[Question 2014a02] Answer 1 (two upgrades, provided by Zhang junrui and Dong Qilin)

[Question 2014a02] Answer 1 (two upgrades, provided by Zhang junrui and Dong Qilin)

To upgrade the original determinant \ (D_n \), consider the following \ (n + 1 \) Order Determinant:

\ [| B | =\begin {vmatrix} 1 &-A_1 &-A_2 & \ cdots &-A _ {n-1} &-a_n \ 0 & 0 & a_1 + A_2 & \ cdots & a_1 + A _ {n-1} & a_1 + a_n \ 0 & A_2 + A_1 & 0 & \ cdots & A_2 + A _ {n-1} & A_2 + a_n \ \ vdots & \ vdots \ 0 & A _ {n-1} + A_1 & A _ {n-1} + A_2 & \ cdots & 0 & A _ {n-1} + a_n \ 0 & a_n + A_1 & a_n + A_2 & \ cdots & a_n + A _ {n-1} & 0 \ end {vmatrix }, \]

Apparently \ (D_n = | B | \). Add the first line of \ (| B | \) to the remaining \ (n \) rows, respectively.

\ [| B | =\begin {vmatrix} 1 &-A_1 &-A_2 & \ cdots &-A _ {n-1} &-a_n \ 1 &-A_1 & A_1 & \ cdots & A_1 & A_1 \ 1 & A_2 &-A_2 & \ cdots & A_2 & A_2 \ vdots & \ vdots \ 1 & A _ {n-1} & A _ {n-1} & \ cdots &-A _ {n-1} & A _ {n-1} \ 1 & a_n &\ cdots & a_n &-a_n \ end {vmatrix }. \]

Once again, consider the order of the above determinant as follows \ (n + 2:

\ [| C | =\begin {vmatrix} 1 & 0 & 0 & 0 & \ cdots & 0 & 0 \ 0 & 1 &-A_1 &-A_2 & \ cdots & -A _ {n-1} &-a_n \-A_1 & 1 &-A_1 & A_1 & \ cdots & A_1 & A_1 \-A_2 & 1 & A_2 &-A_2 &\ cdots & A_2 & A_2 \\\ vdots & \ vdots \-A _ {n-1} & 1 & _ {n-1} & A _ {n-1} & \ cdots &-A _ {n-1} & A _ {n-1} \-a_n & 1 & a_n & \ cdots & a_n &-a_n \ end {vmatrix }, \]

Apparently \ (D_n = | B | = | c | \). Add the first column of \ (| c | \) to the last column of \ (n \) respectively.

\ [| C | =\begin {vmatrix} 1 & 0 & 1 & 1 & \ cdots & 1 & 1 \ 0 & 1 &-A_1 &-A_2 & \ cdots & -A _ {n-1} &-a_n \-A_1 & 1 &-2A_1 & 0 & \ cdots & 0 & 0 \-A_2 & 1 & 0 &-2a_2 &\ cdots & 0 & 0 \ vdots & \ vdots \-A _ {n-1} & 1 & 0 & 0 & \ cdots &-2a _ {n-1} & 0 \-a_n & 1 & 0 & \ cdots & 0 &-2a_n \ end {vmatrix }. \]

The above determine-tion is a typical claw-type deciding factor (refer to example 6th on the 1.2 page of the High-generation White Paper). As long as the non-zero primary diagonal element is used to remove one side of the claw and change it to the upper (lower) the triangular determinant can be used to calculate the value. we chose to remove the claw edges of the first two columns. in the above determinant, column \ (I \) (\ (I = 3,4, \ cdots, N + 2 \)) multiply by \ (-\ frac {1} {2} \) and add the column \ (I \) (\ (I = 3,4, \ cdots, n + 2 \) multiplied by \ (\ frac {1} {2a _ {I-2} \) are added to the second column, available

\ [| C | =\begin {vmatrix} 1-\ frac {n} {2} & \ frac {t} {2} & 1 & \ cdots & 1 & 1 \ frac {s} {2} & 1-\ frac {n} {2} &-A_1 &-A_2 & \ cdots &-A _ {n-1 }&- a_n \ 0 & 0 &-2A_1 & 0 & \ cdots & 0 & 0 \ 0 & 0 &-2a_2 & \ cdots & 0 & 0 \ vdots & \ vdots & \ vdots \ 0 & 0 & 0 & \ cdots &-2a _ {n-1} & 0 \\ 0 & 0 & 0 & 0 & \ cdots & 0 &-2a_n \ end {vmatrix }, \]

Where \ (S = A_1 + A_2 + \ cdots + a_n \), \ (t = \ frac {1} {A_1} + \ frac {1} {A_2} + \ cdots + \ frac {1} {a_n }\). note that the above determining factor is the top triangular deciding factor of the chunk block, so as to obtain

\ [D_n = | c | = (-2) ^ {N-2} \ prod _ {I = 1} ^ na_ I \ Bigg (n-2) ^ 2-\ big (\ sum _ {I = 1} ^ na_ I \ big) \ big (\ sum _ {I = 1} ^ n \ frac {1} {a_ I} \ big) \ Bigg ). \ quad \ Box \]

[Question 2014a02] Answer 1 (two upgrades, provided by Zhang junrui and Dong Qilin)