Power Series for Practical Purposes Ralph Boas The Two

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Power Series for Practical Purposes
Ralph Boas
The Two-Year College Mathematics Journal, Vol. 13, No. 3. (Jun., 1982), pp. 191-195.
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Power Series for Practical Purposes
"Take a bone from a dog-what
remains?"-The
White Queen
Ralph Boas
The author retired from Northwestern University in 1980 with
the title of Professor Emeritus of Mathematics, a title which, he
likes to point out, has nothing to do with merit. He has also
retired as editor of the "American Mathematical Monthly;" the
December 1981 issue has a symbolic portrait of him on the
cover. This article is an outgrowth of his hobby of trying to
encourage people to look at apparently complicated things in a
simple way.
1. Why Taylor series? We hear a great deal about teaching mathematics that is
applicable to "the real world." Consequently I was nonplussed at being told
recently, in all seriousness, that most teachers will reject a calculus text out of hand
if it doesn't discuss the formula for the remainder in a Taylor series.
This claim may well be true; but, if so, most teachers are trying to teach the
wrong course. The Taylor remainder is an important piece of mathematics associated with calculus-important
for proving theorems, that is. So are uniform
convergence and Lebesgue integration, but most teachers realize that these topics
will not be appreciated by the mass of freshmen and sophomores. Why is the Taylor
remainder conceived to be of such central importance?
The conventional answer is that with the remainder formula we can estimate
remainders and so be sure that power series represent the functions we got them
from. This is indeed true for the exponential, sine and cosine functions, for the
binomial series-and not for much of anything else. Indeed, these are practically
the only elementary functions whose successive derivatives are simple enough to
calculate in the first place and then to estimate. Just try to calculate enough
derivatives to estimate the remainder after eight terms for cos(x - sinx). (This
function occurs in the theory of frequency modulation.) You could, of course, use a
computer to evaluate the derivatives, but in that case you probably wouldn't need
the series anyway.
Given that the remainder formula is useful primarily for theoretical purposes,
can we justify the introduction of Taylor series at all? Isn't it now only of academic
interest that
Not long ago, if you wanted a fairly precise approximation to e0.0'982,
you couldn't
do much better than to calculate a few terms of the series with x = 0.01982.
Nowadays, of course, you push seven buttons on the little pocket calculator and get
1.02001772, a result more precise than anything you are likely to need. You can
obtain values of cos(x - sinx) almost as easily.
Nevertheless, many people who apply mathematics do value the ability to write
down several terms of the series of such functions. The usual technique for
cos(x - sinx) would be to form the Maclaurin series for x - sinx, substitute this
series for y in the Maclaurin series for cosy, and rearrange the result as a power
series in x. Authors of calculus books seem to be largely unaware that this is a
completely legitimate process. But, given the existence of calculators and computers,
why should anyone bother to work out the Maclaurin series by any method?
In the first place there are things that calculators don't yet do, or do only after
special programming. One of these is the evaluation of nonelementary functions. I
have seen a calculator that has built-in gamma functions, but, as far as I know,
there is as yet no calculator with built-in Bessel functions or elliptic integrals.
Another operation that is not easy on a calculator is the calculation of a table of
values of an indefinite integral. Still another is making calculations of higher
precision than was built in. You cannot, at least at present, appeal directly to a
calculator for values of
Here J , is the Bessel function
(2) is an elliptic integral, and (3) is a solution of [ l ] x?" + x?' + (x2 - 2)y = 0.
It is relatively easy to start from well-known power series and work out the first
few coefficients of the Maclaurin series of (I), (2) or (3). If one of these functions
turns up in a practical problem, it will probably have to be entered into a computer
to be used for further computation. It is much easier to enter a small number of
coefficients than a whole table of values.
2. Why It works. To get power series for functions like (I), (2) or (3), we need to
know that power series can be differentiated or integrated, multiplied or divided,
and (most importantly) substituted into each other; then, if necessary, rearranged as
power series again, with the same result as if we had worked out the coefficients by
differentiation. Some of the proofs are beyond the scope of a calculus course, but at
least correct statements of the relevant theorems are easily understood. (The same
can be said for other theorems that are used in calculus courses.)
It is most convenient to state the theorems for complex series; but they can be
specialized (although not all the proofs can) to real series by reading "interval of
convergence" for "disk of convergence."
A. Differentiation and integration are permissible in the interior of the
disk of convergence. This is straightforward to prove and is done in many
textbooks.
B.
Multiplication. If the Maclaurin series of f and g both converge for
then their formal product, arranged as a series of ascending powers, has
radius of convergence at least r and represents fg. (Here r is not necessarily the
radius of convergence of either series.)
