## Archive for November, 2008

### Go And Do Likewise

Final 152 Homework (due Wednesday). From the Chapter 3 Review Exercises (pp 195 ff), problems 4, 36, 92, 108, 124, 128, 132, 136, 140, 144, 148, 152. Remarks to follow; I’m late for Learning Center hours. Have a nice holiday!

### 1+2+4+8+…=-1 $\bullet$The p-adics for nonexperts at Division by Zero.

### Trust The Code

I’m teaching a “Mathematical Analysis for Business” course at a Community College. Mostly it’s a quick survey of (the more-or-less standard) Precalculus topics (other than Trig). Way too quick, indeed: ideas taking weeks for similarly-abled students to understand in Precalc are alloted mere days in this course … the old “mile-wide, inch-deep” problem. But that’s another rant. Today I want to talk about what I’ve learned from this course: Annuities and Amortizations and suchlike Business Math (rightly so-called). But really about teaching: I hope to throw some light on some phenomena that could just as well be illustrated with any number of other topics. Here we go.

I posted some calculator code recently for evaluating $a_{n\rceil r} := {{1-(1+r)^{-n}}\over r}$ . This quantity is used in calculating, for example, the “Present Value” of an annuity. Specifically (in our text, whose notations I’ll [mostly] follow just in case any current student of mine should happen to read this post), this is given by $A = Ra_{n\rceil r}$, where $R$ is the periodic payment, $n$ is the number of periodic payments, and $r$ is the periodic rate (unfortunately … there’s some room for improvement here since the text also sometimes uses $r$ for the nominal rate [AKA the APR]). Thus, for example, if I were to have been frightened by a cabal of unscrupulous bankers and lawyers into agreeing to pay $30 every month for the next 33 years (what is in fact the case), and if my bank will pay me 4% (annual) interest on a savings account, compounded monthly (which might be about right; I don’t know and don’t much care), and if I had a big pot of money (unlikely at best), then I could calculate the “Present Value” of my future payments: $Ra_{n\rceil r} = \30a_{(33\cdot 12)\rceil(.04/12)} = \6590.50$ and cover the debt by depositing this amount once-for-all. The text derives the formula for $a_{n\rceil r}$ (pronounced “a-angle-n at r”, by the way)—as well they should … unfortunately, they use “Sigma notation” to do it, and this population of students aren’t prepared to actually read the derivation with understanding (nor, as I’ve already hinted, is there anything like time enough to prepare ’em to read it now). Luckily, the whole thing can be done without Sigmas, so I did that (and, with any luck, at least one student got something out of it since it took an entire lecture). All of the substantial mathematics for these sections of the text is done once the a-angle-n formula is derived; everything else follows from this derivation by very simple algebra (which, yes, does of course count as substantial somewhere … but not here: all of these students knew how to solve linear equations before they ever saw my face or cracked this book). So I say: let’s make this as clear as we know how. For example, since finding $R$ when $A$, $r$, and $n$ are known is obviously accomplished by $R = {A\over{a_{n\rceil r}}}$, let’s put the stress right there instead of, for heck sake, expanding the “angle” symbol to get $R = A[{r\over{1-(1+r)^{-n}}}]$. Now, of course, this represents the same number … but it might not be obviously the same. Anyway, it seems to miss the whole point of having given $a_{n\rceil r}$ a name in the first place! It’s one of the main tricks of the algebraic method: anything you’re going to need to keep referring to again and again in performing calculations should be given a name: “ $x$ stands for the Unknown Quantity” in what seems likely to be the best-known example. And on and on it goes. It’s easy (for students like mine) to understand that if payments are made at the beginning of the pay period, rather than at the end (so-called “annuities due” as opposed to “ordinary annuities”), then the first payment earns no interest and there are $n-1$ more payments after that, so one has $A_d = R + Ra_{(n-1)\rceil r}$ … so why obscure this simple fact by displaying $A_d = R + R[{{1- (1+r)^{-(n-1)}}\over r}]$ instead (as if this were some whole new idea)? It gets worse. The final exam is written by the course co-ordinators; well and good. They’ve also prepared a review packet; better still. Trouble is, there’s a “Formula Sheet” with a whole bunch of the messy formulas and no mention at all of $a_{n\rceil r}$! It begins to appear that somebody’s going out of their way to prevent us from doing this right. ### There I Was … No$#!+ … $\bullet$A war story from Jonathan. $\bullet$“Gay marriage: the category-theoretic perspective” by pozorvlak.

