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  • one year ago

Is it possible for A^2=0 and A is not equal to 0?

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  1. Empty
    • one year ago
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    I'll admit, I'm being ambiguous on purpose here haha.

  2. ganeshie8
    • one year ago
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    \[A=\begin{bmatrix}0&\alpha \\0&0\end{bmatrix}\]

  3. Empty
    • one year ago
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    Alright you win, I can't pull a fast one on you for a nilpotent matrix.

  4. Empty
    • one year ago
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    https://en.wikipedia.org/wiki/Dual_number

  5. ganeshie8
    • one year ago
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    interesting, is that the only matrix form with degree 2 ?

  6. Empty
    • one year ago
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    Yeah, including the transpose I think so. It's interesting because dual numbers are like the complex number representation of a matrix, only instead of giving us rotations they give us shearing transformations instead when we multiply them.

  7. ganeshie8
    • one year ago
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    so they do stuff like transforming a square into rhombus etc is it

  8. Empty
    • one year ago
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    So for contrast to see more vividly: \[a+\epsilon b= a\left[ \begin{array}c 1 & 0\\0 & 1\\\end{array} \right] + b\left[ \begin{array}c 0 & 1\\0 & 0\\\end{array} \right] = \left[ \begin{array}c a & b\\0 & a\\\end{array} \right]\] \[a+i b= a\left[ \begin{array}c 1 & 0\\0 & 1\\\end{array} \right] + b\left[ \begin{array}c 0 & 1\\-1 & 0\\\end{array} \right] = \left[ \begin{array}c a & b\\-b & a\\\end{array} \right]\]

  9. Empty
    • one year ago
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    I just found this now so I don't really know anything much but you can tell by simply multiplying out the rules: \[(a+b \epsilon)(x+y \epsilon) = ax+(bx+ay)\epsilon\] and the part \(-by\) we would normally have is thrown away which is the difference between the rotation and shear! Interesting!

  10. ganeshie8
    • one year ago
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    Okay I have a little experience with transformation matrices, so im trying to relate that with these nilpoint matrices Consider a 2 dimensional point on plane \(z=1\) : \((x,y,1)\) \[\begin{pmatrix}x'\\y'\\1\end{pmatrix}=\begin{bmatrix} a&\color{red}{c}&e\\\color{red}{b}&d&f\\0&0&1\end{bmatrix}\begin{pmatrix}x\\y\\1\end{pmatrix}\] if i remember correctly those red numbers represent the shear effect

  11. Empty
    • one year ago
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    Ahhh perhaps what I've written is confusing for the same reason that if I write out: \[(a+b i)(x+y i) = (ax-by)+(bx+ay)i\] This represents not only a rotation but a stretching. By the same reasoning the equation below is not only strictly shearing, but also extension at the same time. \[(a+b \epsilon)(x+y \epsilon) = ax+(bx+ay)\epsilon\] I'm essentially using their line her to come to this conclusion from the wikipedia article: http://prntscr.com/7o8mzb Specifically in my mind the only difference is the real part (x axis) of the complex numbers is shifted to the left by \(yb\) while in the dual numbers we aren't shifting left by \(yb\) and are instead solely moving upwards (and stretching in this case) but in that article they do a pure shear I believe.

  12. Empty
    • one year ago
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    Specifically I'm interested in grasping the "geometry" section of the article here: https://en.wikipedia.org/wiki/Dual_number#Geometry which should answer our questions about shearing.

  13. Empty
    • one year ago
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    Just playing around right now, I noticed there's an interesting form for powers of dual numbers: \[d=a+b \epsilon\] \[d^n = (a+ b \epsilon)^n = a^n + n a^{n-1}b \epsilon\] Which is just the first two terms of the binomial theorem. I'm sort of just playing around and trying to figure it out since I don't really understand this but it's interesting and new and might have some interesting uses for solving problems since it's so simple to use.

  14. UsukiDoll
    • one year ago
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    don't have an A matrix with 0 entires ...then A^2 won't be equal to 0

  15. UsukiDoll
    • one year ago
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    A^2 = A(A) with matrix multiplication

  16. Empty
    • one year ago
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    I'm talking about this matrix specifically: \[\left[ \begin{array}c 0 & 1\\0 & 0\\\end{array} \right] \left[ \begin{array}c 0 & 1\\0 & 0\\\end{array} \right] = \left[ \begin{array}c 0 & 0\\0 & 0\\\end{array} \right]\]

  17. UsukiDoll
    • one year ago
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    ohhhhhhh

  18. anonymous
    • one year ago
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    I thought it's not possible. I wasn't thinking about matrices.

  19. ganeshie8
    • one year ago
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    nice thats because \(\epsilon^k=0\) for \(k\gt 1\)

  20. Empty
    • one year ago
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    In effect it seems as though I've really calculated more impressive looking haha: \[\left[ \begin{array}c a & b\\0 & a\\\end{array} \right]^n = a^n\left[ \begin{array}c 1 & \frac{nb}{a}\\0 & 1\\\end{array} \right]\]

  21. anonymous
    • one year ago
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    What about: \[ \delta = \begin{bmatrix} 0&0 \\ 1&0 \end{bmatrix} \implies \delta^2= \begin{bmatrix} 0&0 \\ 1&0 \end{bmatrix}^2 =\begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix} = 0 \]And we might think: \[ i = \epsilon - \delta \]

  22. anonymous
    • one year ago
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    Is there any way to compare dual numbers?

  23. anonymous
    • one year ago
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    Actually it seems \(\delta = \epsilon^T\), so we would say: \[ i = \epsilon - \epsilon^T \]

  24. Empty
    • one year ago
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    Interesting one application of dual numbers is that they allow you to do automatic differentiation! Which makes sense I think when you look at them. http://prntscr.com/7o8sqt So specifically I'm guessing because we have \[(a+ b \epsilon)^n = a^n+na^{n-1} b \epsilon\] we subtract a^n and divide by epsilon (well whatever that means). Since these are not invertible. I don't know, that's a good question @wio . One thing I'm noticing is we can possibly make a "dual conjugate" which makes the epsilon term negative but what about if we transpose the matrix, that almost seems to correspond to another operation altogether too.

