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I stumbled upon the first-order taylor series expansion for multivariable functions described by

$$T(x)=f(a)+(x-a)^T\nabla f(a)$$

and I wanted to expand $f(x,y)=\exp(-(x^2+y^2))$ around $(2,1)$ knowing that

$$\begin{align*} \nabla \exp(-(x^2+y^2))&=(-2x\exp(-(x^2+y^2)),-2y\exp(-(x^2+y^2)))^T,\\ f(2,1)&=\exp(-5),\\ \nabla f(2,1) &= (-4\exp(-5),-2\exp(-5))^T. \end{align*}$$

However I was not sure how to compute $T(x)$ because I don't know what expressions will be a vector and what expressions will be a function. How can I compute that?

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The displacement is a vector, and so is the gradient. In your formula, $(x-a)=(x-a_x,y-a_y)=(x-2,y-1)$ and $\nabla f(a)=(f_x(a),f_y(a))$ and you take the dot product between them. In linear term, that just means displacement in x times derivative in x, plus the same thing in y (naturally - in linear term, there is no coupling bewteen dimensions).

In general, the full expansion is

$$f(\vec{r})=\sum_{n=0}^\infty \frac{1}{n!}((\vec{r}-\vec{a})\cdot\nabla)^n f(\vec{a})$$ The parenthesis under the sum is a functional: you multiply and take it to the power before you apply it to the function and evaluate it.

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  • $\begingroup$ Am I right in assuming that $-\exp(-5)(4x+2y-11)$ would be the right polynomial in my case? $\endgroup$ – Christian Ivicevic Jan 27 '16 at 18:50
  • $\begingroup$ Yes, that's correct. $\endgroup$ – orion Jan 27 '16 at 18:56

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