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Mathematical Expression Editor
We study Taylor and Maclaurin series.
We’ve seen that we can approximate functions with polynomials, given that enough
derivative information is available. We have also seen that certain functions can be
represented by a power series. In this section we combine these concepts: If a function
is infinitely differentiable, we show how to represent it with a power series
function.
Let have derivatives of all orders at .
The Taylor series of , centered at is
Setting gives the Maclaurin series of :
Quick: Write down the Taylor series for centered at .
Write down the Taylor series for centered at . -
The difference between a Taylor polynomial and a Taylor series is the former is a
polynomial, containing only a finite number of terms, whereas the latter is a series, a
summation of an infinite set of terms, any number of which (including an infinite
number) may be zero. When creating the Taylor polynomial of degree for a function
at , we needed to evaluate , and the first derivatives of , at . When creating the
Taylor series of , it helps to find a pattern that describes the th derivative of at .
Time for examples!
Compute the Taylor series for centered at .
We’ll start by making a table of
derivatives: Since a repeating pattern has emerged, we see that the Maclaurin series
for is:
Let’s see an example that is not centered at :
Compute the Taylor series for centered at .
We’ll start by making a table of
derivatives:
Since a pattern has emerged, we see that the Taylor series for is:
Finally, sometimes Taylor’s formula may not be the best way to compute the Taylor
series.
Compute the Taylor series for centered at .
If we try to use Taylor’s formula, we
must start by making a table of derivatives:
Hmm. This is getting messy. Let’s try to find the Taylor series via known
power series. We know that setting we now have when and when .
Since we can find the desired power series by integrating. Write with me
since , , and we have our desired power series, which converges with radius of
convergence . However, note the interval of convergence may be different, and it is in
this case. First note that our power series can be written in summation
notation as If or we can see that this sequence is In both cases, the series
converges by the alternating series test. Hence the interval of convergence is
.
Above we implicitly used the following theorem:
The Taylor series centered at of a polynomial in is exactly that polynomial and the
Taylor series of power series centered at is exactly that power series.
This is just saying that if you know a power series for a function, then using Taylor’s
formula will do nothing but give you the power series.
Since we designed Taylor polynomials to approximate functions, you might guess that
the Taylor series of a function is equal to the function (at least on the interval of
convergence for the Taylor series). This is false.
Here is an somewhat unsatisfying example:
Consider:
(a)
Compute the Maclaurin series of .
(b)
Find the radius of convergence.
(c)
Is the Maclaurin series for equal to on the interval of convergence?
We’ll start by making a table of derivatives:
So our Maclaurin series for is: This converges for all values of , and hence
the radius of convergence is , with interval of convergence . However, as
functions.
A more satisfying example is the following:
Let It turns out that is infinitely differentiable everywhere, but all of its derivatives
vanish at . Thus the Maclaurin series for is just The derivatives of “go to ” as goes
to zero faster than any polynomial, and so no polynomial term “detects” that this
function is not the horizontal line .
It is within your power to show that is infinitely differentiable everywhere, and to
prove that . This is quite involved, and we will not do it here. If you have the
gumption, and the willpower, it would make a fantastic exercise.
We will find that “most of the time” they are equal, but we need to consider the
conditions that allow us to conclude this. Taylor’s theorem states that the error
between a function and its th degree Taylor polynomial is : and that where is the
maximum value of on . If goes to for each in an interval as approaches infinity,
we conclude that the function is equal to its Taylor series expansion. This leads us to
our next theorem:
Function and Taylor Series Equality Let have derivatives of all orders at , let , and
let be an interval on which the Taylor series of converges. If for all in containing ,
then on .
We’ll work a representative example of this theorem to see what is going on, the
general case is much the same.
The Maclaurin series for is show that:
To start, note that using the ratio test this
power series has an infinite radius of convergence. We want to know whether for
every real number . Another way of phrasing this is that we want the remainder
to go to zero as goes to infinity. We know that where is the maximum
value of on . We know , since all the derivatives of are just Thus So how
large can be? If is between and , then is between and , so Thus by the
Squeeze Theorem, as proving that the Maclaurin series for converges to
everywhere!
There is good news. A function that is equal to its Taylor series, centered at any
point the domain of , is said to be an analytic function, and most, if not
all, functions that we encounter within this course are analytic functions.
Generally speaking, any function that one creates with elementary functions
(polynomials, exponentials, trigonometric functions, etc.) that is not piecewise
defined is probably analytic. For most functions, we assume the function is
equal to its Taylor series on the series’ interval of convergence and only
check the remainder (as above) when we suspect something may not work as
expected.