POWER SERIES: EXAMPLES AND EXERCISES - MATHS - 2025

A power series consists of a summation of terms in the form of powers of the variable x, or more generally, of xc, where c is a constant real number. In summation notation a series of powers is expressed as follows:

Where the coefficients a _o, a ₁, a ₂ … are real numbers and the series begins at n = 0.

Figure 1. Definition of a power series. Source: F. Zapata.

This series is centered on the value c that is constant, but you can choose that c is equal to 0, in which case the power series simplifies to:

The series start with a _or (xc) ⁰ and a _or x ⁰ respectively. But we know that:

(xc) ⁰ = x ⁰ = 1

Therefore a _o (xc) ⁰ = a _or x ⁰ = a _o (independent term)

The good thing about power series is that functions can be expressed with them and this has many advantages, especially if you want to work with a complicated function.

When this is the case, instead of directly using the function, use its power series expansion, which can be easier to derive, integrate, or work numerically.

Of course everything is conditioned to the convergence of the series. A series converges when adding a certain large number of terms gives a fixed value. And if we add more terms still, we continue to obtain that value.

Functions as Power Series

As an example of a function expressed as a power series, let's take f (x) = e ^x.

This function can be expressed in terms of a series of powers as follows:

and ^x ≈ 1 + x + (x ² /2!) + (x ³ /3!) + (x ⁴ /4!) + (x ⁵ /5!) +…

Where! = n. (n-1). (n-2). (n-3)… and it takes 0! = 1.

We are going to check with the help of a calculator, that indeed the series coincides with the function given explicitly. For example let's start by making x = 0.

We know that e ⁰ = 1. Let's see what the series does:

and ⁰ ≈ 1 + 0 + (0 ² /2!) + (0 ³ /3!) + (0 ⁴ /4!) + (0 ⁵ /5!) +… = 1

And now let's try x = 1. A calculator returns that e ¹ = 2.71828, and then let's compare with the series:

and ¹ ≈ 1 + 1 + (1 ² /2!) + (1 ³ /3!) + (1 ⁴ /4!) + (1 ⁵ /5!) +… = 2 + 0.5000 + 0.1667 + 0.0417 + 0.0083 +… ≈ 2.7167

With only 5 terms we already have an exact match in e ≈ 2.71. Our series has just a little more to go, but as more terms are added, the series certainly converges to the exact value of e. The representation is exact when n → ∞.

If the previous analysis is repeated for n = 2, very similar results are obtained.

In this way we are sure that the exponential function f (x) = e ^x can be represented by this series of powers:

Figure 2. In this animation we can see how the power series gets closer to the exponential function as more terms are taken. Source: Wikimedia Commons.

Geometric series of powers

The function f (x) = e ^x is not the only function that supports a power series representation. For example, the function f (x) = 1/1 - x looks a lot like the well-known convergent geometric series:

It is enough to do a = 1 and r = x to obtain a series suitable for this function, which is centered at c = 0:

However, it is known that this series is convergent for │r│ <1, therefore the representation is valid only in the interval (-1,1), although the function is valid for all x, except x = 1.

When you want to define this function in another range, you simply focus on a suitable value and you are done.

How to find the series expansion of powers of a function

Any function can be developed in a power series centered on c, as long as it has derivatives of all orders at x = c. The procedure makes use of the following theorem, called Taylor's theorem:

Let f (x) be a function with derivatives of order n, denoted as f ⁽ⁿ⁾, which admits a series expansion of powers on the interval I. Her serial development of Taylor is:

So that:

Where R _n, which is the nth term of the series, is called a remainder:

When c = 0 the series is called the Maclaurin series.

This series given here is identical to the series given at the beginning, only now we have a way to explicitly find the coefficients of each term, given by:

However, we must ensure that the series converges to the function to be represented. It happens that not every Taylor series necessarily converges to the f (x) that was had in mind when calculating the coefficients at _n.

This happens because perhaps the derivatives of the function, evaluated at x = c coincide with the same value of the derivatives of another, also at x = c. In this case the coefficients would be the same, but the development would be ambiguous as it is not certain which function it corresponds to.

Fortunately there is a way to know:

Convergence criterion

To avoid ambiguity, if R _n → 0 as n → ∞ for all x in the interval I, the series converges to f (x).

Exercise

- Exercise resolved 1

Find the geometric power series for the function f (x) = 1/2 - x centered at c = 0.

Solution

The given function must be expressed in such a way that it coincides as closely as possible with 1 / 1- x, whose series is known. So let's rewrite numerator and denominator, without altering the original expression:

1/2 - x = (1/2) /

Since ½ is constant, it comes out of the summation, and it is written in terms of the new variable x / 2:

Note that x = 2 does not belong to the domain of the function, and according to the convergence criterion given in the Geometric Power Series section, the expansion is valid for │x / 2│ <1 or equivalently -2 <x <2.

- Exercise solved 2

Find the first 5 terms of the Maclaurin series expansion of the function f (x) = sin x.

Solution

Step 1

First are the derivatives:

-Derivative of order 0: it is the same function f (x) = sin x

-First derivative: (sin x) ´ = cos x

-Second derivative: (sin x) ´´ = (cos x) ´ = - sin x

-Third derivative: (sin x) ´´´ = (-sen x) ´ = - cos x

-Fourth derivative: (sin x) ´´´´ = (- cos x) ´ = sin x

Step 2

Then each derivative is evaluated at x = c, as is a Maclaurin expansion, c = 0:

sin 0 = 0; cos 0 = 1; - sin 0 = 0; -cos 0 = -1; sin 0 = 0

Step 3

The coefficients a _{n are constructed};

a _o = 0/0! = 0; a ₁ = 1/1! = 1; a ₂ = 0/2! = 0; a ₃ = -1 / 3 !; a ₄ = 0/4! = 0

Step 4

Finally the series is assembled according to:

sin x ≈ 0.x ⁰ + 1. x ¹ + 0.x ² - (1/3!) x ³ + 0.x ⁴ … = x - (1/3!)) x ³ +…

Does the reader need more terms? How many more, the series is closer to the function.

Note that there is a pattern in the coefficients, the next non-zero term is ₅ and all those with an odd index are also different from 0, alternating the signs, so that:

sin x ≈ x - (1/3!)) x ³ + (1/5!)) x ⁵ - (1/7!)) x ⁷ +….

It is left as an exercise to check that it converges, the quotient criterion can be used for the convergence of series.

References

1. CK-12 Foundation. Power Series: representation of functions and operations. Recovered from: ck12.org.
2. Engler, A. 2019. Integral Calculus. National University of the Litoral.
3. Larson, R. 2010. Calculation of a variable. 9th. Edition. McGraw Hill.
4. Mathematics Free Texts. Power series. Recovered from: math.liibretexts.org.
5. Wikipedia. Power series. Recovered from: es.wikipedia.org.