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In Exercises $7-12$ , assume the signals listed are solutions of the given difference equation. Determine if the signals form a basis for the solution space of the equation. Justify your answers using appropriate theorems. $$1^{k},(-1)^{k} ; y_{k+4}-2 y_{k+2}+y_{k}=0$$

Since, by the theorem, we have the set H of all solution of the $n ^ { \prime 6 }$ order homogeneous lineardifference equation $y _ { k + m } + a _ { 1 } y _ { k + n - 1 } + \ldots \ldots \ldots \ldots \ldots \ldots + a _ { n - 1 } y _ { k + 1 } + a _ { n } y _ { k } = 0$ For all $k$ is an n-dimensional vector space, the two signals $1 ^ { x }$ and $( - 1 ) ^ { k }$ cannot possibly span a fourth dimensional space and so cannot be a basis for H.

Calculus 3

Chapter 4

Vector Spaces

Section 8

Applications to Difference Equations

Vectors

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we assume that these two signals solve these discreet against on this lunar disc articulation, which is also home agendas. And we need to determine whether they form a basis for that solution. Little space off this occasion? No, by film, 17 on pages 2 50 We have that The dimension off the solution vector space off. The situation equals the number of the order off the Gerson, which is four. And this is greater than two. So which is the number off solutions we have? So these two solutions do not for the basis for all of the solution vector space of this occasion seems there to end their list them. But then the dimension of the

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