March 15, 2004
Dr. Dobb's Math Power Newsletter - March 2004Homer B. Tilton
Expand your base of math knowledge. College Instructor Homer Tilton breaks down barriers from Algebra to Calculus and beyond. From common to complex, Homer makes math concepts exciting for experts and novices alike.
Math Power Official Publication of Math Power ClubIncorporating LIGHT WORK and COMPUTER WRANGLER Math Power Club is an Unofficial Club of Pima CC East Editor and Publisher Emeritus: Homer B. Tilton
South by Due South Gaspard Coriolis was a mid 19th century French civil engineer who discovered a new force which we now call by his name. Today, Coriolis is a force. But references say it is only an apparent force and those same references also say that centrifugal force is only an apparent force. If you have ever rotated a weight on the end of a string 'round and 'round, the outward pull you felt may have seemed real enough -- that's centrifugal force and it's what keeps orbiting satellites from falling to Earth. (The general theory of relativity "geometrizes out" those "apparent" forces, but -- not to "force" the issue -- we might instead postulate an inertial force field to render them real.) If you were to set course due south along the prime meridian (longitude zero degrees) in an airplane from the North Pole, you would soon find yourself drifting westward over the Atlantic as the Earth turned beneath you -- that's the Coriolis force, and again it'd seem real enough. At least a real eastward force would have to be exerted to put you back on track over England. Coriolis force is defined this way in one source:*
The Coriolis force is most often associated with terrestrial navigation as in the airplane example given above; however, it applies not only to rotating spheres but to rotating disks as well; and since our galaxy rotates like a disk, Coriolis force must be accounted for by anyone planning a trip to the center of the galaxy. Frankly my dear, the Coriolis force is a necessary force component if we are to retain Newton's laws. Indeed, the definition of Coriolis effect begins this way in another source:** In a rotating reference frame, Newton's second law of motion is not valid, but it can be made to apply if, in addition to the real forces acting on a body, a (fictitious) Coriolis force and a centrifugal force are introduced. One may come to depend on the centrifugal-force concept. Coriolis force is like that. * Borowski & Borwein, The Harper Collins
Dictionary of Mathematics, 1991, ISBN 0-06-461019-5 George W. Bush, 43rd President of the United States of America, on Wednesday January 14, 2004, did make a Public Announcement declaring his Nation's continuing dedication to mannued space exploration. For the full text of his speech see the President's web site. The scheduled press date for this issue of MATH POWER: 19 Jan '04. See the Special Supplement accompanying this issue, "Begin the Adventure: Achieving Routine Star Travel Within the Bounds of Relativity," an elaboration of "Acceleration" which appeared in the Nov '03 issue. Watch next month for two rebuttals to "Begin the Adventure," both by Jim Malmberg, entitled "Why You Can't Exceed the Velocity of Light" and "Solution of the Constant Acceleration Problem." MAIL MATTERS Response to "Acceleration" (Nov '03 issue) - Note from Dr. David Iadavaia received 12 Dec '03: Hi Homer - Hello, David! - To clarify the point, "Begin the Adventure" appearing in the Special Supplement to this issue reads (p. 9): Those on the ground will see the mass of the ship as increasing without limit; they will also see the jet thrust as getting larger as a steady acceleration of the ship is maintained. There is a profound difference between a particle which
is sent a constant push from stationary coils & electrodes, &
the rocket which is self-propelled. The proof of the pudding is in the
eating. Let's chow down! We need to perform the science experiment dramatized
in "Begin the Adventure," pp. 23-4. The Family Joules When we say "kinetic energy equals 1/2mv2", that presumes a particular frame of reference. In the case of a particle accelerator, the frame is the laboratory; particles are accelerated which then impact a target which is also fixed to the laboratory. We reason that the energy imparted to the target must equal the work performed to accelerate the impacting particle. And if this experiment were performed in a similar laboratory aboard a ship in space accelerating at a steady 1G, that experimenter should get the same result. It is not widely known that relativistic kinetic energy,
Wr=mc2-moc2,*
does not quite match the classical kinetic energy "corrected"
for relativity, Wcr = 1/2mv2
with m=mo/ If you are using SuperCalculator the graphing command
is, with x= Art 2 2 1/sqr(1-x*x)-1'
* This analysis appears at W. G. V. Rosser, An Introduction to the Theory of Relativity, Butterworths, 1964, pp. 181-3. If Mathematica(R) is a monster truck, then SuperCalculator is a utility pickup. Special Calculus One limitation of standard calculus is its reluctance
to accept absolute value as a legitimate operation. However traditionalists
