MathGroup Archive 2000

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Re: Simplifying Problems

  • To: mathgroup at smc.vnet.net
  • Subject: [mg22497] Re: [mg22409] Re: Simplifying Problems
  • From: Jacqueline Zizi <jazi at club-internet.fr>
  • Date: Sun, 5 Mar 2000 00:24:33 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

If you put and evalutate  in 3 different cells:

f [0] = \[Pi];

f [k_; k \[Element] Integers] := 0

f [k_] := Sin [k  \[Pi]] /k

Then you can check that, for example, :
f [3/4 ], f [2/3 ], f [2 ], f [0]  return what you wants.

Then your wanted sum will evaluate well. For example:
f [3/4 ] +  f [2/3 ] + f [2 ] + f [0] + f[3/2]
will give
-2/3+ 2 Sqrt [2]/3+ 3 Sqrt [3]/4 + Pi

Just be carefull that in reality, Sin[0]/0 is not defined, nor is 2/0.
You are free to put what you want, for example Pi for 0.



QUESTION
=========


Hi,

> ------------------------
> First question:
> ------------------------
> I have an expression which has a sum of a number of sinc-like terms.
For
> example,
>
>    f[k] = Sin[k Pi] / k
>
> If I try using simplify with the assumption that k is an integer I get

>
>    In[2]:=
>    Simplify[f[k], k \[Element] Integers]
>
>    Out[2]=
>    0
>
> Although this is true for most integers, it is incorrect for the
integer
> k==0 since f[0] = Pi.  So why is this happening?  I would have
expected it
> to either leave the expression untouched or to give me an If
expression.
etc..........................................................................................



  • Prev by Date: Re: Rearranging a tensor
  • Next by Date: Re: Rearranging a tensor
  • Previous by thread: Re: Re: Simplifying Problems
  • Next by thread: Re: Simplifying Problems