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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: [mg74627] Re: [mg74615] Is this a problem in mathematica?
  • From: Sseziwa Mukasa <mukasa at jeol.com>
  • Date: Thu, 29 Mar 2007 02:26:58 -0500 (EST)
  • References: <200703280644.BAA20508@smc.vnet.net>

On Mar 28, 2007, at 2:44 AM, traz wrote:

> 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?

Minimize just searches for a minimum, if there are two or more minima  
it will only find one at most.  If you want general solutions use  
Solve to find the minima of the first derivative.

In[4]:=
Solve[D[x\^2 + (1 + x^2 - 2)^2, x]== 0, x]
Out[4]=
{{x -> 0}, {x -> -1/Sqrt[2]}, {x -> 1/Sqrt[2]}}

Checking the distance for each point will allow you to reject the  
point at 0 as a minimum.

Regards,

Ssezi


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