  25. Empty
    • one year ago
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    Let's suppose we define \[\epsilon^* = -\epsilon \] \[\epsilon ^T\]and allow us to just put a transpose on it to represent the transposed matrix, there's no other thing we will do. Then if we take the conjugate transpose of a dual number and add it to itself we get a complex number.

  26. Empty
    • one year ago
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    Example: \[(a+b \epsilon)^* = a-b \epsilon \]\[(a+b \epsilon)^T = a+b \epsilon^T\] Combine these operations to make a "Hermitian" operator:\[(a+ \epsilon)^H = a-b \epsilon^T\] Now we add: \[(a+b \epsilon) + (a+ b \epsilon)^H = 2a+bi\] Hmm... good idea @wio!!!

  27. Empty
    • one year ago
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    Interesting thing I've just read is by considering the power series of a function it naturally falls out that: \[f(a+b \epsilon) = f(a)+b f'(a) \epsilon\]

  28. Empty
    • one year ago
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    I wonder if there's a quaternion analog for dual numbers as well

  29. anonymous
    • one year ago
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    \[ \epsilon^T\epsilon = \begin{bmatrix} 0&0 \\ 1&0 \end{bmatrix}\begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix} = \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix} \]Therefore \[ \epsilon \epsilon^T +\epsilon^T\epsilon = 1 \]And we can say: \[ a\epsilon\epsilon^T+b\epsilon+c\epsilon^T+d\epsilon^T\epsilon = \begin{bmatrix} a&b \\ c&d \end{bmatrix} \]From \(\epsilon\) and its transpose, you can get a full \(2\times2\) matrix.

  30. Empty
    • one year ago
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    Interesting, we can get to all 4 parts by algebra and kind of throwing away the matrix in a sense. Hmmm this seems more fundamental than complex numbers in a sense.

  31. anonymous
    • one year ago
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    I don't think there is a single means of generalizing dual numbers. A while back I came up with a way to generalize imaginary numbers that didn't lead to quaternions.

  32. Empty
    • one year ago
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    Hmmm well in this light it almost feels as though we have a companion to the "minus sign" and we are using the "transpose" as sort of another minus sign in a different sense here. Specifically when I said the thing about quaternions I meant the Cayley-wingspanson construction https://en.wikipedia.org/wiki/Cayley%E2%80%93wingspanson_construction

  33. anonymous
    • one year ago
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    What happens when you do \(\epsilon^0\)? Are we saying it is \(1\) or indeterminate?

  34. Empty
    • one year ago
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    When I say more fundamental I also mean that we can see that all rotations are really a specific form of combining two shears. Now that we've decoupled the shears into two parts, it seems like maybe we'd be able to rotate along an ellipse or hyperbola if we desire. According to the wikipedia article \(\epsilon^0 = 1\) from calculating it here: http://prntscr.com/7o8yrw

  35. Empty
    • one year ago
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    I am fine with the idea of interpreting it as the identity matrix I think.

  36. Empty
    • one year ago
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    I feel as though instead of previously we were just throwing complex numbers into a matrix because it happened to work, now it seems to be crucial since we are now giving each entry of the matrix a specific role to play in our algebra. One shears in one axis, one shears in the other axis, one stretches in one axis and the other stretches in the other axis. At least I believe this is what we can now understand a 2x2 matrix to represent.

  37. Empty
    • one year ago
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    Complex numbers are just the case when the shearing is identical and the stretching on the axes is identical which causes circular motion and uniform stretching that commutes with it. I could be wrong on this but this is my hunch, I don't know, this is the direction I'm sorta being dragged along I guess haha weird stuff.

  38. anonymous
    • one year ago
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    Makes me wonder what happens if you do something like: \[ a+b\alpha = \begin{bmatrix} a+b& b \\ 0&a \end{bmatrix} \]It looks like \[ (b\alpha)^2= \begin{bmatrix} b&b \\ 0&0 \end{bmatrix}^2= \begin{bmatrix} b^2&b ^2\\ 0&0 \end{bmatrix}= b^2\alpha \]

  39. anonymous
    • one year ago
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    But might not be closed under multiplication:

  40. Empty
    • one year ago
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    Interesting yeah. \[(aI+b \alpha)^2 = a^2I+2ab \alpha + b^2 \alpha^2\] We can easily and quickly show \[\alpha^2=\alpha\] so \[(aI+b \alpha)^2 = a^2I+(2ab + b^2) \alpha\]

  41. anonymous
    • one year ago
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    I had just computed: \[ (a+b\alpha)(c+d\alpha) = ac+(ad+bc+bd)\alpha \]

  42. Empty
    • one year ago
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    Another thing that I see them saying is that no dual number has an inverse but you can still do division. I think this is essentially the same idea as clifford algebra division of vectors kind of thing but it doesn't exactly translate to matrices as we know them so I'd like to get a better foundation on this operation: \[\frac{a+b \epsilon}{x+ y \epsilon} = \frac{a+b \epsilon}{x+ y \epsilon} \frac{x- y \epsilon}{x- y \epsilon} = \frac{a}{x} + \frac{ bx-ay}{x^2} \epsilon\]

  43. anonymous
    • one year ago
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    \[ (a+b\alpha)^n = \sum_{k=0}^n{n\choose k}a^{n-k}(b\alpha)^k = a^n\alpha^0+\sum_{k=1}^n{n\choose k}a^{n-k}b^k\alpha \]Not sure what \(\alpha^0\) would look like.

  44. Empty
    • one year ago
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    I think in general we can say that for matrices \(A^0 = I\)

  45. Empty
    • one year ago
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    As long as A isn't the zero matrix. The reasoning for why that would be a suitable definition is just by how we define the exponent rules: \(A^{n+m} = A^nA^m\)

  46. anonymous
    • one year ago
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    Actually, that does make sense here: \[ (a+b\alpha)^n = a^n + \left(\sum_{k=1}^n{n\choose k}a^{n-k}b^k\right)\alpha \]Seems similar to what you got for \(n=2\).

  47. anonymous
    • one year ago
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    How would you define modulus or magnitude for dual numbers?

