Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2007
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2007

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: How does Solve works ??

  • To: mathgroup at smc.vnet.net
  • Subject: [mg80250] Re: [mg80183] How does Solve works ??
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Wed, 15 Aug 2007 04:29:15 -0400 (EDT)
  • Reply-to: hanlonr at cox.net

The exponent in expr^(1/2) has to be in parentheses

Clear[n, k];

{n, k} /. 
 FullSimplify[Solve[{e1 == n^2 - k^2, e2 == 2 n*k}, {n, k}][[3]]]

{e2/(Sqrt[2]*Sqrt[Sqrt[e1^2 + e2^2] - e1]), 
 Sqrt[(1/2)*Sqrt[e1^2 + e2^2] - e1/2]}

FullSimplify[% == {(e1 + (e1^2 + e2^2)^(1/2))^(1/2)*Sqrt[1/2],
   (-e1 + (e1^2 + e2^2)^(1/2))^(1/2)*Sqrt[1/2]}, {e1 > 0, e2 > 0}]

True


Bob Hanlon

---- Ravinder K Banyal <r.banyal at neu.edu> wrote: 
> Hi,
> I am just a beginner in Mathematica.
> I have two simple equations as follows:
> e1=n^2-k^2;
> e2=2 n*k;
> If I solve for n and k, I should get
> n = (e1 + (e1^2 + e2^2)^1/2)^1/2 * sqrt(1/2)
> k = (-e1 + (e1^2 + e2^2)^1/2)^1/2 * sqrt(1/2)
> However when I use mathematica (ver 5.1) to solve the above equations I  
> do not get the intended results. 
> Solve[e1 -n^2+k^2 = = 0, e2-2n*k = = 0, {n,k}]
> None of the four roots generated by mathematica matches with above results.
> 
> Any comment please.....
> 
> Regards,
> -ravi
> 
> 
> 
> 



  • Prev by Date: Re: How does Solve works ??
  • Next by Date: Re: Foucault pendulum
  • Previous by thread: Re: How does Solve works ??
  • Next by thread: Re: How does Solve works ??