Let's return to the simple geometric series \begin \sum_^\infty x^n \end where \(x\) is some real number. As we have seen (back in Example 3.2.4 and Lemma 3.2.5), for \(|x| \lt 1\) this series converges to a limit, that varies with \(x\text\) while for \(|x|\geq 1\) the series diverges. Consequently we can consider this series to be a function of \(x\) \begin f(x) &= \sum_^\infty x^n & \text. \end Furthermore (also from Example 3.2.4 and Lemma 3.2.5) we know what the function is. \begin f(x) &= \sum_^\infty x^n = \frac. \end Hence we can consider the series \(\sum_^\infty x^n\) as a new way of representing the function \(\frac\) when \(|x| \lt 1\text<.>\) This series is an example of a power series. Of course, representing a function as simple as \(\frac\) by a series doesn't seem like it is going to make life easier. However the idea of representing a function by a series turns out to be extremely helpful. Power series turn out to be very robust mathematical objects and interact very nicely with not only standard arithmetic operations, but also with differentiation and integration (see Theorem 3.5.13). This means, for example, that \[\begin \frac \left\<\frac\right\> &= \frac \sum_^\infty x^n & \text\\ &= \sum_^\infty \frac x^n & \text\\ &= \sum_^\infty n x^\\ \end\] and in a very similar way \begin \int \frac \, d &= \int \sum_^\infty x^n \, d & \text\\ &= \sum_^\infty \int x^n \, d & \text\\ &= C + \sum_^\infty \frac x^ \end We are hiding some mathematics under the word “just” in the above, but you can see that once we have a power series representation of a function, differentiation and integration become very straightforward. So we should set as our goal for this section, the development of machinery to define and understand power series. This will allow us to answer questions 1 like \begin \text\ e^x &=\sum\limits_^\infty\frac \text < ? >\end Our starting point (now that we have equipped ourselves with basic ideas about series), is the definition of power series.
A series of the form \[ A_0 +A_1(x-c) + A_2(x-c)^2 + A_3 (x-c)^3 + \cdots =\sum_
Because of this, when the limit exists, the quantity
is called the radius of convergence of the series 3 .
All of these possibilities do happen. We give an example of each below. But first, the concept of “radius of convergence” is important enough to warrant a formal definition.
We already know that, if \(a\ne 0\text\) the geometric series \(\sum\limits_^\infty a x^n\) converges when \(|x| \lt 1\) and diverges when \(|x|\ge 1\text<.>\) So, in the terminology of Definition 3.5.3, the geometric series has radius of convergence \(R=1\text<.>\) As a consistency check, we can also compute \(R\) using 3.5.2. The series \(\sum\limits_^\infty a x^n\) has \(A_n=a\text<.>\) So
The series \(\sum\limits_^\infty \frac\) has \(A_n=\frac\text<.>\) So
and \(\sum\limits_^\infty \frac\) has radius of convergence \(\infty\text<.>\) It converges for every \(x\text<.>\)
The series \(\sum\limits_^\infty n! x^n\) has \(A_n=n!\text<.>\) So
and \(\sum\limits_^\infty n! x^n\) has radius of convergence zero 4 . It converges only for \(x=0\text\) where it takes the value \(0!=1\text<.>\)
Comparing the series
\begin &\ 1&+&\ 2x&+&x^2&+&2x^3&+&x^4&+&2x^5&+\cdots \\\\ \end\begin \sum_^\infty A_nx^n=&A_0&+&A_1x&+&A_2x^2&+&A_3x^3&+&A_4x^4&+&A_5x^5&+\cdots \end
\begin A_0&=1&\quad A_1&=2&\quad A_2&=1&\quad A_3&=2&\quad A_4&=1&\quad A_5&=2&\quad\cdots\\ \endand \(\frac>\) does not converge as \(n\rightarrow\infty\text<.>\) Since the limit of the ratios does not exist, we cannot tell anything from the ratio test. Nonetheless, we can still figure out for which \(x\)'s our power series converges.
