Recent content by ShaunDiel

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    Solving for Exact Fit Cubic Polynomial Parameters

    Okay sweet, I'm working this out now, I think you skimmed over the stuff that I worked out because I did a column swap so basically everywhere you have a c it should be a b. But where I am right now, I have: d=(-3/2)-5b c=(8/3)b+124/3 b=(-3/10)-(1/5)d So If I plug them into the...
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    Solving for Exact Fit Cubic Polynomial Parameters

    Right, I understand that much, I just feel like I'm taking the wrong approach completely to starting this. I'm completely screwed now since nothing I did was right
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    Solving for Exact Fit Cubic Polynomial Parameters

    I emailed my prof about the question, asking him if the final cubic: ( (11/8+1019/16)x^3 + x^2 + (8/3+123/3)x+(-5-3/2)=0) Was as incorrect as it looked. He replied with this: "You should get a row of zeros in your equation, giving you a homogeneous solution which is a cubic polynomial and a...
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    Solving for Exact Fit Cubic Polynomial Parameters

    I still don't really understand where I'm supposed to go :confused: I have legitimately no idea
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    Solving for Exact Fit Cubic Polynomial Parameters

    Homework Statement The Attempt at a Solution For Part A) I did the normal best fit straight line stuff, A^τAX=bA^τ etc, resulting in: y= 31/26x -8/3 Now part B) is where I need a hand, I started with just cubing, squaring etc the terms given and doing row ops, which is where I'm at...
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    Matrix Algebra (Recurrences & Diagonalisation)

    Okayy, So I'lll have: [Cn, dn] = 6 (6/11)^n *[3 1] - 3 (-5/11)^n * [1 -4] What do you mean by equate components?
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    Matrix Algebra (Recurrences & Diagonalisation)

    UGH, I'm pretty sure I messed up way back around the start. http://www.wolframalpha.com/input/?i=[[1%2F143*67%2C1%2F143*33]%2C[1%2F143*44%2C-1%2F143*54]] My eigenvalues are right, but 1 vectors wrong.
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    Matrix Algebra (Recurrences & Diagonalisation)

    Oh damn, so it does. So I'll have : [cn dn] = 6*A^n[3 1] -3 *A^n[1 -4] as the expression for cn & dn? or should they be seperate? I'm guessing separate since it asks for a ratio in part b)
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    Matrix Algebra (Recurrences & Diagonalisation)

    Do I just do it the same way as before? Applying A to both sides? So: A[c0,d0] = =6*A[3 1] -3 *A[1 -4]
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    Matrix Algebra (Recurrences & Diagonalisation)

    Okay so I have both general forms for the eigenvectors, how would I write a formula for cn & dn from those though?
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    Matrix Algebra (Recurrences & Diagonalisation)

    Sweet! So does that finish all the diagonalisation for part A?
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    Matrix Algebra (Recurrences & Diagonalisation)

    Would it just be A^2(A[1,-4]))=(-5/11)*A^2[1,-4] A^3[1 -4] = (-5/11)^3[1 -4] ∴A^n[1 -4] = (-5/11)^n[1 -4] ?
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    Matrix Algebra (Recurrences & Diagonalisation)

    Like just multiply them? So if I took A*[1 4] it would be [199/143 -172/143] ??
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    Matrix Algebra (Recurrences & Diagonalisation)

    I just really don't understand what's going on.. I have A[1,-4] = (-5/11)*[1 -4] A[3 1] = (6/11) & [3 1] But what am I trying to do? Get what in terms of what? :s Thanks for being so patient with me I feel like an idiot
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    Matrix Algebra (Recurrences & Diagonalisation)

    I'm sorry, I feel like I'm useless at this.. Do I not have to do the D= P^-1 A P stuff?
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