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Re: RE: greetings and a question!

  • To: mathgroup at smc.vnet.net
  • Subject: [mg83149] Re: [mg83136] RE: [mg83116] greetings and a question!
  • From: DrMajorBob <drmajorbob at bigfoot.com>
  • Date: Tue, 13 Nov 2007 03:36:23 -0500 (EST)
  • References: <33259982.1194871140629.JavaMail.root@m35>
  • Reply-to: drmajorbob at bigfoot.com

The methods of Reduce are mysterious, but I can, at least, verify that the  
result is correct and unique. Start by solving for x and looking at the 
term inside Sqrt:

xRoots = x /. Solve[eq, x];
xDiscriminant = First@Union@Cases[xRoots, Sqrt[z_] :> z, Infinity]

-4 + 4 a - 3 a^2 + 4 b + 2 a b - 3 b^2 + 4 a y - 4 b y - 4 y^2

For x to be Real, this must be >= 0. To have more than one solution, it  
must be > 0, which Reduce says is impossible:

Reduce[xDiscriminant > 0, {a, b, y, z}, Reals]

False

It can be 0, however, at precisely the values you originally posted:

Reduce[xDiscriminant == 0, {a, b, y, z}, Reals]
xRoots /. ToRules@%

a == 1 && b == 1 && y == 0

{-1, -1}

To show that Reduce is correct about that "False" pronouncement above, we  
mimic the above steps, this time solving for y to factor the x  
discriminant:

yRoots = y /. Solve[xDiscriminant == 0, y];
yDiscriminant = Last@Union@Cases[yRoots, Sqrt[z_] :> z, Infinity]

-2 + 2 a - a^2 + 2 b - b^2

As before, the discriminant must be >= 0 to allow a Real y value, and > 0  
to allow more than one.

But

0 == yDiscriminant + (b - 1)^2 + (a - 1)^2 // Expand

True

so a = b = 1 is the only real solution.

Bobby

On Mon, 12 Nov 2007 04:18:15 -0600, Tony Harker <a.harker at ucl.ac.uk> wrote:

> Rearrange the expression as
>
>   (((x + 1) + (a + b - 2)/2)^2 + (y - (a - b)/2)^2) + ((a - 1)^2)/
>   2 + ((b - 1)^2/)2,
>
>  which, if a, b, x and y are real, can only be zero if each term is zero.
> The result follows.
>
>   As to how Mathematica does it -- over to the experts!
>
>    Tony
>
> A.H. Harker
> 112 Cumnor Hill
> Oxford
> OX2 9HY
> UK
>
> ]-> -----Original Message-----
> ]-> From: dimitris [mailto:dimmechan at yahoo.com]
> ]-> Sent: 11 November 2007 08:03
> ]-> To: mathgroup at smc.vnet.net
> ]-> Subject: [mg83116] greetings and a question!
> ]->
> ]-> Hello to all of you!
> ]->
> ]-> Unfortunately, family, working and research issues prevent
> ]-> me from participating to the forum as frequent as I used to.
> ]->
> ]-> Anyway...
> ]->
> ]-> It is my first post since a long time so everybody be patient!
> ]->
> ]-> A student of mine came across the following equation in a
> ]-> mathematical contest:
> ]->
> ]-> In[1]:=
> ]-> eq = x^2 + y^2 + (a + b)*x - (a - b)*y + a^2 + b^2 - a - b + 1==
=0;
> ]->
> ]-> (all variables are assumed real)
> ]->
> ]-> Of course for Mathematica the solution is rather trivial.
> ]->
> ]-> In[1]:=
> ]-> $Version
> ]->
> ]-> Out[1]=
> ]-> "5.2 for Microsoft Windows (June 20, 2005)"
> ]->
> ]-> In[2]:=
> ]-> eq = x^2 + y^2 + (a + b)*x - (a - b)*y + a^2 + b^2 - a - b + 1==
=0;
> ]->
> ]-> In[3]:=
> ]-> Reduce[eq, {a, b, x, y}, Reals]
> ]-> ToRules[%]
> ]-> eq /. %
> ]->
> ]-> Out[3]=
> ]-> a == 1 && b == 1 && x == -1 && y == 0
> ]->
> ]-> Out[4]=
> ]-> {a -> 1, b -> 1, x -> -1, y -> 0}
> ]->
> ]-> Out[5]=
> ]-> True
> ]->
> ]-> Can somebody explain concisely the mathematica concept
> ]-> behind this solution? In fact I would be much obliged if
> ]-> somebody pointed me out how to obtain the result by hand.
> ]-> Also, by curiosity, how Mathematica reaches the result?
> ]->
> ]-> Dimitris
> ]->
> ]->
> ]->
>
>
>



-- 

DrMajorBob at bigfoot.com


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