This first example similar assignment webassign quiz question
Suppose we want to solve the ordinary differential equation
dx
dt= −3x.
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dx1 | |||||||||||
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A = |
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x = | | x1 | andd dtx = | |||||||
−4 1 | 1 | | | dt dx2 |
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5 | x2 | |||||||||||
0 | | | dt dx3 |
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1 | x3 | |||||||||||
| dt | |||||||||||
P = |
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(b) Solve the systems of equations you found in (a). Do the variables y1, y2 and y3 increase or decrease with time?
(c) Rewrite the solution in terms of the original x vector.
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2 |
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Calculus
(b) Give an example of a continuous function f : [0, ∞) → R which is bounded but does not have a maximum value.
(c) Give an example of a continuous function f : (0, 1] → R which is not bounded.
(d) Give an example of a continuous function f : (0, 1] → R which is bounded but does not have a maximum value.(a) f(x) = 1/x, a = 1, ℓ = 1;
[N.B. This first example is similar to Assignment 3 WebAssign Quiz Question 8, except that we are using ε and δ instead of r and s. More importantly, there is another difference: you are not asked to find the largest value of δ that makes the implication true. All you need to do is find some value δ > 0 such that 0 < |x − a| < δ =⇒ |f(x) − ℓ| < ε. Hence there are many possible answers: once you have one answer that works, any smaller value of δ > 0 will also work (why?).](b) f(x) = 3x + 7, a ∈ R is arbitrary but fixed, ℓ = 3a + 7; (c) f(x) = x4, a > 0 is arbitrary but fixed, ℓ = a4;
Question 4. Subspace of a Vector Space
Prove: If U1, U2, · · · , Uk are subspaces of a vector space V , then U1 + U2 + · · · + Uk is a subspace of V that contains all of the subspaces U1, U2, · · · , Uk. Furthermore, each subspace Z of V which contains U1, U2, · · · , Uk must contain U1 + U2 + · · · + Uk. Thus U1 + U2 + · · · + Uk is the smallest subspace of V containing U1, U2, · · · , Uk.