5Cmathrm%7Bd%7Dx%7D%3CM%5Cleft(%20a%2B1%20%5Cright)%20%5En$

另一方面，因为 $f(x)$ 连续，所以可以设 $f(x)%3Em%3E0$ ，那么

$%5Cint_0%5E1%7B%5Cleft(%20a%2Bx%5En%20%5Cright)%20%5Enf%5Cleft(%20x%20%5Cright)%20%5Cmathrm%7Bd%7Dx%7D%3Em%5Cint_%7B1-%5Cfrac%7B2%7D%7Bn%5E2%7D%7D%5E%7B1-%5Cfrac%7B1%7D%7Bn%5E2%7D%7D%7B%5Cleft(%20a%2Bx%5En%20%5Cright)%20%5En%5Cmathrm%7Bd%7Dx%7D%3E%5Cfrac%7Bm%7D%7Bn%5E2%7D%5Cleft(%20a%2B%5Cleft(%201-%5Cfrac%7B2%7D%7Bn%5E2%7D%20%5Cright)%20%5En%20%5Cright)%20%5En$

两边同时开 $n$ 次方取极限可得结论

8.设 $f(x)$ 在 $(-%5Cinfty.%2B%5Cinfty)$ 上可导，且对任意 $x$ 有 $f(x)%3Df(x%2B2k)%3Df(x%2Bb)$ ，其中 $k$ 为正整数， $b$ 为无理数，用 $%5Ctext%7BFourier%7D$ 级数理论证明 $f(x)$ 为常数

听说是某年的竞赛题，不过我没有看答案，下面的方法对不对我也不清楚

设 $f(x)$ 的最小正周期为 $2T$ ，若 $T%3D0$ ，那么依65.1，结论已经成立了，现在设 $T%3E0$ ，那么在 $%5B-T%2CT%5D$ 上

$f%5Cleft(%20x%20%5Cright)%5Csim%20%20%5Cfrac%7Ba_0%7D%7B2%7D%2B%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cleft(%20a_n%5Ccos%20%5Cfrac%7Bn%5Cpi%7D%7BT%7Dx%2Bb_n%5Csin%20%5Cfrac%7Bn%5Cpi%7D%7BT%7Dx%20%5Cright)%7D$

又因为 $f(x)$ 在 $(-%5Cinfty.%2B%5Cinfty)$ 上可导，所以 $f(x)$ 满足李普希兹条件，故 $f(x)$ 收敛于它的傅里叶级数，也就是

$f%5Cleft(%20x%20%5Cright)%20%3D%5Cfrac%7Ba_0%7D%7B2%7D%2B%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%7B%5Cleft(%20a_n%5Ccos%20%5Cfrac%7Bn%5Cpi%7D%7BT%7Dx%2Bb_n%5Csin%20%5Cfrac%7Bn%5Cpi%7D%7BT%7Dx%20%5Cright)%7D%2C%5Cforall%20x%5Cin%20R$

记 $%5Cfrac%7B%5Cpi%7D%7BT%7D%3Dw$ ，因为 $%5Ccos%20%5Cfrac%7Bn%5Cpi%7D%7BT%7Dx%2C%5Csin%20%5Cfrac%7Bn%5Cpi%7D%7BT%7Dx$ 线性无关，所以

$%5Ccos%20wx%3D%5Ccos%20%5Cleft(%20w%5Cleft(%20x%2B2k%20%5Cright)%20%5Cright)%20%3D%5Ccos%20%5Cleft(%20w%5Cleft(%20x%2Bb%20%5Cright)%20%5Cright)%20$

这意味着 $%5Ccos%20wx$ 既以 $2k$ 为周期，又以 $b$ 为周期，而 $%5Ccos%20wx$ 以 $T%3D%5Cfrac%7B2%5Cpi%7D%7Bw%7D$ 为最小正周期，所以存在整数 $N%2CM$ ，使得 $%5Cfrac%7B2%5Cpi%7D%7Bw%7D%3D2Nk%3DMb$ ，那么 $%5Cfrac%7BN%7D%7BM%7D%3D%5Cfrac%7Bb%7D%7B2k%7D$ ，但是左边是有理数，右边是无理数，矛盾

