Suppose 𝐮^(k) and 𝐮^(k) are two solutions to the same iterative system 𝐮^(k+1)=T 𝐮^(k). (a) Supp

Name: Problem 39, Chapter 9: Applied Linear Algebra (Undergraduate Texts in Mathematics)
Uploaded: 2020-04-17T11:25:22-08:00
Duration: 1 min 48 s
Channel: Ankit Gupta
Description: Suppose 𝐮^(k) and 𝐮^(k) are two solutions to the same iterative system 𝐮^(k+1)=T 𝐮^(k). (a) Suppose 𝐮^(k0)=𝐮^(k0) for some k0 ≥0. Can you conclude that these are the same solution: 𝐮^(k)=𝐮^(k) for all k ? (b) What can you say if 𝐮^(k0)=𝐮^(k1) for k0 ≠k1 ?

step 1 For a quadratic equation, kx square plus x plus k equal to 0, given that it has repeated roots,

Question

Suppose $\mathbf{u}^{(k)}$ and $\overline{\mathbf{u}}^{(k)}$ are two solutions to the same iterative system $\mathbf{u}^{(k+1)}=T \mathbf{u}^{(k)}$. (a) Suppose $\mathbf{u}^{\left(k_0\right)}=\overline{\mathbf{u}}^{\left(k_0\right)}$ for some $k_0 \geq 0$. Can you conclude that these are the same solution: $\mathbf{u}^{(k)}=\overline{\mathbf{u}}^{(k)}$ for all $k$ ? (b) What can you say if $\mathbf{u}^{\left(k_0\right)}=\overline{\mathbf{u}}^{\left(k_1\right)}$ for $k_0 \neq k_1$ ?

   Suppose $\mathbf{u}^{(k)}$ and $\overline{\mathbf{u}}^{(k)}$ are two solutions to the same iterative system $\mathbf{u}^{(k+1)}=T \mathbf{u}^{(k)}$.
(a) Suppose $\mathbf{u}^{\left(k_0\right)}=\overline{\mathbf{u}}^{\left(k_0\right)}$ for some $k_0 \geq 0$. Can you conclude that these are the same solution: $\mathbf{u}^{(k)}=\overline{\mathbf{u}}^{(k)}$ for all $k$ ? (b) What can you say if $\mathbf{u}^{\left(k_0\right)}=\overline{\mathbf{u}}^{\left(k_1\right)}$ for $k_0 \neq k_1$ ?

Applied Linear Algebra (Undergraduate Texts in Mathematics)

Peter J. Olver,… 2nd Edition

Chapter 9, Problem 39

Instant Answer

Solved by Expert Ankit Gupta

04/17/2020

Step 1

We know that $\mathbf{u}^{(k+1)} = T \mathbf{u}^{(k)}$ and $\overline{\mathbf{u}}^{(k+1)} = T \overline{\mathbf{u}}^{(k)}$. We are given that $\mathbf{u}^{(k_0)} = \overline{\mathbf{u}}^{(k_0)}$ for some $k_0 \geq 0$. Show more…

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Suppose $\mathbf{u}^{(k)}$ and $\overline{\mathbf{u}}^{(k)}$ are two solutions to the same iterative system $\mathbf{u}^{(k+1)}=T \mathbf{u}^{(k)}$. (a) Suppose $\mathbf{u}^{\left(k_0\right)}=\overline{\mathbf{u}}^{\left(k_0\right)}$ for some $k_0 \geq 0$. Can you conclude that these are the same solution: $\mathbf{u}^{(k)}=\overline{\mathbf{u}}^{(k)}$ for all $k$ ? (b) What can you say if $\mathbf{u}^{\left(k_0\right)}=\overline{\mathbf{u}}^{\left(k_1\right)}$ for $k_0 \neq k_1$ ?

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Key Concepts

Deterministic Iterative Systems

In deterministic iterative systems, each state is uniquely determined by the previous one, so if two sequences have the same state at any index, all future states will be the same. This uniqueness property implies that once the sequence values coincide at a particular index, the entire iteration thereafter must be identical for both solutions.

Periodicity in Iterative Processes

If two solutions of an iterative process coincide at different indices, say at k? and k? with k? ? k?, then the system exhibits a recurrence property. Specifically, the state at k? reappears at k?, suggesting that the sequence is periodic with period equal to the difference k? - k?. This periodicity reflects a deeper invariant property of the transformation governing the iteration.

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