find they are now forced to recognize that One never encounters functions like this clipped parabola in traditional college algebra or calculus courses: y=1/2¦x2-1¦+1/2(x2-1) Graph it for a treat. A reason such functions are not included in traditional texts is that they do not fit the chosen categories: conic sections, polynomial functions, etc. But they do fit other categories long known to, and valued by, engineers. Another self-imposed limitation of standard calculus is the lack of aware- ness that functions of the form Arctrig(trig x) do not reduce to just x. Graph if you would this particular one: Arctan(tan x). You'll see it's a sawtooth wave. Surprise, surprise. The reason, of course, is that Arctan (big "A") is not the full inverse of the tangent function; it is only its principle branch. Standard calculus has other self-imposed limitations as well. One other is its rejection of the Dirac delta function. Others are its non-recognition of infinite numbers and hysteretic numbers. Special calculus embraces such extra- standard tools in a seamless extension of standard calculus. Special calculus is not a threat to standard calculus, it is an extremely valuable extension to it; and there is no reason for the traditional mathematician to fear it or bad-mouth it. Beckmann:* "the delta function, ...the mathematicians proclaimed, was a monstrosity." The x^x connection -In keeping with our efforts to insure that the interface between special calculus and standard calculus is totally seamless the examination of x^x, begun in the Jan issue, is continued here. We observe all the good rules of standard calculus. We shall examine the way in which the negative-x and positive-x branches of Re(x^x) connect. The expression of that function is presented and graphed in the Jan '04 issue, and examination of the graphs there shows that our concern is nontrivial. Two things are shown: (a) For f(x) = Re(x^x), we have that f(0 )
= 1;
(b) For f'(x)=df(x)/dx, f'(0+) and f'(0-)
are equal but negatively infinite.
Since they are beyond the domain of real numbers it is customary to ignore them, justifying that ignorance by saying they "do not exist." However, within the domain of hyperreal numbers they do exist and in special calculus we do not ignore them. We find over and over that something of value is lost when things are ignored. The two promised analyses follow. -- continued -- * Petr Beckmann, A History of 
(Pi), 3rd edition, The Golem Press, 1971; reprinted by Barnes &
Noble Books, 1993, ISBN 0-88029-418-3, p. 112
Q: Why don't inhabitants of Vega eat meat? A: They are Vega-terians. Word of the month - joule - a unit of work or energy equal to the work done by a force of one newton acting through a distance of one meter... Merriam Webster's 10th Collegiate Dictionary Dear Homer - The handout for the prealgebra class that you sent is
great. [Basically the article "A Spiral Nebulosity," Jan '04.]
At this level students not only need to see how to use math every day
-- but they gain confidence when they realize they are already using
it. I may use some of your ideas in my 'teaching tips.' I will send
them to you first for your approval. Is this okay? I just think that
the idea of saying "You already know some math" is good. Dear Jamie - Editorial The prealgebra student comes to the course with a certain capability in arithmetic. Now a new level of abstraction is encountered requiring a huge amount of organization and self discipline. The way that penmanship is taught, as manual exercises with spiraling circles CCW across the page then CW, comes to mind. Something similar might be used with prealgebra to develop the needed skills. The textbook should foster the development of those skills. Some indicators students leave on testsTeaching math (or any subject) is about rewiring a part of the brain in some way; to accomplish that efficiently, the teacher must have some indication of its present wiring. There are these indicators on the student's tests: faint or tiny writing (tentative); bold and large writing (self assured, an "I dare you to teach me something new" attitude); chaotic (very disorganized, confused, troubled); neat & reasoned (ready to learn new math things). ...That, coupled with classroom persona and homework performance. Like the overachiever who always turns in a ton of HW but never on time ("I cannot study new material until I've completely learned all that preceeds" -- a magnet for stress.) Proper presentation Things to do early on:(1) The student must be taught to present proofs properly. A good exercise is simply to have the student copy the examples presented in the book -- repetitively. This can guide the student away from harmful patterns of thinking. (2) The student must be taught how to use the textbook, focusing first on use of the index. Emphasize that the textbook is primarily a reference, and secondarily an aid to learning material it contains. Point out that learning 100 percent of the material may not be a realistic goal and is not required to pull down an "A." Other instructors shudder, and say, "You should not tell the students they don't need to learn everything." But the stress-laden overachiever drives my argument. ...HBT * LIGHT WORK (ISSN 1092-8170) was a companion journal to MATH POWER. Special Calculus is a Seamless Extension of Standard Calculus - continued - Analysis (a) - In the Jan issue it was found that Re(x^x)=¦x¦^x cos 
where =1/2(x-¦x¦).
