Limits and Continuity — reblogging

We have seen many examples of functions earlier. Let us also consider the example of the range of r of a gun pointed at an angle $\theta$ to the horizon. Gun, shell and other conditions remaining the same, we know that for every angle $\theta$ we have a definite range $r(\theta)$ , giving rise to the function r. Suppose we know that the target is at a distance d from the gun and that $r(\theta_{0})=d$ for some $\theta_{0}$ . Then, we point our gun at an angle $\theta_{0}$ to the horizon to hit the target. A little deviation in $\theta$ in likely to cause an error in our hit. But, we also know that if our hit is within a certain distance from the target, then our objective is achieved (depending on the shell). Now, if want the hit to be within a distance $\varepsilon > 0$ from the target, can we adjust the deviation $\delta$ in $\theta$ around $\theta_{0}$ accordingly? Another way of putting the same question is to ask if for a small change in angle $\theta, can one get a small change in range$ latex r(\theta)$? If yes, then we are allowed a little play in aiming our gun. If not, a small play might result in a wide miss. What we are asking is: for every $\varepsilon >0$ , can we find a $\delta >0$ such that $|r(\theta)-r(\theta_{0})|<\varepsilon$ whenever $|\theta -\theta_{0}|<0$ ? We shall come across similar questions in other situations too. For example, perhaps, you might be knowing how to obtain the values of $\pi$ through the series called Madhava-Gregory’s series. That method takes a long time to evaluate the value of $\pi$ correctly to, say, 4 places of decimal. There is also the formula:

$\frac{\pi^{4}}{90}=\frac{1}{1^{4}}+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\ldots+\frac{1}{n^{4}}+\ldots$ .

To find the sum of the above series correct to, say, 4 places of decimal, it is enough to sum the first 25 terms. But the sum gives an approximate value of $\pi^{4}/90$ . To get the value of $\pi$, we need to multiply the sum by $90$ and then extract its fourth root. Now the question is how would the error committed in evaluating $\pi^{4}/90$ be propagated in the subsequent calculations? Put in a different way, let us write $x=\frac{\pi^{4}}{90}$ and $f(x)=(90x)^{1/4}$ , and let $x_{n}$ b/ e the approximate value of $\pi$ calculated by summing the first n terms of the series. So, the approximate value of $\pi$ would be $f(x_{n})$ . This leads naturally to the question: how is the error $|f(x)-f(x_{n})|$ in the value of $f(x)$ related to the error $|x-x_{n}|$ in the value of x? Can we calculate $f(x)$ correct to the desired accuracy by calculating x sufficiently accurately? If not, then this method of calculation is not very useful. If a small perturbation in x causes an abrupt change in $f(x)$ , then we should perhaps think of some other method of calculation. It may be noted that, philosophically, continuity forms the basis of large parts of the experimental sciences where it is tacitly assumed that small errors in measurement will not lead to drastic changes in conclusions.

These ideas lead to the definition of continuity of functions.

Definition. let $f: \Re \rightarrow \Re$ be a function. W say that the function is continuous at $x_{0} \in \Re$ if for every $\varepsilon >0$ , we can find a $\delta>0$ such that $|f(x)-f(x_{0}| , \varepsilon$ whenever $|x-x_{0}|<\delta$ . This is to say that for continuous functions, f, the value of $f(x)$ can be restricted within the interval $(f(x_{0})-\varepsilon, f(x_{0})+\varepsilon)$ by restricting the value of x within $(x_{0}-\delta, x_{0}+\delta)$ . Try to do draw a figure based on this!!

Note that for continuity at $x_{0}$ . we need an interval containing $x_{0}$ to be contained in its domain, and hence, it is enough that the function f has its domain an interval in $[a,b]$ . So we can define continuity of a function $f:[a,b] \rightarrow \Re$ in the same way as above. In the case of the end points a and b, we can only talk of the right and left continuity, respectively.

More later,

Nalin Pithwa

Mathematics Hothouse