In conclusion, our series converges if and only if \(|x| \lt 1\text\) and so has radius of convergence \(1\text<.>\)
Lets construct a series from the digits of \(\pi\text<.>\) Now to avoid dividing by zero, let us set
Since \(\pi = 3.141591\dots\)
\begin A_0 = 4 \quad A_1 = 2 \quad A_2 = 5 \quad A_3 = 2 \quad A_4 = 6 \quad A_5 = 10 \quad A_6 = 2 \quad \cdots \end
Consequently every \(A_n\) is an integer between 1 and 10 and gives us the series
\[ \sum_^\infty A_n x^n = 4 + 2x + 5x^2 + 2x^3 + 6x^4 +10 x^5 + \cdots \nonumber \]
The number \(\pi\) is irrational 5 and consequently the ratio \(\frac>\) cannot have a limit as \(n\rightarrow\infty\text<.>\) If you do not understand why this is the case then don't worry too much about it 6 . As in the last example, the limit of the ratios does not exist and we cannot tell anything from the ratio test. But we can still figure out for which \(x\)'s it converges.
In conclusion, our series converges if and only if \(|x| \lt 1\text\) and so has radius of convergence \(1\text<.>\)
Though we won't prove it, it is true that every power series has a radius of convergence, whether or not the limit \(\lim\limits_
Let \(\sum\limits_^\infty A_n (x-c)^n\) be a power series. Then one of the following alternatives must hold.
Consider the power series
\begin \sum_^\infty A_n (x-c)^n. \end
The set of real \(x\)-values for which it converges is called the interval of convergence of the series.
Suppose that the power series \(\sum\limits_^\infty A_n (x-c)^n\) has radius of convergence \(R\text<.>\) Then from Theorem 3.5.9, we have that
To reiterate — while the radius convergence, \(R\) with \(0 \lt R \lt \infty\text\) tells us that the series converges for \(|x-c| \lt R\) and diverges for \(|x-c| \gt R\text\) it does not (by itself) tell us whether or not the series converges when \(|x-c|=R\text\) i.e. when \(x=c\pm R\text<.>\) The following example shows that all four possibilities can occur.
Let \(p\) be any real number 7 and consider the series \(\sum_^\infty \frac\text<.>\) This series has \(A_n= \frac\text<.>\) Since
the series has radius of convergence \(1\text<.>\) So it certainly converges for \(|x| \lt 1\) and diverges for \(|x| \gt 1\text<.>\) That just leaves \(x=\pm 1\text<.>\)
We are told that a certain power series with center \(c=3\text\) converges at \(x=4\) and diverges at \(x=1\text<.>\) What else can we say about the convergence or divergence of the series for other values of \(x\text\)
We are told that the series is centerd at \(3\text\) so its terms are all powers of \((x-3)\) and it is of the form
A good way to summarise the convergence data we are given is with a figure like the one below. Green dots mark the values of \(x\) where the series is known to converge. (Recall that every power series converges at its center.) The red dot marks the value of \(x\) where the series is known to diverge. The bull's eye marks the center.
Can we say more about the convergence and/or divergence of the series for other values of \(x\text\) Yes!
Let us think about the radius of convergence, \(R\text\) of the series. We know that it must exist and the information we have been given allows us to bound \(R\text<.>\) Recall that
We have been told that
We still don't know \(R\) exactly. But we do know that \(1\le R\le 2\text<.>\) Consequently,
The following figure provides a resume of all of this convergence data — there is convergence at green \(x\)'s and divergence at red \(x\)'s.
Notice that from the data given we cannot say anything about the convergence or divergence of the series on the intervals \((1,2]\) and \((4,5]\text<.>\)
One lesson that we can derive from this example is that,
Just as we have done previously with limits, differentiation and integration, we can construct power series representations of more complicated functions by using those of simpler functions. Here is a theorem that helps us to do so.
