MathGroup Archive 2007

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

Search the Archive

Re: Very Strange Behaviour about something Very Simple

  • To: mathgroup at
  • Subject: [mg77193] Re: Very Strange Behaviour about something Very Simple
  • From: Steven Siew <stevensiew2 at>
  • Date: Tue, 5 Jun 2007 06:51:57 -0400 (EDT)
  • References: <f40h9q$6j9$>

On Jun 4, 6:08 pm, "Apostolos E. A. S. Evangelopoulos"
<a.e.a.evangelopou... at> wrote:
> Hello all!
> The following results appear extremely weird and I shall definitely appreciate ideas about why this might be happening!
> I am asking for a solution of the following equation:
> Solve[8R^3/(3h)-h^2/3\[Equal](8R^3+2h^3-3h^3)/(3h), h]
> The result is {}, meaning -as far as I know- that there are no solutions.
> Equivalently, I ask for the following:
> True &&8R^3/(3h)-h^2/3\[Equal](8R^3+2h^3-3h^3)/(3h),
> and, instead of getting a `False' statement, which would be consistent with the previous output, I get the same line in equation form, i.e.
> -h^2/3 + (8*R^3)/(3*h) == (-h^3 + 8*R^3)/(3*h),
> meaning that -again as far as I know- there exists a finite number of particular vaules of R and h satisfying this equality.
> And here comes the incredible bit:
> Taking the right hand side of the above just one small step further and splitting the fraction into two, one easily observes that the above is an identity! Indeed, Mathematica will respond as follows (and correctly this time):
> Solve[8R^3/(3h)-h^2/3\[Equal]8R^3/(3h)-h^2/3, h]
> Output: {{}}
> True && 8R^3/(3h) - h^2/3 == 8R^3/(3h) - h^2/3
> Output: True
> Ultimately, in asking for a simplification of either of the two sides (even from much more complex but equivalent forms) Mathematica always gives the same output and verifies identity. But does this mean I always have to be so careful and simplify everything as much as I can before manipulating it at all!?
> What more can I say...
> I eagerly await your view on this!
> Cheers,
> Apostolos

In[1]:=  eq1 = 8*(R^3/(3*h)) - h^2/3 == (8*R^3 + 2*h^3 - 3*h^3) /

Out[1]=  -(h^2/3) + (8*R^3)/(3*h) == (-h^3 + 8*R^3)/(3*h)

In[2]:=  FullSimplify[eq1]

Out[2]= True

What you have inputted is an identity. Which simplifies to   1 == 1

  • Prev by Date: Re: Evaluate, Replace, and Hold
  • Next by Date: Re: Re: pure function to generate a list of integrals]
  • Previous by thread: Re: Very Strange Behaviour about something Very Simple
  • Next by thread: Re: Very Strange Behaviour about something Very Simple