Re: Simplifying Finite Sums With A Variable # of Terms
- To: mathgroup at smc.vnet.net
- Subject: [mg21771] Re: [mg21744] Simplifying Finite Sums With A Variable # of Terms
- From: "Tomas Garza" <tgarza at mail.internet.com.mx>
- Date: Thu, 27 Jan 2000 22:56:52 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
Wretch [arc at astro.columbia.edu] wrote:
> Hello. I'm a user of Mathematica 3.0. A simple example of
> what I'd like to do is as follows:
>
> Suppose you have the finite series
>
> S[i_]=Sum[x[k],{k,1,i}]
>
> which is equal to x[1]+x[2]+...+x[i], where the total # of
> terms i is left variable.
>
> I'd like mathematica to calculate the difference
>
> S[N]-S[N-1] = x[N]-x[0] .
>
> I've tried commands like
>
> Expand[S[N]-S[N-1]] and Simplify[S[N]-S[N-1]] ,
>
> but mathematica doesn't simplify it as you would expect.
> It basically does nothing. I suspect that it needs some
> sort of clarification as to the nature of N (i.e. it's a
> positive integer), but I'm not sure. Is there an easy
> way for me to do what I'd like?
To start with, there is no way to incorporate x[0] into your results, since
it is not included in the definition of S (the iterator therein is {k,1,i}).
Then, using e.g. CumulativeSums in the Statistics`DataManipulation` AddOn to
define the sums,
In[1]:=
<< Statistics`DataManipulation`
In[2]:=
s[q_List] := CumulativeSums[q]
In[3]:=
z[q_List] := Prepend[Drop[s[q], -1], 0]
so that, for any list q, the difference s[q] - z[q] will give the list
{S[1], S[2] - S[1],S[3] - S[2],..., S[k] - S[k-1]} = q. Take, for example,
In[4]:=
q = {a, b, c, d}
Out[4]
{a, b, c, d}
In[5]:= s[u]
Out[5]=
{a, a + b, a + b + c, a + b + c + d}
In[6]:=
s[q] - z[q]
Out[6]=
{a, b, c, d}
Tomas Garza
Mexico City