- #1
jeebs
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I have 3 basis vectors:
[tex] e_1 = sin^2(x), e_2 = cos^2(x), e_3 = sin(x)cos(x) [/tex]
I am told that the combination rule is just normal addition, and that the differential operator is defined by Dp(x) = p'(x).
My task is to show that [tex] D = \left(\begin{array}{ccc}0&0&-1\\0&0&1\\2&-2&0\end{array}\right)[/tex] in this basis. So, what I've done is calculated the derivative of each of the basis vectors:
[tex] \frac{d}{dx}e_1 = 2cos(x)sin(x) = 2e_3 ... \frac{d}{dx}e_2 = -2cos(x)sin(x) = -2e_3 ... \frac{d}{dx}e_3 = cos^2(x) - sin^2(x) = e_2 - e_1 [/tex]
Now I'm looking at this vector p(x) being transformed into its 1st derivative p'(x). If I'm not mistaken, we can write p(x) as some arbitrary linear combination of the basis vectors, like:
[tex] p(x) = p_1e_1 + p_2e_2 + p_3e_3 = \sum_k p_ke_k = \left(\begin{array}{c}p_1e_1 \\ p_2e_2 \\ p_3e_3 \end{array}\right)... p'(x) = p_1\frac{d}{dx}e_1 + p_2\frac{d}{dx}e_2 + p_3\frac{d}{dx}e_3 = \left(\begin{array}{c}p_1e_1' \\ p_2e_2' \\ p_3e_3' \end{array}\right) [/tex]
and I need to somehow find the coefficients Dij for [tex] p'(x) = Dp(x) = \left(\begin{array}{ccc}D_{11}&D_{12}&D_{13}\\D_{21}&D_{22}&D_{23}\\D_{31}&D_{32}&D_{33} \end{array}\right) \left(\begin{array}{c}p_1e_1 \\ p_2e_2 \\ p_3e_3 \end{array}\right) = \left(\begin{array}{c}D_{11}p_1e_1 + D_{12}p_2e_2 + D_{13}p_3e_3\\ D_{21}p_1e_1 + D_{22}p_2e_2 + D_{23}p_3e_3 \\ D_{31}p_1e_1 + D_{32}p_2e_2 + D_{33}p_3e_3 \end{array}\right) = \left(\begin{array}{c}p_1e_1' \\ p_2e_2' \\ p_3e_3' \end{array}\right) [/tex]
My problem is that I cannot quite figure out how I can get from what I currently know to having the coefficients of the D matrix. I know this is probably trivial but I've been sat staring at this for ages now...
For instance, take, the first component of p'(x). we have [tex] p_1e_1' = D_{11}p_1e_1 + D_{12}p_2e_2 + D_{13}p_3e_3 = 2p_1e_3[/tex]. Since the coefficients pi are arbitrary, it means we can set D11 = D12 = 0, hence p1e1' = D13p3e3 therefore D13 = p1e1' / p3e3, and since the pi factors are arbitrary, they can be set = 1.
thus D13 = e1' / e3, but this does not give me the D13 = -1 that I require - it gives me D13 = 2e3 / e3 = 2.
What am I doing wrong here? I notice there is a 2 in the matrix, but it's in the wrong corner - D31 rather than D13. I can't see where my mistake is though, unless that switching of corners is just a fluke.
[tex] e_1 = sin^2(x), e_2 = cos^2(x), e_3 = sin(x)cos(x) [/tex]
I am told that the combination rule is just normal addition, and that the differential operator is defined by Dp(x) = p'(x).
My task is to show that [tex] D = \left(\begin{array}{ccc}0&0&-1\\0&0&1\\2&-2&0\end{array}\right)[/tex] in this basis. So, what I've done is calculated the derivative of each of the basis vectors:
[tex] \frac{d}{dx}e_1 = 2cos(x)sin(x) = 2e_3 ... \frac{d}{dx}e_2 = -2cos(x)sin(x) = -2e_3 ... \frac{d}{dx}e_3 = cos^2(x) - sin^2(x) = e_2 - e_1 [/tex]
Now I'm looking at this vector p(x) being transformed into its 1st derivative p'(x). If I'm not mistaken, we can write p(x) as some arbitrary linear combination of the basis vectors, like:
[tex] p(x) = p_1e_1 + p_2e_2 + p_3e_3 = \sum_k p_ke_k = \left(\begin{array}{c}p_1e_1 \\ p_2e_2 \\ p_3e_3 \end{array}\right)... p'(x) = p_1\frac{d}{dx}e_1 + p_2\frac{d}{dx}e_2 + p_3\frac{d}{dx}e_3 = \left(\begin{array}{c}p_1e_1' \\ p_2e_2' \\ p_3e_3' \end{array}\right) [/tex]
and I need to somehow find the coefficients Dij for [tex] p'(x) = Dp(x) = \left(\begin{array}{ccc}D_{11}&D_{12}&D_{13}\\D_{21}&D_{22}&D_{23}\\D_{31}&D_{32}&D_{33} \end{array}\right) \left(\begin{array}{c}p_1e_1 \\ p_2e_2 \\ p_3e_3 \end{array}\right) = \left(\begin{array}{c}D_{11}p_1e_1 + D_{12}p_2e_2 + D_{13}p_3e_3\\ D_{21}p_1e_1 + D_{22}p_2e_2 + D_{23}p_3e_3 \\ D_{31}p_1e_1 + D_{32}p_2e_2 + D_{33}p_3e_3 \end{array}\right) = \left(\begin{array}{c}p_1e_1' \\ p_2e_2' \\ p_3e_3' \end{array}\right) [/tex]
My problem is that I cannot quite figure out how I can get from what I currently know to having the coefficients of the D matrix. I know this is probably trivial but I've been sat staring at this for ages now...
For instance, take, the first component of p'(x). we have [tex] p_1e_1' = D_{11}p_1e_1 + D_{12}p_2e_2 + D_{13}p_3e_3 = 2p_1e_3[/tex]. Since the coefficients pi are arbitrary, it means we can set D11 = D12 = 0, hence p1e1' = D13p3e3 therefore D13 = p1e1' / p3e3, and since the pi factors are arbitrary, they can be set = 1.
thus D13 = e1' / e3, but this does not give me the D13 = -1 that I require - it gives me D13 = 2e3 / e3 = 2.
What am I doing wrong here? I notice there is a 2 in the matrix, but it's in the wrong corner - D31 rather than D13. I can't see where my mistake is though, unless that switching of corners is just a fluke.
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