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how will the third equation here help me to get H

Anonymous Thu 17 Oct 01:22:56 2019 No.11065870 View Reply Original Report

Quoted By: >>11065878 >>11065917 >>11066184 >>11066198

Anonymous

Anonymous Thu 17 Oct 2019 01:25:22 No.11065878 Report

Quoted By: >>11065892

>>11065870
what is this

Anonymous

Anonymous Thu 17 Oct 2019 01:29:40 No.11065892 Report

Quoted By:

>>11065878
https://en.wikipedia.org/wiki/Discretization

Anonymous

Anonymous Thu 17 Oct 2019 01:33:41 No.11065907 Report

Quoted By:

>>11056713

Anonymous

Anonymous Thu 17 Oct 2019 01:37:24 No.11065917 Report

Quoted By:

>>11065870
>approximation

Definition.

Anonymous

Anonymous Thu 17 Oct 2019 03:33:24 No.11066184 Report

Quoted By:

>>11065870
Suppose we have the matrix equation
$X(t) = I + \int_0^t \! AX(\tau) \, \mathrm{d}\tau$
then, by the existence and uniqueness theorems (use Gronwall’s lemma), and because A doesn’t change w.r.t. t, we have
$G(t) = X(t) = e^{At}$
So, assuming A is invertible, we have
$A^{-1}G(t)B - A^{-1}B = \int_0^t \! e^{A\tau} \, \mathrm{d}\tau B = H(t)$
$A^{-1}G(t)B - A^{-1}B = H(t)$

Anonymous

Anonymous Thu 17 Oct 2019 03:40:49 No.11066198 Report

Quoted By:

>>11065870
well? just integrate e^(A lambda)
H = int_0^T (I + A lambda + 1/2 A^2 lambda^2 + ...) d lambda B, the sum converges uniformly on the bounded interval 0 to T, so we may swap the sum and the integral to get (T + AT^2/2! + A^2T^3/3! + ...)B = A^{-1}(AT + A^2T^2 / 2 + ...)B = A^{-1}(e^{AT} - I)B = A^{-1}GB - A^{-1}B.
Like the dude above said.

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