  48. anonymous
    • one year ago
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    I guess it's really asking what your inner product would be.

  49. anonymous
    • one year ago
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    I think that imaginary numbers would use the determinate for the inner product.

  50. Empty
    • one year ago
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    To find out what an inverse dual matrix representation would be, I calculated out multiplying a matrix with values and plugged it in to get: \[\left[ \begin{array}c \frac{1}{x} & -y\\0 & \frac{1}{x}\\\end{array} \right]\] is the inverse of \[\left[ \begin{array}c x & y\\0 & x\\\end{array} \right]\] Which supposedly doesn't exist but now I'm curious if multiplying these gives us 1. Inner product is a good question, I don't know. Since we can turn a dual into a complex number we might be able to just hijack that

  51. anonymous
    • one year ago
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    But for dual numbers, the determinate seems to give \(\langle a+b\epsilon, a+b\epsilon\rangle = a^2\).

  52. anonymous
    • one year ago
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    For the alpha concept, it would seem determinate inner product would give \[ \langle a+b\alpha, a+b\alpha\rangle = a^2+ab \]

  53. Empty
    • one year ago
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    I guess I don't know if I even understand these things geometrically yet in the "dual plane". While I can visualize complex number multiplication I can't quite visualize dual multiplication yet, so I guess that's what's to look into now for me.

  54. anonymous
    • one year ago
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    Hmmm \[(a+b\alpha)^{-1}=\begin{bmatrix} a+b&b\\0&a \end{bmatrix}^{-1} =\frac{1}{a^2+ab}\begin{bmatrix} a&-b\\0&a+b \end{bmatrix}^{-1} =\frac{(a+b)+-b\alpha}{a^2+ab} \]

  55. Empty
    • one year ago
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    So while \(e^{i t} = \cos t + i \sin t\) it appears that we have \(e^{\epsilon t} = 1+ \epsilon t\) which is supposedly some kind of parabolic rotation. I'm not sure, I'm just about to download and skim this paper: http://arxiv.org/abs/0707.4024

  56. Empty
    • one year ago
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    See @wio I said earlier I was studying probability and now I'm onto this already haha

  57. Empty
    • one year ago
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    Interesting, another alternative is that we can use a different matrix other than i or epsilon to create a hyperbola.

  58. anonymous
    • one year ago
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    \[ e^{\epsilon t} = \sum_{n=0}^\infty \frac{(\epsilon t)^n}{n!} = 1 + \epsilon t + \sum_{n=2}^\infty \frac{(\epsilon t)^n}{n!} \]Where \[ \sum_{n=2}^\infty \frac{(\epsilon t)^n}{n!} = \sum_{k=0}^\infty \frac{(\epsilon t)^{k+2}}{(k+2)!} = \sum_{k=0}^\infty \frac{0\cdot (\epsilon t)^k}{(k+2)!} = 0 \]

  59. Empty
    • one year ago
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    All of the shapes traced out by the "rotations" are defined by this single function: \[x^2-i^2y^2=1\] So if i is the traditional imaginary number we have \[x^2+y^2=1\] if i is our new dual epsilon we have: \[x^2=1\] and if we have this other alternative hyperbolic i we have \[x^2-y^2=1\] I wonder what matrix they're using for this hyperbolic i that squares to 1 but isn't 1.

  60. Empty
    • one year ago
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    Probably \[\left[ \begin{array}c 0 & 1\\1 & 0\\\end{array} \right]\]

  61. Empty
    • one year ago
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    This is illuminating: http://prntscr.com/7o9ctw

  62. anonymous
    • one year ago
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    One way to generalize dual numbers: \[ \epsilon_3 = \begin{bmatrix} 0&1&0 \\ 0&0&1 \\ 0&0&0 \end{bmatrix}, (\epsilon_3)^2 = \begin{bmatrix} 0&0&1 \\ 0&0&0 \\ 0&0&0 \end{bmatrix} \]We have: \[ (\epsilon_3)(\epsilon_3)= (\epsilon_3)^2 \\ (\epsilon_3)^3 = (\epsilon_3)^2(\epsilon_3) = 0 \]This is similar to my previous generalization of imaginary numbers, where tried to start with an identity: \[ (i_3)^3 = -1 \]It turned out that \(i_3\) was just a 6th of a rotation around the unit circle, if you projected onto the imaginary plane.

  63. anonymous
    • one year ago
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    complex plane

  64. anonymous
    • one year ago
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    However, the 'quaternion' approach to it might be more along the lines of: \[ \epsilon = \begin{bmatrix} 0&0&1&1 \\ 0&0&1&1 \\ 0&0&0&0 \\ 0&0&0&0 \end{bmatrix} \]Or \[ \epsilon = \begin{bmatrix} 0&0&0&1 \\ 0&0&0&0 \\ 0&0&0&0 \\ 0&0&0&0 \end{bmatrix} \]

  65. Empty
    • one year ago
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    The amazing parabolic sine and cosine functions lol, and they satisfy the hilarious \[e^{\epsilon t} = cosp(t) + \epsilon sinp(t) = 1+\epsilon t\] \[sinp(x) = x=\frac{e^{\epsilon x}-e^{-\epsilon x}}{2 \epsilon} \]\[cosp(x)=1=\frac{e^{\epsilon x}+e^{-\epsilon x}}{2 }\] \[\frac{d}{dx} sinp(x) = cosp(x)\]\[\frac{d}{dx} cosp(x) = 0\] So at least it feels natural in a stupid way.

  66. Empty
    • one year ago
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    Ah interesting I see what you're saying. I think I am sort of seeing how pauli spin matrices might just be how we write quaternions using matrices and complex numbers or something, I'm not sure.

