The Journal of Online Mathematics and Its Applications, Volume 7 (2007)
An Evolutionary Textbook, Håkan Lennerstad, David Erman, Maria Salomonsson
The following are translations from Swedish to English of sample dialogues at the website Calculus with dialogues. For students in the course, participation in dialogues was compulsory. As is seen below, many dialogues refer to topics at particular pages in the book that was used.
Student A: Sometimes I'm a bit confused over all these integral concepts.: =) In the Mathematical Continuation on page 517, one can read a little bit about double integrals. Then I wonder if the double integral corresponds to what one learned to solve in Mathematics D in high school, when one calculates the area between two different curves (functions)?
Student B: No, it is not the area between two different curves (function). It is the volume above the x,y-plane. You do not only integrate in x-direction, but also in y-direction (dxdy). V=(double integral)f(x,y)dxdy. A little difficult to show when you cannot draw it. But maybe it becomes a little more comprehensible.
Student C: At the bottom of page 311 it says that "the class of differentiable function at an interval (a,b) is denoted by C1(a,b)". On page 297 it is said that "the set of functions that are continuous on the interval [a,b] is denoted by C[a,b]". what does C mean?
Student D: Here C is the set of continuous functions, it is not a constant.
Student E: What happens if the exponent or a base is not an integer?
Teacher A: If the base is not an integer but the exponent is an integer there are no problems. The exponent gives the number of factors, so the factors (the base) does not have to be an integer. For example, (0.5)^3 = 0.5 x 0.5 x 0.5 (here "x" is multiplication!). Exponents of the form 1/integer we can handle using inverses, see page 171. Then we can handle exponents of the form p/q by composition, see bottom of page 172. Then it is possible to handle irrational exponents with a limiting procedure. For example: 2^e can be calculated from 2^2, 2^2.7, 2^2.71, 2^718, ..., which converges.
Student F: Periodic decimal expansion means simply that groups of numbers "reoccur" among the decimals of a number. For example,. in 4.123456123456123456..., the block 123456 reoccurs again and again, so it is periodic. It would be very difficult to say how you see that a number does not have a periodic decimal expansion-- the periodic expansion may be very complicated. Assume that we want to see if Pi has a periodic expansion--we can check the first 1000 decimals and see that there is no "pattern", but maybe there is a pattern that repeats after 1500 decimals? How do you prove that? I think I will let Teacher A take that question?
Teacher A: Exactly, in a periodic decimal expansion a set of numbers reoccur. But not just in any way. The example of Student F is good, where 123456 reoccurs. But the number 3.123012300123000123000012300000123... is not periodic, where 123 reoccurs with every time with more zeros in between. There has to be a fixed part of the decimal expansion that reoccurs constantly, and nothing else, from a certain decimal and forward. That a number has a periodic expansion is actually the same thing as that it is rational, i.e. a ratio of two integers, see Completion K2.4 to the book. It is not easy to show that Pi is not rational (i.e., does not have a periodic decimal expansion). I do not know the proof. It is much easier to prove that the square of 2 is irrational, since by squaring gives integers. This proof is classical and can be found in the book (page 32).. It took a long time until someone managed to prove that Pi and e are irrational. Proofs can be found at http://mcraeclan.com/mathhelp/BasicNumberIrrationalityProofs.htm . The proof of that e is irrational is not so difficult, almost a part of the course. It uses Completion C8.3 to the book.
Student G: I get stuck in Exercise 12.12d. Anyone have a clue? Thank you!
Teacher A: If we are to do partial integration, we need that int(1/sin^2 x)dx = -cosx/sinx + C (test this by taking derivatives of the right side!). This is a relative to the more famous d/dx tan x = 1/cos^2 x, i.e. that int(1/cos^2 x)dx = tanx + C. This gives int(x/sin^2 x)dx = (Partial Integration) = -xcosx/sinx + int(cosx/sinx)dx. Now we may substutitute y=sinx, which gives dy=cosxdx, so tha last integral is only int(1/y)dy = ln|y| + C = ln|sinx| + C. And what is left is to put the pieces together.