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 Posted: Tue, 25th Jan 2011 17:15 Post subject: Small math problem |
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Just a small problem that I encouter on a regular basis.
Let's say I have a term like
-x^3-x^2+2x
Now good old google can tell me I can convert this into
-x(x-1)(x+2)
I can get to -x(x^2+x-2) but how on earth am I supposed to see/know that (x^2+x-2)= (x-1)(x+2)?
And that's even a rather easy example but still I don't get it.
I suppose there's some very easy method of doing this but somehow I think I never heard of one in class.
p.s.
If you wanna know what I do that for.
-x^3-x^2+2x is the characteristic polynomial of a matrix and I'm trying to find the eigenvalues and eigenvectors.
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garus
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| Posted: Tue, 25th Jan 2011 17:19 Post subject: |
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snip
Last edited by garus on Tue, 27th Aug 2024 21:42; edited 1 time in total
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| Posted: Tue, 25th Jan 2011 17:23 Post subject: |
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NEVERMIND
| sabin1981 wrote: | | Fuck you troll. Fuck you and your entire aids-infested family. Get cancer and die. Slowly. |
Last edited by Mussolinka on Tue, 25th Jan 2011 17:26; edited 1 time in total
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| Posted: Tue, 25th Jan 2011 17:24 Post subject: |
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you ned to find the values where the term equals zero.
In this case, the first one is easy. obviously if x=0, f(x)=0.
Now divide: -x^3-x^2+2x : -x = x^2+x-2
you can now use pq-formula to find the last two. (1; -2).
hope that helps
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| Posted: Tue, 25th Jan 2011 18:05 Post subject: |
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@Reklis basically what garus said. Can you give the complete problem?
"Quantum mechanics is actually, contrary to it's reputation, unbeliveably simple, once you take the physics out."
Scott Aaronson | chiv wrote: | | thats true you know. newton didnt discover gravity. the apple told him about it, and then he killed it. the core was never found. |
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| Posted: Tue, 25th Jan 2011 18:33 Post subject: |
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satz(gruppe) von vieta. google it.
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| Posted: Wed, 26th Jan 2011 00:16 Post subject: |
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in order to use polynomial division you'll have to guess the first root. however you might not get lucky with the root (example: x=1,-1,0,2,-2) do you use any theorem or do you just guess your way forward?
there's a special theorem that you could use to find the roots, if you need it i can dig it up from my book and paste it with an example. but usually with problems like that they're nice and give you something even in order to do it all without a calculator.
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| Posted: Wed, 26th Jan 2011 00:27 Post subject: |
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@Razacka2
Not using any theorem.
I just pulled the idea to use the polynomial division out of my ass.
I just figure that they won't use something that would cause trouble.
If you can give me another way to solve this, let me hear it, it's just the only way I could come up with to solve this at all.
| sabin1981 wrote: | Now you're just arguing semantics. Getting fucked in the ass with a broom stale is an "improvement" over getting stabbed in the eye with a fork  |
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| Posted: Wed, 26th Jan 2011 01:14 Post subject: |
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Here  http://mathbin.net/58018
You do have degeneracy issue, but that's not that bad. Also Mathematica is great for checking if you have the right answer
"Quantum mechanics is actually, contrary to it's reputation, unbeliveably simple, once you take the physics out."
Scott Aaronson | chiv wrote: | | thats true you know. newton didnt discover gravity. the apple told him about it, and then he killed it. the core was never found. |
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| Posted: Wed, 26th Jan 2011 13:36 Post subject: |
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No that I understand how to get the eigenvalues I still struggle with the eigenvectors.
So if lambda = 5 it looks like this
http://mathbin.net/58034
and I might be stupid but I can't solve that. All I get is 0 = 0 and that isn't right :/
| sabin1981 wrote: | | Now you're just arguing semantics. Getting fucked in the ass with a broom stale is an "improvement" over getting stabbed in the eye with a fork |
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| Posted: Wed, 26th Jan 2011 13:50 Post subject: |
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It's zero because of degeneracy. You have
2a_1-11a_2+2a_3=0 that means that you can arbitrarily find vectors that will satisfy a_1=11/2a_2-a_3, and you'll have your eigenvector. You can arbitrarily choose one to be say, 0 and see what'll follow.
In Mathematica I get:ž
{-1, 0, 1}
{11, 2, 0}
{0, 0, 0}
As eigenvectors
"Quantum mechanics is actually, contrary to it's reputation, unbeliveably simple, once you take the physics out."
Scott Aaronson | chiv wrote: | | thats true you know. newton didnt discover gravity. the apple told him about it, and then he killed it. the core was never found. |
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| Posted: Wed, 26th Jan 2011 14:18 Post subject: |
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Derp. I feel sort of stupid right now, thanks again dingo.
