Date: Sep 19, 2007 10:54 AM Author: Dave L. Renfro Subject: [ap-calculus] Neat Taylor series application I know it's way too early for anyone to be

worrying about Taylor series yet, but I came

across something this morning that many of you

teaching BC calculus may find useful to look

at later, when you get to Taylor series. If

you've taught BC calculus more than a couple

of times, you probably have a collection of

"applications of Taylor series" in a folder

(evaluating limits where L'Hopital's rule is

very difficult, best quadratic and cubic and

so on approximations at a point, various

things you can obtain by differentiating or

integrating Taylor series expansions, etc.),

but what follows is one that you probably

haven't seen before.

This is a problem in Clement V. Durell and

Alan Robson's "Advanced Trigonometry" [Dover

Publications, 1930/2003], Exercise V.f #20

on p. 102.

Using Taylor expansions, determine the "near x = 0"

growth rate ordering for the following collection

of 6 functions:

sin(sin x)

sin(arc-tan x)

tan(tan x)

tan(arc-sin x)

arc-sin(arc-sin x)

arc-tan(arc-tan x)

Each of these functions is increasing in a

sufficiently small interval about x = 0

(a non-maximal such interval is, for example,

-0.8 < x < 0.8). Moreover, the linear approximations

for each of these functions at x = 0 is L(x) = x.

[Technical note: It is possible for a function to

be everywhere differentiable, have a linear

approximation L(x) = x at x = 0, and yet not be

increasing on any interval containing x = 0.

An example is f(x) = x + (x^2)*sin(1/x^2).]

Therefore, linear approximations at x = 0

do not give us enough information to order

these functions. However, by using Taylor

expansions, we _can_ obtain enough information

about their growth rates at x = 0 to put

them in order.

All we need are these expansions:

sin(x) = x - (1/6)x^3 + ...

tan(x) = x + (1/3)x^3 + ...

arc-sin(x) = x + (1/6)x^3 + ...

arc-tan(x) = x - (1/3)x^3 + ...

Note: If the trig. and arc-trig. expansion symmetry

catches your interest, apply "reversion of series"

to y = x + Bx^3 + ... (i.e. constant and quadratic

coefficients are zero, linear coefficient is 1).

http://mathworld.wolfram.com/ReversionofSeries.html

http://books.google.com/books?q=reversion-of-series

It's now easy to get the first nonzero correction

terms for each of the linear expansions of the 6

functions above. For example,

tan[arc-sin x] = tan[x + (1/6)x^3 + ...]

= [x + (1/6)x^3 + ...] + (1/3)*[x + (1/6)x^3 + ...]^3

= [x + (1/6)x^3 + ...] + (1/3)*[x^3 + ...]

= x + (1/2)x^3 + ...

Do this for the other functions and then arrange

them from least to greatest rate of increase

at x = 0:

1. arc-tan(arc-tan x) = x - (2/3)x^3 + ...

2. sin(arc-tan x) = x - (1/2)x^3 + ...

3. sin(sin x) = x - (1/3)x^3 + ...

4. arc-sin(arc-sin x) = x + (1/3)x^3 + ...

5. tan(arc-sin x) = x + (1/2)x^3 + ...

6. tan(tan x) = x + (2/3)x^3 + ...

You can see this graphically with a graphing

calculator by graphing all 6 of these functions

(USE RADIAN MODE) along with y = x. Use a window

of -a < x < a for a = 0.6, 0.7, or 0.8 (experiment

some) and, if you don't have a "zoom-fit" option,

use -b < y < b for b between 0.8 and 1.7 (depending

on what you used for 'a'). Of course, the graph

will look better if you use a CAS such as MAPLE

or Mathematica . . .

Dave L. Renfro

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Course related websites:

http://apcentral.collegeboard.com/calculusab

http://apcentral.collegeboard.com/calculusbc

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