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ganeshie8
 one year ago
Find the matrix for linear transformation
\(T : \mathbb{P}_n \to \mathbb{P}_n\), \[T f(x) =f(x) 2f'(x)+3f''(x)\]
ganeshie8
 one year ago
Find the matrix for linear transformation \(T : \mathbb{P}_n \to \mathbb{P}_n\), \[T f(x) =f(x) 2f'(x)+3f''(x)\]

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\(\mathbb{P}_n\) is the set of polynomials of degree not greater than \(n\)

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.2\[f(x)=a_0=a_1x+a_2x^2+...a_nx^n=\\\begin{bmatrix} a_0 &a_1 &a_2 &... &a_n \end{bmatrix} \begin{bmatrix} 1\\ x\\ x^2\\ x^3\\ ...\\ x^n \end{bmatrix}\\f'=0+a_1 +2a_2x+3a_3x^2+...=\\\begin{bmatrix} a_0 &a_1 &a_2 &... &a_n \end{bmatrix} \begin{bmatrix} 0\\ 1\\ 2x\\ 3x^2\\ ...\\n x^n1 \end{bmatrix}\]

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.2\[f''=0+0+2a_2+6a_3x+12a_4x^2+...=\\\begin{bmatrix} a_0 &a_1 &a_2 &... &a_n \end{bmatrix} \begin{bmatrix} 0\\ 0\\ 2\\ 6x\\12x^2\\ ...\\n(n1)x^{n2} \end{bmatrix}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4how do we get the transformation matrix ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4doing \(f(x) 2f'(x)+3f''(x)\) gives a vector, not matrix right ?

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.2for f\[\begin{bmatrix} 1\\ x\\ x^2\\ x^3\\ x^4\\ x^5\\ ... \end{bmatrix}\] for f'\[\begin{bmatrix} 0\\ 1\\ 2x\\ 3 x^2\\ ... \end{bmatrix}\] for f'' \[\begin{bmatrix} 0\\0\\2\\ 6x\\ 12 x^2\\ ... \end{bmatrix}\] find f2f'+3f'' vector and multiply by \[\begin{bmatrix} a_0 &a_1 &a_2 &... &a_n \end{bmatrix}\] it gives you matrix

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4hmm \(f2f'+3f'\)' gives \(n+1\) x \(1\) vector premultiplying \(\begin{bmatrix} a_0 &a_1 &a_2 &... &a_n \end{bmatrix}\) gives 1x1 matrix (scalar)

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.2this the way to wirte matrix form \[\begin{bmatrix} 0\\ 1\\ 2x\\ 3x^2\\... \end{bmatrix}=\begin{bmatrix} 0 &0 &0 &0 &... \\ 1&0 &0 &0 &... \\ 0&2 &0 &0 &... \\ 0&0 &3 &0 &... \\ 0 &0 &0 &4 &... \end{bmatrix}\begin{bmatrix} 1\\ x\\ x^2\\ x^3\\ x^4\\ ... \end{bmatrix}\]

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.2that was f' we do like this for f'' then you have sum of some matrix to find the transformation

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4I think that should work because now the vector has correct exponents ordering

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4matrix for \(Tf(x)=f(x)\) is \[\begin{bmatrix} 1&0&0&0&\cdots&0&0\\ 0&1&0&0&\cdots&0&0\\ 0&0&1&0&\cdots&0&0\\ 0&0&0&1&\cdots&0&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&0&0\\ 0&0&0&0&\cdots&1&0\\ 0&0&0&0&\cdots&0&1\\ \end{bmatrix}\] matrix for \(Tf(x)=f'(x)\) is \[\begin{bmatrix} 0&1&0&0&\cdots&0&0\\ 0&0&2&0&\cdots&0&0\\ 0&0&0&3&\cdots&0&0\\ 0&0&0&0&\cdots&0&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&n1&0\\ 0&0&0&0&\cdots&0&n\\ 0&0&0&0&\cdots&0&0\\ \end{bmatrix}\] matrix for \(Tf(x)=f''(x)\) is \[\begin{bmatrix} 0&0&2\cdot 1&0&\cdots&0&0\\ 0&0&0&3\cdot 2 &\cdots&0&0\\ 0&0&0&0&\cdots&0&0\\ 0&0&0&0&\cdots&(n1)(n2)&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&0&n(n1)\\ 0&0&0&0&\cdots&0&0\\ 0&0&0&0&\cdots&0&0\\ \end{bmatrix}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4i don't really see a more simpler way to express the final matrix hmm