  67. Empty
    • one year ago
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    Out of curiosity can we extend Fermat's Last theorem to matrices? \[A^n+B^n=C^n\]

  68. anonymous
    • one year ago
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    \[ e^{\alpha t} = \sum_{n=0}^\infty \frac{(\alpha t)^n}{n!} = \sum_{n=0}^\infty \frac{\alpha^n t^n}{n!} = 1+ \sum_{n=1}^\infty \frac{\alpha t^n}{n!} \]We can can start that \(n\) back at zero by adding and subtracting \(\alpha\):\[ 1+ \sum_{n=1}^\infty \frac{\alpha t^n}{n!} + \alpha t^0 - \alpha t^0 = 1 - \alpha + \alpha\sum_{n=0}^\infty \frac{t^n}{n!} = 1-\alpha + \alpha e^{t} \]In summary:\[ e^{\alpha t} =1-\alpha + \alpha e^{t} \]

  69. Empty
    • one year ago
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    This is very illuminating to me since the difference between these three matrices is simply the bottom left value being -1, 0 or 1. http://prntscr.com/7o9h0g It almost seems like we should consider all of these simultaneously with a value in the bottom left that we can change spaces with.

  70. anonymous
    • one year ago
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    Which could also be said as: \[ e^{\alpha t} = 1 + (e^t-1)\alpha \]So we could say: \[ \cos\alpha(t) = 1 \\ \sin\alpha(t) = e^t-1 \\ \]Very weird result.

  71. anonymous
    • one year ago
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    Well, not completely sure about that last one but

  72. Empty
    • one year ago
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    \[\left[ \begin{array}c 0 & 1\\s & 0\\\end{array} \right]^2 = s\left[ \begin{array}c 1& 0\\0 & 1\\\end{array} \right]\] So we have this matrix where the bottom left entry is the value of the square. 0, 1, or -1. How are you coming to that result wio?

  73. anonymous
    • one year ago
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    I am just using the weird identity: \[ e^{\alpha t} = \cos_\alpha(t) + \alpha \sin_\alpha(t) = 1+(e^t-1)\alpha \]

  74. anonymous
    • one year ago
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    Though I'm not sure what the underlying assumptions of that identity are.

  75. anonymous
    • one year ago
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    It's more of a definition though, so...

  76. Empty
    • one year ago
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    well as long as you define sine and cosine by this structure: \[sinp(x) = x=\frac{e^{\epsilon x}-e^{-\epsilon x}}{2 \epsilon} \]\[cosp(x)=1=\frac{e^{\epsilon x}+e^{-\epsilon x}}{2 }\] I think you will in a good position.

  77. Empty
    • one year ago
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    I'll try to play around with the alpha values a bit to see if I get to it as well

  78. anonymous
    • one year ago
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    That structure? Why?

  79. Empty
    • one year ago
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    It produces an even and an odd function that add to the exponential function because: \[f(x) = \frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}=g(x)+h(x)\] and we have the identities that g(x)=g(-x) and h(x)=-h(-x) like we expect from a sine or cosine.

  80. Empty
    • one year ago
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    Also I'm noticing we might be able to generalize your matrix a bit: \[\left[ \begin{array}c s & 1\\0 & 0\\\end{array} \right]^2=s\left[ \begin{array}c s & 1\\0 & 0\\\end{array} \right]\] or in other words \[\alpha_s^2=s \alpha_s\] So you have some control over what the square is, you can make s=0, 1, or -1 for different interesting but also easier to track properties. Something to play with.

  81. anonymous
    • one year ago
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    Let's see \[ e^{-\alpha t} = 1 + (e^{-t}-1)\alpha \]So \[ e^{\alpha t}+e^{-\alpha t} = 2 + \alpha\bigg(e^t-1 + e^{-t}-1\bigg) = 2+\alpha (e^t+e^{-t}-2) \]So we have: \[ \frac{e^{\alpha t}+e^{-\alpha t} }{2} = 1+\cosh(t) \]For the other one: \[ e^{\alpha t}- e^{-\alpha t} =\alpha\bigg(e^t - e^{-t}\bigg) = \]So \[ \frac{e^{\alpha t}- e^{-\alpha t}}{2\alpha} = \sinh(t) \]

  82. anonymous
    • one year ago
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    Hmmm, the only issue thought is that I know division by \(\alpha\) is problematic, just as it is for \(\epsilon\).

  83. Empty
    • one year ago
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    It is weird I'll agree with you hmm.\[\frac{a+b \alpha}{x+ y \alpha} = \frac{a+b \alpha}{x+ y \alpha} \frac{a-b \alpha}{x- y \alpha} = \frac{ax}{x^2-y^2} + \alpha \frac{bx-ay-by}{x^2-y^2}\]

  84. anonymous
    • one year ago
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    For \(a+b\epsilon\) we have division problems at \(a=0\) (regardless of \(b\)), and for \(a+b\alpha\) we have division problems at \(a=-b\). You could say that for complex we have division problems at \(a^2+b^2=0\) for obvious reasons.

  85. Empty
    • one year ago
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    Yeah, I am not sure how I feel about "dividing by a matrix" but somehow it seems to just sorta work even though it definitely has problems going on.

  86. anonymous
    • one year ago
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    I think we have to say that at these places, the magnitude ought to be defined as \(0\).

  87. anonymous
    • one year ago
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    Everything we have done so far can be generalized as "Using a \(2\times 2\) matrix to define multiplication for a \(2\) component vector."

  88. Empty
    • one year ago
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    Hahaha yeah basically.

  89. Empty
    • one year ago
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    But I think we have some interesting relationship that the matrices satisfy a relation like this: \[A^2=sA\] or \[A^2=sI\] So there's some sort of structure here. Hmmm. It's sorta all nonsense in a way though unless we can get something out of it I think.

  90. anonymous
    • one year ago
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    We could even do Hadamard matrix for this: \[ a+bH = \begin{bmatrix} a+b & a+b \\ a+b & a-b \end{bmatrix} \]It appears that \(H^2 = 2I\), so: \[ (a+bH )(c+dH ) = (ac+2bd)+(ad+bc)H \]

  91. anonymous
    • one year ago
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    That one quantum computing tutorial said something about Unitary matrices or something being special.

  92. Empty
    • one year ago
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    Oh interesting. I am sort of getting a new idea from this. Yeah a unitary matrix is basically just the complex version of an orthogonal matrix. An orthogonal matrix is the inverse of its transpose while a unitary matrix is the inverse of its hermitian, which is the conjugate transpose. The DFT uses this for instance.

