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MathGroup Archive 2002

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Re: Taylor Series Expansions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg32381] Re: Taylor Series Expansions
  • From: "Allan Hayes" <hay at haystack.demon.co.uk>
  • Date: Thu, 17 Jan 2002 02:23:34 -0500 (EST)
  • References: <a23g46$9mq$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Joe,

    Select[Normal[Series[Exp[x y], {x, 0, 2}, {y, 0, 2}]],
      Exponent[#,x]+Exponent[#,y]<=2&]

            1+x y

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

    Module[{a,b},
      Select[a+b+Normal[Series[Exp[x y], {x, 0, 2}, {y, 0, 2}]],
          Exponent[#,x]+Exponent[#,y]<=2&]-a-b]

            1+x y

--
Allan

---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565


"Joe Helfand" <jhelfand at wam.umd.edu> wrote in message
news:a23g46$9mq$1 at smc.vnet.net...
> 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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