Mr. Fegur
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 Apr1 awarded Popular Question Jan24 awarded Yearling Jan24 revised Discrete Kalman Filter deleted 12 characters in body Jan24 revised Discrete Kalman Filter added 139 characters in body Jan24 answered Discrete Kalman Filter Dec10 answered Which vectors $b$ in $\mathbb{R}^3$ does the matrix equation have a solution? Dec10 answered linear systems? Oct16 awarded Supporter Oct10 comment Discover where Bob is sleeping using hidden Markov chains The only reason that the probability matches what happens on night 1 is because I have assumed a uniform probability to start with. If you don't choose those, you won't get the same result. I only did night 1 because the procedure for night 2 is identical. I'm leaving it to you to use what I have provided to calculate the result for night 2. Oct10 comment Discover where Bob is sleeping using hidden Markov chains $p(x1=A|y0,y1)$ is the probability that Bob is in house A at time 1 given satellite measurements at time 0 AND time 1. I have not calculated $p(x2=A|y0,y1,y2)$, which is taking into account satellite measurements at time 2. I guess I should say that time 2 means night 2. Oct10 answered Discover where Bob is sleeping using hidden Markov chains Sep24 awarded Autobiographer May29 comment Principle of superposition for linear systems Yes I do. However, I thought that since the first system obeys homogeneity and scaling, and so does the second system, there must be some form of the superposition principle holding for this system too. Maybe I'm wrong but I don't see why it fails. May29 comment Principle of superposition for linear systems u is a known function of time, the independent variable. It looks like a linear ODE with varying coefficients. May29 asked Principle of superposition for linear systems Mar18 asked Numerical integration of function where the input data is not sampled uniformly in time Nov1 accepted Continuous map from $D^{n}$ to $S^{n}$ Sep22 comment Rotation Matix: Going from 3D to 2D by using norm Ok clearly I need to pause before I hit the enter button. In my last comment, I said "I am mostly interested in having $w_{3}$ unchanged after transformation from $B′$ to $A′$.", but what I actually meant was that I am interested in having $w_{3}$ equal to $v_{3}$ after transformation from $B'$ to $A'$, where $v_{3}$ as defined above is $v_{3}=r_{1}w_{1}+r_{2}w_{2}+r_{3}w_{3}$. Sorry for the confusion and thanks for bearing with me. Sep22 revised Rotation Matix: Going from 3D to 2D by using norm added 195 characters in body Sep22 comment Rotation Matix: Going from 3D to 2D by using norm Oh I see what you are asking now. Say $F_{B}$ has basis vectors $i_{B}$, $j_{B}$, $k_{B}$, and we write $v = w_{1}i_{B} + w_{2}j_{B}+w_{3}k_{B}$. Now when I compress $w_{1}$ and $w_{2}$ to $||w_{1},w_{2}||$, the resulting vector $\begin{bmatrix}||w_{1},w_{2}|| & w_{3}\end{bmatrix}$ is not in the $B$ frame. Well, can I just assume it is in some new frame $B'$ with basis $i_{B'}$, $k_{B'}$, $k_{B'}=k_{B}$ and similarly for $A'$, and so $v|_{n} = ||w_{1},w_{2}||i_{B'} + w_{3}k_{B'}$? I am mostly interested in having $w_{3}$ unchanged after transformation from $B'$ to $A'$.Thanks!