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I know of two different formulas for the determination of the Quadratic Taylor Polynomial:

  • One with the Multi Index Notation:
    $$f(x_0 + \zeta) = \sum_{\vert\alpha\vert \le 2} \frac{1}{\alpha!} \, \partial^{\alpha} \, f(x_0) \, \zeta^{\alpha}$$

  • One without:
    $$ f(x_0 + \zeta) = f(x_0) + \sum_{i=1}^{n} \partial_{i} \, f(x_0) \, \zeta_i + \frac{1}{2}\sum_{i,\,j=1}^{n} \partial_{i} \, \partial_{j} \, f(x_0) \, \zeta_i \, \zeta_j$$

Let's pretend $f = f(x,y)$.

I'm curious about how often the term containing $\partial_{x} \partial_{y}$ appears. With the Multiple Index Notation this would be the tuple (1, 1), therefore the term appears exactly once $\frac{1}{2} \partial_{x} \partial_{y} \,f(x_0) \zeta_x \, \zeta_y$. Sadly the second sum in the later formula contains both of the cases ($i = 1, j = 2$ and $i = 2, j = 1$), which results in two terms $\frac{1}{2} \partial_{x} \partial_{y} \, f(x_0) \zeta_x \, \zeta_y$.

Which is the right calculation of the Quadratic Taylor Polynomial? And what am I doing wrong/missing in my "wrong" formula?

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  • $\begingroup$ In multi index rotation $\alpha!=1$ when $\alpha =(1,1)$ $\endgroup$
    – N74
    Aug 1 '16 at 20:45
  • $\begingroup$ @N74 Thank you! Seem's like I missed this one.. If you'd like to post this comment in form of a short answer I would be happy to accept it! $\endgroup$
    – user326377
    Aug 1 '16 at 22:09
  • $\begingroup$ I thought it was too short for an answer... but it seemed to work. $\endgroup$
    – N74
    Aug 1 '16 at 22:23
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In multi index notation $\alpha !=1$ when $\alpha=(1,1)$

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Not sure what to tell you; the cleanest way to remember it is, with column vector $\zeta,$ $$ f( x_0 + \zeta) \approx f(x_0) + \nabla f(x_0) \cdot \zeta + \frac{1}{2} \zeta^T H \zeta, $$ where $H$ is the symmetric Hessian matrix of second partial derivatives, evaluated at $x_0.$ Oh, $\nabla f$ is the gradient vector of $f.$

I cannot tell whether this is part of your difficulty, but mixed partials commute once the function is smooth enough. That is a quick application of Fubini's Theorem.

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  • $\begingroup$ When you evaluate your formula, does it contain the term $\frac{1}{2} \partial_x \partial_y \, f(x_0) \zeta_x \zeta_y$ or $\partial_x \partial_y \, f(x_0) \zeta_x \zeta_y$? $\endgroup$
    – user326377
    Aug 1 '16 at 20:08
  • $\begingroup$ @Herickson I think you should work it out yourself for the example $f(x,y) = 5 x^2 + 4xy + 7 y^2, $ at the origin, although the Hessian is constant. $\endgroup$
    – Will Jagy
    Aug 1 '16 at 20:13

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