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Re: Difference of numbers in the list

  • To: mathgroup at
  • Subject: [mg45922] Re: Difference of numbers in the list
  • From: Bill Rowe <readnewsciv at>
  • Date: Thu, 29 Jan 2004 05:36:49 -0500 (EST)
  • Sender: owner-wri-mathgroup at

On 1/28/04 at 5:19 AM, mtmtk2003 at (mtmtk) wrote:

>I have this table:

>Table[{((6.67/10^8)*157.85)/(0.225 - x)^2, (Pi^2/420^2)*x}, {x, 0.,
>0.007245, 0.001}]

>I wanted to plot the difference of each pair but the only solution
>I could find was to first have two tables A and B then take A-B and
>try to plot that.

>A = Table[{((6.67/10^8)*157.85)/(0.225 - x)^2}, {x, 0., 0.007245,

>B = Table[{(Pi^2/420^2)*x}, {x, 0, 0.007245, 0.001}]

>A - B

>{{0.00020797224691358022}, {0.0002097773366920289},
>{0.00021160751897569133}, {0.00021346324689854517},
>{0.00021534498386961064}, {0.0002172532038538137},
>{0.00021918839166184726}, {0.00022115104324936126}}

>But ListPlot will give an error for this :

>" not a list of numbers or pairs of numbers."

>When I delete the curly braces around the numbers then ListPlot

>But this is too complicated for plotting the difference of two sets
>of numbers.

>How can this be done simply? Thanks for any help.

There are a number of ways this can be solved. First, you could get rid of the curly braces in your definition of A and B. That is defining

A = Table[((6.67/10^8)*157.85)/(0.225 - x)^2, {x, 0., 0.007245,0.001}]

B = Table[(Pi^2/420^2)*x, {x, 0, 0.007245, 0.001}]

Now both A and B will be vectors as will be A-B and ListPlot will not complain.

However, to me the simplest is to work directly with the initial table. That is

data=Table[{((6.67/10^8)*157.85)/(0.225 - x)^2, (Pi^2/420^2)*x}, {x, 0.,0.007245, 0.001}];

But using your definition of A and B


will work just fine. This is nothing more than removing one level of curly braces which you describe as "too complicated". I wonder what you are doing when you "delete the curly braces" that you find "too complicated".

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