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Topic: [ap-calculus] Neat Taylor series application
Replies: 0

 Dave L. Renfro Posts: 2,165 Registered: 11/18/05
[ap-calculus] Neat Taylor series application
Posted: Sep 19, 2007 10:54 AM

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
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

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
====
Course related websites:
http://apcentral.collegeboard.com/calculusab
http://apcentral.collegeboard.com/calculusbc
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