He radius of convergence of a power series is the radius of the circle of convergence to which the series converges. Said circle extends from the value that cancels the base of the powers to the closest singularity of the function associated to the series.
All analytical function f(z) has associated a power series about a non-singular point, called Taylor series:
Where to is the center of the circle of convergence, z the independent variable of the function and the cn are coefficients related to the derivatives of the function F on the point z=a.
convergence radius r is a positive real number that defines the region:
|z–a| < r
where the series converges. Outside this region, the series diverges, that is, it takes infinite values. When the radius of convergence is infinite, then the series converges in the entire complex plane.
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How is the radius of convergence determined?
For a series to be convergent, it is necessary that the absolute value of the successive terms decrease when the number of terms is very large. In mathematical form it would be expressed as follows:
Using the properties of the limits in the previous expression, we obtain:
Here r is the radius of convergence and |z–a| < r is the open boundary circle in the complex plane where the series converges. In case the value to and the variable z are real numbers, then the open interval of convergence on the real axis will be: (a – r, a + r).
Taylor series
The Taylor series of a function f(x) around a value to in which the function has infinitely many derivatives, is a power series defined as:
In the environment | x–a | < rwith r as the radius of convergence of the series, we have that the Taylor series and the function f(x) match.
On the other hand, the radius of convergence r is the distance from the point to and the singularity x’s closest to the point tobeing the singular points those values where the limit of the function tends to infinity.
That is to say, when x → x’s so f → ±∞.
examples
Example 1
Be S(x) the power series given by the following expression:
S(x) = 1 – x + x2– x3+ x4– …….+(-1)n ⋅ xn + ….
To determine the region where the series converges, we calculate the quotient between the (nth + 1) term and the (nth) term:
The absolute value of the above quotient is |x| and its limit when n → ∞ it is also |x|.
For the series to be convergent it is necessary that:
Then the radius of convergence of this series is r=1since it converges for values of x that are at a distance less than 1 from the center x = 0.
Example 2
We want to find the Taylor series of the function f(x) = 1 / (1 + x) around the point x=0 and determine its radius of convergence.
To find the series we take the successive derivatives of the function f(x), of which we will show the first three:
Taking into account that the zero order term of the Taylor series is:
f(0)=1,
The first order: f'(0)/1!
Second order:
f»(0)/2!
Third order:
f»'(0)/3!
And so on, we have that the Taylor series of the given function is:
f(x) = 1 – x + x2 – x3 + x4 – …….+(-1)n ⋅ xn + ….
Which coincides with the power series studied in Example 1.
We have already said that the radius of convergence of a Taylor series is the distance from the center of the series expansion, which in our case is the value x=0 up to the first singularity of the function f(x).
Since our function has a singularity (i.e., an infinity) at x = -1the distance between the value -1 and the expansion center 0 is |-1 – 0| = 1we conclude that the radius of convergence of the Taylor series is 1.
This result fully coincides with that obtained in example 1 by another method.
The fact that the zone of convergence of the Taylor series is the open interval (-1, 1) implies that the function and the series coincide in this interval, but not outside it.
This is shown in figure 2, where 41 terms have been taken from the Taylor series, drawn by the solid blue line, while the original function is shown by the dashed red line.
solved exercises
– Exercise 1
Consider the same function f(x) = 1 / (1 + x) from example 2, but this time we are asked to find the Taylor series of said function around the point a = 1.
Solution
We find the successive coefficient terms of the series, beginning with the independent term that is f(1) = ½.
The next coefficient that corresponds to the first order term is:
f'(1)/1! = -¼
The second order is:
f»(1)/2! = 2/(23 2!)
The third-order coefficient follows:
f»'(1)/3! = -6 / (24 3!)
And so on. The Taylor series will be:
Sf(x) = ½ – 1/22 (x-1) + 1/23(x-1)2 – 1/24 (x-1)3 + 1/25 (x-1)4– …..
– Exercise 2
Find the radius of convergence of the previous series
Solution
We write the nth term and the nth term plus one:
We calculate the quotient of these two terms which is shown below in simplified form:
The absolute value of the previous expression is taken, obtaining:
| x–1 | / 2
However, for the series to be convergent, it is necessary for the previous quantity to be strictly less than unity, that is:
| x–1 | < 2
Which tells us that the radius of convergence around the value x=1 is:
r = 1
On the other hand, the previous expression is equivalent to the double inequality:
-2 < x – 1 < +2
If we add +1 to each of the three members of the previous expression we obtain:
-1 < x < 3
What is the interval of convergence of the series?
Figure 1 shows the original function and the Taylor series of that function around the point x=1. In the figure it can be verified that the series coincides with the function in the neighborhood of the point x=1, but within the radius of convergence.
References
CK-12 Foundation. Power Series: representation of functions and operations. Retrieved from: ck12.org.
Engler, A. 2019. Integral Calculus. National University of the Coast.
Larson, R. 2010. Calculus of a variable. 9na. Edition. McGraw Hill.
Mathematics Free Texts. power series. Retrieved from: math.liibretexts.org.
Wikipedia. Power series. Recovered from: es.wikipedia.org.
Wikipedia. Radius of convergence. Retrieved from: en.wikipedia.org