Gonna solve some more fo those so that I can be sure that I really got it now.
| sabin1981 wrote: | | Now you're just arguing semantics. Getting fucked in the ass with a broom stale is an "improvement" over getting stabbed in the eye with a fork |
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| Posted: Wed, 26th Jan 2011 14:21 Post subject: |
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(x^2+x-2)
(x^2 -x + 2x - 2) | add +x -x
(x(x-1) + 2(x-1) | extract x and 2 | or if pro extract directly (x-1) from previous
(x-1)(x+2) | extract (x-1)
thus (x^2+x-2) = (x-1)(x+2)
9-th grade stuff i usually can't remember lol
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| Posted: Wed, 26th Jan 2011 15:52 Post subject: |
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@dingo_d
I fear I'm still struggling a bit
How do you get this?
| Quote: | | 2a_1-11a_2+2a_3=0 that means that you can arbitrarily find vectors that will satisfy a_1=11/2a_2-a_3 |
If I solve 2a_1-11a_2+2a_3=0
I get
a_1 = (11a_2-2a_3)/2?
But I guess I'm doing it wrong :/
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| Posted: Wed, 26th Jan 2011 16:20 Post subject: |
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Well you put \lambda=5 in the matrix and multiply that matrix by the column vector (a_1, a_2, a_3). You get 3 equations (the second is 0=0) and from 1st you get that.
And you get the same thing as me  you can decompose the numerator:
"Quantum mechanics is actually, contrary to it's reputation, unbeliveably simple, once you take the physics out."
Scott Aaronson | chiv wrote: | | thats true you know. newton didnt discover gravity. the apple told him about it, and then he killed it. the core was never found. |
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| Posted: Wed, 26th Jan 2011 16:32 Post subject: |
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Is the null vector defined as an eigenvector?
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| Posted: Wed, 26th Jan 2011 16:41 Post subject: |
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| Atropa wrote: | | Is the null vector defined as an eigenvector? |
That's what was kinda strange to me. I think that by definition the eigenvectors should be non zero vectors.
"Quantum mechanics is actually, contrary to it's reputation, unbeliveably simple, once you take the physics out."
Scott Aaronson | chiv wrote: | | thats true you know. newton didnt discover gravity. the apple told him about it, and then he killed it. the core was never found. |
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| Posted: Wed, 26th Jan 2011 17:05 Post subject: |
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So long ago since I worked with this :/. It's propably some general feature that a degenerate matrix haven't got eigenvectors which span the whole space.
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| Posted: Wed, 26th Jan 2011 17:14 Post subject: |
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I mean, degeneracy is a normal occurrence in quantum mechanics, but to have null vector as eigen vector wouldn't give any information of the system...
"Quantum mechanics is actually, contrary to it's reputation, unbeliveably simple, once you take the physics out."
Scott Aaronson | chiv wrote: | | thats true you know. newton didnt discover gravity. the apple told him about it, and then he killed it. the core was never found. |
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| Posted: Wed, 26th Jan 2011 17:24 Post subject: |
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Aren't alot of quantum operators unbounded? I guess we really don't care about all the energies or states at any rate.
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| Posted: Wed, 26th Jan 2011 18:14 Post subject: |
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my brain exploded upon entering this topic.
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| Posted: Wed, 26th Jan 2011 18:39 Post subject: |
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| Atropa wrote: | | Aren't alot of quantum operators unbounded? I guess we really don't care about all the energies or states at any rate. |
True, in nuclear physics, you often do not need every state, and Hamiltonian that has 100000 terms won't give you much insight in the problem 
"Quantum mechanics is actually, contrary to it's reputation, unbeliveably simple, once you take the physics out."
Scott Aaronson | chiv wrote: | | thats true you know. newton didnt discover gravity. the apple told him about it, and then he killed it. the core was never found. |
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| Posted: Wed, 26th Jan 2011 20:27 Post subject: |
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| Reklis wrote: | | dingo_d wrote: | Well you put \lambda=5 in the matrix and multiply that matrix by the column vector (a_1, a_2, a_3). You get 3 equations (the second is 0=0) and from 1st you get that.
And you get the same thing as me you can decompose the numerator:
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I guess I just read that wrong in your last post because that's what I got so I think I finally got how this works.
But it works the same for a non degeneracy case, right?
So if I get two or three different eigenvalues I gotta substitute the eigenvalues for lambda and solve the set of equation? |
Yep. Only in non degenerate case you'd get a set of equations which you'll be able to solve and normalize.
"Quantum mechanics is actually, contrary to it's reputation, unbeliveably simple, once you take the physics out."
Scott Aaronson | chiv wrote: | | thats true you know. newton didnt discover gravity. the apple told him about it, and then he killed it. the core was never found. |
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| Posted: Wed, 26th Jan 2011 21:36 Post subject: |
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Really the only equations you should remember are:
det(A-yI) = 0
Av=yv
A = matrix
y = eigen value. v = eigen vector
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| Posted: Wed, 9th Feb 2011 19:07 Post subject: |
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Let's dig this one up again for gits and shiggles.
I am trying to show that
f: [0,1] -> |R , x -> sqrt(x) is a continuous function.
I fiddled around with the whole epsilon-delta shabang but didn't get anything that looked remotely correct.
So NFOHump math crowd.
Show me the money! Err. Way I mean. Show me the way.
| sabin1981 wrote: | | Now you're just arguing semantics. Getting fucked in the ass with a broom stale is an "improvement" over getting stabbed in the eye with a fork |
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| Posted: Wed, 9th Feb 2011 19:25 Post subject: |
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Ugh real math :/. Why would any one doubt that sqrt(x) is continuous?
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garus
VIP Member
Posts: 34197
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| Posted: Wed, 9th Feb 2011 19:34 Post subject: |
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