復習筆記Day107：華中科技大學2023數學分析參考答案（上）

2023-02-23 23:38 作者:間宮_卓司 0人讀過 | 我要投稿

之前在專欄里面說了，可能很長一段時間不會更新數分高代相關的內容了，但是復試的內容里面還有數分高代，所以我不得不花點時間復健一下數分高代，復健的方式大概就是寫幾套考研真題，然后再把課本和之前的筆記看一遍，然后我要側重練習計算題，這次的高代就是吃了計算能力太差和基礎不扎實的虧（我想了半天線性方程組要怎么解···）

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

這題可以通過直接通分來暴力求解，不過如果注意到極限 $%5Cunderset%7Bx%5Crightarrow%200%7D%7B%5Clim%7D%5Cleft(%20%5Cfrac%7B1%7D%7Bx%5E2%7D-%5Cfrac%7B1%7D%7B%5Csin%20%5E2x%7D%20%5Cright)%20$ 存在的話，可以減少計算量

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

現在分別來計算最后一式中兩項的極限，對于第一項，因為當 $x%5Crightarrow%200$ 時，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

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

從而

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

對于第二項，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.計算 $%5Cbegin%7Bequation%7D%0A%5Cint_0%5E%7B%2B%5Cinfty%7D%20e%5E%7B-%5Cfrac%7B1%7D%7B4%20s%7D%7D%20s%5E%7B-%5Cfrac%7B3%7D%7B2%7D%7D%20e%5E%7B-s%7D%20%5Cmathrm%7B~d%7D%20s%0A%5Cend%7Bequation%7D$

這題還是比較哈人的，一開始我覺得它和90.2有點像，但是好像又不太一樣，經過一番嘗試后，我發(fā)現通過換元法可能可以把夾在兩個 $e$ 中間的 $s$ 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 $%5Calpha%3D-2$ （注意上下限要互換，上面那個式子只在 $%5Calpha%3E0$ 的時候是對的）

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

進而

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

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

3.判斷積分 $%5Cbegin%7Bequation%7D%0A%5Cint_0%5E%7B2%20%5Cpi%7D%20e%5E%7B-x%5E2%7D%20%5Ccos%20x%20%5Cmathrm%7B~d%7D%20x%0A%5Cend%7Bequation%7D$ 的正負，并證明你的結論

這題我一開始想了很長時間，用積分第二中值定理搞了半天都沒有搞出來，后來我把被積函數的圖像畫了出來

雖然我知道 $e%5E%7B-x%5E2%7D$ 收斂的很快，但是沒想到這么快，所以我覺得實際上證明這個結論并不需要很精確的估計，隨便估計一下就夠了

所以現在先來估計 $%5Cint_0%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%7Be%5E%7B-x%5E2%7D%7D%5Ccos%20x%5Cmathrm%7Bd%7Dx$

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

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

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

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

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.設 $%5Cbegin%7Bequation%7D%0AI%5Cleft(x_0%2C%20x_1%5Cright)%3D%5Ciint_%7B%5CSigma%7D%20%5Cfrac%7Be%5E%7B-x%5E%5Calpha%7D%7D%7B%5Csqrt%7By%5E2%2Bz%5E2%7D%7D%20%5Cmathrm%7B~d%7D%20y%20%5Cmathrm%7B~d%7D%20z%0A%5Cend%7Bequation%7D$ ，其中 $%5CSigma$ 為拋物面 $x%3Dy%5E2%2Bz%5E2$ 與平面 $x%3Dx_0%2Cx%3Dx_1$ 所圍立體表面的內側， $%5Calpha%3E0$ ， $x_1%3Ex_0%3E0$ ，求極限

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

做換元 $y%3Dr%5Ccos%20%5Ctheta%20%2Cz%3Dr%5Csin%20%5Ctheta%20%2Cx%3Dr%5E2$ ，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)證明：方程 $%5Cbegin%7Bequation%7D%0A(x%2B1)%5E%7Bx%2B1%7D%3De%20x%5Ex%0A%5Cend%7Bequation%7D$ 有唯一正根

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

所以只要證明 $f%5Cleft(%20x%20%5Cright)%20%3D%5Cleft(%20x%2B1%20%5Cright)%20%5Cln%20%5Cleft(%20x%2B1%20%5Cright)%20-x%5Cln%20x-1$ 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 $f(x)$ 是單調遞增的，進而 $f(0%5E%2B)%3D-1%3C0$ ， $f(1)%3D%5Cln4-1%3E0$ ，所以有唯一正根

(2)設 $p(x)%3Dx-x%5E2$ ， $f(x)$ 是二階連續(xù)可微函數，證明對任意非負整數 $k$ ，成立

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

這題考察的是歐拉-麥克勞林公式，謝惠民上面就有，一開始看到取整函數的時候可能會被嚇到，但是只需要注意到

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

也就沒那么嚇人了

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

再次使用第一行的技巧，可得

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

這就證明了結論

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

沒有想法，甚至懷疑是題目記錯了，湊不出定積分的形式

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

這題好像是陳紀修上面的例題

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

(2)設 $%5Calpha%2C%5Cbeta$ 是任意非零實數，對正整數 $n$ ，證明：

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

其中

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

這題是史濟懷上面的例題，一方面，在收斂范圍內，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

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

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

因為b站的專欄一個只能放100張圖片，而一個公式算一張圖片，所以剩下的題目只能放到下去了

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