Definition of a Power Series
We now investigate a generalization of polynomials. The polynomial
below is a fifth degree polynomial
3x5 - 2x4 + 5x3 + 4x - 1
instead of a fifth degree polynomial, we consider a polynomial of infinite
Definition of a Power Series
Let f(x) be the function represented by the series
Then f(x) is called a power series function.
More generally, if f(x) is represented by the series
Then we call f(x) a power series centered at x =
The power series
centered at 0.
centered at -2.
The Radius of Convergence
Next we want to investigate the domain of power series. Recall to find
the domain, we ask what values of x can the
function handle? This is particularly important with power series, since
infinite series often do not converge. It would be an insurmountable task
to plug in each value of x and see if the series
converges for that value. Fortunately, we have the following theorem.
a power series centered at c
then only the following three are possibilities for the domain of f.
domain is the value c
domain is all real numbers.
exists a real number R
such that all values of x
| x - c | < R
are in the domain and values that satisfy
| x - c | > R
are not in the domain.
is called the radius of convergence of f.
To compute the radius of convergence, we use the ratio test.
Example: Find the radius of convergence of
We use the Ratio Test:
|x - 3| < 2
1 < x < 5
1/2(5 - 1) = 2
the radius of convergence is 2
Find the radius of convergence of
Taylor and Maclaurin Series
Since power series are functions, a natural question to ask is, "Can our
everyday functions be represented as power series?" Also, "Given
a power series, can we find an everyday function that is equivalent to the power
The following definition helps to answer these questions.
The Taylor Series
centered at x = c is
= 0, then the series is called
the Maclaurin series for f.
We use the notation f
(n) to denote
the nth derivative of f.
Find the McLaurin Series for
f(0) = 1
f '(0) = 0 f ''(0) = -1 f (3)(0) = 0
= 1 f (5)(0) = 0 f (6)(0) =
-1 f (7)(0) = 0
= 1 f (9)(0) = 0
f (10)(0) =
-1 f (11)(0) = 0
Hence we have the series
1 - x2 /2+ x4/4! - x6/6! +
x8/8! - x8/8! + ...
We see that the series is
Exercises Find the Taylor series expansion for
sin(x) centered at x = 0
ln(x) centered at x = 0
Differentiation and Integration of Power Series
Since a power series is a function, it is natural to ask if the function is
continuous, differentiable or integrable. The following theorem
answers this question.
Suppose that a function is given by the power series
f(x) =San(x - c)n
and that the interval of convergence is
(c - R,c + R) (plus possible endpoints)
then f(x) is continuous, differentiable, and integrable on that interval
(not necessarily including the endpoints). To obtain the derivative or the integral of
f(x) we can pass the derivative or integral through the S. In other
Furthermore, the radius of convergence for the derivative and integral is R.
Consider the series
f(x) = Sxn
by the GST this series converges for |x| <
1, hence the center of convergence is 0 and the radius is
By the above theorem
f '(x) = Snxn-1
has center of convergence 0 and radius of convergence
1 also. We can also say that
also has center of convergence 0 and radius of convergence
satisfies the differential equation
y'' + xy' - y = 0
We can also use substitution to find power series.
Find the Maclaurin series for
1 - x2
Substituting x2 for x
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