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I have a function that (I believe) correctly takes the multivariate Taylor series expansion about the origin for some expression (first argument), in some variables (second argument, list), to different orders (third argument, list).

My method for taking the multivariate series expansion follows the answer provided here. I take the series expansion to the highest order requested in the list (unless the highest order is infinity, in which case the next-highest is used), then I truncate the series by adding a "big-O" for each variable at its requested order plus one. If the requested order of a variable is infinity, no expansion is performed on that variable.

mySeriesMWE[exp_, x_List, o_List] := Module[
  {pos, ow, xw, xDummy},
  pos = Position[o, Infinity];
  If[(Length[o] < 1) || (Length[pos] == Length[o]),
  ow = o // Delete[#, pos] &;
  xw = x // Delete[#, pos] &;
  (exp /.
         Thread[xw -> xDummy*xw] //
        Series[#, {xDummy, 0, Max[ow]}] & //
       Normal) /.
     {xDummy -> 1} //
    # + 
      Sum[O[xw[[i]]]^(ow[[i]] + 1), {i, 1, Length[xw]}] & //

Usually, the results are as-expected, but there is a test-case that provides a result that is unintuitive, but I think, technically correct: when the expression is $x y z$ and we compare the expansion at the origin for the powers $\{1,1,3\}$ and $\{1,1,\infty\}$.

mySeriesMWE[x y z, {x, y, z}, {1, 1, 3}]
mySeriesMWE[x y z, {x, y, z}, {1, 1, Infinity}]
x y z

It was surprising to me that taking the third power in $z$ gave a better approximation (in fact, the entire expression) than not taking the expansion in $z$ at all! I checked in the general case, however, and I think this result is correct:

mySeriesMWE[f[x, y, z], {x, y, z}, {1, 1, 3}] // Expand
mySeriesMWE[f[x, y, z], {x, y, z}, {1, 1, Infinity}]
f[0,0,0]+z (f^(0,0,1))[0,0,0]+1/2 z^2 (f^(0,0,2))[0,0,0]+1/6 z^3 (f^(0,0,3))[0,0,0]+y (f^(0,1,0))[0,0,0]+y z (f^(0,1,1))[0,0,0]+1/2 y z^2 (f^(0,1,2))[0,0,0]+x (f^(1,0,0))[0,0,0]+x z (f^(1,0,1))[0,0,0]+1/2 x z^2 (f^(1,0,2))[0,0,0]+x y (f^(1,1,0))[0,0,0]+x y z (f^(1,1,1))[0,0,0]
f[0,0,z]+y (f^(0,1,0))[0,0,z]+x (f^(1,0,0))[0,0,z]

The last term in the first output is $x y z$ when the expression is $x y z$, and I believe both expansions match the definition of a multivariate Taylor series about the origin, throwing away the proper higher-order terms.

My first question is: is the function valid for orders less-than infinity? (I believe the answer is "yes," but I have never seen a multivariate Taylor series expansion truncated in certain orders.)

My second question is: is the function's behavior correct when the requested order of a variable is infinity?

My third question is: is there a better way to write the function such that the variable we do not wish to expand in is somehow better-preserved?


Here's a better way, based on a suggestion by John Sidles. The idea is to "penalize" higher-order terms by weighting the replacement of xDummy with an exponent that depends (unintuitively: inversely) on the requested orders of expansion in that variable. Here is the modified function.

mySeriesMWE[exp_, x_List, o_List] := Module[
   {pos, ow, xw, n, xDummy, xDummyPower, xDummyList, testList},
   pos = Position[o, Infinity];
   If[(Length[o] < 1) || (Length[pos] == Length[o]),
    ow = o // Delete[#, pos] &;
    xw = x // Delete[#, pos] &;
    n = Length[xw];
    xDummyList = Array[xDummy^(1 + Max[ow] - ow[[#]]) &, n];
    testList = Array[If[#1 == #2, ow[[#1]], 0] &, {n, n}];
    xDummyPower = 
        testList[[j, k]]*(1 + Max[ow] - ow[[k]]),
        {k, 1, n}
       {j, 1, n}
    (exp /.
           Table[xw[[i]] -> xDummyList[[i]]*xw[[i]], {i, 1, n}] //
                 Series[#, {xDummy, 0, xDummyPower}] & //
         Normal) /.
       {xDummy -> 1} //
      # + Sum[O[xw[[i]]]^(ow[[i]] + 1), {i, 1, n}] & //

