MathGroup Archive 2009

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Don't understand the response

  • To: mathgroup at smc.vnet.net
  • Subject: [mg97625] Re: [mg97587] Don't understand the response
  • From: DrMajorBob <btreat1 at austin.rr.com>
  • Date: Tue, 17 Mar 2009 05:00:03 -0500 (EST)
  • References: <17032823.1237030239609.JavaMail.root@m02>
  • Reply-to: drmajorbob at bigfoot.com

You didn't see fit to give the value of outB, so I'll first conjure one  
that duplicates your output:

outB = RandomInteger[{1, 5}, {3, 2}];
outB[[1 + 1 ;; 3, 1]] = 0;
outB

{{2, 3}, {0, 4}, {0, 1}}

(1 <= 4) And (outB[[1 + 1 ;; 3, 1]] ==
    ConstantArray[0, {3 - (1 + 1) + 1}])

And True^2

Now to deconstruct that:

1 <= 4

True

and also

(outB[[1 + 1 ;; 3, 1]] == ConstantArray[0, {3 - (1 + 1) + 1}])

True

so In[3308] is equivalent to

True And True

And True^2

(multiplying three symbols, two of which are the same)

You could have entered, instead:

True ~And~ True

True

or

(1 <= 4) ~And~(outB[[1 + 1 ;; 3, 1]] ==
    ConstantArray[0, {3 - (1 + 1) + 1}])

True

or

(1 <= 4) && (outB[[1 + 1 ;; 3, 1]] ==
    ConstantArray[0, {3 - (1 + 1) + 1}])

True

Bobby

On Mon, 16 Mar 2009 04:24:33 -0500, dglawler <dglawler at comcast.net> wrote:

> Don't understand the result below.  Can someone gently unpack this for  
> me?
> True^2??
>
>
> In[3308]:= (1 <= 4) And (outB[[1 + 1 ;; 3, 1]] ==
>    ConstantArray[0, {3 - (1 + 1) + 1}])
>
> Out[3308]= And True^2
>
> Later.........Dennis
>
>



-- 
DrMajorBob at bigfoot.com


  • Prev by Date: Re: inverse of calculation
  • Next by Date: Re: Different results with FourierTransform[]
  • Previous by thread: Re: Don't understand the response
  • Next by thread: Re: collecting certain powers