A power series consists of a summation of terms in the form of powers of the variable x, or more generally, of x-c, where c is constant real number. In summation notation, a series of powers is expressed as follows:
∑an (x -c)n = aor + to1 (x - c) + atwo (x - c)two + to3 (x - c)3 +… + An (x - c)n
Where the coefficients aor, to1, totwo… They are real numbers and the series starts at n = 0.
This series is focused on value c which is constant, but you can choose which c is equal to 0, in which case the power series simplifies to:
∑an xn = aor + to1 x + atwo xtwo + to3 x3 +… + An xn
The series begin with toor(x-c)0 Y toorx0 respectively. But we know that:
(x-c)0= x0 = 1
Therefore toor(x-c)0 = toorx0 = toor (independent term)
The good thing about power series is that with them you can express functions 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 expansion in power series is used, 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.
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As an example of a function expressed as a power series, let's take f (x) = ex.
This function can be expressed in terms of a series of powers as follows:
andx ≈ 1 + x + (xtwo / 2!) + (X3 / 3!) + (X4 / 4!) + (X5 / 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 e0 = 1. Let's see what the series does:
and0 ≈ 1 + 0 + (0two / 2!) + (03 / 3!) + (04 / 4!) + (05 / 5!) +… = 1
And now let's try with x = 1. A calculator shows that and1 = 2.71828, and then let's compare with the series:
and1 ≈ 1 + 1 + (1two / 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 has just a little more to do, 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 to n = 2 very similar results are obtained.
In this way we are sure that the exponential function f (x) = ex can be represented by this series of powers:
The function f (x) = ex it 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:
∑a.rn = a / 1 - r
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, por lo tanto la representación es válida únicamente en el intervalo (-1,1), aunque la función sea válida para todo x, excepto x=1.
When you want to define this function in another interval, you simply focus on a suitable value and you are done..
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 in the interval I. Its development in taylor series it is:
So that:
f (x) = f (c) + f '(c) (x-c) + f "(c) (x-c)two / 2 + f "(c) (x-c)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 must be ensured 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 ton.
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 as it is not certain which function it corresponds to..
Fortunately there is a way to know:
Convergence criterion
To avoid ambiguity, if Rn → 0 when n → ∞ for all x in the interval I, the series converges to f (x).
Find the Geometric Power Series for the Function f (x) = 1/2 - x centered at c = 0.
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. Therefore let's rewrite numerator and denominator, without altering the original expression:
1/2 - x = (1/2) / [1 - (x / 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 section Geometric power series, the expansion is valid for │x / 2│< 1 o equivalentemente -2 < x < 2.
Find the first 5 terms of the Maclaurin series expansion of the function f (x) = sin x.
Derivatives are first found:
-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
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
The coefficients a are constructedn;
toor = 0/0! = 0; to1 = 1/1! = 1; totwo = 0/2! = 0; to3 = -1 / 3 !; to4 = 0/4! = 0
Finally the series is assembled according to:
sin x ≈ 0.x0 + 1. x1 + 0 .xtwo - (1/3!) X3 + 0.x4… = X - (1/3!)) X3 +...
Does the reader need more terms? How many more, the series gets 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 the 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 quotient criterion for series convergence.
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