Assume that the functions \(f(x)\) and \(g(x)\) are given by the power series
\[ f(x) = \sum_^\infty A_n (x-c)^n \qquad g(x) = \sum_^\infty B_n (x-c)^n \nonumber \]
for all \(x\) obeying \(|x-c| \lt R\text<.>\) In particular, we are assuming that both power series have radius of convergence at least \(R\text<.>\) Also let \(K\) be a constant. Then
for all \(x\) obeying \(|x-c| \lt R\text<.>\)
In particular the radius of convergence of each of the six power series on the right hand sides is at least \(R\text<.>\) In fact, if \(R\) is the radius of convergence of \(\sum\limits_^\infty A_n (x-c)^n\text\) then \(R\) is also the radius of convergence of all of the above right hand sides, with the possible exceptions of \(\sum\limits_^\infty [A_n+B_n]\, (x-c)^n\) and \(\sum\limits_^\infty KA_n\, (x-c)^n\) when \(K=0\text<.>\)
The last statement of Theorem 3.5.13 might seem a little odd, but consider the following two power series centerd at \(0\text\)
\begin \sum_^\infty 2^n x^n & \text < and >\sum_^\infty (1-2^n) x^n. \end
The ratio test tells us that they both have radius of convergence \(R=\frac\text<.>\) However their sum is
\begin \sum_^\infty 2^n x^n + \sum_^\infty (1-2^n) x^n &= \sum_^\infty x^n \end
which has the larger radius of convergence \(1\text<.>\)
A more extreme example of the same phenomenon is supplied by the two series
\begin \sum_^\infty 2^n x^n & \text < and >\sum_^\infty (-2^n) x^n. \end
They are both geometric series with radius of convergence \(R=\frac\text<.>\) But their sum is
\begin \sum_^\infty 2^n x^n + \sum_^\infty (-2^n) x^n &= \sum_^\infty (0)x^n \end
which has radius of convergence \(+\infty\text<.>\)
We'll now use this theorem to build power series representations for a bunch of functions out of the one simple power series representation that we know — the geometric series
Find a power series representation for \(\frac\text<.>\)
Solution: The secret to finding power series representations for a good many functions is to manipulate them into a form in which \(\frac\) appears and use the geometric series representation \(\frac = \sum_^\infty y^n\text<.>\) We have deliberately renamed the variable to \(y\) here — it does not have to be \(x\text<.>\) We can use that strategy to find a power series expansion for \(\frac\) — we just have to recognize that \(\frac\) is the same as \(\frac\) if we set \(y\) to \(x^2\text<.>\)
This is a perfectly good power series. There is nothing wrong with the power of \(x\) being \(2n\text<.>\) (This just means that the coefficients of all odd powers of \(x\) are zero.) In fact, you should try to always write power series in forms that are as easy to understand as possible. The geometric series that we used at the end of the first line converges for
\begin |y| \lt 1 \iff \big|x^2\big| \lt 1 \iff |x| \lt 1 \end
So our power series has radius of convergence \(1\) and interval of convergence \(-1 \lt x \lt 1\text<.>\)
Find a power series representation for \(\frac\text<.>\)
Solution: This example is just a more algebraically involved variant of the last one. Again, the strategy is to manipulate \(\frac\) into a form in which \(\frac\) appears.
Now use Theorem 3.5.13 twice
The geometric series that we used in the second line converges when
\begin |y| \lt 1 &\iff \big|\frac\big| \lt 1\\ & \iff |x|^2 \lt 2 \iff |x| \lt \sqrt \end
So the given power series has radius of convergence \(\sqrt\) and interval of convergence \(-\sqrt \lt x \lt \sqrt\text<.>\)
Find a power series representation for \(\frac\) with center \(3\text<.>\)
Solution: The new wrinkle in this example is the requirement that the center be \(3\text<.>\) That the center is to be \(3\) means that we need a power series in powers of \(x-c\text\) with \(c=3\text<.>\) So we are looking for a power series of the form \(\sum_^\infty A_n(x-3)^n\text<.>\) The easy way to find such a series is to force an \(x-3\) to appear by adding and subtracting a \(3\text<.>\)
Now we continue, as in the last example, by manipulating \(\frac\) into a form in which \(\frac\) appears.
The geometric series that we used in the second line converges when
\begin |y| \lt 1 & \iff \Big|\frac\Big| \lt 1\\ & \iff |x-3| \lt 2\\ & \iff -2 \lt x-3 \lt 2\\ & \iff 1 \lt x \lt 5 \end
So the power series has radius of convergence \(2\) and interval of convergence \(1 \lt x \lt 5\text<.>\)
In the previous two examples, to construct a new series from an existing series, we replaced \(x\) by a simple function. The following theorem gives us some more (but certainly not all) commonly used substitutions.
