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3F→:1/(1-ax)^n = Σ[r=0~∞] C(r+n-1,r) (ax)^r,證明自己找09/12 16:57
1F→:1就是錯的,2是對的,readcount用完了當然先解開mutex09/12 16:53
3F→:如果你是沒看到證明就不想背公式的人09/12 16:37
4F→:去找本有證明的課本,或是google..09/12 16:38
25F→:就只是distinct不代表線性獨立而已09/10 12:57
3F→:[T-λΙ]β = [I](T-λΙ)[I] = [I](T)[I] - λ[I](I)[I]09/07 11:00
4F→: = [I](T)[I] - λI = [T]β-λΙ, [I]是基底變換09/07 11:02
5F推:至於相同特徵值不知道你是問哪個,T跟[T]β ?09/07 11:08
6F→:第一行可以看出λ為T的特徵值時 |T-λΙ|= 0 = |[T]β-λΙ|09/07 11:10
7F→:所以λ也是[T]β的特徵值09/07 11:12
4F→:推矩陣≠行列式08/31 03:54
1F→:這個解法其實就是 B = P([2 1 3]^t)08/29 22:26
2F→:A^100 = P(D^100)(P^-1),所以(A^100)B = P(D^100)[2 1 3]^t08/29 22:27
3F→:其中 P = [v1 v2 v3],觀念都是一樣的08/29 22:30
4F→:就是把要乘的向量分解成eigenvector的線性組合08/29 22:31
5F→:利用 (A^n)x = (λ^n)x,然後再把算完後的向量合起來而已08/29 22:32
8F→:(A^100)B = P(D^100)(P^-1)P([2 1 3]^t) =P(D^100)[2 1 3]^t08/29 22:42
9F→:B = P([2 1 3]^t)就是B = 2(v1)+(v2)+3(v3)08/29 22:43
10F→:你說的是[2 1 3]^t = (P^-1)B,就是B以v1v2v3為basis的座標08/29 22:47
2F→:e1 = [1 0 0 ...], e2 = [0 1 0 ...] 依此類推08/29 13:16
2F→:題庫那些都不要買,沒補習的話去買本筆記吧08/18 23:06
1F→:不然多出來的bit怎麼辦,當然還要多一層啊07/15 07:41