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復(fù)習(xí)筆記 Day120

2023-07-13 18:44 作者:間宮_卓司  | 我要投稿

120.1 設(shè)數(shù)列%5C%7By_n%5C%7D滿足y_%7Bn%2B1%7D%3Day_%7Bn%7D%5E%7B2%7D%2B2by_n%2Bc,其中a%2Cb%2Cc%2Cy_0都是實數(shù)且a%5Cne0,問何時數(shù)列%5C%7By_n%5C%7D保持有界?

y'_n%3Day_n%2Bb%2Cc'%3Dac-b%5E2%2Bb,那么y_%7Bn%2B1%7D%5E%7B'%7D%3Dy_%7Bn%7D%5E%7B'2%7D%2Bc',所以接下來只需要討論y_%7Bn%2B1%7D%3Dy_%7Bn%7D%5E%7B2%7D%2Bc這種特殊情況就好了,記f(x)%3Dx%5E2%2Bc

(i)若f(x)%3Dx沒有實數(shù)解,則對任意y_0%5C%7By_n%5C%7D無界

此時y_%7Bn%2B1%7D-y_n%3Dy_%7Bn%7D%5E%7B2%7D-y_n%2Bc%3E0,又因為%5C%7By_n%5C%7D沒有不動點(diǎn),所以y_n單調(diào)的趨于正無窮

接下來討論f(x)%3Dx有實數(shù)解的情況,為了便于討論,引入一些記號,記

f_1%5Cleft(%20x%20%5Cright)%20%3D-%5Csqrt%7Bx%5E2-c'%7D%2Cf_2%5Cleft(%20x%20%5Cright)%20%3D%5Csqrt%7Bx%5E2-c'%7D%2Cx%5Cin%20%5Cleft%5B%20c'%2C%2B%5Cinfty%20%5Cright)%20

分別為f(x)在0左端和右端的反函數(shù),x_1%2Cx_2為方程f(x)%3Dx的兩個實數(shù)解(可以相等)

(ii)若y_0%3Cf_1(x_2)y_0%3Ex_2,則%5C%7By_n%5C%7D單調(diào)趨于正無窮

直接解方程可知

x_1%3D%5Cfrac%7B1-%5Csqrt%7B1-4c%5E2%7D%7D%7B2%7D%2Cx_2%3D%5Cfrac%7B1%2B%5Csqrt%7B1-4c%5E2%7D%7D%7B2%7D

y_0%3Cf_1(x_2),則y_1%3Df(y_0)%3Ex_2,所以只需討論y_0%3Ex_2的情況即可

因為y_1-y_0%3Dy_%7B0%7D%5E%7B2%7D-y_0%2Bc%3E0,遞推可得

%5Cbegin%7Baligned%7D%0A%09y_%7Bn%2B1%7D-y_n%26%3Df%5Cleft(%20y_n%20%5Cright)%20-f%5Cleft(%20y_%7Bn-1%7D%20%5Cright)%5C%5C%0A%09%26%3Df'%5Cleft(%20%5Cxi%20%5Cright)%20%5Cleft(%20y_n-y_%7Bn-1%7D%20%5Cright)%5C%5C%0A%09%26%3Ef'%5Cleft(%20x_2%20%5Cright)%20%5Cleft(%20y_n-y_%7Bn-1%7D%20%5Cright)%5C%5C%0A%09%26%3E%5Ccdots%5C%5C%0A%09%26%3E%5Cleft(%20f'%5Cleft(%20x_2%20%5Cright)%20%5Cright)%20%5En%5Cleft(%20y_1-y_0%20%5Cright)%5C%5C%0A%5Cend%7Baligned%7D

f'%5Cleft(%20x_2%20%5Cright)%20%3D1%2B2%5Csqrt%7B1-4c%5E2%7D%3E0,因為在x_2的右端沒有不動點(diǎn),所以只能有%5C%7By_n%5C%7D單調(diào)趨于正無窮

(iii)若y_0%5Cin%20%5Cleft%5B%20f_1%5Cleft(%20x_2%20%5Cright)%20%2Cx_2%20%5Cright%5D%20%5Cleft%5B%20f%5Cleft(%200%20%5Cright)%20%2Cf%5Cleft(%20x_2%20%5Cright)%20%5Cright%5D%20%5Csubset%20%5Cleft%5B%20f_1%5Cleft(%20x_2%20%5Cright)%20%2Cx_2%20%5Cright%5D%20,則%5C%7By_n%5C%7D保持有界

y_0%5Cin%20%5Cleft%5B%20f_1%5Cleft(%20x_2%20%5Cright)%20%2Cx_2%20%5Cright%5D%20,則

y_1%5Cin%20f%5Cleft(%20%5Cleft%5B%20f_1%5Cleft(%20x_2%20%5Cright)%20%2Cx_2%20%5Cright%5D%20%5Cright)%20%5Csubset%20%5Cleft%5B%20f%5Cleft(%200%20%5Cright)%20%2Cf%5Cleft(%20x_2%20%5Cright)%20%5Cright%5D%20%5Csubset%20%5Cleft%5B%20f_1%5Cleft(%20x_2%20%5Cright)%20%2Cx_2%20%5Cright%5D%20