  93. Empty
    • one year ago
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    \[M=\left[ \begin{array}c a & b\\c & d\\\end{array} \right]=a\left[ \begin{array}c 1 & 0\\0 & 0\\\end{array} \right]+b\left[ \begin{array}c 0 & 1\\0 & 0\\\end{array} \right]+c\left[ \begin{array}c 0 & 0\\1 & 0\\\end{array} \right]+d\left[ \begin{array}c 0 & 0\\0 & 1\\\end{array} \right]\] \[M=aA+bB+cC+dD\] So now if I square this I'm going to get a more in depth understanding of how each part influences the square rather than looking at our specific cases like we've been doing.

  94. Empty
    • one year ago
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    \[M= \left[ \begin{array}c A & B & C & D\\\end{array} \right]\left[ \begin{array}c a\\ b\\ c\\ d\\\end{array} \right]\] So now I've got a matrix of matrices.... Maybe this isn't the right path to take on this haha. Nevermind.

  95. anonymous
    • one year ago
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    \[ H^n = \begin{cases} 2^{k}I & n=2k\\ 2^{k}H & n=2k+1 \end{cases} \]I think the power of \(n\) is important, since we can use it in the MacLaurin series of \(e^x\).

  96. Empty
    • one year ago
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    Yeah I think you're right about that. Splitting it up like you've done is also going to make it nice for thinking about the sine and cosine functions

  97. anonymous
    • one year ago
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    \[ e^{tH} = \sum_{n=0}^\infty \frac{(tH)^n}{n!} = \sum_{k=0}^\infty \frac{t^{2k}}{(2k)!} + H\sum_{k=0}^\infty \frac{t^{2k+1}}{(2k+1)!} \]Hmmm, not quite trignometric, since we don't have the \(-1\) coefficient.

  98. anonymous
    • one year ago
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    Oh wow. \[ e^{tH} = \cosh(t) + H\sinh(t) \]

  99. anonymous
    • one year ago
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    That is funky

  100. anonymous
    • one year ago
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    I mean the fact that \(\cos_H (t) = \cosh\). Seems really coincidental.

  101. Empty
    • one year ago
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    Haha weird, the ghost of Hadamard knew his name would coincide with Hyperbola or something XD

  102. anonymous
    • one year ago
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    Okay, so here is my opinion. If you have some algebra: \[ a+bA \]where \(A\) is some \(2\times2\) matrix, then what if we define it so that: \[ a+bA = re^{A\theta} = r\bigg(\cos_A(\theta) + A\sin_A(\theta)\bigg) \]Somehow this makes sent to me because \(\theta\) is an angle and \(r\) is a magnitude. I suppose there are still some issues though, because multiplication of \(a+bA\) won't always give you some sort of rotational result. Maybe this pattern is only worthwhile for certain cases like complex numbers.

  103. Empty
    • one year ago
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    No I think you're right, I was thinking something along the same lines, that for every matrix we essentially can define a pair of sine and cosine functions. I am just wondering if we can perhaps extend this to nxn matrices and instead of creating even and odd functions (even is the 2 of 2x2 matrices) so we can make other like "mod 3" functions for our 3x3 matrices for instance.

  104. UsukiDoll
    • one year ago
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    You guys should publish a novel on this lol xD

  105. Empty
    • one year ago
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    Haha we just have to publish, writing the novel we already did it lol

  106. anonymous
    • one year ago
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    Anyway, for rules mentioned before: \[ A^2=cA\implies A^n = c^{n}A \\ A^2 = cI \implies A^n = \begin{cases} c^kI &n=2k \\ c^kH &n=2k+1 \end{cases} \]

  107. Empty
    • one year ago
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    slight correction:\[ A^2=cA\implies A^n = c^{n-1}A \\ \]

  108. anonymous
    • one year ago
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    Whoops, that looks right.

  109. anonymous
    • one year ago
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    For that first case \[ e^{At} = \sum_{n=0}^\infty \frac{(At)^n}{n!} =A\sum_{n=0}^\infty \frac{c^{n-1}t^n}{n!} = \frac{A}{c}e^{ct} \]

  110. anonymous
    • one year ago
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    For the second case, I think I'm getting: \[ e^{At} = \cosh(ct) + A\sinh(ct) \]

  111. anonymous
    • one year ago
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    Okay, the first case I think I was a bit too hand wavy though... let me double check.

  112. Empty
    • one year ago
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    Now that I'm thinking about these generalizations I thought why not make the matrix square to a different matrix altogether? But then I thought, what about if we use group theory? Let's instead just use group theory which already has multiplication defined and allows us to do the same exact things, but we're no longer confined to what a matrix multiplication gives us.

  113. anonymous
    • one year ago
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    Can you find matrix such that: \[ A\neq I \land A^3=I \]I'm not sure, but my gut feeling is we need at least a \(3\times3\) to actually get that.

  114. UsukiDoll
    • one year ago
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    so we have to find... a matrix that makes this true so we can't have Matrix A equal to an identity and have A^3 = to an identity matrix

  115. UsukiDoll
    • one year ago
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    so A can't be an identity matrix but using matrix multiplication A(A)(A) that result should come up to an identity matrix

  116. Empty
    • one year ago
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    Yeah good point, just work it out with variables and solve the system of equations... but who wants to do that? haha

  117. anonymous
    • one year ago
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    We're talking about a very ugly system of equations.

  118. Empty
    • one year ago
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    True, but we might be able to simplify it. I think tensor notation makes each equation simpler to look at in a sense. \[A^i_jA^j_kA^k_l = \delta ^i_l\] Or in terms of individual components (in linear algebra terms) where n is the row and m is the column of the entry: \[a_{n1}a_{11}a_{1m}+a_{n1}a_{12}a_{2m}+a_{n2}a_{21}a_{1m}+a_{n2}a_{22}a_{2m}=\delta_{nm}\] So this represents our 4 equations. Plug in n and m equal to 1 or 2 to get them.