There is some ambiguity in choosing the order of expansion in xDummy. I chose a conservative value (including extra terms) and truncated with O in the usual manner, in case an obvious term was missed in the expansion (which does happen).

share|improve this question
It may work better if you replace Max by Total. –  Daniel Lichtblau Apr 21 '14 at 19:33
@DanielLichtblau, I like your suggestion. However, it yields more terms than I want. With it, when I request $O(x^n), O(y^m), O(z^p)$, I throw-away terms like those that include x^i when $i>n$, etc. But, say the expression is $x y z$ and ${n,m,p}={1,1,3}$. I would get a term with x*y*z^3, which is really of higher-order. That's why I'm taking Max, although I can't really say which method is "right." (Can you, or somebody else?) Thanks for the comment! –  Rico Picone Apr 21 '14 at 22:24
There is actually a pretty good solution for this, shown to me by John Sidles. I'm going to see if he posts his answer, otherwise I'll do so. –  Rico Picone Apr 22 '14 at 1:30
Maybe you want a weighted total degree bound? Could get that using the lcm of the vector of exponents. For each variable x, change it to x^(lcm/max_x_weight). Then use the method from the posts you linked to, with lcm as the total degree. Last, undo the power transformations. –  Daniel Lichtblau Apr 22 '14 at 13:49
@DanielLichtblau, I just saw your comment. This sounds analogous to the edit I posted, but I'll give it a try, anyhow. Thanks. –  Rico Picone Apr 22 '14 at 19:28

1 Answer 1

up vote 1 down vote accepted

This was what I had in mind (does not handle infinites in the order spec though). I'm not at all convinced that it's what you want but it might give some ideas for coding.

weightedMultivariateSeries[f_, vars_List, ord_List] /; 
  Length[vars] === Length[ord] && 
   VectorQ[ord, IntegerQ[#] && # > 0 &] := Module[
  {t, n = Length[ord], lcm = LCM @@ ord, newf},
  newf = f /. Thread[vars -> vars^(lcm/ord)];
  newf = newf /. Thread[vars -> t*vars];
  Normal[Series[newf, {t, 0, Max@(lcm/ord)}]] /. t -> 1 /. 
   Thread[vars -> vars^(ord/lcm)]
share|improve this answer
This is close to how I implemented your suggestion. I'll test yours and accept it as the answer, as soon as I can. –  Rico Picone Apr 23 '14 at 18:55
I think instead of {t, 0, Max@(lcm/ord)}, you need {t, 0, lcm}. Right? –  Rico Picone Apr 23 '14 at 22:56
I'm not sure. It's not clear to me what is the balance between individual degrees and total degree. I suspect what you may need is another integer input to give the total degree of your series, with ord used to reweight individual variable powers. –  Daniel Lichtblau Apr 23 '14 at 23:35
If I use {t, 0, lcm}, I get pretty much exactly what I expect in the test-cases I've used. I do have a "bump" variable input that I use to add to (or subtract from) the order, {t, 0, lcm+bump}, in case I want to see more or fewer terms. However, I can't find a case in which bump=0 isn't "correct" in the sense of giving only (and all) the terms that wouldn't be of higher-order, considering mixing (see the example I gave in the comments on my original question). –  Rico Picone Apr 24 '14 at 0:15
I accept your answer with the qualification that in my application I think it should be {t, 0, lcm}, as discussed above. Others can tweak it as-desired. I would up-vote it, too, but I don't yet have the minimum rep. –  Rico Picone Apr 25 '14 at 18:19

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