We know that
lim x^x = 1 x->0+because for x>0, x^x = exp(x ln x) = exp(x) + x thus lim Re(x^x) = lim ¦x¦^x = lim ¦x¦^(Analysis (b) - Taking Re'(x^x) = d/dx (¦x¦^x cosgives Re'(x^x) = -¦x¦^x sinUsing dand d/dx(¦x¦^x) = ¦x¦^x(1+ln¦x¦)we have
Re'(x^x) = ¦x¦^x {-1/2
With
lim ¦x¦^x = 1 x->0and limwe have finally that lim Re'(x^x) = ln¦x¦ x->0QED Send for the book, Special Calculus. Write via U.S. mail to Homer B. Tilton/ 8401 E. Desert Steppes Dr./ Tucson, AZ 85710-4207. Front matter is free. The book is not. MATH POWER is on www.ddj.com at "Dr.Dobb's MATH POWER Newsletter" Copy GWBASIC.EXE OK.MOMHe then tagged the last byte by affixing "LastByte" to it using this command: Echo LastByte >>OK.MOMHe then loaded the modified OK.MOM file into DEBUG with DEBUG OK.MOMand searched for "LastByte" with this DEBUG command: S ssss:0000 FFFF "LastByte"where ssss is 1000 + CS (Norman had previously found CS with the DEBUG R command.) This returned ssss:3BE0. He backed up a paragraph and launched this command: D ssss:3BD0 L20giving this two-line return: 2E9E:3BE0 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................ 2E9E:3BE0 4C 61 73 74 42 79 74 65-0D 0A 00 B0 0D E8 6E 1B LastByte......n.So the last byte in Smart BASIC is clearly 00, the ASCII nul character. In fact the last 16 bytes are that. BASIC apparently uses this as some kind of storage buffer. He made a printout with the PrntScrn key for posterity. Leaving DEBUG with the Q command, he then deleted OK.MOM with the DEL command so it would not clutter things up and lead to possible future confusion. Mission accomplished. Written reader comments are invited on all material. Those intended for my attention must be submitted by US Mail to my Tucson address. All such comments are subject to being published unless requested otherwise. They may also be subject to editing. -HBT The hard copy version of MATH POWER is published as a shareletter; that means you are permitted to make not-for-profit copies of it for distribution to your colleagues and students. MATH POWER is published monthly. It is published and edited by Homer B. Tilton under the auspices of Pima Community College, East Campus, 8181 E. Irvington Rd., Tucson AZ 85709-4000. Editorial Assistant, Jo Taylor. All material is copyright Homer B. Tilton unless otherwise noted. A limited number of copies may be made at educational institutions for internal use of faculty and students. For more extended copying or to request additional copies contact the Editor at the above address. Letters and editorial material are welcome. All submitted material may be published in MATH POWER, and edited, unless specifically requested otherwise. Your newletter is interesting to read and consider. In the July '03 issue, your puzzle challenge, "Puzzle me one more time," p. 5, caught my eye and I thought I'd give it a shot. I am currently a software developer with a graduate physics background (who probably took more math than anything else!). I have always enjoyed math and I have a 9-year old son who is following in my footsteps. Thanks for the mental stimulation! Marty Troyer, Redmond, OR Dear Marty - Thank you for your letter. Your proof, indicated above by three asterisks, appears below as a separate article. It is pretty much like what Norman came up with appearing in the Aug '03 issue, p. 1. Kudos to you! ...HBT by Marty Troyer  (x3-i*x+3),
to find exact expressions for the value of i (=n) and the value of x (= )
for which f(x,i) borders on the complex vs.real.
My first thought was to determine the condition for the non-real solutions
by determining when f(x,i) would have a negative argument. After
that, to find the tangent to the curve where it appears to just touch
the x-axis when we use the critical value n for i.
