In the previous section we started looking at writing down a
power series representation of a function.
The problem with the approach in that section is that everything came
down to needing to be able to relate the function in some way to
and while there are many functions out there that can be
related to this function there are many more that simply can’t be related to
this.
So, without taking anything away from the process we looked
at in the previous section, what we need to do is come up with a more general method
for writing a power series representation for a function.
So, for the time being, let’s make two assumptions. First, let’s assume that the function does in fact have a power series representation
about ,
Next, we will need to assume that the function, ,
has derivatives of every order and that we can in fact find them all.
Now that we’ve assumed that a power series representation
exists we need to determine what the coefficients, cn, are. This is
easier than it might at first appear to be.
Let’s first just evaluate everything at . This gives,
So, all the terms except the first are zero and we now know
what c0 is. Unfortunately, there isn’t any other value of
x that we can plug into the function
that will allow us to quickly find any of the other coefficients. However, if we take the derivative of the
function (and its power series) then plug in we get,
and we now know c1.
Let’s continue with this idea and find the second
derivative.
So, it looks like,
Using the third derivative gives,
Using the fourth derivative gives,
Hopefully by this time you’ve seen the pattern here. It looks like, in general, we’ve got the following
formula for the coefficients.
This even works for n=0
if you recall that and define .
So, provided a power series representation for the function about exists the Taylor Series for about
is,
Taylor Series
If we use ,
so we talking about the Taylor Series about ,
we call the series a Maclaurin
Series for or,
Maclaurin Series
Before working any examples of Taylor Series we first need
to address the assumption that a Taylor Series will in fact exist for a given
function. Let’s start out with some
notation and definitions that we’ll need.
To determine a condition that must be true in order for a
Taylor series to exist for a function let’s first define the nth degree Taylor polynomial of as,
Note that this really is a polynomial of degree at most n!
If we were to write out the sum without the summation notation this
would clearly be an nth degree polynomial. We’ll see a nice application of Taylor polynomials in the
next section.
Notice as well that for the full Taylor Series,
the nth degree Taylor polynomial is just the
partial sum for the series.
Next, the remainder
is defined to be,
So, the remainder is really just the error between the function and the nth degree Taylor polynomial
for a given n.
With this definition note that we can then write the
function as,
We now have the following Theorem.
Theorem
In general showing that is a somewhat difficult process and so we will
be assuming that this can be done for some R
in all of the examples that we’ll be looking at.
Now let’s look at some examples.
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Example 1 Find
the Taylor Series for about .
Solution
This is actually one of the easier Taylor Series that
we’ll be asked to compute. To find the
Taylor Series for a function we will need to determine a general formula for . This is one of the few functions where this
is easy to do right from the start.
To get a formula for all we need to do is recognize that,
and so,
Therefore, the Taylor
series for about x=0
is,
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Example 2 Find
the Taylor Series for about .
Solution
There are two ways to do this problem. Both are fairly simple, however one of them
requires significantly less work.
We’ll work both solutions since the longer one has some nice ideas
that we’ll see in other examples.
Solution 1
As with the first example we’ll need to get a formula for . However, unlike the first one we’ve got a
little more work to do. Let’s first
take some derivatives and evaluate them at x=0.
After a couple of computations we were able to get general
formulas for both and . We often won’t be able to get a general
formula for so don’t get too excited about getting that
formula. Also, as we will see it won’t
always be easy to get a general formula for .
So, in this case we’ve got general formulas so all we need
to do is plug these into the Taylor Series formula and be done with the
problem.
Solution 2
The previous solution wasn’t too bad and we often have to
do things in that manner. However, in
this case there is a much shorter solution method. In the previous section we used series that
we’ve already found to help us find a new series. Let’s do the same thing with this one. We already know a Taylor Series for about and in this case the only difference is
we’ve got a “-x” in the exponent
instead of just an x.
So, all we need to do is replace the x in the Taylor Series that we found in the first example with “-x”.
