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MathGroup Archive 2004

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Re: Re: New user: Abs[] problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg50123] Re: [mg50107] Re: [mg50079] New user: Abs[] problem
  • From: Murray Eisenberg <murray at math.umass.edu>
  • Date: Mon, 16 Aug 2004 06:45:29 -0400 (EDT)
  • Organization: Mathematics & Statistics, Univ. of Mass./Amherst
  • References: <200408140550.BAA15327@smc.vnet.net> <200408150714.DAA12689@smc.vnet.net>
  • Reply-to: murray at math.umass.edu
  • Sender: owner-wri-mathgroup at wolfram.com

Oops, I miscopied the original poster's equation -- changed the first 
minus to a plus.  So ignore my post (except perhaps as suggestions for 
investigating such situations).

Murray Eisenberg wrote:

> Yes, sometimes Mathematica cannot find (all) solutions.  But in this 
> case, BELIEVE!  Mathematica DID solve it: there are no solutions!
> 
> You can see this, e.g., by graphing your piecewise-linear function:
> 
>    Plot[x + 1 + Abs[2x - 4] + Abs[5 - x], {x, -10, 10}];
>    Plot[x + 1 + Abs[2x - 4] + Abs[5 - x], {x, 0, 5}];
> 
> (The second Plot is just to convince you that the minimum really is 
> strictly positive.)
> 
> Or, you can consider the 4 cases:
> 
>    case1 = x >= 2 && 5 >= x;
>    case2 = x >= 2 && 5 < x;
>    case3 = x < 2 && 5 >= x;
>    case4 = x < 2 && 5 < x;
> 
> The first case gives the closed interval from 2 to 5; the second and 
> third simplify...
> 
>    Reduce[case2, x]
> x>5
>    Reduce[case3, x]
> x<2
> 
> ... and the fourth case gives the empty set:
> 
>    Reduce[case4, x]
> False
> 
> (You could have done that much without Mahematica, of course.)
> Now look at your function of x on each of these intervals:
> 
>    expr = x + 1 + Abs[2x - 4] + Abs[5 - x];
> 
>    lin1 = Simplify[expr, case1]
> 2(1+x)
>    lin2 = Simplify[expr, case2]
> 4 (-2 + x)
>    lin3 = Simplify[expr, case3]
> -2 (-5 + x)
> 
> Now try to solve the resulting linear equation in each case:
> 
>    Solve[lin1 == 0, x]
> {{x->-1}}
>    Solve[lin2 == 0, x]
> {{x->2}}
>    Solve[lin3 == 0, x]
> {{x->5}}
> 
> In each case, the line meets the x-axis at a point outside the interval 
> on which the corresponding line segment over the interval in question is 
> the domain of the piece.  QED
> 
> 
> Edson.Brusque at smc.vnet.net wrote:
> 
>>Hello,
>>
>>     I'm starting Electrical Engineering and using Mathematica for 
>>helping me in Calculus e Algebra studies.
>>
>>     I'm trying to solve this equation on Mathematica:
>>         x + 1 - |2x - 4| + |5 - x| = 0
>>
>>     on the Math* notebook I'm typing:
>>         Solve[x + 1 - Abs[2x - 4] + Abs[5 - x] == 0,x]
>>
>>     but only got an empty output: {{}}
>>
>>     Someone please can help me get Math* to solve this?
> 
> 

-- 
Murray Eisenberg                     murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower      phone 413 549-1020 (H)
University of Massachusetts                413 545-2859 (W)
710 North Pleasant Street            fax   413 545-1801
Amherst, MA 01003-9305


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