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Learning:
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A Cloud Chamber of the Mind

March 2006 Mathematics Notebook

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An Example of a "Learning Process" Journal

Thursday March 23, 2006
Learning Log Number 21
7:00 am Ballina NSW Australia

7:00 am Ballina NSW Australia

This is a continuation of my notes for "Calculus".

Yesterday I made some handwritten notes for chapter 1. This acted as a form of review for the entire chapter. Hopefully I will be able to maintain a more regular pattern of mathematics during the next few months.

Here is a copy of my notes:

Calculus is the mathematics of change. The key concept is that of a limit. Another important idea is that of continuity. Underlying both ideas are the concepts of function and, surprisingly, number. And intertwined among all of this is the idea of infinity. Much of the relationship between calculus, number and infinity is best discussed in courses such as analysis, functions of a real variable, functions of a complex variable and number theory.

Aside: If complex variables are the extension of number to two dimensions, what are the extensions to 3, 4, ..., k dimensions?

One should try to keep in mind at least 4 different approaches when considering a mathematics problem.

1. Numerical (tables of values)
2. Graphical (use software such as Mathematica)
3. Analytic (algebra, logic)
4. Language (use your own words).

A limit (of a function) may fail to exist for 3 main reasons:

1. The behavior of the function near the point may differ when approaching from the left and the right.
2. The function may be unbounded near the point.
3. The function may oscillate near the point.
Also, there may be complexities about the definition of the function near the point.

The Dirichlet function: f(x) = 0 if x is rational, = 1 if x is irrational, has NO limit at any real number c.

Peter Gustav Dirichlet (1805 - 1859), German, made fundamental contributions to our understanding of number.

Augustine-Louis Cauchy (1789 - 1857), French, made important contributions to calculus, including the epsilon-delta notation.

Many functions (e.g. polynomial functions, rational functions with a non-zero denominator, radical functions, and trigonometric functions) can have their limits evaluated by direct substitution. If one cannot evaluate the limit of a rational function because it has an indeterminate form (i.e. 0/0) then try either cancelling a like factor from both numerator and denominator, or by rationalizing the numerator and then cancelling a like factor.

Evaluating expressions involving trigonometric functions can be very tricky because of possible indeterminate forms.

A function f is continuous at c if f(c) is defined and .

The following problem was giving me difficulty the last time I looked at it:

105. Consider the function .

(a) Find the domain of f. f is clearly undefined for x = 0. Since the secant is equal to 1/cos x, the function is also undefined whenever the cosine is equal to 0 (i.e. whenever c is a multiple of p/2)

(b) Looking at the graph for this function that Mathematica draws, the domain appears to be valid for all x since there are no error messages. On the other hand the graph clearly approaches plus and minus infinity at many points.

(c) The limit of f as x approaches 0 appears to be 1/2.

(d) BUT I am not able to show this analytically!

Now to try again.

I am still not able to work this out. I think it may involve using a trigonometric identity that I do not have at my fingertips. Time to move on.

8:00 am That is all for the moment. Now to get ready for the day.

11:00 am It is raining outside, a good time for another mathematics session.

Exercises For Section 1.4 [p. 75 - 78]

1. Think About It. (a) The continuity is destroyed at x = c because the graph of the function approaches a different value from the right than from the left as x approaches c.
(b) The continuity is destroyed at x = c because the function is not defined at f(c).
(c) The continuity is destroyed at x = c because the limit of f as x approaches c is not equal to f(c).
(d) The continuity is destroyed at x = c because the limit of f as x approaches from the left is not equal to f(c).

I like problems that begin with Think About It.

2. Writing. A removable discontinuity is one which consists of a single point whereas a nonremovable discontinuity is one where the limit from at least one side approaches to positive or negative infinity.
(a) is a function with a nonremovable discontinuity at x = 2.

Here is the graph of this function:

It took awhile to find a way to copy this image directly from Mathematica. Here are the steps:

1. In Mathematica, click on the image to select it.

2. File -> Save As Special -> HTML. This will save all the images within a file folder in the location you have specified.

3. Locate the appropriate image within the file folder and move it to the appropriate location within the Dreamweaver file structure.

4. Display the image in the normal way within Dreamweaver.

5. Delete all of the unwanted files that came over with the Save command.



(b) has a removable discontinuity at x = - 2.

Here is the graph of this function:

(c) 12:00 PM Lunch 12:35 PM

The function has both a removable discontinuity at x = -2 and a nonremovable discontinuity at x = 2.

Here is the graph of this function:

I also like problems that begin with Writing.

3. Think About It. All one need do is draw a curve that moves to the right and ends at the point (3, 1) but does not include that point, and then draw another curve that moves to the left at end at the point (3, 0) but does not include that point. The function represented by the curve is not continuous at x = 3 because the two limits do not coincide.

4. Think About It. If the functions f and g are continuous for all real x, f + g is also always continuous for all real x. However f/g may not always be continuous since there may be points where g is zero and at those points the ratio will be undefined.

5. - 10. easy

11. 1/10. Mathematica agrees.

12. -1/4. Mathematica agrees.

13. - infinity. Mathematica agrees.

Here are the results using Mathematica (Web, Mathematica). 1:25 PM

Reflection: I am comfortable with the ideas, but not with some of the exercises.

Difficulties: