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Re: Bug with Series[] - help wanted

The order notation O[1/r] is part of how the output of Series
is formatted.  It does not closely correspond to the internal
form of a series.

In[1]:= $VersionNumber

Out[1]= 5.2

In[2]:= s=Series[1/(r+1), {r, Infinity,3}]

1    -2    -3     1 4
- - r   + r   + O[-]
r                 r

Look at the internal structure of the Series output:

In[3]:= FullForm[s]

SeriesData[r, DirectedInfinity[1], List[1, -1, 1], 1, 4, 1]

Turn it into a polynomial.  Note that the O term goes away.

In[4]:= Normal[s]

  -3    -2   1
r   - r   + -

In[5]:= FullForm[%]

Plus[Power[r, -3], Times[-1, Power[r, -2]], Power[r, -1]]

I hope this is helpful.

Bruce Miller
Technical Support
Wolfram Research, Inc.
support at

On Mar 30, 2006, at 4:29 AM, GidiL wrote:

> Hello all!
> I would really appreciate some help.
> Mathematica has a built-in function called Series:
> Series[x,x_0,n],
> which allows to expand functions into power series where x is the
> variable, x_0 is the point about which we expand, and n is the desired
> order of expansion.
> It also allows x_0 to be infinity, which is very useful when one needs
> multiple expansion (as in gravitational waves and electromagnetic
> radiation).
> Although is allows to expand series in terms of 1/x and writes O(1/x)
> (and its powers) in the output, it does not allow this to be entered in
> the input. To convince yourselves, try it out. Enter in the input,
> e.g., 1+ 1/r+ (O(1/r))^2, and it will tell you that 1/r is not a
> variable.
> If, on the other hand, you entered:
> In=Series[1/(r+1), {r, Infinity,3}]//Simplify
> you will get
> Out=1/r-(1/r)^2+(1/r)^3+(O(1/r))^4
> which shows that the output is possible, but the input is impossible.
> Can anyone offer some assistance?
> Thanks,
> Gideon

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