  119. ganeshie8
    • one year ago
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    \[A=\begin{bmatrix}1&0 \\0&e^{i5 \pi/6}\end{bmatrix}\]

  120. anonymous
    • one year ago
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    I can easily find a solution for \(3\times 3\) though. \[ A = \begin{bmatrix} 0&1&0 \\ 0&0&1 \\ 1&0&0 \end{bmatrix} \]Then \[ A^3=I \]And I think: \[ A_c = \begin{bmatrix} 0&1&0 \\ 0&0&1 \\ c&0&0 \end{bmatrix} \implies (A_c)^3 = cI \]

  121. anonymous
    • one year ago
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    @ganeshie8 In our context, the matrix you have provided technically would be something along the lines of a \(4\times 4\), once expanded to real numbers.

  122. ganeshie8
    • one year ago
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    Ohk I see, I'm not able to follow the conversation completely as it is way too above my head lol

  123. ganeshie8
    • one year ago
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    does this work http://www.wolframalpha.com/input/?i=%7B%7B1%2C2%7D%2C%7B-3%2F2%2C-2%7D%7D%5E3+

  124. anonymous
    • one year ago
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    Yeah, that could work.

  125. Empty
    • one year ago
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    Ahhh how did you think up either of these matrices, I'm impressed @ganeshie8 !

  126. ganeshie8
    • one year ago
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    I don't think I let wolfram think lol

  127. anonymous
    • one year ago
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    That matrix would seem to work like: \[ A^n = \begin{cases} I &n=3k \\ A &n=3k+1 \\ A^2 &n=3k+2\end{cases} \]

  128. anonymous
    • one year ago
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    Maybe ganeshie can help identify the MacLaurin series we get?

  129. anonymous
    • one year ago
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    \[ e^{At} = \sum_{k=0}^\infty \frac{t^{3k}}{(3k)!}+A\sum_{k=0}^\infty \frac{t^{3k+1}}{(3k+1)!}+A^2\sum_{k=0}^\infty \frac{t^{3k+2}}{(3k+2)!} \]

  130. anonymous
    • one year ago
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    http://www.wolframalpha.com/input/?i=sum+t%5E%7B3n%7D%2F%283n%29%21++for+n+%3D+0+to+infty For the first one. This looks messy.

  131. Empty
    • one year ago
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    \[e^{At}=f(t)+A g(t)+ A^2 h(t)\] where we have \(\omega = e^{i 2 \pi /3}\) and analgously we have our ENTIRELY REAL FUNCTIONS satisfying these "even odd" mod 3 symmetries: \[f(t)=f( \omega t) = f( \omega ^2 t) \\ g(t)=\omega g( \omega t) = \omega^2 g(\omega^2 t) \\ h(t) = \omega^2 h(\omega t) = \omega h (\omega^2 t)\] And additionally we can represent: \[f(t) = \frac{e^{At}+e^{\omega At} + e^{\omega^2 A t}}{3}\] and we have two more corresponding ones, where when we divide by 3 in the denominator we add A and A^2 to match just like we would for complex sine we have to add an i.

  132. Empty
    • one year ago
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    Maybe I'm off a bit, since I believe f'(t)=g(t) and g'(t)=h(t) and h'(t)=f(t), but I think what I've said is mostly correct, I am sorta just saying these from memory I didn't actually calculate any of this right now, but I've done this particular thing before.

  133. anonymous
    • one year ago
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    I think the link I gave will give you \(f(t)\), and it's messy.

  134. anonymous
    • one year ago
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    I'm not sure though if there is a unique matrix solution to the equations anymore, though.

  135. Empty
    • one year ago
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    Now I feel like we should just throw away matrices and use group theory since the entries of the matrix are starting to hinder us when we really aren't using them for anything.

  136. anonymous
    • one year ago
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    But without matrices, it's hard to check out results.

  137. Empty
    • one year ago
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    What's to check?

  138. anonymous
    • one year ago
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    Matrices allow easy definitions of inverses.

  139. Empty
    • one year ago
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    Groups allow us to arbitrarily define inverses and we can insure that it's closed simultaneously if we want

  140. anonymous
    • one year ago
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    If \(A^3 = 1\), then how would you even go about \(A^{-1}\)?

  141. Empty
    • one year ago
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    It's its own inverse in that case. |dw:1435918078087:dw|

  142. anonymous
    • one year ago
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    You're saying \(A^2 = A^{-1}\)?

  143. Empty
    • one year ago
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    No sorry, here so from this cayley table |dw:1435918188833:dw| I should have just drawn out the whole table. In order to check that every element has a unique inverse we simply check to see that there is only one identity element per column and per row.

  144. anonymous
    • one year ago
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    I think there is a good chance that you can find a matrix which represents any sort of variable you come up with.

  145. anonymous
    • one year ago
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    Hmmm

  146. Empty
    • one year ago
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    I don't doubt it, but I don't think we particularly gain anything from playing with elements in the matrices except longer calculations. Plus it looks like we have multiple different types of matrices that end up obeying the same rules, so it doesn't feel very unique and this shifts the focus a bit I guess. I don't know, just something to think about.

  147. Empty
    • one year ago
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    We now have the ability to define noncommutative structures and other larger weirder things I think while it still being manageable since we're looking it up in a chart rather than carrying out multiplication on nxn matrices which becomes difficult for n=3 and higher. Suppose we want \(A^{8}=I\) I think we would be in trouble.

  148. anonymous
    • one year ago
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    Hmm, to be really generalized for the 2 component case, we can define: \[ A^2 = aI + bA \]For the \(n\) component case , we can define: \[ \bigg( \sum_{k=0}^{n-1} a_kA^k \bigg)\bigg( \sum_{k=0}^{n-1} b_kA^k \bigg)=\bigg( \sum_{k=0}^{n-1} c_kA^k \bigg) \]For coefficients as vectors \(\mathbf a\) and \(\mathbf b\) we say \(\mathbf c =f(\mathbf a, \mathbf b)\).

  149. Empty
    • one year ago
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    Ultimately it seems like we're really just using these matrices to define how we multiply "complexish numbers" in our vector space and plugging them into power series to see if we get a cool set of "rotation" functions.

  150. anonymous
    • one year ago
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    I suppose there might be some kinds of \(f\) which can't be achieved through matrix multiplication.

  151. Empty
    • one year ago
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    I'm sorta not sure how to raise A to the third power from this definition: \(A^2 = aI+bA \implies A^3 = ?\)

  152. Empty
    • one year ago
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    The convolution you've written there is interesting to me though, I would like to understand this more I don't quite know what you're saying but I think I like where it's going.

