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# Continuity as accuracy

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Continuity can be understood by considering the accuracy of a measurement.

Suppose we have a collection of square tiles of various sizes and we need to find the area $A$ of each in order to be able to compute how many we need to cover the whole floor.

Of course, we simply measure the side, $x$, of each tile. We know from geometry that $$A=x^2,$$ so we can always compute: $$x=10\Rightarrow A=100,$$ (in inches) etc.

But what if the measurement isn't fully accurate? What if there is always some error? It's never $x=10$ but $$x=10 \pm .3.$$ As a result the computed value of the area -- what we care about -- will also have some error! Consider, the area won't be just $A=100$ but $$A=(10 \pm .3)^2.$$ Therefore, $$A=10^2 \pm 2 \cdot 10 \cdot .3 +.3^2$$ $$=100.09 \pm 6.$$ The result means that the actual area must be within the interval $(94.09,106.09)$.

Suppose however that we can always improve the accuracy of the measurement of the side of the tile $x$ -- as much as we like. The question is, can we also improve the accuracy of the computed value of $A$ -- to our satisfaction?

What do we mean by that? Suppose $x=10$. The above computation shows that if the desired accuracy of $A$ is $\pm 5$, we haven't achieved it with the given accuracy of measurement $x$, which is of $\pm .3$. We can easily show however that $\pm .2$ would solve the problem: $$A=(10 \pm .2)^2=10^2 \pm 2 \cdot 10 \cdot .2 +.2^2 .$$ $$=100.04 \pm 4.$$ It follows that the real area must be within $4.04$ from $100$.

Let's rephrase this problem in order to solve it for all possible values of desired accuracy of $A$.

Suppose the measurement of the side is $a$ and, therefore, the assumed area $A$ is $a^2$. Now suppose we want the accuracy of $A$ to be some small value $\epsilon >0$, what accuracy $\delta$ of $x$ do we need to accomplish that?

Suppose the actual length is $x$ and, therefore, the actual area is $A=x^2$. Then we want to ensure that $A$ is within $\epsilon$ from $a^2$ by making sure that $x$ is within $\delta$ from $a$. What that $\delta$ should be?

To rephrase algebraically, we want to find $\delta$ such that $$|x-a| <\delta \Rightarrow |x^2-a^2|<\epsilon.$$

Why is this even possible? Because $x^{2}$ depends continuously on $x$.

Here's the definition suggested by the above discussion: $f$ is continuous at $x=a$ if for any $\epsilon >0$ there is $\delta >0$ such that $$|x-a| <\delta \Rightarrow |f(x)-f(a)|<\epsilon.$$

Exercise. Prove that $f(x)=x^2$ is continuous at $x=0,x=1,x=a$.

To further illustrate this idea, consider a different situation. Suppose we don't care about the area anymore; we just want to fit these tiles in a strip of $10$ inches. We take a tile and if it fits it is used; otherwise it is discarded.

So, we still get a measurement $a$ of the side of the tile but our real interest is whether it's less or more than $10$.

As above, we don't know $x$ exactly; it's always $5.0 \pm 0.5$ or $a \pm \delta$. Here $\delta$ is the accuracy of measurement of $x$. The situation is much simpler than before. For example, if the length is measured as $11$, we need the accuracy $\delta=1$ or better to make the determination. it's similar for length $9$.

But what if the measurement is exactly $10$? Even if we can improve the accuracy, i.e., $\delta$, as long as $\delta > 0$, we don't always know whether $x$ is larger or smaller than $10$.

Let's define a function $f$ such that

• $f(x)=1$ (pass) if $x \le 10$ and
• $f(x)=-1$ (no pass) if $x>10$.

Algebraically, we can use this: $$f(x) = -\text{sign}(x -10).$$

Suppose we need the accuracy of $y = f(x)$ to be $\epsilon = 0.5$. Can we achieve this by decreasing $\delta$? In other words, can we find $\delta$ such that $$|x-10|<\delta \Rightarrow |f(x)-1|<\epsilon.$$

Of course not: $$x>10 \Rightarrow |f(x)-1|=|-1-1|=2.$$ The reason to be quoted is that $f$ is discontinuous at $x=10$.

Exercise. Carry out this kind of analysis for: the total test score vs. the corresponding letter grade.

Thus, the idea of continuity of the dependence of $y$ on $x$ is:

We can ensure the desired accuracy of $y$ by increasing the accuracy of $x$.

Exercise. In addition to being continuous, $f(x)=x^2$ is also differentiable. How does that help with accuracy? Hint: there is simpler dependence between $\epsilon$ and $\delta$.