This shall be my last lecture that I will give in this tutorial. This shall be a continuation of the last lecture. Now since we have an understanding of what a sequence and a series are we can talk about a very important topic that appears a lot in analysis related areas, power series.

Definition:Apower series(centered at zero) is an "infinite polynomial" of the form . The domain isdefinedto be the values for such that the power series converges. And the value isdefinedto be the convergent value.

The above definition is good because it always converges at least at one point that is . Note, the are the terms of some sequence.

Here is a more general definition.

Definition:Apower seriescentered at is .

Again it converges at least at one point, .

There is one interesting theorem about infinite series which is not even proved in the standard Calculus III course.

Theorem:Given a power series, then the series converges at: a single point (its center), some interval, or the entire number line.

When I say "some interval" I mean to say an interval one of the forms: . For example, a power series can coverge for . Another example, it can converge for . But it cannot converge for .The symbol represents "union" it means , because it is not one complete interval, that is, it is with missing which does not make it complete, and hence violates the theorem. When we sayinterval of convergencewe means the interval for which the series converges, it can happen that the series converges for the entire number line (that is for all ) in that case we shall say the interval of convergence is infinite. And it can also happen that a power series converges at a single point (the center) and we shall say the interval of convergence is zero.

Example 73:Given the power series find the interval of convergence. The typical way of doing these problems is through the ratio test, because the ratio test is useful for dealing with exponents and factorials. Thus we have a series where are the terms of the sequence. Now we find the ratio of these terms. But before we do that I will mention one important point. The ratio test that we had the condition that the terms of the sequence are non-zero for sufficiently large , that is the condition for it to work. In the case where we cannot use the ratio test because it is always zero, and hence, never non-zero for sufficiently large . But we can ignore that case because it is the center and we know the series converges at the center. Thus, it is safe to assume . Now we use ratio test,

Cancel, now no matter what you choose you get a limit of zero and by the ratio test this shows convergence. Thus, the interval of convergence is infinite, or you may want to write .

Example 74:Given the power series find the interval of convergence. Again without any fear we can use the ratio test without considering the case (division by zero) first because we know it converges for it is its center. The terms of the sequence for some fixed are thus we need to evaluate the limit . But the limit of is one, thus the final limit we get is . We know by the ratio test that if the limit is strictly less than 1 then we have convergence. Thus the power series converges. And for the power series diverges. But what about , remember that is the "bad point", the point where the ratio test is inconlusive. Thus, we check each point seperately. The solution to is . Thus, if then the power series evaluated at that point is , this is the infamous harmonic series we had before, it diverges. If then we have this is the alternating harmonic series which we know converges. Thus, the series converges for and , expressing the absolute value as interval we have , this is the interval of convergence. Note we did not get any other interval as promised by the theorem.

Example 75:Given the power series find the interval of convergence. Again we need to consider the case but as we already seen we can ignore that without any fear. This is non-zero for suffiently large we can use the ratio test, . The limit as always diverges to unless is zero. Thus, in this case the interval of convergence is zero, that is, only converges at its center .

If you are smart you should realize what I am about to say. If exists then there are two cases. When the limit is zero, in that case the interval of convergence is infinite. The limit is not zero (it must be positive because of absolute value). In that case, for convergence, must satisfy where is the limit of . (And also check endpoints). In this case we have an interval. The final case is when diverges to thus the interval of convergence is zero. Thus, we proved the theorem, because the cases are the full line, the interval, or a point, theorem proved! But if you are Hacker smart then, no, it is not as simple as I just said. Because in the last case when it does not exist, does not mean it must diverge to infinity (remember the sequence limit does not exists but it does not increase without bound). Thus, we completely do not know what the interval is because the limit does not exist nor diverge to . But there is something else that can happen, the ratio test cannot be used, if if never non-zero for sufficiently large . Thus, the theorem is powerful because it appliesevento the case when the ratio limit does not exists nor diverges to and in the case where the ratio test cannot be used.

The nice thing about power series is that they are easy to differenciate. We can do it term by term. Though it might seem obvious, this is not the case. The same thing about integration.

Example 76:If is a power series function. Then, term by term differenciation says, .

Theorem:If then

The power series, as mentioned, represents a function whose domain is the interval of convergence and function value is the convergent value. The question that is of interest to us is given some arbitraty function can we find a power series representation for the function? Let us assume that has a power series representation,

We will assume is infinitely differenciable at meaning we can take the derivative again and again.

Evaluate at to get,

Next, take the derivative,

Evaluate at to get,

Next, take the derivative again,

Evaluate at to get,

Next, take the derivative again,

Evaluate at to get,

.

The pattern is clear, in general we get, for ,

Thus,

This, show thatifan infinitely differenciable function is expressible as a power series at then such a representation must be unique.

(Note the in the exponent does not mean power, it means derivative).

Definition:If is infinitely differenciable at then theTaylor Series(centered at ) is defined by,

.

