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Re: indeterminate and infinity expressions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg67613] Re: indeterminate and infinity expressions
  • From: Bill Rowe <readnewsciv at earthlink.net>
  • Date: Sun, 2 Jul 2006 06:27:32 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

On 7/1/06 at 5:11 AM, merser at image.dk (Søren Merser) 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}

>Depending on the value of #, I get Infinity and Inderterminate
>results.

This makes sense because you are evaluating the expression at 0,1.

The first term of the expression when evaluated at 0 reduces to Log[0] which evaluates to -Infinitiy.

The second term of the expression when evaluated will reduce to 0 times -Infinity which is indeterminate.

>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?

There are other ways to do the same thing. For example I could do something like

In[10]:=
List @@ (Log[#1/yhat] + (1 - #1)*
       Log[(1 - #1)/(1 - yhat)] & )[x] /. yhat -> 0.123

Out[10]=
{(1-x) log(1.14025 (1-x)),log(8.13008 x)}

followed by

In[11]:=
Cases[Flatten[MapThread[#1 /. x -> #2 & , {%, {0, 1}}]], 
  _?NumericQ]

Out[11]=
{0.131248,2.09557}

I don't clainm this is simpler, but it does avoid the manual splitting of the expression you did.

But I am curious as to what you hope to accomplish since the result certainly isn't the value of the full expression at 0 or 1.
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