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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?

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  • $\begingroup$ What do you mean by "theory"? It sounds like perhaps you mean "theorem"? $\endgroup$ – Namaste Feb 24 '14 at 20:58
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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 $\,\ g(x,x) = x^{n-1} + \cdots + x^{n-1} = n x^{n-1} = f'(x).$

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  • $\begingroup$ Yes, my answer always has a polynomial form ... prior to being evaluated to a single value. $\endgroup$ – annoying_squid Feb 24 '14 at 22:06
  • $\begingroup$ 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. $\endgroup$ – Transfinite Numbers Oct 29 '16 at 22:12
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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.

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  • $\begingroup$ Is there a way to determine that the property of a problem is best described using limits as opposed to absolute values? $\endgroup$ – annoying_squid Feb 24 '14 at 22:03
  • $\begingroup$ 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. $\endgroup$ – naslundx Feb 24 '14 at 22:03

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