### Mathman’s Silver Hammer $\bullet$Carnival of Mathematics #44, by Edmund Harriss. $\bullet$The inimitable C. Shalizi debunks some Statistics Abuse. (Cosma’s an old-school keep-it-simple HTML-by-hand kinda guy; shrink the window to the size of a column for easy reading.) $\bullet$J. Fisher on “The Appeal to Common Practice” (sometime this month).

### Age Before Beauty $\bullet$A long-running conversation: “Geometry before Algebra?” at Continuities. $\bullet$The 198th Carnival of Education is hosted by a math teacher this week.

### They Are Standing Still

Here’s some calculator code for yesterday’s formulae:
AANGLE
:Prompt N,R
:(1-(1+R)^-N)/R $\rightarrow$A
:Disp A, A $\rightarrow$Frac

To produce this code, first enter the programming interface (push the PRGM button). We’re creating a new program, so select NEW and enter its name (AANGLE, say). Hit enter; we’re editing the new program. Now type PRGM again (this puts us in the PRGM submenu of the programming interface); select I/O (with the “arrow” keys; I/O stands for “Input/Output”) and then select Prompt. The symbols N and R are produced with the ALPHA key in the usual way; the comma is above the 7. The rest of the code follows easily (Disp is of course also found in the I/O submenu).

Now we can forget the formula for $a_{n\rceil r}$ (and never mind the tables in the back of the book). The first thing to notice about this procedure is how doggone easy it is. Indeed, as far as I know, it’s much easier than it would be on a Windows box. It was this easy in DOS (the predecessor to Windows) since BASIC was standard on DOS boxes. Several generations of computer coders have a great deal to answer for.

### Saving The Appearances

Yesterday we loaded programs AANGLE and SANGLE, used in computing annuities. For the record, one has $a_{n\rceil r} := {{1-(1+r)^{-n}}\over r}$ and $s_{n\rceil r} := {{(1+r)^n -1}\over r} (= (r+1)^n a_{n\rceil r})$. I’ve gotta go mark papers; if the actual rant on this material ever materializes, it won’t be till tomorrow at the soonest.

### Ramifications $\bullet$Rolfe Schmitt has bifurcated his blog activity. $\bullet$I’ll be looking at T.R. Mayes’ “Time Value Mathematics” pages with my Math 130 class soon (so here’s the bookmark).

### Today’s Lemniscate Problem

… is essentially an example from the text, first of all. So hopefully, that’ll inspire some reading (nobody had the right slope on this one). Here are some other remarks.

In Calc III (here’s an old blog by me), one considers polar co-ordinates (so skip the rest of this paragraph if you prefer not to think about ’em yet). Our curve $(x^2+y^2)^2 = 50xy$ “translates” into $r^4 = 50(r\cos(\theta))(r\sin(\theta))$, so we can put $r =5\sqrt{2\cos\theta\sin\theta}$ into a grapher. The “Draw” feature can then be used to produce the desired tangent line ( $(x,y) = (2,4)$ gives us $\theta =\arctan({4\over2})$; input this at the prompt); this feature also produces the display ${{dy}\over{dx}}=.181818$ (or words to that effect); being translated (via the “Frac” feature, say), one has y’=2/11.

The Calc I version: differentiating both sides of the given equation gives $D_x[(x^2+y^2)^2] = D_x[50xy]$ $2(x^2+y^2)\cdot D_x[x^2+y^2] = 50x\cdot D_x[y] + D_x[50x]\cdot y$ $2(x^2+y^2)(2x + 2yD_x[y])=50xD_x[y] + 50y\,.$
One now easily isolates the $D_x[y]$ (or ${{dy}\over{dx}}$) terms, “factors out” ${{dy}\over{dx}}$ (and “cancels” a 2): dividing by the other factor on both sides of the equation gives ${{dy}\over{dx}} = {{25y - (x^2+y^2)\cdot 2x}\over{(x^2+y^2)\cdot 2y - 25x}}$. Finally, substituting the given values gives ${2\over 11}$—by Calc I methods (and without a calculator). “Plug in” on the point-slope form; done.

• ## (Partial) Contents Page

Vlorbik On Math Ed ('07—'09)
(a good place to start!)