Now, that we know how to represent function as power series
we can now talk about at least a couple of applications of series.
There are in fact many applications of series, unfortunately
most of them are beyond the scope of this course. One application of power series (with the
occasional use of Taylor Series) is in the field of Ordinary Differential
Equations when finding Series Solutions to
Differential Equations. If you are
interested in seeing how that works you can check out that chapter of my
Differential Equations notes.
Another application of series arises in the study of Partial
Differential Equations. One of the more
commonly used methods in that subject makes use of Fourier Series.
Many of the applications of series, especially those in the
differential equations fields, rely on the fact that functions can be
represented as a series. In these
applications it is very difficult, if not impossible, to find the function
itself. However, there are methods of
determining the series representation for the unknown function.
While the differential equations applications are beyond the
scope of this course there are some applications from a Calculus setting that
we can look at.
Example 1 Determine
a Taylor Series about for the following integral.
To do this we will first need to find a Taylor Series
about for the integrand. This however isn’t terribly difficult. We already have a Taylor Series for sine
about so we’ll just use that as follows,
We can now do the problem.
So, while we can’t integrate this function in terms of
known functions we can come up with a series representation for the integral.
This idea of deriving a series representation for a function
instead of trying to find the function itself is used quite often in several
fields. In fact, there are some fields
where this is one of the main ideas used and without this idea it would be very
difficult to accomplish anything in those fields.
Another application of series isn’t really an application of
infinite series. It’s more an
application of partial sums. In fact,
we’ve already seen this application in use once in this chapter. In the Estimating
the Value of a Series we used a partial sum to estimate the value of a
series. We can do the same thing with
power series and series representations of functions. The main difference is that we will now be using
the partial sum to approximate a function instead of a single value.
We will look at Taylor
series for our examples, but we could just as easily use any series
representation here. Recall that the nth degree Taylor Polynomial of f(x) is given by,
Let’s take a look at example of this.
Example 2 For
the function plot the function as well as ,
and on the same graph for the interval [-4,4].
Here is the general formula for the Taylor polynomials for cosine.
The three Taylor
polynomials that we’ve got are then,
Here is the graph of these three Taylor polynomials as well as the graph of
As we can see from this graph as we increase the degree of
polynomial it starts to look more and more like the function itself. In fact by the time we get to the only difference is right at the
ends. The higher the degree of the Taylor polynomial the
better it approximates the function.
Also the larger the interval the higher degree Taylor polynomial we
need to get a good approximation for the whole interval.
Before moving on let’s write down a couple more Taylor polynomials from
the previous example. Notice that
because the Taylor series for cosine doesn’t
contain any terms with odd powers on x
we get the following Taylor
These are identical to those used in the example. Sometimes this will happen although that was
not really the point of this. The point
is to notice that the nth degree Taylor
polynomial may actually have a degree that is less than n. It will never be more
than n, but it can be less than n.
The final example in this section really isn’t an
application of series and probably belonged in the previous section. However, the previous section was getting too
long so the example is in this section.
This is an example of how to multiply series together and while this
isn’t an application of series it is something that does have to be done on
occasion in the applications. So, in
that sense it does belong in this section.
Example 3 Find
the first three non-zero terms in the Taylor Series for about .
Before we start let’s acknowledge that the easiest way to
do this problem is to simply compute the first 3-4 derivatives, evaluate them
plug into the formula and we’d be done.
However, as we noted prior to this example we want to use this example
to illustrate how we multiply series.
We will make use of the fact that we’ve got Taylor Series
for each of these so we can use them in this problem.
We’re not going to completely multiply out these
series. We’re going to do enough of the
multiplication to get an answer. The
problem statement says that we want the first three non-zero terms. That
non-zero bit is important as it is possible that some of the terms will be
zero. If none of the terms are zero
this would mean that the first three non-zero terms would be the constant
term, x term, and x2 term. However, because some might be zero let’s
assume that if we get all the terms up through x4 we’ll have enough to get the answer. If we’ve assumed wrong it will be very easy
to fix so don’t worry about that.
Now, let’s write down the first few terms of each series
and we’ll stop at the x4
term in each.
Note that we do need to acknowledge that these series
don’t stop. That’s the purpose of the
“ ” at the end of each. Just for a second however, let’s suppose
that each of these did stop and ask ourselves how we would multiply each
out. If this were the case we would
take every term in the second and multiply by every term in the first. In other words, we would first multiply
every term in the second series by 1, then every term in the second series by
x, then by x2 etc.
By stopping each series at x4 we have now guaranteed that we’ll get all terms
that have an exponent of 4 or less. Do
you see why?
Each of the terms that we neglected to write down have an
exponent of at least 5 and so multiplying by 1 or any power of x will result in a term with an
exponent that is at a minimum 5.
Therefore, none of the neglected terms will contribute terms with an
exponent of 4 or less and so weren’t needed.
So, let’s start the multiplication process.
Now, collect like terms ignoring everything with an
exponent of 5 or more since we won’t have all those terms and don’t want them
either. Doing this gives,
There we go. It
looks like we over guessed and ended up with four non-zero terms, but that’s
okay. If we had under guessed and it
turned out that we needed terms with x5
in them all we would need to do at this point is go back and add in those
terms to the original series and do a couple quick multiplications. In other words, there is no reason to
completely redo all the work.