（这出现了伪证吗？而且其中求傅里叶级数完全是多余的）

9.设二元函数 $f(x%2Cy)$ 在 $(x_0%2Cy_0)$ 的某邻域 $U$ 内有定义，且在 $U$ 内存在偏导数。证明：若

$f_x%5Cleft(%20x%2Cy%20%5Cright)%20%2Cf_y%5Cleft(%20x%2Cy%20%5Cright)%20$ 都在 $(x_0%2Cy_0)$ 可微，则 $f_%7Bxy%7D%5Cleft(%20x_0%2Cy_0%20%5Cright)%20%3Df_%7Byx%7D%5Cleft(%20x_0%2Cy_0%20%5Cright)%20$

这道题的证明思路类似于课本上证明若 $f_%7Bxy%7D$ 和 $f_%7Byx%7D$ 连续，则 $f_%7Bxy%7D%3Df_%7Byx%7D$

依题意，有

$%5Cbegin%7Baligned%7D%0A%09%26%5Cleft(%20f%5Cleft(%20x%2B%5CDelta%20x%2Cy%20%2B%5CDelta%20y%5Cright)%20-f%5Cleft(%20x%2B%5CDelta%20x%2Cy_0%20%5Cright)%20%5Cright)%20-%5Cleft(%20f%5Cleft(%20x_0%2Cy_0%2B%5CDelta%20y%20%5Cright)%20-f%5Cleft(%20x_0%2Cy_0%20%5Cright)%20%5Cright)%5C%5C%0A%09%26%3D%5CDelta%20x%5Cleft(%20f_x%5Cleft(%20x_0%2B%5Ctheta%20%5CDelta%20x%2Cy_0%2B%5CDelta%20y%20%5Cright)%20-f_x%5Cleft(%20x_0%2B%5Ctheta%20%5CDelta%20x%2Cy_0%20%5Cright)%20%5Cright)%5C%5C%0A%09%26%3D%5CDelta%20x%5Cleft(%20f_x%5Cleft(%20x_0%2Cy_0%20%5Cright)%20%2Bf_%7Bxx%7D%5Cleft(%20x_0%2Cy_0%20%5Cright)%20%5Ctheta%20%5CDelta%20x%2Bf_%7Bxy%7D%5Cleft(%20x_0%2Cy_0%20%5Cright)%20%5CDelta%20y%2B%20%5Cright.%5C%5C%0A%09%26%5Cleft.%20%2Bo%5Cleft(%20%5Csqrt%7B%5CDelta%20x%5E2%2B%5CDelta%20y%5E2%7D%20%5Cright)%20-f_x%5Cleft(%20x_0%2Cy_0%20%5Cright)%20%2Bf_%7Bxx%7D%5Cleft(%20x_0%2Cy_0%20%5Cright)%20%5Ctheta%20%5CDelta%20x%2Bo%5Cleft(%20%5CDelta%20x%20%5Cright)%20%5Cright)%5C%5C%0A%09%26%3D%5CDelta%20x%5CDelta%20yf_%7Bxy%7D%5Cleft(%20x_0%2Cy_0%20%5Cright)%20%2Bo%5Cleft(%20%5Csqrt%7B%5CDelta%20x%5E2%2B%5CDelta%20y%5E2%7D%20%5Cright)%5C%5C%0A%5Cend%7Baligned%7D$

同理 $%5Cbegin%7Baligned%7D%0A%09%26%5Cleft(%20f%5Cleft(%20x%2B%5CDelta%20x%2Cy%2B%5CDelta%20y%20%5Cright)%20-f%5Cleft(%20x%2Cy_0%2B%5CDelta%20y%20%5Cright)%20%5Cright)%20-%5Cleft(%20f%5Cleft(%20x_0%2B%5CDelta%20x%2Cy_0%20%5Cright)%20-f%5Cleft(%20x_0%2Cy_0%20%5Cright)%20%5Cright)%5C%5C%0A%09%26%3D%5CDelta%20x%5CDelta%20yf_%7Byx%7D%5Cleft(%20x_0%2Cy_0%20%5Cright)%20%2Bo%5Cleft(%20%5Csqrt%7B%5CDelta%20x%5E2%2B%5CDelta%20y%5E2%7D%20%5Cright)%5C%5C%0A%5Cend%7Baligned%7D$