Re: Accuracy and Precision

*To*: mathgroup at smc.vnet.net*Subject*: [mg36861] Re: [mg36853] Accuracy and Precision*From*: Andrzej Kozlowski <andrzej at tuins.ac.jp>*Date*: Mon, 30 Sep 2002 03:02:59 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

Well, first of of all, your using SetAccuracy and SetPrecision does nothing at all here, since they do not change the value of a or b. You should use a = SetAccuracy[a, Infinity] etc. But even then you won't get the same answer as when you use exact numbers because of the way you evaluate f. Here is the order of evaluation that will give you the same answer, and should explain what is going on: f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121* b^4 - 2) + 5.5*b^8 + a/(2*b), Infinity]; a = 77617.; b = 33096.; a = SetAccuracy[a, Infinity]; b = SetAccuracy[b, Infinity]; f 54767 -(-----) 66192 Andrzej Kozlowski Toyama International University JAPAN On Sunday, September 29, 2002, at 03:55 PM, Peter Kosta wrote: > Could someone explain what is going on here, please? > > In[1]:= > a = 77617.; b = 33096.; > > In[2]:= > SetAccuracy[a, Infinity]; SetAccuracy[b, Infinity]; > SetPrecision[a, Infinity]; SetPrecision[b, Infinity]; > > In[4]:= > f := 333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8 + > a/(2*b) > > In[5]:= > SetAccuracy[f, Infinity]; SetPrecision[f, Infinity]; > > In[6]:= > f > > Out[6]= > -1.1805916207174113*^21 > > In[7]:= > a = 77617; b = 33096; > > In[8]:= > g := (33375/100)*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + > (55/10)*b^8 + a/(2*b) > > In[9]:= > g > > Out[9]= > -(54767/66192) > > In[10]:= > N[%] > > Out[10]= > -0.8273960599468214 > > Thanks, > > PK > > >