A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

ganeshie8

  • one year ago

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

  • This Question is Closed
  1. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    \(\mathbb{P}_n\) is the set of polynomials of degree not greater than \(n\)

  2. amoodarya
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[f(x)=a_0=a_1x+a_2x^2+...a_nx^n=\\\begin{bmatrix} a_0 &a_1 &a_2 &... &a_n \end{bmatrix} \begin{bmatrix} 1\\ x\\ x^2\\ x^3\\ ...\\ x^n \end{bmatrix}\\f'=0+a_1 +2a_2x+3a_3x^2+...=\\\begin{bmatrix} a_0 &a_1 &a_2 &... &a_n \end{bmatrix} \begin{bmatrix} 0\\ 1\\ 2x\\ 3x^2\\ ...\\n x^n-1 \end{bmatrix}\]

  3. amoodarya
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[f''=0+0+2a_2+6a_3x+12a_4x^2+...=\\\begin{bmatrix} a_0 &a_1 &a_2 &... &a_n \end{bmatrix} \begin{bmatrix} 0\\ 0\\ 2\\ 6x\\12x^2\\ ...\\n(n-1)x^{n-2} \end{bmatrix}\]

  4. amoodarya
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    almost solved!

  5. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    how do we get the transformation matrix ?

  6. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    doing \(f(x) -2f'(x)+3f''(x)\) gives a vector, not matrix right ?

  7. amoodarya
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    for 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 f-2f'+3f'' vector and multiply by \[\begin{bmatrix} a_0 &a_1 &a_2 &... &a_n \end{bmatrix}\] it gives you matrix

  8. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    hmm \(f-2f'+3f'\)' gives \(n+1\) x \(1\) vector premultiplying \(\begin{bmatrix} a_0 &a_1 &a_2 &... &a_n \end{bmatrix}\) gives 1x1 matrix (scalar)

  9. amoodarya
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    this the way to wirte matrix form \[\begin{bmatrix} 0\\ 1\\ 2x\\ 3x^2\\... \end{bmatrix}=\begin{bmatrix} 0 &0 &0 &0 &... \\ 1&0 &0 &0 &... \\ 0&2 &0 &0 &... \\ 0&0 &3 &0 &... \\ 0 &0 &0 &4 &... \end{bmatrix}\begin{bmatrix} 1\\ x\\ x^2\\ x^3\\ x^4\\ ... \end{bmatrix}\]

  10. amoodarya
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    that was f' we do like this for f'' then you have sum of some matrix to find the transformation

  11. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    I think that should work because now the vector has correct exponents ordering

  12. imqwerty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    '.'

  13. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    matrix 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 \[\begin{bmatrix} 0&1&0&0&\cdots&0&0\\ 0&0&2&0&\cdots&0&0\\ 0&0&0&3&\cdots&0&0\\ 0&0&0&0&\cdots&0&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&n-1&0\\ 0&0&0&0&\cdots&0&n\\ 0&0&0&0&\cdots&0&0\\ \end{bmatrix}\] matrix for \(Tf(x)=f''(x)\) is \[\begin{bmatrix} 0&0&2\cdot 1&0&\cdots&0&0\\ 0&0&0&3\cdot 2 &\cdots&0&0\\ 0&0&0&0&\cdots&0&0\\ 0&0&0&0&\cdots&(n-1)(n-2)&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&0&n(n-1)\\ 0&0&0&0&\cdots&0&0\\ 0&0&0&0&\cdots&0&0\\ \end{bmatrix}\]

  14. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    i don't really see a more simpler way to express the final matrix hmm

  15. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    for 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 \[\begin{bmatrix} 1&\color{blue}{-2\cdot}1&\color{red}{3\cdot }2\cdot 1&0&\cdots&0&0\\ 0&1&\color{blue}{-2\cdot}2&\color{red}{3\cdot }3\cdot 2 &\cdots&0&0\\ 0&0&1&\color{blue}{-2\cdot}3&\cdots&0&0\\ 0&0&0&1&\cdots&\color{red}{3\cdot }(n-1)(n-2)&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\color{blue}{-2\cdot}(n-1)&\color{red}{3\cdot }n(n-1)\\ 0&0&0&0&\cdots&1&\color{blue}{-2\cdot}n\\ 0&0&0&0&\cdots&0&1\\ \end{bmatrix}\]

  16. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    You 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 \(x^1\) ): \[\Large D_{nm}^{(k)} = \delta_{n,(m-k)} \binom{n+k}{n} \] 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 \(f(x)= px^2+qx+r\) $$A= \begin{bmatrix} 1 & a & a^2 \\ 0 & 1 & 2a \\ 0 & 0 & 1 \end{bmatrix}$$ $$A_{ij}=\binom{i}{j}a^{j-i}$$ anywho I thought you might think this is interesting.

  17. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    So using this matrix D I put we can write: \[Ty = y-2y'+3y''\]\[Ty=D^{(0)}y-2D^{(1)}y +3D^{(2)}y\]\[T=I-2D^{(1)}+3D^{(2)}\] 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.

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wow this has yo be some really advanced math

  19. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    to*

  20. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    That looks neat, pretty sure you meant \[\Large D_{nm}^{(k)} = \delta_{n,(m-k)} \binom{n+k}{n}\color{red}{k!}\]

  21. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    \[Ty = y-2y'+3y''\]\[Ty=D^{(0)}y-2D^{(1)}y +3D^{(2)}y\]\[T=I-2D^{(1)}+3D^{(2)}\] This is indeed compact, was just wondering if we can do any operations with the kronecker delta symbol and maybe further simplify or something..

  22. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Ahhh 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.

  23. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Something 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.

  24. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.