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4for the overall transformation, I think il end up with matrix for \(Tf(x)=f(x)\color{blue}{2\cdot}f'(x)+\color{red}{3\cdot }f''(x)\) is \[\begin{bmatrix} 1&\color{blue}{2\cdot}1&\color{red}{3\cdot }2\cdot 1&0&\cdots&0&0\\ 0&1&\color{blue}{2\cdot}2&\color{red}{3\cdot }3\cdot 2 &\cdots&0&0\\ 0&0&1&\color{blue}{2\cdot}3&\cdots&0&0\\ 0&0&0&1&\cdots&\color{red}{3\cdot }(n1)(n2)&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\color{blue}{2\cdot}(n1)&\color{red}{3\cdot }n(n1)\\ 0&0&0&0&\cdots&1&\color{blue}{2\cdot}n\\ 0&0&0&0&\cdots&0&1\\ \end{bmatrix}\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.1You can represent the entries of the derivative matrix using the Kronecker delta and the binomial coefficients by starting at the top left entry being indexed with 0,0 instead of 1,1 (which is more natural because \(x^0\) is also the first entry of the vector multiplying it, not \(x^1\) ): \[\Large D_{nm}^{(k)} = \delta_{n,(mk)} \binom{n+k}{n} \] So for instance, k=0 corresponds to the identity matrix, \[\Large D_{nm}^{(0)} = \delta_{nm} \] Or let's look at this entry on the second derivative which represents the map from \(x^4\) to \(12 x^2\): \[\Large D_{24}^{(2)} = \delta_{22} \binom{4}{2} =12\] Cool, also I found this other matrix the other day too that might be related through the Taylor series since it represents shifting the function f(x) to f(x+a), that is if f(x) can be represented by the same vector you're using. Part of A for just \(f(x)= px^2+qx+r\) $$A= \begin{bmatrix} 1 & a & a^2 \\ 0 & 1 & 2a \\ 0 & 0 & 1 \end{bmatrix}$$ $$A_{ij}=\binom{i}{j}a^{ji}$$ anywho I thought you might think this is interesting.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1So using this matrix D I put we can write: \[Ty = y2y'+3y''\]\[Ty=D^{(0)}y2D^{(1)}y +3D^{(2)}y\]\[T=I2D^{(1)}+3D^{(2)}\] So I guess that's one way to write T down, I don't know if that's quite what you like or not but I think it's a little better.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wow this has yo be some really advanced math

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4That looks neat, pretty sure you meant \[\Large D_{nm}^{(k)} = \delta_{n,(mk)} \binom{n+k}{n}\color{red}{k!}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\[Ty = y2y'+3y''\]\[Ty=D^{(0)}y2D^{(1)}y +3D^{(2)}y\]\[T=I2D^{(1)}+3D^{(2)}\] This is indeed compact, was just wondering if we can do any operations with the kronecker delta symbol and maybe further simplify or something..

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Ahhh yeah I somehow left that out haha, oh well you got it. As it stands, I don't think there's a whole lot more we can do. Each one of those D matrices is going to be its own separate band at least, so together it will form one fairly predictable banded matrix.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Something kind of interesting I'm seeing is, each of these D matrices are linearly independent of each other and all these D matrices form a vector space.
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