This is a much shorter method of arriving at the same
answer so don’t forget about using previously computed series where possible
(and allowed of course).
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To this point we’ve only looked at Taylor Series about (also known as Maclaurin Series) so let’s take
a look at a Taylor Series that isn’t about . Also, we’ll pick on the exponential function
one more time since it makes some of the work easier. This will be the final Taylor Series for
exponentials in this section.
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Example 4 Find
the Taylor Series for about .
Solution
Finding a general formula for is fairly simple.
The Taylor Series is then,
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Okay, we now need to work some examples that don’t involve
the exponential function since these will tend to require a little more work.
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Example 5 Find
the Taylor Series for about .
Solution
First we’ll need to take some derivatives of the function
and evaluate them at x=0.
In this example, unlike the previous ones, there is not an
easy formula for either the general derivative or the evaluation of the
derivative. However, there is a clear
pattern to the evaluations. So, let’s
plug what we’ve got into the Taylor
series and see what we get,
So, we only pick up terms with even powers on the x’s.
This doesn’t really help us to get a general formula for the Taylor
Series. However, let’s drop the zeroes
and “renumber” the terms as follows to see what we can get.
By renumbering the terms as we did we can actually come up
with a general formula for the Taylor Series and here it is,
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This idea of renumbering the series terms as we did in the
previous example isn’t used all that often, but occasionally is very
useful. There is one more series where
we need to do it so let’s take a look at that so we can get one more example
down of renumbering series terms.
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Example 6 Find
the Taylor Series for about .
Solution
As with the last example we’ll start off in the same
manner.
So, we get a similar pattern for this one. Let’s plug the numbers into the Taylor
Series.
In this case we only get terms that have an odd exponent
on x and as with the last problem
once we ignore the zero terms there is a clear pattern and formula. So renumbering the terms as we did in the
previous example we get the following Taylor Series.
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We really need to work another example or two in which f(x) isn’t
about .
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Example 7 Find
the Taylor Series for about .
Solution
Here are the first few derivatives and the evaluations.
Note that while we got a general formula here it doesn’t
work for . This will happen on occasion so don’t worry
about it when it does.
In order to plug this into the Taylor Series formula we’ll
need to strip out the term first.
Notice that we simplified the factorials in this
case. You should always simplify them
if there are more than one and it’s possible to simplify them.
Also, do not get excited about the term sitting in front
of the series. Sometimes we need to do
that when we can’t get a general formula that will hold for all values of n.
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Example 8 Find
the Taylor Series for about .
Solution
Again, here are the derivatives and evaluations.
Notice that all the negative signs will cancel out in the
evaluation. Also, this formula will
work for all n, unlike the previous
example.
Here is the Taylor Series for this function.
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Now, let’s work one of the easier examples in this
section. The problem for most students
is that it may not appear to be that easy (or maybe it will appear to be too
easy) at first glance.
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Example 9 Find
the Taylor Series for about .
Solution
Here are the derivatives for this problem.
This Taylor series will terminate after . This will always happen when we are finding
the Taylor Series of a polynomial.
Here is the Taylor Series for this one.
When finding the Taylor Series of a polynomial we don’t do
any simplification of the right hand side.
We leave it like it is. In
fact, if we were to multiply everything out we just get back to the original
polynomial!
While it’s not apparent that writing the Taylor Series for
a polynomial is useful there are times where this needs to be done. The problem is that they are beyond the
scope of this course and so aren’t covered here. For example, there is one application to
series in the field of Differential Equations where this needs to be done on
occasion.
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So, we’ve seen quite a few examples of Taylor Series to this
point and in all of them we were able to find general formulas for the
series. This won’t always be the
case. To see an example of one that
doesn’t have a general formula check out the last example in the next section.
Before leaving this section there are three important Taylor
Series that we’ve derived in this section that we should summarize up in one
place. In my class I will assume that
you know these formulas from this point on.