  153. anonymous
    • one year ago
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    \[ A^3 = aA + b(aI+bA) = ab + (a+b^2)A \]

  154. anonymous
    • one year ago
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    I'm trying to generalize, but hmmm

  155. Empty
    • one year ago
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    Interesting your definition of A^2 means we can write: \(A= \frac{1}{b}A^2 - \frac{a}{b}I\)

  156. anonymous
    • one year ago
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    Okay, the most general I would ever want to get is to say we're looking for multiplication functions, which can be described as \(f:\mathbb R^d \times \mathbb R^d \to \mathbb R^d\). Obviously that will keep things closed.

  157. anonymous
    • one year ago
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    But for \(d=2\), and working in a vector space, the only thing we really need to define is \(A^2\) because algebra will solve the rest.

  158. Empty
    • one year ago
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    Sounds good to me. I feel like from this perspective we can define multiplication between any of our "complexish" numbers to be anything from rotations, to shear, to whatever and then we can just go ahead and define the group theory elements to match what we want.

  159. anonymous
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @wio \[ A^3 = aA + b(aI+bA) = ab + (a+b^2)A \] \(\color{blue}{\text{End of Quote}}\) For complex numbers, we would say \((a,b) = (-1,0)\), and so: \[ i^3 = (-1)(0)+((-1)+(0)^2)i = -i \]

  160. Empty
    • one year ago
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    That way we can simply calculate stuff with algebra knowing it obeys our rules.

  161. anonymous
    • one year ago
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    For dual numbers \((a,b) = (0,0)\), and so: \[ \epsilon^3 = 0 \]For \(\alpha\), we'd say \((a,b) = (0,1)\), and we get: \[ \alpha^3 = (0)(1) + ((0)+(1)^1)\alpha = \alpha \]

  162. anonymous
    • one year ago
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    Hmmm, interesting.

  163. anonymous
    • one year ago
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    If we consider only cases where \((a,b) \in \{-1,0,1\}^2\), that gives us about \(9\) combinations we can mess with.

  164. anonymous
    • one year ago
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    Hmm, interesting.

  165. Empty
    • one year ago
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    Hmmm I'm trying to find some interesting multiplication definition.

  166. anonymous
    • one year ago
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    Using: \[ A^2 = a+bA \]as a generalization, then: \[ (c+dA)(f+gA) = (cf+adg) + (cg+df+bdg)A \]Hmmm, it looks overly general, but I've been thinking something...

  167. anonymous
    • one year ago
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    If we want to do \[ \frac{c+dA}{f+gA} \]All we need is to find a way to get rid of that \(A\) in the denominator.

  168. anonymous
    • one year ago
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    I think the only way this can be fun is if we have something we want to try to find.

  169. anonymous
    • one year ago
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    We have found \(\sin_i\) and \(\cos_i\) for the different conic sections, so that is off the table.

  170. Empty
    • one year ago
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    Ellipse or a torus is what I've been searching for, but I think I am too tired to think up something that'll do that.

  171. anonymous
    • one year ago
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    But those are 3d objects, not conic sections

  172. anonymous
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    And ellipse corresponds to original trig functions

  173. anonymous
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    Well, hmmm, I suppose circle is a type of ellipse so...

  174. Empty
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    I think an ellipse might be nice to be able to have operations that let us rotate the ellipse itself like this |dw:1435921176282:dw| Also what's stopping us from doing more than one? We can have something like quaternions with multiple parts interacting.

  175. Empty
    • one year ago
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    One idea is that whatever manages the ellipses will collapse down into complex numbers

  176. Empty
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    But ellipses can stretch and rotate, something circles can't do. I mean you can rotate a circle, but it's meaningless.

  177. Empty
    • one year ago
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    I don't know, hmm.

  178. anonymous
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    Okay, now that I think about it... \(i\) corresponds to the circle, a class of an ellipse. \(\epsilon\) corresponds to a specific class of parabola. And then \(H\) corresponded to a class of hyperbola.

  179. anonymous
    • one year ago
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    And \(H\) would be like \((2,0)\), I think. That is: \[ H^2 = (2) + (0)H = 2 \]

  180. Empty
    • one year ago
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    Yeah, actually since these are all sorta just separate sections of a conic maybe we can do a linear combination of these to get an ellipse?

  181. anonymous
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    So clearly we are getting conic sections for \((x,0)\) configurations.

  182. Empty
    • one year ago
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    I never really even bothered to try to calculate something like: \[(x+yi+z \epsilon)^2\]

  183. anonymous
    • one year ago
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    Conic sections have a certain property, eccentricity or something.

  184. Empty
    • one year ago
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    Yeah I don't know either haha, I just know that elliptic integrals are hard and finding arclength, there's a possibility that what we're doing gets us something interesting to play with there.

  185. anonymous
    • one year ago
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    https://en.wikipedia.org/wiki/Eccentricity_(mathematics)

  186. anonymous
    • one year ago
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    For a circle, we use \((-1,0)\). For a hyperbola we use \((2,0)\). For a parabola we use \((0,0)\). Maybe it is just \((x-1,0)\) where \(x\) is the eccentricity. Just a hypothesis.

  187. Empty
    • one year ago
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    That sounds like a great idea to me

  188. anonymous
    • one year ago
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    Eccentricity of an ellipse is \(\sqrt{1-b^2/a^2}\).

  189. Empty
    • one year ago
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    I'm too tired to really learn anything new right now but it'll give me something to think about later

  190. anonymous
    • one year ago
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    Yeah, maybe another day we can experiment.

  191. anonymous
    • one year ago
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    Come up with some way to test hypothesis

  192. anonymous
    • one year ago
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    And perhaps figure out how \((0,y)\) changes things up.

  193. anonymous
    • one year ago
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    It's clear that \(\alpha\) gave us some clear insight into how that might work.

  194. Empty
    • one year ago
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    Yeah definitely, I don't think we should throw away matrices they helped us quite a bit in concrete ways of determining what products _should_ be. For instance right now I'm just calculating \(\epsilon * i \) and \(i*\epsilon \) as something to explore with my brain later lol.

