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:


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 $2$:


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?


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.

  • 1
    $\begingroup$ Thank you very much!! Your solution is elegant. simple and general. $\endgroup$ – matheorem Nov 22 '12 at 8:03
  • $\begingroup$ Does this trick work in this way for any expansion order? $\endgroup$ – granular bastard Jan 7 at 21:51
  • $\begingroup$ @granularbastard Yes, it does. $\endgroup$ – Jens Jan 9 at 4:05

Here's an attempt:

multiTaylor[f_, {vars_?VectorQ, pt_?VectorQ, n_Integer?NonNegative}] :=
                   Sum[Nest[(vars - pt).# &, (D[f, {vars, \[FormalK]}]
                       /. Thread[vars -> pt]), \[FormalK]]/\[FormalK]!,
                       {\[FormalK], 0, n}, Method -> "Procedural"]

multiTaylor[f[x, y], {{x, y}, {0, 0}, 2}]
   f[0, 0] + y*Derivative[0, 1][f][0, 0] + x*Derivative[1, 0][f][0, 0] +
   (y*(y*Derivative[0, 2][f][0, 0] + x*Derivative[1, 1][f][0, 0]) +
    x*(y*Derivative[1, 1][f][0, 0] + x*Derivative[2, 0][f][0, 0]))/2
  • $\begingroup$ I admire your mathematica skill. Your answer is right. But I think the following Jens's solution is better, it is simple and more enlightening. What do you think? Thank you all the same!! $\endgroup$ – matheorem Nov 22 '12 at 8:21
  • 1
    $\begingroup$ No problem; you should always pick the answer that is most helpful to you and give that answer the acceptance, after all. I wanted to write something a bit more general (i.e. something that can also handle three, four... variables), so mine is a little bit complicated. $\endgroup$ – J. M.'s ennui Nov 22 '12 at 8:24
  • $\begingroup$ Thanks for the function – it's very useful to me. One slight problem I just noticed occurs when f contains the variable k, which will be replaced by some integer after the expansion. To improve you function, you might want to wrap it in Module[{k}, (* function code *)] to protect the local variable. $\endgroup$ – David Zwicker Jul 30 '15 at 20:07
  • $\begingroup$ @David, you're right; I'll incorporate the fix later. Thank you! $\endgroup$ – J. M.'s ennui Jul 30 '15 at 22:02
  • $\begingroup$ This function should be implemented in Mathematica $\endgroup$ – Pedro Harunari Sep 9 '19 at 20:16

Jens answered this nicely. But my preferred way of thinking about this question is via re-scaling (essentially using perturbation theory) instead of dummy variables.

For example, to generate the series expansion, re-scale all variables by s and expand by series coercion:

(f[x,y] /. Thread[{x,y} -> s {x,y}]) + O[s]^3

TraditionalForm output

This form is generally useful—and although one can use Normal, often working with the series is what you really end up wanting to do.

  • $\begingroup$ Thank you so much! O[s] is doing like magic. I am wondering how O is defined, it behaves not like a normal function. $\endgroup$ – matheorem Apr 16 '18 at 0:36
  • $\begingroup$ Not magic. If you add a function to a series expansion, the only way for it to make sense is to automatically coerce the function into a series (about the same expansion point). $\endgroup$ – TheDoctor Apr 22 '18 at 6:04
  • $\begingroup$ If you add 1.0 to Pi, what you would expect should happen. You might consider this to be "magic" too but surely such automatic coercion is what you want to happen... $\endgroup$ – TheDoctor Apr 22 '18 at 6:16
  • $\begingroup$ Finally, I should have pointed out that Series automatically generates O. For example, if you enter Series[f[x],{x,0,3}] then you'll see O[x] in the output. $\endgroup$ – TheDoctor Apr 22 '18 at 6:18

Another possibility is to strip the "too high" terms with a rule:

ser = Series[f[x, y], {x, x0, 2}, {y, y0, 2}];
Normal[ser] /.
 Derivative[m__][f][args__] /; Plus[m] > 2 :> 0

Mathematica graphics


If f is dependent on a variable k, the above definition leads to a conflict. It is better to rename k in a less common name such as var.

multiTaylor[f_, {vars_?VectorQ, pt_?VectorQ, n_Integer?NonNegative}] :=
                   Sum[Nest[(vars - pt).# &, (D[f, {vars, var}] /. Thread[vars -> pt]),var]/var!,
                       {var, 0, n}, Method -> "Procedural"]

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