For all values of the radicand x3-i*x+3 < 0 we will get non-real
solutions for f(x,i). We will look at the point where this function
goes to zero to find the limiting value for i, which we've called n. To
solve the cubic, I used the substitution x=z+n/(3z) to transform it into
(z^3)2 + 3*z^3 + (n/3)^3 = 0(after multiplying thru by another z3) This is now in the form of a quadratic equation in z3, so the solution looks like this: z^3 = 0.5*(-3These solutions are non-real when 9-4(i/3)^3<0;so: n = (3^5/22)^(1/3)Next, we want to find where it crosses (or just touches in this case) the x-axis. If we find where the slope is zero in that region, we should be able to determine that point on the x-axis. I chose to implicitly differentiate y2 = x3-n*x+3 instead of first solving for y, since this will yield the same result, but more easily. I got: 2y*y' = 3x2 - nAnd so y' = (3x2-n)/(2y) = 0or 3x2 - n = 0This results in zeroes at x= (n/3),
after discarding the negative result.
Thus
and after simplification, Summarizing, exact expressions for and n are:
MY SOLUTION NORMAN'S SOLUTION BOTH SOLUTIONS
I think that gets at the heart of the solution, even if it may not be completely rigorous. I hope you and Norman approve! In Norman's proof which I later saw in the Aug '03 issue, he saw no need to solve the cubic; and he quaintly used "intrinsically" for implicitly, referring to differentiation. Is that a carry-over from the old-school? Where is the point that has no sign? When class had convened, Norman held up two blank report covers; one green, one red. "Let me have your attention!" he reminded the class. He then over-lapped the two sheets about 1/4 inch, green on the right and red on the left, with the boundary joining the two colors vertical to lie on a y-axis to be. "It is important they overlap, not just butt up against each other. Okay?" He made a construct by sticking the two covers together in that position with chewing gum and stuck the construct up on the board. "Let this construct be cut from quadrants I and II of the Cartesian plane, with corners at say, (10,0), (10,20), (-10,20), and (-10,0). Here; I'll sketch that system around it. Okay?" And he did. It looked like this. 
The border between red and green lined up perfectly with the invisible y axis. "Green points within the construct have positive x; red points have negative x. All points have positive y," Norman said. "I've passed out the same kind of sheets* so you can all participate, okay?" He hastily added, "But no chewing gum, okay?" "Every x-point in the construct is either green or red. That is, every x-point is either positive or negative. There is no point without a sign." "Now consider the line y=10 cutting horizontally across the construct, centrally, half way up. Don't draw it, just consider it. Okay? Consider it to be a gedanken number line like the x-axis. It is an invisible number line which you are to reason and think about." Norman asked everyone to point to x=0 on that invisible number line. Everyone confidently pointed to the center of the construct. "The question now is, what is the sign of that zero point, okay? Remember, every point on the construct is either plus or minus. Okay?" ** Student Ruby: "But Professor Nerdnick, doesn't zero have no sign?" Norman: "Okay, ah, Roo-bee. Address that concern in your write-up. This is a thinking kind of question; not an ask-your-professor-for-the-answer kind." Everyone was asked, as part of their homework, to write a statement of the problem along with their evaluation and ultimate answer, and turn in their write-up at the start of the next class meeting. They were to write only that on one side of a clean sheet of paper with their name and the date clearly and prominently displayed in the upper right-hand portion. Norman promised to see that the best writeups were published in a future issue of MATH POWER. You in the reading audience are also invited to send in your answer. Send it by U.S. mail to Prof. Homer B. Tilton / Math Dept. / Pima Community College EC / 8181 E. Irvington Rd. / Tucson, AZ 85709-4000. Be sure to include your name and address. The best two or three writeups (possibly more) will appear in a future issue of MATH POWER. Don't disappoint us by ignoring this challenge. ...Okay?
* Norman had before-hand cut several sheets of red and of
green paper into quarters for passing out.
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v
[a real number! Ed.] for a particle of mass m moving with velocity
v with respect to a rotating frame of reference with ANGULAR VELOCITY
2).
Still, someone on the ship will see the ship's mass as unchanging. This
is all in accord with relativity. To speak of the ship's KE as measured
by someone on the ship would be a non sequitur; and to think of KE as
something a body simply "has" in varying amount would not
be correct. Since velocity is relative as opposed to being absolute,
so is kinetic energy. 
Wr/m
dx/x=Ln¦x¦+C,
for instance, and the biomass pretty much just hit the fan. A question
now is, how does one formally differentiate Ln¦x¦ so as to
get 1/x. Special calculus shows us how. 
(1-sgn x) sin 
3.931112... .
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