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calculus and analysis - Multivariable Taylor expansion does not work as expected


The basic multivariable Taylor expansion formula around a point is as follows:



$$ f(\mathbf r + \mathbf a) = f(\mathbf r) + (\mathbf a \cdot \nabla )f(\mathbf r) + \frac{1}{2!}(\mathbf a \cdot \nabla)^2 f(\mathbf r) + \cdots \tag{1}$$


In Mathematica, as far as I know, there is only one function, Series that deals with Taylor expansion. And this function surprisingly doesn't expand functions in the way the above multivariable Taylor expansion formula does. What I mean is that the function Series doesn't produce a Taylor series truncated at the right order.


For example, if I want to expand $f(x,y)$ around $(0,0)$ to order $2$, I think I should evaluate the following Mathematica expression:


Normal[Series[f[x,y],{x,0,2},{y,0,2}]]

But the result also gives order $3$ and order $4$ terms. Of course, I can write the expression in the following way to get a series truncated at order $2$:


Normal[Series[f[x,y],{x,0,1},{y,0,1}]]

but in this way I lose terms like $x^2$ and $y^2$, so it is still not right.


The formula $(1)$ gives each order in each term, so if the function Series would expand a function in the way formula $(1)$ does, there will be no problem.



I am disappointed that the Mathematica developers designed Series as they did. Does anyone know how to work around this problem?



Answer



It's true that the multivariable version of Series can't be used for your purpose, but it's still pretty straightforward to get the desired order by introducing a dummy variable t as follows:


Normal[Series[f[(x - x0) t + x0, (y - y0) t + y0], {t, 0, 2}]] /. t -> 1


$(x-\text{x0}) (y-\text{y0}) f^{(1,1)}(\text{x0},\text{y0})+\frac{1}{2} (x-\text{x0})^2 f^{(2,0)}(\text{x0},\text{y0})+(x-\text{x0}) f^{(1,0)}(\text{x0},\text{y0})+(y-\text{y0}) f^{(0,1)}(\text{x0},\text{y0})+\frac{1}{2} (y-\text{y0})^2 f^{(0,2)}(\text{x0},\text{y0})+f(\text{x0},\text{y 0})$



The expansion is done only with respect to t which is then set to 1 at the end. This guarantees that you'll get exactly the terms up to the total order (2 in this example) that you specify.


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