Find P_300 Given P_299: Initial State Vector Homework

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To find the state vector P_300 given P_299=[.1 .1 .8] and the transition matrix T, the calculation involves multiplying P_299 by T. The transition matrix is defined as T = [[.3 .4 .3], [.5 0 .5], [.5 .5 0]]. Instead of calculating the inverse of T raised to the 299th power, the direct multiplication P_300 = P_299 * T can be used to obtain the state at time t=300. This approach simplifies the problem by leveraging the properties of matrix multiplication for state transitions. The discussion emphasizes that knowing P_299 and the transition matrix is sufficient to compute P_300 directly.
alexcc17
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Homework Statement


Suppose that the initial vector P0 is not given and instead we are given that at time t=299 the state vector is P_299=[.1 .1 .8].
Than find P_300, the state vector at time t=300


Homework Equations


I also know from the problem it is referencing that the transition matrix is:

[.3 .4 .3]
T= [.5 0 .5]
[.5 .5 0]


The Attempt at a Solution



I thought I would just set up the equation as such:

P_299=P0 * T^299

and just take the inverse of T^299 and multiply it by P_299 to get the P0, then use that to find P_300, but my graphing calculator won't even calculate the 299th power of the transition matrix, so there must be another way to do this.

Any help would be great!
 
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I don't see any need to worry about P0. T describes how the state changes from one timestamp to the next, right? And you know the state at time 299. And you want the state at time 300? Either I'm missing some subtlety or you are missing the obvious.
 
Would it just be p_299*T to get p_300?
 
Yes, that's what an exponent means isn't it? T300v= T(T299v)
 
Wasn't sure if that applied to matrices. Thanks
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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