Re: Q about using Round.
- To: mathgroup at smc.vnet.net
- Subject: [mg20609] Re: Q about using Round.
- From: Cheng Lee Lung <eeacheng at cityu.edu.hk>
- Date: Thu, 4 Nov 1999 02:13:27 -0500
- Organization: City University of Hong Kong
- References: <7v3clp$607@smc.vnet.net> <7vfe0u$bcu$7@dragonfly.wolfram.com>
- Sender: owner-wri-mathgroup at wolfram.com
Wen-Feng Hsiao wrote: > OK! I am sorry. My post is not readable. > About my question, Lichtblau has kindly answered it for me. It is because > the Map procedure will map the function to all parts of an expression. > Thus, after mapping, the Sqrt[Pi] would be equal to Round[Pi]^Round[1/2] > => 3^0 =1. > > Wen-Feng > > In article <7v3clp$607 at smc.vnet.net>, d8442803 at student.nsysu.edu.tw > says... > > Hi, > > > > The following procedures are an example of 'Round' in the Mathematica > > help. It is strange that the round of square-root of pi is 1. It is > > supposed to be 2. Why? > > > > Wen-Feng > > > > -=-=-=- > > In[7]:= > > NumericRound[x_] := MapAll[If[NumericQ[#], Round[#], #] &, x] > > In[8]:= > > NumericRound[Sqrt[Pi] + (E + a) x + (4.3 - 8.6i )x^3] // InputForm > > Out[8]//InputForm= > > 1 + (3 + a)*x + (4 - 9*i)*x^3 > > > > (* When changing it into the following , I obtain the desired result. > > Why? *) Try N[Sqrt[Pi]]. LL