For A S calculate I+tA+ 0.904837 (-2-6) X 10 -0.200334 X 1.10517 t2 The actual value of e014, to six decimal places, is you plug t = 0.1 into the approximation with no rounding.) X 2 4² +4³: 6 1-t 0 L X 0.904837 -0.200334 t² etAI+tA+ 2 0 1.105170] +3 -A² + What do you get when -A³? (Give six decimal places 6
For A S calculate I+tA+ 0.904837 (-2-6) X 10 -0.200334 X 1.10517 t2 The actual value of e014, to six decimal places, is you plug t = 0.1 into the approximation with no rounding.) X 2 4² +4³: 6 1-t 0 L X 0.904837 -0.200334 t² etAI+tA+ 2 0 1.105170] +3 -A² + What do you get when -A³? (Give six decimal places 6
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.2: Exponents And Radicals
Problem 92E
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solve using matrix exponentials: For A=[(-1,0),(-2,1)] calculate I+tA+t^2/2 A^2 + t^3/6 A^3: [(_,0),(t(-t^2-6) / 3,_)] The actual value of e^(0.1A), to six decimal places, is [(0.904837, 0),(-0.200334, 1.105170)]. What do you get when you plug t = 0.1 into the approximation e^(tA)~~I+tA+t^2/2 A^2 + t^3/6 A^3? (Give six decimal places with no rounding.)
Transcribed Image Text:For A
-1 01
-21 calculate I + A+
2-F
1-2-6)
0.904837
-0.200334
X
10
The actual value of e014, to six decimal places, is
you plug t = 0.1 into the approximation etAI+tA+
with no rounding.)
1.10517
+3
224² +4²³:
6
X
1+t-
0
X
0.904837
0
-0.200334 1.105170
t²
+3
24²
What do you get when
A³? (Give six decimal places
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