  195. Empty
    • one year ago
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    In some sense we're sorta thinking in terms of instead of functions of a complex variable, more like functions of a 2x2 matrix variable.

  196. anonymous
    • one year ago
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    Looking at the eccentricity of a hyperbola, I think my hypothesis was wrong. Hadamard seemed to give us a standard hyperbola, so its eccentricity should be: \[ \sqrt{1+\frac {1^2}{1^2}} = \sqrt{2} \]And \(\sqrt{2} + 1 \neq 2\).

  197. anonymous
    • one year ago
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    I'm going to bed, but now I have a goal.

  198. Empty
    • one year ago
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    In my mind I'm imagining the conic section. If you cut it one way you get a circle and if you cut it the other way you get a hyperbola. I'd like to rotate that plane that we're cutting through to somewhere inbetween there. Actually... I think I know how to do it, because earlier we had: \[\left[ \begin{array}c 0 & 1\\s & 0\\\end{array} \right]\] where s=1 representing a hyperbola, s=0 representing a parabola, and s=-1 representing a circle. And I know that this is like the same spacing, so if -1<s<0 is what we use, we will be on an ellipse I think.

  199. anonymous
    • one year ago
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    It's clear to me that the matrix you have presented corresponds to \((s,0)\). Hmm

  200. anonymous
    • one year ago
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    Is there something to correspond to \((0, t)\), now I wonder...

  201. Empty
    • one year ago
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    Haha well perhaps we have chosen the transpose arbitrarily, let's just choose the lower triangular version of this maybe?

  202. anonymous
    • one year ago
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    I guess it would be going back to: \[ A^2 = tA \]We know for \(t=1\) we can use: \[ \begin{bmatrix}1&1\\0&0\end{bmatrix} \]

  203. anonymous
    • one year ago
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    Since that is what was used for \(\alpha\).

  204. Empty
    • one year ago
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    Oh right

  205. Empty
    • one year ago
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    Woah absolutely fascinating. Instead of calling them real and imaginary parts they call it bosonic and fermionic directions for real and \(\epsilon\) directions. This is used for particle physics. http://prntscr.com/7ob3pm

  206. anonymous
    • one year ago
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    I think I found it:\[ A = \begin{bmatrix} 1&1 \\ t-1&t-1 \end{bmatrix} \implies A^2 = tA \]

  207. UsukiDoll
    • one year ago
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    This is probably the longest thread I've ever been on xD!

  208. anonymous
    • one year ago
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    I wonder if there is a matrix such that \[ A^2 = sI + tA \]If we can find such a matrix, then matrix would be just fine.

  209. anonymous
    • one year ago
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    It's kinda funny, because ultimately what we have here is: \[ A^2 -tA-sI \]Which is like a quadratic equation for matrices.

  210. anonymous
    • one year ago
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    This is an ugly solution, but: \[ A = \bigg(\frac{t+\sqrt{t^2+4s}}{2} \bigg) I \implies A^2 = sI +tA \]But since \(A=cI\), it sort of doesn't count. I wonder if there is another solution.

  211. anonymous
    • one year ago
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    the reason it lets you do automatic differentiation is because dual numbers capture a property of nilpotency used by Fermat, Newton, etc. that \(\varepsilon^2=0\) (recall Fermat's https://en.wikipedia.org/wiki/Adequality ); this is used in functional languages to explicitly extend real computational functions to dual numbers and 'compute' their derivatives in a way unlike symbolic or numerical differentiation. it's mainly used in machine learning contexts to come up with derivatives of complex functions (like a feedforward neural net) for error propagation in learning algorithms fyi

  212. anonymous
    • one year ago
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    $$f(a+b\varepsilon)=f(a)+\varepsilon f'(a)+\frac12\varepsilon^2 f''(a)+\dots=f(a)+\varepsilon f'(a)$$this is kinda the same idea behind the trick used in numerical differentiators to evaluate derivatives of \(x\) at \(x+i\) for smaller error terms

  213. anonymous
    • one year ago
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    ps the 'hyperbolic' numbers you were talking about are just the split-complex numbers with \(j^2=1\)

  214. anonymous
    • one year ago
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    that is the most straightforward way of encoding split-complex numbers in \(M_2(\mathbb{R})\) https://en.wikipedia.org/wiki/Split-complex_number#Matrix_representations

  215. anonymous
    • one year ago
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    and as you noticed \(j^2=1\) ends up causing \(|a+bj|=1\) to giv ethe unit hyperbola (a^2-b^2=j\) and its paramterized by \(\exp(bj)=\cosh(b)+j\sinh(b)\)

  216. anonymous
    • one year ago
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    the problem with an ellipse is that there is no 'unit' ellipse, but we can define an ellipse as \(a^2+k^2b^2=1\) which just reduces to \(a+bw\) where \(w\) is some scaled kind of \(i\) (specifically \(w=ki\)), so it ends up giving rise to the same algebra \(\mathbb{C}\) as just \(i\) alone -- boring. let's rehash the different types of conics: 1) ellipses (using \(i^2=-1\)) 2) parabolas (using \(\varepsilon^2=0\)) 3) hyperbolas (using \(j^2=1\)) ... note the correspondence which is actually a consequence of the limitations we face when trying to come up with two-dimensional unital algebras over the reals https://en.wikipedia.org/wiki/Hypercomplex_number#Two-dimensional_real_algebras

  217. Empty
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @wio I wonder if there is a matrix such that \[ A^2 = sI + tA \]If we can find such a matrix, then matrix would be just fine. \(\color{blue}{\text{End of Quote}}\) Yes because the Cayley-Hamilton theorem says every square matrix satisfies its own characteristic equation. That is to say we can replace A with \(\lambda\) which represent the eigenvalues of A and solve. \[\lambda^2 = s+t \lambda\] The solution to this quadratic has two roots so we can throw it into a 2x2 diagonal matrix: \[\left[ \begin{array}c \lambda_+ & 0\\0 & \lambda_-\\\end{array} \right]\] So this is a nice trick for making symmetric, diagonal matrices that obey any polynomial equation.

  218. nincompoop
    • one year ago
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    great post

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