Re: [線代] 矩陣
※ 引述《semmy214 (黃小六)》之銘言:
: https://imgur.com/a/FtSTD
: 問一下e^Dt右上角 te^2t 怎來的 3Q
D不是對角矩陣
D^n = [2^n a_n]
[ 0 2^n]
a_n - 2a_(n-1) = 2^(n-1)
a_n = c1 + d2^n + f_n, c = const
f_n = (n - 1)2^(n - 1)
a_1 = 1, a_2 = 4 => a_n = 2^(n - 1) + (n - 1)2^(n - 1)
= n2^(n - 1)
00
=> [exp(Dt)]_12 = Sigma (1/k!)a_kt^k
k=0
00
= Sigma (1/(k-1)!)2^(k - 1) t^(k)
k=1
00
= Sigma (1/k!)2^k t^(k+1)
k=0
= texp(2t)
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