Taking a derivative is essentially the correct way to divide by zero - based on taking the limit.
If you can take the limit, you can divide. If there is no unique limit, you can't.
The notation
essentially divides by 0 n times.
Of course, that's not the only case in which you actually can divide by 0.
In case of
, you can look at the taylor polynomial of
which is
}^k x^{1+2k}}{{\left(1+2k\right)}!})
which, in turn, you can easily divide by
to obtain the series:
}^k x^{2k}}{{\left(1+2k\right)}!})
.
In doing so, you'll find, that, for x=0, the function returns 1.
Another method to figure out a division by zero is via L'Hôpital's rule:
Take two completely generic functions
)
and
)
.
Let's assume, at a point
,
=0)
.
Thus
}{g\left(x_0\right)})
would be a division by 0.
However, if this is possible, a value can be found at some point by looking at
}{g^'\left(x_0\right)})
.
If that still results in a division by 0 (i.e.
=0)
), you just have to keep going. At some point, the n-th derivative
} \left( x_0 \right) \neq 0)
. and the division
}\left(x_0\right)}{g^{\left(n\right)}\left(x_0\right)})
can be done.
The value you get in that case will be precisely the value you'll get for the very first division
.
Technically, to be more exact, you'd need to use limits.
But to cut a long story short, it already is known how to properly divide by 0. You just need to keep track of extra information, namely the behavior of the two functions you divide as the denominator approaches 0. In case of two constant functions, with, say,
=1)
and
, this will just result in a directed infinity.
So really, when people say, dividing by 0 is undefined, what they really mean is, dividing two mere numbers, not functions, by 0, is underdefined. More information can (but doesn't have to) resolve the problems. If the problems remain, you get a singularity. Most singularities, however, can be dealt with pretty well. (Polynomial singularities, namely)
Exponential singularities are rarer but also more problematic.
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March 30, 2010, 08:22:26 PM
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