遞推可得%5C%7By_n%5C%7D保持有界

(iv)若%5Cleft%5B%20f%5Cleft(%200%20%5Cright)%20%2Cf%5Cleft(%20x_2%20%5Cright)%20%5Cright%5D%20%5Csubset%20%5Cleft%5B%20f_1%5Cleft(%20x_2%20%5Cright)%20%2Cx_2%20%5Cright%5D%20不成立,則%5C%7By_n%5C%7D只能在一個零測集上保持有界

%5C%7By_n%5C%7D保持有界,則一定有y_n%5Cin%20%5Cleft%5B%20f_1%5Cleft(%20x_2%20%5Cright)%20%2Cx_2%20%5Cright%5D%20成立。并且此時有%5Cleft%5B%20f_1%5Cleft(%20x_2%20%5Cright)%20%2Cx_2%20%5Cright%5D%20%5Csubset%20%5Cleft%5B%20f%5Cleft(%200%20%5Cright)%20%2Cf%5Cleft(%20x_2%20%5Cright)%20%5Cright%5D%20,所以f_1(x)%2Cf_2(x)%5Cleft%5B%20f_1%5Cleft(%20x_2%20%5Cright)%20%2Cx_2%20%5Cright%5D%20上有定義,那么對有界的%5C%7By_n%5C%7D,成立

y_0%3Df_%7Bi_1%7D%5Cleft(%20y_1%20%5Cright)%20%3D%5Ccdots%20%3Df_%7Bi_1%7D%5Cleft(%20f_%7Bi_2%7D%5Cleft(%20%5Ccdots%20f_%7Bi_n%7D%5Cleft(%20y_n%20%5Cright)%20%5Cright)%20%5Cright)%20

其中i_k%3D1%2C2。接下來對于確定的i_k,去研究p_n%5Cleft(%20x%20%5Cright)%20%3Df_%7Bi_1%7D%5Cleft(%20f_%7Bi_2%7D%5Cleft(%20%5Ccdots%20f_%7Bi_n%7D%5Cleft(%20y_n%20%5Cright)%20%5Cright)%20%5Cright)%20,記

p_%7Bn%2Ck%7D%5Cleft(%20x%20%5Cright)%20%3Df_%7Bi_k%7D%5Cleft(%20f_%7Bi_%7Bk%2B1%7D%7D%5Cleft(%20%5Ccdots%20f_%7Bi_n%7D%5Cleft(%20y_n%20%5Cright)%20%5Cright)%20%5Cright)%20,那么

%5Cbegin%7Baligned%7D%0A%09p_%7Bn%7D%5E%7B'%7D%5Cleft(%20x%20%5Cright)%20%26%3Df_%7Bi_1%7D%5E%7B'%7D%5Cleft(%20p_%7Bn%2C1%7D%5Cleft(%20x%20%5Cright)%20%5Cright)%20p_%7Bn%2C1%7D%5E%7B'%7D%5Cleft(%20x%20%5Cright)%5C%5C%0A%09%26%3D%5Ccdots%5C%5C%0A%09%26%3D%5Cprod_%7Bk%3D1%7D%5E%7Bn-1%7D%7Bf_%7Bi_k%7D%5E%7B'%7D%5Cleft(%20p_%7Bn%2Ck%2B1%7D%5Cleft(%20x%20%5Cright)%20%5Cright)%7D%5C%5C%0A%5Cend%7Baligned%7D

再根據(jù)%5Cleft(%20g%5E%7B-1%7D%5Cleft(%20y%20%5Cright)%20%5Cright)%20'%3D%5Cfrac%7B1%7D%7Bg'%5Cleft(%20x%20%5Cright)%7D%2Cg%5Cleft(%20x%20%5Cright)%20%3Dy,可知

f_%7Bi_k%7D%5E%7B'%7D%5Cleft(%20p_%7Bn%2Ck%2B1%7D%5E%7B%7D%5Cleft(%20x%20%5Cright)%20%5Cright)%20%3D%5Cfrac%7B1%7D%7Bf'%5Cleft(%20f_%7Bi_k%7D%5Cleft(%20p_%7Bn%2Ck%2B1%7D%5E%7B%7D%5Cleft(%20x%20%5Cright)%20%5Cright)%20%5Cright)%7D%3D%5Cfrac%7B1%7D%7Bf'%5Cleft(%20p_%7Bn%2Ck%7D%5E%7B%7D%5Cleft(%20x%20%5Cright)%20%5Cright)%7D,所以

