Math 1b. Calculator Policy

Calculators are not used in this course, they are not needed to solve any homework or exam problems, and their use is not permitted on exams or on homework. If you want to know why this policy being used, please read the rest of this handout.

Part of the reason for the vast range of applications of calculus (and mathematics in general) is that a handful of general concepts can be used to describe many different kinds of phenomena. However, the ability to use general principles in practical settings depends upon understanding how to properly apply such principles and what kinds of difficulties might arise. This understanding is something which cannot be told to you by a machine. Once you understand a fundamental idea, it is usually a simple matter to learn how to use a calculator or a computer to apply the idea in a concrete numerical setting. The purpose of this course is to teach you the hard part: understanding the concepts and the methods of computation.

Computers and calculators are wonderful and powerful tools for carrying out difficult numerical tasks, but if you are unable to think about and work with concepts in the absence of a machine, then your ability to intelligently confront new problems will be severely restricted. It is for this reason that calculators are not permitted in this course.

The following example illustrates the right way and wrong way to use a calculator:

Problem: The graphs of y = x^1/10 and $y = 10\ln(x)$ intersect for some x₀ between 1 and 2. Do the graphs ever intersect for $x \ge 2$ ?

Bogus Solution: Graph on a calculator, and it looks like x^1/10 is always smaller than $10\ln(x)$ for $x \ge 2$ and x^1/10 seems to grow more slowly as well (try it!). Thus, the graphs never cross for $x \ge 1$ .

This solution is wrong for several reasons. First, although it is true that $x^{1/10} < 10\ln(x)$ over the range of $x \ge 2$ that you are likely to consider on a graphing calculator, and moreover the rate of growth of x^1/10 is less than that of $10\ln(x)$ over such ranges as well, the conclusion is actually incorrect. Since the range of a calculator is limited, the fact that the calculator fails to show something does not imply that it can't happen, even if the problem (such as the above one) seems to only involve ``small'' numbers. Two examples are the ``facts'' that the graphs do not cross for $x \ge 2$ and that lower graph, y = x^1/10, is growing more slowly than the higher one, $y = 10\ln(x)$ . How do we know that these trends will continue forever? It turns out that they don't!

Let's now consider the bogus solution in more detail. It is certainly true that if x^1/10 is smaller than $10\ln(x)$ at some point, such as at x = 2, and always grows more slowly beyond this point, then the graphs never cross again. This is is the type of conceptual explanation that is expected when you are justifying an answer, so the bogus solution would have been fine if the hypothesis about growth rates were correct, as the calculator appears to suggest. Is is true that for large x, the function x^1/10 always grows more slowly than $10\ln(x)$ ? Let's compute the derivatives!

$\begin{displaymath}{d\over dx} x^{1/10} = {1\over 10}x^{-9/10},\,\,\,\,\, {d\over dx} (10\ln(x)) = {10\over x}.\end{displaymath}$

Is $(1/10)x^{-9/10} \le 10/x$ always true for $x \ge 1$ ? Multiplying through by 10x, our inequality is equivalent to $x^{1/10} \le 100$ , or $x \le 100^{10} = 10^{20}$ . No wonder the calculator graph showed what it did: you have to go all the way to x = 10²⁰ = 100000000000000000000 to see the growth rate of x^1/10 overtake that of $10\ln(x)$ . Aha, so perhaps something does eventually happen ...

Solution without Calculator: Make the change of variable z = x^1/10 to get rid of the fractional exponents, so we are considering the functions z and $10\times\ln(z^{10}) = 100\ln(z)$ , for $z \ge 1$ . From basic facts about the rate of growth of polynomials and logarithms, we know that polynomials always grow much faster than logarithms when the variable is large. Thus, $z > 100 \ln(z)$ for large z. Since we have $z < 100\ln(z)$ for z = 2, by the intermediate value theorem there is a solution to $z = 100\ln(z)$ with z > 2 > 2^1/10. Thus, there is a solution to $x^{1/10} = 10\ln(x)$ with x^1/10 > 2^1/10 (i.e., with x > 2).

Can we solve it with a Calculator? It is possible to use a calculator to correctly solve the problem (instead of appealing to the intermediate value theorem), but we will now see that such a solution requires the same insight that we needed to solve the problem without the calculator, namely a clever choice of change of variable. The use of a calculator does not enable us to abandon the need to think about what we are doing.

We saw by a consideration of growth rates in the bogus solution that we need to consider x beyond 100¹⁰ = 10²⁰. In order to bring such large numbers ``into range'' on the calculator, it is easier to graph functions in terms of the variable z = x^1/10 rather than x. For example, when x = 100¹⁰ (a very large number) we have z = x^1/10 = 100(a much smaller number). Since x = z¹⁰, we see that $10\ln(x) = 10\ln(z^{10}) = 10\times 10\ln(z) = 100\ln(z)$ . Thus, we want to graph z and $100\ln(z)$ and see if there is an intersection for $z \ge 1$ . Put another way, we are trying to solve the equation $z = 100\ln(z)$ , or $z/\ln(z) = 100$ , with z large. Graphing $y(z) = z/\ln(z)$ , the calculator finds an approximate z₀ = 647.27751 where y(z₀) = 100. Since x = z¹⁰, raising z₀ to the 10th power gives the approximate x-coordinate $z_0^{10} \sim 1.3 \times 10^{28}$ of a point of intersection of the original graphs y = x^1/10 and $y = 10\ln(x)$ . This coordinate is out of the range in which a student would look.

This looks rather complicated (and it is, compared to the conceptual solution without the calculator), but the essential idea of the solution (change of variable) is still the same. The moral of the story is that the calculator is very good at finding explicit numerical solutions, but in order to avoid errors, one often needs the same level of understanding that is required if one tries to avoid the use of a machine.

Thomas J. Brennan
2000-01-31