P (x) = a0+a1x + a2x2+a3x3+· · · ,
wherea0, a1, a2, a3 etc. are constants. Such series regularly occur in physics and engineering and are useful because, for|x| < 1, the later terms in the series may become very small and be discarded. For example the series
P (x) = 1 + x + x2+x3+· · · ,
although in principle infinitely long, in practice may be simplified ifx happens to have a value small compared with unity. To see this note that P (x) for x = 0.1 has the following values: 1, if just one term is taken into account; 1.1, for two terms; 1.11, for three terms; 1.111, for four terms, etc. If the quantity that it represents can only be measured with an accuracy of two decimal places, then all but the first three terms may be ignored, i.e. whenx = 0.1 or less
P (x) = 1 + x + x2+ O(x3)≈ 1 + x + x2.
This sort of approximation is often used to simplify equations into manageable forms. It may seem imprecise at first but is perfectly acceptable insofar as it matches the experimental accuracy that can be achieved.
The symbols O and ≈ used above need some further explanation. They are used to compare the behaviour of two functions when a variable upon which both functions depend tends to a particular limit, usually zero or infinity (and obvious from the context). For two functionsf(x) and g(x), with g positive, the formaldefinitions of the above symbols are as follows:
(i) If there exists a constantk such that|f| ≤ kg as the limit is approached thenf = O(g).
(ii) If as the limit ofx is approached f/g tends to a limit l, where l= 0, then f≈ lg. The statement f ≈ g means that the ratio of the two sides tends to unity.
4.5.1 Convergence of power series
The convergence or otherwise of power series is a crucial consideration in practical terms. For example, if we are to use a power series as an approximation, it is clearly important that it tends to the precise answer as more and more terms of the approximation are taken. Consider the general power series
P (x) = a0+a1x + a2x2+· · · .
Using d’Alembert’s ratio test (see subsection 4.3.2), we see thatP (x) converges absolutely if
ρ = lim
n→∞
an+1 an x
= |x| limn→∞
an+1 an
< 1.
Thus the convergence ofP (x) depends upon the value of x, i.e. there is, in general, a range of values ofx for which P (x) converges, an interval of convergence. Note that at the limits of this rangeρ = 1, and so the series may converge or diverge.
The convergence of the series at the end-points may be determined by substituting these values ofx into the power series P (x) and testing the resulting series using any applicable method (discussed in section 4.3).
4.5 POWER SERIES
Determine the range of values of x for which the following power series converges:
P (x) = 1 + 2x + 4x2+ 8x3+· · · .
By using the interval-of-convergence method discussed above, ρ = lim
n→∞
2n+1 2n x
= |2x|,
and hence the power series will converge for|x| < 1/2. Examining the end-points of the interval separately, we find
P (1/2) = 1 + 1 + 1 +· · · , P (−1/2) = 1 − 1 + 1 − · · · .
ObviouslyP (1/2) diverges, while P (−1/2) oscillates. Therefore P (x) is not convergent at either end-point of the region but is convergent for−1 < x < 1.
The convergence of power series may be extended to the case where the parameterz is complex. For the power series
P (z) = a0+a1z + a2z2+· · · , we find thatP (z) converges if
ρ = lim
n→∞
an+1 an
z
= |z| limn→∞
an+1 an
< 1.
We therefore have a range in|z| for which P (z) converges, i.e. P (z) converges for values ofz lying within a circle in the Argand diagram (in this case centred on the origin of the Argand diagram). The radius of the circle is called the radius of convergence: ifz lies inside the circle, the series will converge whereas ifz lies outside the circle, the series will diverge; if, though, z lies on the circle then the convergence must be tested using another method. Clearly the radius of convergenceR is given by 1/R = limn→∞|an+1/an|.
Determine the range of values of z for which the following complex power series converges:
P (z) = 1−z 2+z2
4 −z3 8 +· · · .
We find thatρ =|z/2|, which shows that P (z) converges for |z| < 2. Therefore the circle of convergence in the Argand diagram is centred on the origin and has a radiusR = 2.
