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MathGroup Archive 2007

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Re: Very Strange Behaviour about something Very Simple

  • To: mathgroup at smc.vnet.net
  • Subject: [mg77177] Re: Very Strange Behaviour about something Very Simple
  • From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
  • Date: Tue, 5 Jun 2007 06:43:37 -0400 (EDT)
  • Organization: The Open University, Milton Keynes, UK
  • References: <f40h9q$6j9$1@smc.vnet.net>

Apostolos E. A. S. Evangelopoulos 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

If you suspect an identity, if it is usually better to check it first 
with Simplify. Also, SolveAlways and Reduce will return the expected 
result without any additional manipulation.

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

Out[1]=
True

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

Out[2]=
{{}}

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

Out[3]=
True

Regards,
Jean-Marc


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