您在這裡

首頁 » Brief Calculus » Limits and continuity

Another perspective on indeterminate forms

12 十一月, 2015 - 11:22
Available under Creative Commons-ShareAlike 4.0 International License.

An expression like 0/0, called an indeterminate form, can be thought of in a different way in terms of infinitesimals. Suppose I tell you I have two infinitesimal numbers d and e in my pocket, and I ask you whether d/e is finite, infinite, or infinitesimal. You can't tell, because d and e might not be infinitesimals of the same order of magnitude. For instance, if e=37d, then d/e=1/37 is finite; but if e=d^2, then d/e is infinite; and if d=e^2, then d/e is infinitesimal. Acting this out with numbers that are small but not infinitesimal,

\begin{align*} \frac{.001}{.037} &= \frac{1}{37}\\ \frac{.001}{.000001}&=1000 \\ \frac{.000001}{.001}&=.001 \end{align*}

On the other hand, suppose I tell you I have an infinitesimal number d and a finite number x, and I ask you to speculate about d/x. You know for sure that it's going to be infinitesimal. Likewise, you can be sure that x/d is infinite. These aren't indeterminate forms.

We can do something similar with infinite numbers. If H and K are both infinite, then H-K is indeterminate. It could be infinite, for example, if H was positive infinite and K=H/2. On the other hand, it could be finite if H=K+1. Acting this out with big but finite numbers,

\begin{align*} 1000-500 &=500 \\ 1001-1000&=1 \end{align*}

Example

If H is a positive infinite number, is \sqrt{H+1}-\sqrt{H-1} finite, infinite, infinitesimal, or indeterminate?

Trying it with a finite, big number, we have\begin{matrix} \sqrt{1000001}-\sqrt{999999}\\ =1.00000000020373\times 10^{-3} \end{matrix} 

: H=1/d
d^-1
: sqrt(H+1)-sqrt(H-1)
d^1/2+0.125d^5/2+...

For convenience, the first line of input defines an infinite number H in terms of the calculator’s built-in infinitesimal d. The result has only positive powers of d, so it’s clearly infinitesimal.

More rigorously, we can rewrite the expression as \sqrt{H}\left ( \sqrt{1+1/H}-\sqrt{1-1/H} \right ). Since the derivative of the square root function \sqrt{x} evaluated at x=1 is 1/2, we can approximate this as
\sqrt{H}\left [ 1+\frac{1}{2H}+...-\left ( 1-\frac{1}{2H}+... \right ) \right ]\\ \begin{align*} &= \sqrt{H}\left [ \frac{1}{H}+... \right ]\\ &=\frac{1}{\sqrt{H}} \end{align*} which is infinitesimal.

  • 瀏覽次數：2948