# cancelling out before evaluation of variable

I'm been working on a theory, though my math is weak. Let's say I've managed to determine that I can arrive at an answer A by always using the formula BCD / D. Of course this evaluates to BC after canceling out D. However sometimes D can be zero 0 which results in an undefined answer. My question is theoretical in nature: Are there any mathematical theories that permit for the D's to cancel out even if D is zero?

• What do you mean by "theory"? It sounds like perhaps you mean "theorem"? – Namaste Feb 24 '14 at 20:58

Yes, if one knows that the answer has polynomial form then one can perform such cancellations. As a simple example, if we wish to solve $$\, x f(x) = x^2\,$$ and we know the solution $$\,f\,$$ is a polynomial in $$\,x\,$$ then the solution is $$\,f(x) = x\,$$ This can lead to very efficient solutions in less trivial contexts. For example, see this slick proof of Sylvester's determinant identity $$\rm\, det (I+AB)=det(I+BA)\,$$ that proceeds by universally cancelling $$\rm\ det\, A\$$ from the $$\rm\, det\,$$ of $$\rm \ (1+A\ B)\, A\, =\, A\, (1+B\ A),\,$$ thus trivially eliminating the "apparent singularity" at $$\rm\ det\, A\, =\, 0.\,$$ Further discussion is here.

As another example, one can algebraically define derivatives of polynomials by a formula involving universal cancellation. By the Factor Theorem we know that $$\,x-y\mid f(x)-f(y)\,$$ in $$\,R[x,y]\,$$ for any ring $$\,R.\,$$ Let the quotient be the polynomial $$\,g(x,y)\in R[x,y].\,$$ Then one easily shows using linearity that the derivative of $$\,f(x)\,$$ w.r.t. $$\,x\,$$ is $$\,f'(x) = g(x,x),\,$$ i.e.

$$\begin{eqnarray}{}& g(x,y)\ &=&\ \frac{f(x)-f(y)}{x-y}\ \in\ R[x,y]\\ \Rightarrow\ & g(x,x)\ &=&\ f'(x)\ \in\ R[x] \end{eqnarray}$$

For example, $$\,f(x) = x^n$$ $$\,\Rightarrow$$ $$\,g(x,y) = \dfrac{(x^n\!-y^n)}{(x\!-\!y)} = x^{n-1}\! + x^{n-2}y+\cdots+xy^{n-2}\!+y^{n-1}$$

therefore $$\,\ g(x,x) = x^{n-1} + \cdots + x^{n-1} = n x^{n-1} = f'(x).$$

• Yes, my answer always has a polynomial form ... prior to being evaluated to a single value. – annoying_squid Feb 24 '14 at 22:06
• By taylor expansion it seems to work always, see for example here: mathforum.org/kb/… , but then when I try f(x)=e^x I cannot do it practically. – Transfinite Numbers Oct 29 '16 at 22:12

Simple answer is that $\frac{ab}{b} = a$ whenever $b \not=0$. If $b=0$, the expression simply has no defined value.

A longer answer is that $\frac{f(x)}{g(x)}$ may yet be defined even when $f(a) \to 0$ and $g(a) \to 0$ for some $x=a$. The classic example is

$$\lim_{x \to 0} \frac{\sin{x}}{x} = 1$$

Even though both numerator and denominator tend to 0. Of course, inserting $x=0$ giving $\frac{\sin{0}}{0}$ is nonsense.

• Is there a way to determine that the property of a problem is best described using limits as opposed to absolute values? – annoying_squid Feb 24 '14 at 22:03
• I'm not sure I can give a general description. If you post more details on what you are trying to calculate, I can try to help. – naslundx Feb 24 '14 at 22:03