On this circle we must test the convergence by substituting the value ofz into P (z) and considering the resulting series. On the circle of convergence we can writez = 2 exp iθ.
Substituting this intoP (z), we obtain P (z) = 1−2 expiθ
2 +4 exp 2iθ 4 − · · ·
= 1− exp iθ + [exp iθ]2− · · · ,
which is a complex infinite geometric series with first term a = 1 and common ratio
r =− exp iθ. Therefore, on the the circle of convergence we have P (z) = 1
1 + expiθ.
Unlessθ = π this is a finite complex number, and so P (z) converges at all points on the circle|z| = 2 except at θ = π (i.e. z = −2), where it diverges. Note that P (z) is just the binomial expansion of (1 +z/2)−1, for which it is obvious thatz =−2 is a singular point.
In general, for power series expansions of complex functions about a given point in the complex plane, the circle of convergence extends as far as the nearest singular point. This is discussed further in chapter 24.
Note that the centre of the circle of convergence does not necessarily lie at the origin. For example, applying the ratio test to the complex power series
P (z) = 1 +z− 1
2 +(z− 1)2
4 +(z− 1)3 8 +· · · ,
we find that for it to converge we require|(z − 1)/2| < 1. Thus the series converges for z lying within a circle of radius 2 centred on the point (1, 0) in the Argand diagram.
4.5.2 Operations with power series
The following rules are useful when manipulating power series; they apply to power series in a real or complex variable.
(i) If two power seriesP (x) and Q(x) have regions of convergence that overlap to some extent then the series produced by taking the sum, the difference or the product ofP (x) and Q(x) converges in the common region.
(ii) If two power series P (x) and Q(x) converge for all values of x then one series may be substituted into the other to give a third series, which also converges for all values ofx. For example, consider the power series expansions of sin x and exgiven below in subsection 4.6.3,
sinx = x−x3 3!+x5
5! −x7 7!+· · · ex= 1 +x +x2
2! +x3 3!+x4
4! +· · · ,
both of which converge for all values of x. Substituting the series for sin x into that forexwe obtain
esinx= 1 +x +x2 2! −3x4
4! −8x5 5! +· · · , which also converges for all values ofx.
If, however, either of the power seriesP (x) and Q(x) has only a limited region of convergence, or if they both do so, then further care must be taken when substituting one series into the other. For example, suppose Q(x) converges for all x, but P (x) only converges for x within a finite range. We may substitute
4.5 POWER SERIES
Q(x) into P (x) to obtain P (Q(x)), but we must be careful since the value of Q(x) may lie outside the region of convergence forP (x), with the consequence that the resulting seriesP (Q(x)) does not converge.
(iii) If a power seriesP (x) converges for a particular range of x then the series obtained by differentiating every term and the series obtained by integrating every term also converge in this range.
This is easily seen for the power series
P (x) = a0+a1x + a2x2+· · · ,
which converges if|x| < limn→∞|an/an+1| ≡ k. The series obtained by differenti-atingP (x) with respect to x is given by
dP
dx =a1+ 2a2x + 3a3x2+· · · and converges if
|x| < limn→∞
nan (n + 1)an+1
= k.
Similarly the series obtained by integratingP (x) term by term,
P (x) dx = a0x +a1x2 2 +a2x3
3 +· · · , converges if
|x| < lim
n→∞
(n + 2)an
(n + 1)an+1
= k.
So, series resulting from differentiation or integration have the same interval of convergence as the original series. However, even if the original series converges at either end-point of the interval, it is not necessarily the case that the new series will do so. The new series must be tested separately at the end-points in order to determine whether it converges there. Note that although power series may be integrated or differentiated without altering their interval of convergence, this is not true for series in general.
It is also worth noting that differentiating or integrating a power series term by term within its interval of convergence is equivalent to differentiating or integrating the function it represents. For example, consider the power series expansion of sinx,
sinx = x−x3 3!+x5
5! −x7
7!+· · · , (4.14)
which converges for all values of x. If we differentiate term by term, the series becomes
1−x2 2! +x4
4!−x6 6! +· · · , which is the series expansion of cosx, as we expect.