Warning, do not think that this says that there is the power series, it is simply a definition. Given a function we define its Taylor series to be that series. Of course, what we are really are hoping for is that the Taylor series of a function and the function are the same, meaning the Taylor series represents the function. There is a way to show that given a infinitely differenciable function to show it matches with its Taylor series. But since this procedure is a little bit too advanced we willassumethat it exists. But we will do more than assume, we will assume it exists on its interval of convergence. Meaning, we will take a function, find its Taylor series, assume it matches the function, find the interval of convergence, and state that this power series represents the given function on the given interval.

Example 77:We will find the power series for . Again, we willassumethat the Taylor series is the unique power series centered at .

...

Thus,

.....

Thus, the coefficients in the Taylor series are,

.....

Thus,

We are almost finished. Remember I said to check the interval of convergence. By using the techinque with the ratio test shown before we find that is the interval of convergece. Thus, the power series for the functiononlyworks for . And for other values it fails. Again, we did not actually show that there is a power series, we just assumed, because that is a little too advanced right now.

The above series is extremely important, it is called theinfinite geometric series. For example, we can use if in the following manner. What does represent? We can write, by definition of decimals,

Thus, by the formula, the sum is, .

Example 78:We know that on , thus, the derivative of both sides, . Multiply by through to get,

. This tells us that, (note )

.

Example 79:We will find the power series for the exponential function centered at . Thus, the function that we are working with is . To find the coefficients for the power series we take derivatives and evaluate.

.

.....

Thus,

.....

Thus, the coefficients in the Taylor series are,

.....

Thus, we have,

.

Now, we have to find the interval for which this formula is valid, that is we need to find the interval of convergence. We done this example in the previous lecture. We find that is the interval of convergece. Thus, it works for all . Specifically when we have a beautify equation, . We can use this to approximate , by just taking a few terms in this infinite series we get an awfully close approximation.

Example 80:We will do a different example this time, . The problem with this one is that ideally the nicest looking series is when the center is zero, as we had before. The problem is that is not infinitely differenciable at (the center) because it is not even defined there! Thus, we need to chose a different center for which the function is defined. We cannot chose for the same reason thus we will chose . Thus, we have,

.....

Thus,

.....

Thus, the coefficients in the Taylor series are,

.....

Thus, we have,

.

By using the ratio test we find that to converge we need thus . Checking the endpoints we find that leads to negative harmonic series, which diverges to , and that leads to alternating harmonic series which converges. Thus, the interval of convergence is . This is the interval on which this power series works. Specifically when we have,

. A beautiful formula, it also shows what the sum of the alternating harmonic series is.

Unlike the power series for this one converges too slowly, meaning many many terms are needed to obtain accuracy. The trick to determine whether it is a fast or slow is so look at how quickly it gets small, since the exponential series gets small very quickly because factorials grow fast it is reasonable to say it converges quickly, and over here denominators get smaller slowly thus it is reasonable to say the series does not converge quickly enough. Of course, this trick does not always work but you can look out for it.

There are two extremely important series that appear a lot, the sine and cosine. The standard way how this is done in Calculus III class is by the use of the following facts and . However, mathematicians do not really consider the derivations in Calculus I class to be full of rigor. And hence mathematiciansdefinesine and cosine as follows:

.

Note, by taking the derivatives term by term we obtain the fundamental derivative identities for sine and cosine. Since we done several Taylor series expansions it will be an excerise to find using the derivative identities above. Furthermore, show that the interval of convergence for both is .

If you wish to read the next section, it would help to know this definition.

Definition:A function is analytic at some point, means there exists a power series centered at the point having non-zero interval of convergence. Thus, for example, is analytic everywhere, you can center the series anywhere.

Application to Differencial Equations*

We can use the above concepts of power series to solve the differencial equation on some non-zero open interval ,

.

The idea here is toassumethat there exists a function which is analytic at (say zero, for simplicity) that solves the differencial equation on some .

This is one of the more difficult ones because usually we divide by the expression in front to get,

.

But the problem is that that can lead to division by zero, and we do not know if we can do that.

When you will study differencial equations you will see that there is an entire theory that explains when and when not an analytic solution exists (a solution expressible as a power series). Since this is not a lecture on differencial equations we will just assume, if our assumption leads to a contradiction we will know that it was false. Thus, the assumption that we are making is that the solution is analytic, that is,

Thus,

Substitute that into the differencial equation,

Multiply through,

To add these together we will evalute the second summation at and the third at ,

Combine, term by term,

Thus,

Because, power series' are unique we have that each term must be zero,

for

for .

Thus, for .

We also found that .

Hence the only one which was non-zero was .

In fact, it is arbitraty because it ends up cancelling out.

Thus, we have that, is a solution.

Meaning any line passing through the origin is a solution. In fact, it works out! Substitute it into the differencial equation to confirm the solution. But are there any more solutions? If they are then they cannot be analytic at zero because we found all. It turns out that is a solution, but the reason why we did get it is because is not analytic at , it is not even defined there! There is a method for finding that other solution but that does not interest us because we just looked for analytic solutions.

~~~

Excerises

1)Find the power series for and as explained above.

2)Find the power series for . (Hint: Use the logarithm identities).

3)Find the power series for be integrating term by term the infinite geometric series.

Note, and integrate.

4)Find the power series for centered at .

5*)Solve the differencial equation, . It happens to be a slightly different result than the one we did together.

6)What does represent as a rational number?