This is a special case of the theorem that the formal product of two absolutely
convergent series converges (absolutely) to the product of the sums of the series. In
fact, we have
IzI < r,
x
00
n=O
x
00
a,zn.
x
00
k
bmzm= ckzk, where ck =
ajbk_,.
m=O
k=O
j=O
Hence we can calculate the coefficients in the product series or program them for a
computer to calculate.
C. Division. If the Maclaurin series of f and g converge for lzl < r and
g(z) 0 for 0 < lzl < r, then if the Maclaurin series for f is divided by the
Maclaurin series for g by long division (as if the series were polynomials), the
resulting series represents f / g for lzl < r.
Again, r is not necessarily the radius of convergence of either series.
If h = f / g then f = gh. Write h(z) = C;='=,cnzn,multiply h by g by using the
formula under B, and solve for the cn recursively. We can then see by induction that
these c, are exactly what one gets by long division.
For example, C does not allow us to divide the series for cosz by the series for
sinz near z = 0; fortunately, since cotz does not have a Maclaurin series.
+
D. Substitution. The following theorem, although not the most general possible, covers many cases that are likely to arise in practice. Curiously enough, it is
absent from most calculus books and is not discussed adequately in most modern
textbooks on complex analysis. It was rather difficult to locate a formal proof (see,
however, [2] or [3]); I give one in 03 in case anyone wants to see it. It involves
nothing deeper than the theorem that a uniformly convergent series of analytic
functions can be differentiated term by term.
Substitution theorem. Let f(w) be represented by the series C;='=,anwnfor
lzl < r, and let I g(0)I < s. Then the
Maclaurin series of F(z) = f(g(z)) can be obtained by substituting w = C?=',,bkzk
into C;='=,a,wn and rearranging the result as a power series in t.
IwI < S; let g(z) be represented by C:='=,bkzk for
The resulting series represents F(z) in any disk lzl < t in which F is analytic;
there is such a disk because F is analytic at 0. In practice, f and g are often
elementary functions and the radius of convergence of the Maclaurin series of F
can be found by inspection.
Notice particularly what the theorem does not permit. For example, we cannot
get a Maclaurin series for ln(1 - cosz) by substituting the Maclaurin series for cos z
into the series for ln(1 - w), because the composite function is not even defined at
z = 0. The theorem does not let us try because the Maclaurin series for ln(1 - w)
has radius of convergence s = 1 but cosO = 1 is not less than s. We must also be
careful to avoid the mistake of substituting the Maclaurin series of g into the
Maclaurin series of f when g is analytic at z = a but g(a) is outside the disk of
convergence of the Maclaurin series of f. For example, consider a branch of
(1 - 2 c o ~ z ) ' / ~The
. Maclaurin series of (1 - 2w)'l2 converges only for Iwl < f , but
cosO = 1; even though (1 - 2 c 0 s z ) ' / ~ is analytic at z = 1, we cannot substitute
w = cosz into a divergent Maclaurin series and expect to get a meaningful result
when z is near 0.
3. Proof of the substitution theorem. Since F(z) is ,analytic at 0 it has a
Maclaurin series C h n z n .On the other hand, when I g(z)l < s we have
by the multiplication theorem. What we want to show is that
Now CA,zn is uniformly convergent in any closed disk lzl < r where F is analytic
and hence can be differentiated repeatedly in a neighborhood of 0. Setting z = 0, we
get
Now C+,wm converges uniformly for Iwl < s , < s and so C+,[g(z)lm converges
uniformly provided I g(z)l < s l . This will be the case if lzl is sufficiently small and s ,
is sufficiently close to s, since I g(0)I < s. Hence C+,[g(z)lm, although not a power
series, can also be differentiated term by term; consequently
setting z
=0
we get
This process can be continued to yield (5).
This shows that the series for f(g(z)) can be rearranged into a power series when
z is sufficiently close to 0; this series then represents F(z) in the largest disk, center
at 0, in which F is analytic.
REFERENCES
I. E. Kamke, Differentialgleichungen, Losungsmethoden und Losungen, vol. 1, 3rd ed., Leipzig,
Akademische Verlagsgesellschaft, 1944, p. 450, (2.208).
2. A. I. Markushevich, Theory of Functions of a Complex Variable, vol. I (translated and edited by R.
A. Silverman), Prentice-Hall, Englewood Cliffs, N.J., 1965, p. 433.
3. W. F. Osgood, Lehrbuch der Funktionentheorie, vol. 1, 5th ed., Teubner, Leipzig, 1928, p. 362.
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Intuition Out to Sea
William A. Leonard, California State University, Fullerton, CA
Using only intuition and without writing anything, try to answer the following
question:
A rope, attached to boat B and passing over pulley C, is drawn to the
right one foot at A . Does the boat move more than one foot, less than
one foot, or exactly one foot to the right?
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