A **power series** consists of a sum 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 power series is expressed as follows:

*∑an (x -c)n = ao + a1 (x – c) + a2 (x – c)2 + a3 (x – c)3 + …+ an (x – c)n*

Where the coefficients ao, a1, a2… are real numbers and the series starts at n = 0.

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

*∑an xn = ao + a1 x + a2 x2 + a3 x3 + …+ an xn*

The strings start with *ao*(xc)0 y *ao*x0 respectively. But we know that:

(xc)0=x0 = 1

Therefore *ao*(xc)0 = *ao*x0 = *ao* (independent term)

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

When this is the case, instead of using the function directly, its power series expansion is used, which may 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 results in a fixed value. And if we add even more terms, we still get that value.

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**Functions as Power Series**

As an example of a function expressed as a power series, let us take *f(x)** = former*.

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

ex ≈ 1 + x + (x2 / 2!) + (x3 / 3!) + (x4 / 4!) + (x5 / 5!) + …

Where! = no. (n-1). (n-2). (n-3)… and 0 is taken! = 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 e0 = 1. Let’s see what the series does:

e0 ≈ 1 + 0 + (02 / 2!) + (03 / 3!) + (04 / 4!) + (05 / 5!) + … = 1

And now let’s try *x = 1*. A calculator shows that *e1 = 2.71828*and then we compare with the series:

e1 ≈ 1 + 1 + (12 / 2!) + (13 / 3!) + (14 / 4!) + (15 / 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 is just a little further away, but as more terms are added, the series certainly converges to the exact value of *and*. The representation is exact when *n → ∞*.

If the above analysis is repeated for *n=2* very similar results are obtained.

Thus we are sure that the exponential function *f(x) = former* It can be represented by this power series:

**geometric power series**

The function *f(x) = former *is not the only function that admits a power series representation. For example, the function *F*(*x) =1 / 1 – x* looks a lot like the familiar *convergent geometric series*:

*∑a.rn = a / 1 – r*

It is enough to make a = 1 and r = x to obtain a suitable series 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 interval, you simply center it on a suitable value and you’re done.

**How to find the power series expansion of a function**

Any function can be expanded into a power series centered at 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 *no*denoted as *f(n)*which admits a power series expansion in the interval *Yo*. Its development in *taylor series* is:

So that:

*f(x) = f(c) + f´(c) (xc) + f´´(c) (xc)2 /2 + f´´´(c) (xc)3 /6 +…Rn*

Where Rn, which is the nth term of the series, is called *residue*:

When c = 0 the series is called *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, it is necessary to 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 *still*.

This happens because perhaps the derivatives of the function, evaluated in *x=c *coincide with the same value of the derivatives of another, also in *x=c*. In this case, the coefficients would be the same, but the development would be ambiguous since there is no certainty of which function it corresponds to.

Fortunately there is a way to know:

**convergence criterion**

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

**Exercise**

**– Solved exercise 1**

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

**Solution**

We must express the given function 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) / [1 – (x/2)]

Since ½ is constant, it leaves the summation, and this 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 section *geometric power series*the expansion is valid for │x/2│< 1 or equivalently -2 < x < 2.

**– Solved exercise 2**

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

**Solution**

**Step 1**

First find 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)´´´ = (-sin x)´ = – cos x

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

**Step 2**

Then each derivative is evaluated at x = c, as it 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 an are constructed;

ao = 0 / 0! = 0; a1 = 1 / 1! = 1 ; a2 = 0 / 2! = 0; a3 = -1/3!; a4 = 0 / 4! = 0

**Step 4**

Finally the series is assembled according to:

sin x ≈ 0.x0 + 1. x1 + 0 .x2 – (1/3!)x3 + 0.x4… = x – (1/3!))x3 + …

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 a5 and all those with odd index are also different from 0, alternating signs, so that:

sin x ≈ x – (1/3!))x3 + (1/5!))x5 – (1/7!))x7 + ….

It is left as an exercise to check that it converges, you can use the *ratio criterion* for the convergence of series.

**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.