p_%7Bn%7D%5E%7B'%7D%5Cleft(%20x%20%5Cright)%20%3D%5Cprod_%7Bk%3D1%7D%5E%7Bn-1%7D%7B%5Cfrac%7B1%7D%7Bf'%5Cleft(%20p_%7Bn%2Ck%7D%5E%7B%7D%5Cleft(%20x%20%5Cright)%20%5Cright)%7D%7D%3D%5Cprod_%7Bk%3D1%7D%5E%7Bn-1%7D%7B%5Cfrac%7B1%7D%7B2p_%7Bn%2Ck%7D%5Cleft(%20x%20%5Cright)%7D%7D

接下來對%5Cprod_%7Bk%3D1%7D%5E%7Bn-1%7D%7Bp_%7Bn%2Ck%7D%5Cleft(%20x%20%5Cright)%7D進(jìn)行估計,因為f%5Cleft(%20p_%7Bn%2Ck%7D%5Cleft(%20x%20%5Cright)%20%5Cright)%20%3Dp_%7Bn%2Ck%7D%5E%7B2%7D%5Cleft(%20x%20%5Cright)%20%2Bc%3Dp_%7Bn%2Ck%2B1%7D%5Cleft(%20x%20%5Cright)%20,化簡可得

%5Cfrac%7Bx_2-p_%7Bn%2Ck%2B1%7D%5Cleft(%20x%20%5Cright)%7D%7Bx_2-p_%7Bn%2Ck%7D%5Cleft(%20x%20%5Cright)%7D%3Dx_2%2Bp_%7Bn%2Ck%7D%5E%7B%7D%5Cleft(%20x%20%5Cright)%20

再遞推一步可得

%5Cfrac%7Bx_2-p_%7Bn%2Ck%2B1%7D%5Cleft(%20x%20%5Cright)%7D%7Bx_2-p_%7Bn%2Ck%7D%5Cleft(%20x%20%5Cright)%7D%3Dx_2%2Bp_%7Bn%2Ck%7D%5E%7B%7D%5Cleft(%20x%20%5Cright)%20%3D2x_2-x_%7B2%7D%5E%7B2%7D%2Bp_%7Bn%2Ck-1%7D%5E%7B2%7D%5Cleft(%20x%20%5Cright)%20

那么

%5Cprod_%7Bk%3D2%7D%5E%7Bn-1%7D%7B%5Cleft(%20%5Cfrac%7Bx_2-p_%7Bn%2Ck%2B1%7D%5Cleft(%20x%20%5Cright)%7D%7Bx_2-p_%7Bn%2Ck%7D%5Cleft(%20x%20%5Cright)%7D%2Bx_%7B2%7D%5E%7B2%7D-2x_2%20%5Cright)%20%3D%5Cprod_%7Bk%3D2%7D%5E%7Bn-1%7D%7Bp_%7Bn%2Ck-1%7D%5E%7B2%7D%5Cleft(%20x%20%5Cright)%7D%7D

容易證明

%5Cleft(%20%5Cprod_%7Bk%3D1%7D%5En%7B%5Cleft(%20a_k%2Bb_k%20%5Cright)%7D%20%5Cright)%20%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D%3E%5Cleft(%20%5Cprod_%7Bk%3D1%7D%5En%7Ba_k%7D%20%5Cright)%20%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D%2B%5Cleft(%20%5Cprod_%7Bk%3D1%7D%5En%7Bb_k%7D%20%5Cright)%20%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D

其中a_k%2Cb_k%3E0,所以

%5Cprod_%7Bk%3D2%7D%5E%7Bn-1%7D%7Bp_%7Bn%2Ck-1%7D%5E%7B2%7D%5Cleft(%20x%20%5Cright)%7D%3E%5Cleft(%20x_%7B2%7D%5E%7B2%7D-2x_2%2B%5Cleft(%20%5Cfrac%7Bx_2-p_%7Bn%2Cn%7D%5Cleft(%20x%20%5Cright)%7D%7Bx_2-p_%7Bn%2C2%7D%5Cleft(%20x%20%5Cright)%7D%20%5Cright)%20%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D%20%5Cright)%20%5En

注意%5Cleft%5B%20f_1%5Cleft(%20x_2%20%5Cright)%20%2Cx_2%20%5Cright%5D%20%5Csubset%20%5Cleft%5B%20f%5Cleft(%200%20%5Cright)%20%2Cf%5Cleft(%20x_2%20%5Cright)%20%5Cright%5D%20蘊(yùn)含了x_2%3E2


復(fù)習(xí)筆記 Day120的評論 (共 條)

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