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Re: Q about using Round.


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? *)


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