Assume that the function \(f(x)\) is given by the power series
\[ f(x) = \sum_^\infty A_n x^n \nonumber \]
for all \(x\) in the interval \(I\text<.>\) Also let \(K\) and \(k\) be real constants. Then
\begin f\big(Kx^k\big) &= \sum_^\infty A_nK^n\, x^ \end
whenever \(Kx^k\) is in \(I\text<.>\) In particular, if \(\sum_^\infty A_n x^n\) has radius of convergence \(R\text\) \(K\) is nonzero and \(k\) is a natural number, then \(\sum_^\infty A_nK^n\, x^\) has radius of convergence \(\root\of\text<.>\)
Find a power series representation for \(\frac\text<.>\)
Solution: Once again the trick is to express \(\frac\) in terms of \(\frac\text<.>\) Notice that
Note that the \(n=0\) term has disappeared because, for \(n=0\text\)
\[ \frac x^n = \frac x^0 = \frac 1= 0 \nonumber \]
Also note that the radius of convergence of this series is one. We can see this via Theorem 3.5.13. That theorem tells us that the radius of convergence of a power series is not changed by differentiation — and since \(\sum_^\infty x^n\) has radius of convergence one, so too does its derivative.
Without much more work we can determine the interval of convergence by testing at \(x=\pm 1\text<.>\) When \(x=\pm 1\) the terms of the series do not go to zero as \(n \to \infty\) and so, by the divergence test, the series does not converge there. Hence the interval of convergence for the series is \(-1 \lt x \lt 1\text<.>\)
Notice that, in this last example, we differentiated a known series to get to our answer. As per Theorem 3.5.13, the radius of convergence didn't change. In addition, in this particular example, the interval of convergence didn't change. This is not always the case. Differentiation of some series causes the interval of convergence to shrink. In particular the differentiated series may no longer be convergent at the end points of the interval 8 Similarly, when we integrate a power series the radius of convergence is unchanged, but the interval of convergence may expand to include one or both ends, as illustrated by the next example.
Find a power series representation for \(\log (1+x)\text<.>\)
Solution: Recall that \(\frac\log (1+x) = \frac\) so that \(\log(1+t)\) is an antiderivative of \(\frac\) and
Theorem 3.5.13 guarantees that the radius of convergence is exactly one (the radius of convergence of the geometric series \(\sum_^\infty (-t)^n\)) and that
\begin \log(1+x) = \sum_^\infty (-1)^n\frac> \qquad\text\quad -1 \lt x \lt 1 \end
When \(x=-1\) our series reduces to \(\sum_^\infty \frac\text\) which is (minus) the harmonic series and so diverges. That's no surprise — \(\log(1+(-1)) =\log 0=-\infty\text<.>\) When \(x=1\text\) the series converges by the alternating series test. It is possible to prove, by continuity, though we won't do so here, that the sum is \(\log 2\text<.>\) So the interval of convergence is \(-1 \lt x\le 1\text<.>\)
Find a power series representation for \(\arctan x\text<.>\)
Solution: Recall that \(\frac\arctan x = \frac\) so that \(\arctan t\) is an antiderivative of \(\frac\) and
Theorem 3.5.13 guarantees that the radius of convergence is exactly one (the radius of convergence of the geometric series \(\sum_^\infty (-t^2)^n\)) and that
When \(x=\pm 1\text\) the series converges by the alternating series test. So the interval of convergence is \(-1\le x\le 1\text<.>\) It is possible to prove, though once again we won't do so here, that when \(x=\pm 1\text\) the series \(\sum_^\infty (-1)^n\frac>\) converges to the value of the left hand side, \(\arctan x\text\) at \(x=\pm 1\text<.>\) That is, to \(\arctan(\pm 1)=\pm \frac<\pi>\text<.>\)
The operations on power series dealt with in Theorem 3.5.13 are fairly easy to apply. Unfortunately taking the product, ratio or composition of two power series is more involved and is beyond the scope of this course 9 . Unfortunately Theorem 3.5.13 alone will not get us power series representations of many of our standard functions (like \(e^x\) and \(\sin x\)). Fortunately we can find such representations by extending Taylor polynomials 10 to Taylor series.
