I have been reviewing my analysis course and there is something that seems odd to me. My professor has wrote that the linear approximation for a function of two variables is :
$ x = x_{0} + h$
$ y = y_{0} + k$
$ h = x - x_{0}$
$ k = y - y_{0}$
$$ f(x_{0}+h,y_{0}+k) = f(x_{0},y_{0}) + \nabla{f}(x_{0},y_{0})*\begin{pmatrix}
h \\
k \\
\end{pmatrix} +\text{negligible}.$$
Then, he has explained that in dimension $n$, the formula is generalised as follows :
$$ f(x^{0}+H) = f(x^{0}) + \nabla{f}(x^{0})*H + \frac{1}{2}*H^{T}*\nabla{^2f}(x^0)*H + \text{negligible}.$$
With $x^0$ being : $\begin{pmatrix} x_{1}^{0} \\ ... \\ x_{n}^{0} \\ \end{pmatrix}$ and $ H = \begin{pmatrix} h_{1} \\ ... \\ h_{n} \\ \end{pmatrix} $
I am a bit confused because he gave us an example for a function $(x,y)\mapsto x^2+y^2x^2y+4$ to compute : $$f(1+0.0001,2+0.0006)$$ and he has used the generalised expression instead of the first one.
Will the result be different ? Because the expressions seem different...
After reviewing my course again and looking at some stuff over the internet I come to the following conclusions : An approximation can be computed using the n-order approximation and in my coures, we have covered the first and second order approximation.
If you are dealing with a one-variable function :
1st order approximation would be $ f(x_{0}+h) = f(x_{0})+f'(x_{0})*h + 0(h^2) $
with h the small floating number you want to add to a random integer
2nd order approximation would be $ f(x_{0}+h) = f(x_{0})+f'(x_{0})*h + \frac{1}{2}*f''(x_{0})*h^2 $
If you are dealing with a two-variables function :
1st order approximation would be : $ f(x_{0}+h,y_{0}+k) = f(x_{0},y_{0}) + \nabla{f}(x_{0},y_{0})* \begin{pmatrix} h \\ k \\ \end{pmatrix} +\text{negligible}. $
if you are dealing with a n-variables function :