It is time to consider functions other than polynomials. In particular it is time to give a
mathematically acceptable definition of functions like e^{x},sin
(x)
and cos
(x)
. It has been
assumed these functions are known from beginning calculus but this is a pretence. Most
students who take calculus come through it without a complete understanding of the circular
functions. This is because of the reliance on plane geometry in defining them. Fortunately,
these functions can be completely understood in terms of power series rather than wretched
plane geometry. The exponential function can also be defined in a simple manner using power
series.
Definition 8.1.1Let
{ak}
_{k=0}^{∞}be a sequence of numbers. The expression,
∞∑ k
ak(x− a) (8.1)
k=0
(8.1)
is called a Taylor series centered at a.This is also called a power series centered at a.It isunderstood that x and a ∈ F, that is, either ℂ or ℝ.
In the above definition, x is a variable. Thus you can put in various values of x and ask
whether the resulting series of numbers converges. Defining D to be the set of all values of x
such that the resulting series does converge, define a new function f defined on D having
values in F as
∞∑ k
f (x) ≡ ak(x− a) .
k=0
This might be a totally new function one which has no name. Nevertheless, much can be
said about such functions. The following lemma is fundamental in considering the
form of D which always turns out to be of the form B
(a,r)
along with possibly
some points, z such that
|z − a|
= r. First here is a simple lemma which will be
useful.
Lemma 8.1.2lim_{n→∞}n^{1∕n} = 1.
Proof: It is clear n^{1∕n}≥ 1. Let n^{1∕n} = 1 + e_{n} where 0 ≤ e_{n}. Then raising both sides to
the n^{th} power for n > 1 and using the binomial theorem,
n ( )
n = (1+ e )n = ∑ n ek ≥ 1+ ne + (n (n− 1)∕2)e2
n k=0 k n n n
2
≥ (n(n − 1)∕2)en
Thus
0 ≤ e2≤ ----n--- = --1--
n n (n− 1) n− 1
From this the desired result follows because
| |
||n1∕n − 1|| = en ≤ √-1--.■
n − 1
Theorem 8.1.3Let∑_{k=0}^{∞}a_{k}
(x − a)
^{k}be a Taylorseries. Then there existsr ≤ ∞ such that the Taylor series converges absolutely if
|x − a|
< r. Furthermore,if
|x − a|
> r, the Taylor series diverges.If λ < r then the Taylor series convergesuniformly on the closed disk
|x − a|
≤ λ.
Proof: Note
| |
lim sup ||a (x− a)k||1∕k = lim sup |a |1∕k |x − a|.
k→∞ k k→ ∞ k
Then by the root test, the series converges absolutely if
| |
lim sup ||a (x − a)k||1∕k = lim sup |a |1∕k|x − a| ≤ λlim sup |a |1∕k ≤ λ-< α < 1
k→ ∞ k k→∞ k k→ ∞ k r
It follows that for all k large enough and such x,
| |
||ak(x − a)k||
< α^{k}. Then by the Weierstrass
M test, convergence is uniform. ■
Note that the radius of convergence r is given by
lim sup |a |1∕kr = 1
k→ ∞ k
Definition 8.1.4The number in the above theorem is called the radius ofconvergence and the set on which convergence takes place is called the disc ofconvergence.
Now the theorem was proved using the root test but often you use the ratio test to
find the interval of convergence. This kind of thing is typical in math so get used
to it. The proof of a theorem does not always yield a way to find the thing the
theorem speaks about. The above is an existence theorem. There exists an interval of
convergence from the above theorem. You find it in specific cases any way that is most
convenient.
Example 8.1.5Find the disc of convergence of the Taylor series∑_{n=1}^{∞}