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Symbolic math
Message #1 Posted by Tizedes Csaba on 11 Aug 2003, 5:17 p.m.

Hello!

Try to solve for 'x' the following equation:

3*x+1/(x-5)=15+1/(x-5)

The MAPLE, MATHEMATICA and HP48SX (solve with QUAD) give me the bad answer (x=5). (The "good" answer is 'no solution'.)

Who understand it...!?

Csaba

      
Re: Symbolic math
Message #2 Posted by Wayne Brown on 11 Aug 2003, 5:52 p.m.,
in response to message #1 by Tizedes Csaba

I would imagine it is rearranging it as:

3*x=15+(1/(x-5)-1/(x-5))

and simplifying that to:

3*x=15

which it then solves to get x=5. (I'm just guessing here.) Obviously it doesn't try to go back and substitute x=5 into the original equation and solve it, else it would get a divide-by-zero error. Hence the admonition of my algebra teachers over 30 years ago: Always go back and check your answer!

BTW, Erable 3.201 on my 48GX gives the same answer: x=5.

            
Well...
Message #3 Posted by Jim L on 11 Aug 2003, 6:48 p.m.,
in response to message #2 by Wayne Brown

Undefined-approaches infinity = Undefined-approaches infinity?

Seems like a pretty good "machine answer". Needs a person to put it in context though.

      
Re: Symbolic math
Message #4 Posted by Larry Corrado, USA (WI) on 11 Aug 2003, 5:56 p.m.,
in response to message #1 by Tizedes Csaba

If you try to find a solution by inserting values and testing, the 1/(x-5) terms on each side "blow up" when x approaches 5. However, if you first algebraically cancel out those denominators (by multiplying by (x-5)), you can solve it normally (and get x = 5). To be honest, I don't know which solution ("no soultion" or x = 5) a mathematician would call correct.

Don't know if this is helpful or not... Larry

      
Re: Symbolic math
Message #5 Posted by Speck on 11 Aug 2003, 9:55 p.m.,
in response to message #1 by Tizedes Csaba

On the TI-92:

Solve(3*x+1/(x-5)=15+1/(x-5),x)

Returns x=5

The TI-85 numeric solver returns X=5.00000000001, but does so with a sign change error before final resolve.

As a direct algebraic problem, x=5 doesn't really work. However, if the machine is internally doing some sort of a limit process, then the returned answer x=5 is quite correct, as it would never actually reach the discontinuity at x=5. If something like Newton's method (or bisection, or one of the better methods) is being used, then there is some sort of stop condition where, based on a preset tolerance level, the algorithm figures it's "close enough," and returns the found answer to the user, even though that answer may appear to be exact. The machine doesn't just magically pick x=5, plug it in and see what happens. You could just as easily do that on your own. The machine has to use a method more suited to it's type of "thinking." The advice posted above about always checking your answer is definately good advice. The computer can never be better than the person using it.

Unfortunately, this doesn't look like one of the problems where the conjugate trick works, either.

Speck.

      
Re: Symbolic math
Message #6 Posted by J. Osugi (E. U.) on 12 Aug 2003, 1:32 a.m.,
in response to message #1 by Tizedes Csaba

 > Try to solve for 'x' the following equation:
 > 3*x+1/(x-5)=15+1/(x-5)
 >
 > The MAPLE, MATHEMATICA and HP48SX (solve with QUAD) give me the bad answer (x=5). (The "good" answer is 'no solution'.)
 >
 > Who understand it...!?

I wasn't able to get a symbolic result with the HP48GX (expan, colct and isol fail to isolate the 'x' variable). The numeric solver (root) instead works and returns 5.

The functions solve(), nsolve(), and zeros() of the TI89 <EG> return 5 as well.

Mupad Light and Math Cad return 5, too.

I suppose it's a matter of how the function behaves when the value of x tends to 5: both the terms 1/(x-5) tend to the same order of infinite and elide each other, so that x=5 is the actual "good" solution. In other words: evaluate both the terms of your equation, 3*x+1/(x-5) and 15+1/(x-5) for x=4.9, 4.99, 4.999, and so forth, or for x=5.1, 5.01, 5.001, and so forth. It should be self explaining. Just my two Eurocents. :-)

            
Re: Symbolic math
Message #7 Posted by Thibaut.be on 12 Aug 2003, 10:25 a.m.,
in response to message #6 by J. Osugi (E. U.)

I agree with previous comments. When it goes about solving an undefined equation, then you need to calculate the limit for f(x) when x tends to the value of the indertimenation, which is here x-5 = 0, or x=5. The limit clearly tends to 5.

      
My really problem...
Message #8 Posted by Tizedes Csaba on 12 Aug 2003, 4:10 p.m.,
in response to message #1 by Tizedes Csaba

Thanks for your answers, opinions!

I think, the really problem is that, if a program works symbolically, why dont examine the denominators?

And if the solver works numerically why give back x=5? (And not 'division by zero' error...) In this case the 'good' solution is a number near 5 (where 1/(x-5) is properly big number (=:N) considering 3*x or 15 AND 3*x+N equal with 15+N for 10, 12 or many digits.)

Csaba


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