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MathGroup Archive 2002

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Re: Taylor Series Expansions

  • To: mathgroup at
  • Subject: [mg32381] Re: Taylor Series Expansions
  • From: "Allan Hayes" <hay at>
  • Date: Thu, 17 Jan 2002 02:23:34 -0500 (EST)
  • References: <a23g46$9mq$>
  • Sender: owner-wri-mathgroup at


    Select[Normal[Series[Exp[x y], {x, 0, 2}, {y, 0, 2}]],

            1+x y

And to allow for the expansion being 0 or only one term

      Select[a+b+Normal[Series[Exp[x y], {x, 0, 2}, {y, 0, 2}]],

            1+x y


Allan Hayes
Mathematica Training and Consulting
Leicester UK
hay at
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565

"Joe Helfand" <jhelfand at> wrote in message
news:a23g46$9mq$1 at
> Wow!
>     I have definitely come to the right place.  Thanks for all the
> responses.  Using the Map built in function solved my problem (it still
> took a bit, so you can imagine what I was dealing with).  Here is
> something else which I have wasted some time on not knowing as much
> about Mathematica as I should.  It has to do with multi-variable Taylor
> series expansion.  Mathematica has a built in Series function.  But when
> you use this for multi-variable functions, it doesn't do quite what I'd
> expect.  Let's say I have a function for two fariables, and I want to
> expand to 2nd order.  When I use Series, it expands each varible to
> second order, but includes the cross terms, which I want to belong to a
> 4th order expansion.  For example:
> In[1172]:=
> Normal[Series[Exp[x y], {x, 0, 2}, {y, 0, 2}]]
> Out[1172]=
> \!\(1 + x\ y + \(x\^2\ y\^2\)\/2\)
> But what I really want is just 1 + x y, where if I go to fourth order,
> then I'll take the x^2 y^2 / 2.  I had to take some time to write some
> sloppy Taylor series expansion functions that did what I wanted.  Is
> there a way to get around this problem or do you have any suggestions?
> Thanks Again,
> Joe

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