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Assume a vector-valued function, for example ${\bf f}=(f_1, f_2)$, where

$$f_1(x,y)= x^2+3xy$$ $$f_2(x,y)= 2xy+y^2$$

(here f is column vector, x, y are variables)

Assume that each $f_i$ is a polynomial with degree at most 2, and thus we can write (in vector form) that:

$${\bf f}({\bf x}) = Q({\bf x},{\bf x})+L{\bf x}+{\bf c}$$ where $\bf f$ and $\bf x$ are both vectors, and $Q({\bf x},{\bf x})$ is the "quadratic" part,$L{\bf x}$ is the linear part, and c is a constant vector.

It seems to me that $f({\bf x}+1)= f(1)+f'(1){\bf x}+Q({\bf x},{\bf x})$ where $f'({\bf x})$ is the Jacobian matrix.

It looks like Taylor expansion, but I do not know exactly how to prove, except for stragihtforward calculation. Does anyone help me out?

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Added Tex formatting. You can use the $ signs to do so. – nbubis Apr 16 '12 at 16:23

You could approach this as follows: First of all, note that you can rewrite $Q$ as $x^T M x$ for some appropriate matrix $M$. Without loss of generality, you may choose $M$ to be symmetric.

With the convention that $1$ denotes the vector made up of $1$ entries, we have: $$\begin{array}{r&l} f(x+1) &= (x+1)^T M (x+1) + L (x+1) + c\\ &= x^T M x + x^T M 1 + 1^T M x + 1 M 1 + L x + L 1 + c\\ &= (1^T M 1 + L 1 + c) + 2 \cdot 1^T M x + L x + x^T M x\\ &= f(1) + 2 \cdot 1^T M x + L x + Q(x,x). \end{array}$$

Since $\nabla f(x) = 2 M x+ L$, your claim follows.

Alternatively, you can, as you indicated, just apply the multi-dimensional Taylor formula with development point $x$. Using the matrix representation of $Q$ will make this quite easy.

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Thanks. A might stupid question, I can see that $\nabla f(x) = 2Mx +L$, but how do you derive this? – user29271 Apr 16 '12 at 21:51
Just direct calculation: $\frac{\partial}{\partial x_i} x^T M x = \frac{\partial}{\partial x_i} \sum_{i,j} x_i M_{i,j} x_j = 2 M_{i,i} x_{i} + \sum_{j < i} M_{i,j} x_j + \sum_{j > i} M_{i,j} x_j = 2 \cdot e_i M x$, $\nabla Lx = L$ using linearity and hence $\nabla f(x) = 2Mx + L$ by linearity. – Johannes Kloos Apr 17 '12 at 9:38

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