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MathGroup Archive 2007

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Re: Is this a problem in mathematica?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74639] Re: Is this a problem in mathematica?
  • From: "Dana DeLouis" <dana.del at gmail.com>
  • Date: Thu, 29 Mar 2007 02:33:08 -0500 (EST)

> Minimize[minDist, y == 1 + x^2, {x, y}]

> but x also has another answer: +1/Sqrt[2].

 

Here's an alternative using Derivatives.

 

Plot[x^2 + 1, {x, -2, 2}, 

   AspectRatio -> Automatic, 

   GridLines -> Automatic]; 

 

pt = {0, 2}; 

{x, 1 + x^2} - pt

 

{x, -1 + x^2}

 

Distance Equation:  I used Simplify to remove Abs[ ]

 

d = Simplify[Norm[%], x >= 0]

 

Sqrt[1 - x^2 + x^4]

 

Take the Derivative...

 

D[d, x]

 

(-2*x + 4*x^3)/(2*Sqrt[1 - x^2 + x^4])

 

Solve[% == 0, x]

 

{{x -> 0}, {x -> -(1/Sqrt[2])}, {x -> 1/Sqrt[2]}}

 

There are 3 possible solutions.  However, with x->0, the distance is not the
shortest:

 

d /. %

 

{1, Sqrt[3]/2, Sqrt[3]/2}

 

Therefore, the remaining 2 solutions would be the shortest.  Ie..

{x -> -(1/Sqrt[2])}, {x -> 1/Sqrt[2]}

 

-- 

HTH   :>)

Dana DeLouis

Windows XP & Mathematica 5.2

 

 

"traz" <t_raz at yahoo.com> wrote in message news:eud3g1$l0t$1 at smc.vnet.net...

> Let's say I wanna solve this problem:

> 

> Determine point(s) on y = x^2 +1 that are

> closest to (0,2).

> 

> So in mathematica:

> 

> minDist = (x - 0)^2 + (y - 2)^2;

> Minimize[minDist, y == 1 + x^2, {x, y}]

> 

> Output will give you: x -> -1/Sqrt[2], y -> 3/2

> 

> but x also has another answer: +1/Sqrt[2]. Is this a problem in
mathematica or can my code be changed to output the other value of x for the
minimum distance?

> 


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