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Homework Help: How to calculate derivative of a vector valued function

  1. Nov 26, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex] \vec f_1 : \bf R^n \rightarrow R^1 [/tex], [tex]\vec f_1(\vec x) = \|\vec x\|^2[/tex], calculate [tex]\vec f_1'(\left[ \begin{array}{c} 1 \\\vdots\\ 1 \end{array} \right])[/tex]

    2. Relevant equations

    3. The attempt at a solution
    I calculated the derivative as follow:
    [tex]\vec f_1(\vec x) =\|\vec x \|^2[/tex], [tex]\vec f_1(\vec x) = \sum \limits_{i=1}^n x_i^2[/tex].
    [tex]\vec f_1(\vec x + \vec h) = \sum \limits_{i = 1}^n (x_i + h_i)^2 = \sum \limits_{i=1}^n (x_i^2 + 2x_i h_i + h_i^2) = \sum \limits_{i=1}^n x_i^2 + \sum \limits_{i=1}^n 2x_i h_i + \sum \limits_{i=1}^n h_i^2[/tex].
    Thus, we have:
    [tex]\vec f_1(\vec x + \vec h) - \vec f_1(\vec x) = \sum \limits_{i=1}^n (2x_i h_i +h_i^2)[/tex]

    Since according to the definition of the derivative of vector valued function (Rudin Analysis Page 212):
    If there exists a linear transformation [tex] A [/tex] such that
    [tex] \lim \limits_{\vec h to 0} \frac{|\vec f(\vec x + \vec h) - \vec f(\vec x) - A\vec h |}{|\vec h|} = 0 [/tex],
    then [tex] \vec f [/tex] is differentiable at [tex] \vec x [/tex], and
    [tex] \vec f'(\vec x) = A [/tex]

    I defined a linear transformation as: [tex] A = 2 \vec x^T [/tex], such that we may have [tex] A \vec h = \sum \limits_{i = 1}^n 2x_i h_i [/tex]. Then since [tex]\lim \limits_{\vec h to 0} \frac{|\sum \limits_{i = 1}^n h_i^2|}{|\vec h|} = \lim \limits_{\vec h to 0} |\vec h| = 0 [/tex], [tex] A = 2 \vec x^T = \vec f'(\vec x) [/tex], and then plug [tex] \vec x = \begin{pmatrix} 1 \\ \vdots \\1 \end{pmatrix} [/tex] into the expression [tex] \vec f'(\vec x) [/tex].

    Is this solution correct? Thanks!
  2. jcsd
  3. Nov 26, 2008 #2


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    A function from Rn to R is essentially a function f such that [itex]f(x_1, x_2, x_3,\cdot\cdot\cdot, f_n)[/itex] is a number for each [itex](x_1, x_2, \cdot\cdot\cdot, x_n)[/itex] Its derivative can represented by the vector
    [tex]<\frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2}, \cdot\cdot\cdot, \frac{\partial f}{\partial x_n}>[/tex]
    the gradient of the function.

    [itex]||x||^2= x_1^2+ x_2^2+ \cdot\cdot\cdot+ x_n^2[/itex]
    The gradient is [itex]2x_1+ 2x_2+ \cdot\cdot\cdot+ 2x_n[/itex]

    Evaluate that at [itex]x_1= x_2= x_3= \cdot\cdot\cdot= x_n= 1[/itex].

    To answer your specific question,
    [tex]\lim \limits_{\vec h to 0} \frac{|\sum \limits_{i = 1}^n h_i^2|}{|\vec h|}[/tex]
    is 0 but that omly tells you the derivative at (0, 0, ..., 0), not at (1, 1, 1,...,1).
    What is
    [tex]\lim \limits_{\vec h to 0} \frac{|\sum \limits_{i = 1}^n (1+h_i)^2- 1|}{|\vec h|}[/tex]?
  4. Nov 26, 2008 #3
    Thanks for replying!
    [tex]\lim \limits_{\vec h \rightarrow 0} \frac{|\sum \limits_{i=1}^n h_i^2|}{|\vec h|}[/tex] was obtained from [tex] \lim \limits_{\vec h \rightarrow 0} \frac{|\vec f(\vec x + \vec h) - \vec f(\vec x) - A\vec h|}{|\vec h|} [/tex]. Since I defined the linear transform at [tex]\vec x = \left[\begin{array}{c} x_1 \\ \vdots \\ x_n \end{array}\right][/tex] as [tex] A = \left(\begin{array}{ccc} 2x_1 & \ldots & 2x_n \end{array}\right) [/tex], I got [tex] A\vec h = \sum \limits_{i = 1}^n 2x_i h_i[/tex].

    Plug this [tex] A\vec h[/tex] back to the definition of derivative, I got:
    [tex] \lim \limits_{\vec h \rightarrow 0} \frac{|\vec f(\vec x + \vec h) - \vec f(\vec x) - A\vec h|}{|\vec h|} = \lim \limits_{\vec h \rightarrow 0} |\vec h| = 0. [/tex].

    I think I have shown that the linear transformation [tex]A[/tex] does exist at [tex]\vec x[/tex], so according to the definition [tex] \vec f'(\left[\begin{array}{c} 1 \\ \vdots \\ 1 \end{array}\right]) = \left(\begin{array}{ccc} 2 & \ldots & 2 \end{array}\right)[/tex].

    But I just was confused by the fact that the function [tex] \vec f(\vec x)[/tex] actually maps [tex] \bf R^n [/tex] to [tex] \bf R^1 [/tex], but the result of my calculation implies that [tex] \vec f'(\vec x)[/tex] maps [tex] \bf R^n [/tex] to itself? Am I wrong somewhere? Or I still didn't really understand the definition?
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