Recent content by Sergey S
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Diagonalizing matrix in a vector equation
I agree, I really should read a book on numerical methods. I am aware that my questions are not anything I can't find in books, and I am sorry for taking everyone's time.- Sergey S
- Post #12
- Forum: Calculus and Beyond Homework Help
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S
Diagonalizing matrix in a vector equation
I see. Yes I can do that. The reason I went for diagonalization instead of just solving the equation in the first place was because my ##y## vector is ridiculously long and I was hoping to find a way to avoid touching it as far as possible. But I can still use Cramer's rule to solve the...- Sergey S
- Post #9
- Forum: Calculus and Beyond Homework Help
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S
Diagonalizing matrix in a vector equation
Yeah I got it. That solved almost all of my problems - now I know that T exists and how to find it. At this point I just wonder if there is a more computation-friendly method than finding inverse of A; maybe this equation can yield something - SDS^{-1}x=y? I am also curious if I can isolate Dx...- Sergey S
- Post #6
- Forum: Calculus and Beyond Homework Help
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The right amount of Resources for a self-study attempt
I went through lectures of MIT 18.02 multivariable calculus by Denis Auroux a month ago, I think it is a very accessible course. They have all the lectures uploaded to youtube, it makes it pretty easy to follow. It assumes background in single variable calculus though.- Sergey S
- Post #10
- Forum: Science and Math Textbooks
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S
The right amount of Resources for a self-study attempt
Walter Lewin uses a little bit of Calculus in his courses, especially in 8.03 from what I have seen of it. If you are not comfortable with notation like \nabla \cdot B and \nabla \times E you may want to add a book on Multivariable Calculus into the cart.- Sergey S
- Post #8
- Forum: Science and Math Textbooks
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If 0<x<y and 0<a<b then ax<yb proof.
That is probably a poor way to do it, but here is what I would have done with that problem. First rewrite it in this form: x<y, a<b, where x,y,a,b are positive; Prove by-ax>0 Let's introduce a function f(y,b)=by-ax. We can say that: \lim_{y\rightarrow x, b\rightarrow a} {f(y,b)}=0 Then...- Sergey S
- Post #18
- Forum: Calculus and Beyond Homework Help
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S
Diagonalizing matrix in a vector equation
So this works for any invertible A. It almost seems too good to be true. Thank you very much! Yes I am aware of that, but I can assume that A is diagonalizable. I see that I can write diagonal matrix D using eigenvalues of A; I still don't know how I can isolate Dx in the left side of the...- Sergey S
- Post #4
- Forum: Calculus and Beyond Homework Help
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S
Diagonalizing matrix in a vector equation
Homework Statement I have a vector equation of a form: Ax=y where A is a square n \times n matrix and y, x are vectors. I want to rewrite it in a new from: Dx=Ty where D is a diagonal matrix, and T is some square n \times n matrix. The question is, is it possible to find such D and T...- Sergey S
- Thread
- Matrix Vector
- Replies: 12
- Forum: Calculus and Beyond Homework Help
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Derivative of a rotating unit vector
Thank you D H! The idea of planar rotations in four dimensions is kind of hard to digest (well, even with 3-d for me it takes effort to visualize some operations). I will probably have to do the differentiation for the 4-d case to get the feel of it, but now that I know the answer it should be...- Sergey S
- Post #5
- Forum: Calculus and Beyond Homework Help
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S
Derivative of a rotating unit vector
I have spend some time thinking about it. First, for 3-dimensional case we can use well known result from classical mechanics and say that \frac{d\vec{r}}{dt}=\vec{ω}\times\vec{r}. I don't know if the result holds for the general case. Well, at least now I know how to prove that result I've...- Sergey S
- Post #3
- Forum: Calculus and Beyond Homework Help
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Derivative of a rotating unit vector
I think this is a textbook-style question, if I am wrong, please redirect me to the forum section where I should have posted this. This is my first time here, so I am sorry if I am messing it up. Homework Statement We have an n-dimensional vector \vec{r} with a constant length \|\vec{r}\|=1...- Sergey S
- Thread
- Derivative Rotating Unit Unit vector Vector
- Replies: 4
- Forum: Calculus and Beyond Homework Help