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Re: [Q] Solve?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg23420] Re: [Q] Solve?
*From*: "Allan Hayes" <hay at haystack.demon.co.uk>
*Date*: Sun, 7 May 2000 21:18:05 -0400 (EDT)
*References*: <8f29a1$uf@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
James:
Your equation is equivalent to
eq1 = Sqrt[4. + 2.25*z^2] == -2 + 0.525*z;
The difficulty lies in how we interpret Sqrt[4. + 2.25*z^2].
If x != 0 the there are exactly two solution to w^2 = x . Each of them is
the negative of the other. Mathematica interprets Sqrt[x] to be the one with
argument in the range (-Pi/2, Pi/2], this is called the principal square
root of x.
The other one is therefore - Sqrt[x]. To make Mathematica allow this we use
the equation
eq2 = -Sqrt[4. + 2.25*z^2] == -2 + 0.525*z;
Now, If there is a solution to either eq1 or eq2, then, on squaring both
sides of the equation, we see that it will satisfy
eq3 = 4. + 2.25*z^2 == (-2 + 0.525*z)^2;
Mathematica solves this
sln = Solve[eq3, z]
{{z -> -1.06363}, {z -> 0.}}
And, although
eqn1 /. sln
{False, False}
we do have
eq2 /. sln
{True, True}
One way out of this kind of difficulty is to rewrite the equation as a pair
of equations
eqX = {w == -2 + 0.525*z, w^2 == 4. + 2.25*z^2};
Then we get
slnX = Solve[eqX, {z, w}]
{{z -> -1.06363, w -> -2.5584},
-16
{z -> 6.76707 10 , w -> -2.}}
Tidy up:
slnX = Chop[slnX]
{{z -> -1.06363, w -> -2.5584}, {z -> 0, w -> -2.}}
Now we have
eqX /. slnX
{{True, True}, {True, True}}
--
Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565
"James" <research at proton.csl.uiuc.edu> wrote in message
news:8f29a1$uf at smc.vnet.net...
>
>
> Hi!
>
> I come across a problem that I don't understand.
> I generate systems of equations
> and try to solve them using mathematica.
> For example,
>
> ---------------------------------------------------------------
> In[1]= Solve[((4/2)*( 1.50*z+ 2.00)+(4-4/2)*
> Sqrt[( 1.50*z+ 0.00)^2+ 4.00]) +
> ((4/2)*( 1.50*z+ 2.00)+(4-4/2)*
> Sqrt[( 1.50*z+ 0.00)^2+ 4.00])- 8.10*z==0, z]
>
> Out[1]= {}
> ---------------------------------------------------------------
>
> But mathematica can't solve it, simply returing empty.
> This equation can be solved, though, as you see.
> I restarted mathematica to run again, but it doesn't help.
>
>
> The weird thing is some equations are solved, like the below.
> ---------------------------------------------------------------
> In[2]= Solve[((4/2)*( 1.50*z+ 2.00)+(2-4/2)*
> Sqrt[( 1.50*z+ 0.00)^2+ 4.00]) +
> ((4/2)*( 1.50*z+ 2.00)+(3-4/2)*
> Sqrt[( 1.50*z+ 0.00)^2+ 4.00])- 8.10*z==0, z]
>
> Out[2]= {{z->13.4976}}
> ---------------------------------------------------------------
>
> What's wrong with it and how can I solve it?
> I need to use mathematica, because the number of the systems
> of equations are large.
> (Please disregard any redundunt terms in the equations..
> they are generated by a program.)
> With that, when does mathematica return '{}' as an answer?
> Thank you.
>
>
> --- James, newbie in mathematica.
>
>
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