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| Math 1b03 project 1 |
Ush.s |
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11-01-2011 05:25 AM |
11-28-2011 at 10:23 PM
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#46
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Quote:
Originally Posted by Ush.s
for number 5 I got
V = x[p, p+q, 1]^T =S
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so? solve for x= 1/(2p+q+1).
p.s did u read chapter 10.5? if dont, do it
Last edited by Differential : 11-28-2011 at 10:26 PM.
Ush.s
says thanks to Differential for this post.
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11-28-2011 at 10:26 PM
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#47
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.. and the steady state vector then becomes
1/(2p+q+1) * [p, q+p, 1]^T
when plugging in 0.02 and 0.28 for p and q respectively, I get a steady state vector of
[1/66, 5/22, 25/33]^T
and this does not satisfy
(I-A)*[1/66, 5/22, 25/33]^T = 0
so.. I'm confused o_o
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11-28-2011 at 10:28 PM
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#48
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Quote:
Originally Posted by Ush.s
.. and the steady state vector then becomes
1/(2p+q+1) * [p, q+p, 1]^T
when plugging in 0.02 and 0.28 for p and q respectively, I get a steady state vector of
[1/66, 5/22, 25/33]^T
and this does not satisfy
(I-A)*[1/66, 5/22, 25/33]^T = 0
so.. I'm confused o_o
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use original one, not (I-A) |
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11-28-2011 at 10:31 PM
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#49
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@Bug: Facebook? o___o Is there a 1B03 HW group or something?
@omgmt: YOU'RE IN PNB! AND TAKING MATH 1B03! Would that mean you also have Cognition not on the 17th but the 19th? Am I not alone?  |
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11-28-2011 at 10:31 PM
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#50
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original what? (λI-A), where λ = 1 ?
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11-28-2011 at 10:33 PM
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#51
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Quote:
Originally Posted by Ush.s
original what? (λI-A), where λ = 1 ?
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Initial dynamical system i.e V(t+1)=AV(t)
p.s lamda always 1 for steady state
Last edited by Differential : 11-28-2011 at 10:38 PM.
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11-28-2011 at 10:48 PM
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#53
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Quote:
Originally Posted by Ush.s
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oh...
http://imageshack.us/f/832/21859393.png/
p.s why so much hate for math? lol
Ush.s
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11-28-2011 at 10:58 PM
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#54
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Quote:
Originally Posted by Faer
@omgmt: YOU'RE IN PNB! AND TAKING MATH 1B03! Would that mean you also have Cognition not on the 17th but the 19th? Am I not alone? |
LOL yupp! I'm so glad we get an extra two days to study for cog, cuz seriously I need it. That course is killing my average so bad. Plus I'm trying to work on this math project while doing the cog. paper... fun times . . . O__e |
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11-28-2011 at 11:11 PM
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#55
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Quote:
Originally Posted by Differential
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I don't understand anything after out[17]
out[17] is just the same as my values for q
but i dont see where in[19] comes from
=[
btw, I totally owe you a beer or two, thanks soooo much for helping me out =)
nvm
I made another stupid mistake
was typing "32000" instead of "13200" >.<
ughhh
Last edited by Ush.s : 11-28-2011 at 11:25 PM.
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11-28-2011 at 11:11 PM
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#56
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Quote:
Originally Posted by Faer
@Bug: Facebook? o___o Is there a 1B03 HW group or something?
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http://www.facebook.com/groups/254177191292039/
__________________
Honours Chemistry '15
Faer
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11-28-2011 at 11:39 PM
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#57
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hello;
This is what I am doing when I attempt to find X1, X2 and X3:
I computed the eigenvalues and eigenvectors form A:
eigenvalues:
lambda1 = 1
lambda2 = (-1/5)
lambda3 = (-1/10)
v1 = {(1/50),(3/10),1}
v2 = {(-1/10),(-9/10),1}
v3 = {(-1/5),(-4/5),1}
I am assuming that to compute X1 X2 and X3 I would multiply by the factor which would make the first value in each of the vectors equal to 1. I have done that and my calculation is equal to:
X1 = 50
X2 = -10
X3 = -5
However when I try and use these values in question 4, I know it's not right. I'm ending up with negative values (you can't have negative components), and moreover I'm not getting the vector to add up to a total of 13200 components. Can someone point me in the right direction. I am almost 100% sure I computed the eigenvalues correctly and I don't believe what I'm doing is a major problem, it's problem a slight calculation error.
I'm thinking it might be because i didn't multiply the state vector of {0,0,13200} into my original matrix A (i am not sure if that is what you are supposed to do, purely speculating). Help would be much appreciated. Thanks a lot.
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11-28-2011 at 11:41 PM
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#58
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Quote:
Originally Posted by Ush.s
I don't understand anything after out[17]
out[17] is just the same as my values for q
but i dont see where in[19] comes from
=[
btw, I totally owe you a beer or two, thanks soooo much for helping me out =)
nvm
I made another stupid mistake
was typing "32000" instead of "13200" >.<
ughhh
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I am muslim and I dont drink alcohol, thanks though. And as for help, I dont see any reason not to help since I already handed in and others getting same mark doesnt affect my mark anyways |
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11-29-2011 at 12:03 AM
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#59
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Quote:
Originally Posted by redwine
hello;
This is what I am doing when I attempt to find X1, X2 and X3:
I computed the eigenvalues and eigenvectors form A:
eigenvalues:
lambda1 = 1
lambda2 = (-1/5)
lambda3 = (-1/10)
v1 = {(1/50),(3/10),1}
v2 = {(-1/10),(-9/10),1}
v3 = {(-1/5),(-4/5),1}
I am assuming that to compute X1 X2 and X3 I would multiply by the factor which would make the first value in each of the vectors equal to 1. I have done that and my calculation is equal to:
X1 = 50
X2 = -10
X3 = -5
However when I try and use these values in question 4, I know it's not right. I'm ending up with negative values (you can't have negative components), and moreover I'm not getting the vector to add up to a total of 13200 components. Can someone point me in the right direction. I am almost 100% sure I computed the eigenvalues correctly and I don't believe what I'm doing is a major problem, it's problem a slight calculation error.
I'm thinking it might be because i didn't multiply the state vector of {0,0,13200} into my original matrix A (i am not sure if that is what you are supposed to do, purely speculating). Help would be much appreciated. Thanks a lot.
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200[1,15,50]+400[1,4,-5]-100[1,9,-10]=[500,3700,9000] ?
redwine
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11-29-2011 at 12:08 AM
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#60
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i think you switched the 100 and 400 in that 
its the other way around I believe |
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