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Thread: linear approximation of mapping

  1. #1
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    linear approximation of mapping

    question says to find the linear approximation of the given mapping at the indicated point


    $\displaystyle f(x,y) = (xe^y + cos(y), x, x+e^y); (1,0) $

    is this related to the tangent plane to a graph

    just take it and use it in this formula

    $\displaystyle z = f(x_{0}, y_{0}) + \frac{\partial f}{\partial x}(x_{0},y_{0})(x-x_{0}) + \frac{\partial f}{\partial y}(x_{0},y_{0})(y-y_{0})$
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    Re: linear approximation of mapping

    Yes, and then substitute the point in.
    Thanks from Jonroberts74
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    Re: linear approximation of mapping

    actually leads me to another question

    $\displaystyle f(x,y)$ has 3 outputs

    how does that work with a point in 2-space?

    $\displaystyle z= (2,1,2) + (1,1,1)(x-1)+(1,0,1)(y-0)$ ??
    Last edited by Jonroberts74; Jun 27th 2014 at 09:12 PM.
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    Re: linear approximation of mapping

    A linear function, from $\displaystyle R^n$ to $\displaystyle R^m$ can be represented by an n by m matrix.

    In particular the linear approximation to $\displaystyle f(x, y)= (xe^y+ cos(y), x, x+ e^y)$ at any point, (x, y), has as "slope"
    $\displaystyle \begin{bmatrix}e^y & xe^y- sin(y) \\ 1 & 0 \\ 1 & e^y\end{bmatrix}$

    At (1, 0) that is $\displaystyle \begin{bmatrix} 1 & 1 &\\ 1 & 0 \\ 1 & 1\end{bmatrix}$

    Since f(1, 0)= (1+ 1, 1, 1+ 1)= (2, 1, 2), the approximation to the function at (1, 0) is
    $\displaystyle \begin{bmatrix}2 \\ 1 \\ 2\end{bmatrix}+ \begin{bmatrix} 1 & 1 &\\ 1 & 0 \\ 1 & 1\end{bmatrix}\begin{bmatrix}x-1 \\ y \end{bmatrix}$
    $\displaystyle = \begin{bmatrix}2+ (x- 1)+ y\\ 1+ (x- 1) \\ 2+ (x- 1)+ y\end{bmatrix}= \begin{bmatrix}x+ y+ 1 \\ x\\ x+ y+ 1\end{bmatrix}$.

    That is exactly your $\displaystyle z= (2, 1, 2)+ (1, 1, 1)(x- 1)+ (1, 0, 1)(y- 0)$
    Last edited by HallsofIvy; Jun 28th 2014 at 06:35 AM.
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