184 reputation
6
bio website
location Michigan
age 30
visits member for 1 year, 11 months
seen 11 hours ago

Hi I am the village idiot. I like to ask dumb questions.

Good day.


Oct
16
awarded  Supporter
Oct
10
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.
Oct
10
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.
Oct
10
answered Discover where Bob is sleeping using hidden Markov chains
Sep
24
awarded  Autobiographer
May
29
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.
May
29
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.
May
29
asked Principle of superposition for linear systems
Mar
18
asked Numerical integration of function where the input data is not sampled uniformly in time
Nov
1
accepted Continuous map from $D^{n}$ to $S^{n}$
Sep
22
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.
Sep
22
revised Rotation Matix: Going from 3D to 2D by using norm
added 195 characters in body
Sep
22
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!
Sep
22
comment Rotation Matix: Going from 3D to 2D by using norm
Yes. Each frame has associated with it three basis unit vectors. So you can say FA has associated with it iA, jA, and kA, and FB has associated with it iB, jB, and kB. I am trying to compress two of them by usage of the norm.
Sep
22
asked Rotation Matix: Going from 3D to 2D by using norm
Mar
24
revised Continuous map from $D^{n}$ to $S^{n}$
added 754 characters in body
Mar
24
revised Continuous map from $D^{n}$ to $S^{n}$
added 122 characters in body
Mar
24
asked Continuous map from $D^{n}$ to $S^{n}$
Mar
17
accepted Quotient space definition
Mar
17
comment Quotient space definition
I guess I am getting too hung up on visualization.