Re: indeterminate and infinity expressions
- To: mathgroup at smc.vnet.net
- Subject: [mg67615] Re: indeterminate and infinity expressions
- From: "David W.Cantrell" <DWCantrell at sigmaxi.org>
- Date: Sun, 2 Jul 2006 06:27:39 -0400 (EDT)
- References: <e85es4$k8e$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Søren_Merser <merser at image.dk> wrote:
> I need to evaluate the expression below:
>
> In[1]:= Log[(#/yhat)] + (1-#) Log[(1-#)/(1-yhat)] & /@ {0,1} /.
> yhat->123 Out[1]= {Indeterminate, -Infinity}
First, Søren, if you really want others to help you, please be much more
careful in your posting. There were several errors.
Evaluating your In[1], exactly as you gave it, returns
{-Infinity, Indeterminate}, which is perfectly correct.
Your claimed Out[1] had the list in the wrong order.
But looking at your In[2], which is missing /. BTW, it seems that your
In[1] was missing an initial # and that yhat was to have been replaced
by .123, rather than 123 .
To get what you want, use Limit. Here's my suggestion:
In[8]:= Limit[x Log[(x/yhat)] + (1-x) Log[(1-x)/(1-yhat)], x->#]
& /@ {0,1} /. yhat -> .123
Out[8]= {0.131248,2.09557}
David
> Depending on the value of #, I get Infinity and Inderterminate results.
>
> To get the wanted answer I've to split the expression into two parts
>
> In[2]:= ans1 = # Log[#/yhat] & /@ {0, 1} yhat -> .123
> Out[2]= {Indeterminate, 2.09557}
>
> In[3]:= ans2 = (1-#) Log[(1-#)/(1-yhat)] & /@ {0, 1} /. yhat -> .123
> Out[3]= {0.131248, Indeterminate}
>
> In[4]:= Max[#] & /@ ({ans1, ans2} /. Indeterminate -> 0)
> Out[4]= {2.09557, 0.131248}
>
> Now, is there an easier way to do it?
>
> Regards Soren
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