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Unit VII Power Series
UNIT VII
Power Series
7.1 Power Series
Definition 7.1 A series of the form is called a power series in x and
is called a power series in x a, where a .
Now assume that x 0 = 1 x , so that
= + + + . . .
If = 0 n N, then the power series is a polynomial of degree at most N.
Example 1 x2 3 + 2x = , where = 3, = 2, = 1 and = 0 n 3.
and are examples of power series.
Note that: Every power series in x a, converges for x = a a .
Example 2 Show that converges only for x = 0.
Solution For x 0, = = .
Hence diverges for x 0.
Therefore, converges only for x = 0.
Example 3 Show that converges for every number x.
Solution For x 0, = = = 0.
Therefore, converges for every number x.
Example 4 Show that converges for 1.
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Solution For x 0, = = .
Hence by generalized ratio test converges for 1 and x 0.
Therefore converges for 1.
Moreover; , where x 0, is a geometric series and for 1, = .
Lemma 7.1
a) If converges, then converges absolutely for
.
b) If diverges, then diverges for .
Theorem 7.1 let be a power series. Then exactly one of the following
conditions hold.
a) converges only for x = 0.
b) converges for all x .
c) There is a number R 0 such that converges for
R and diverges for R.
Note that: R is called the radius of convergence of .
If converges only for x = 0, then we let R = 0 and if converges for x , then
we let R = .
The set of all values of x for which converges is called the interval of convergence
Example 5 Find the interval of convergence of .
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Solution For x 0, = = .
Hence converges for and for x = , diverges.
Therefore ( , ) is the interval of convergence of .
Example 6 Determine the interval of convergence of the series
a) b) c) d)
Solutions a) For x 0,
= = . Thus converges for
1.
For x = 1 we get and both of which diverge.
Therefore, ( 1, 1) is the interval of convergence of .
b) For x 0, = = .
Thus converges for 1.
For x = 1, converges and for x = 1, diverges.
Therefore, ( 1, 1] is the interval of convergence of .
c) For x 0, =
= = .
Thus converges for .
For x = , = converges
and for x = , = = diverges.
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Therefore, ( , ] is the interval of convergence of .
d) For x 2, , = =
.
Thus converges for 1 1 x 3.
For x = 1, = which converges and for x = 3,
= which converges.
Therefore, [1, 3] is the interval of convergence of .
Remark: The radius of convergence of is given by
R =
Differentiation of Power Series
Theorem 7.2 (Differentiation theorem for Power Series)
Let be a power series with radius of convergence R 0.
Then has the same radius of convergence and
= = for
R.
Example 7 Show that = .
Solution For x 0, = = 0.
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Thus converges for x and = = =
.
Therefore = x.
From the result of the above example we get:
f (x) = f (x) x, where f (x) = .
Since f (0) = 1, then we get f (x) = ex for all x.
Therefore ex = x.
Moreover = , e – x = and = etc.
Note that: Theorem 7.2 states that and have the same radius
of convergence but the interval of convergence of these series may not be the same.
Example 8 Consider .
Let f (x) = .
R = = 1 0 and diverges for x = 1 and converges for x = 1.
Hence [1, 1) is the interval of convergence of and f (x) = converges for 1
and diverges for 1.
Thus (1, 1) is the interval of convergence of .
Therefore, the two series have different interval of convergences.
Theorem 7.3 Suppose a power series has radius of convergence R 0. Let
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f (x) = for R x R.
Then f has derivatives of all orders on ( R, R) and f (n) (0) = n! cn for n 0.
Consequently
f (x) = for R x R.
Corollary 7.3.1 Let R 0 and suppose and be power series that
converge for R x R.
If = for R x R, then cn = bn for each n 0.
Integration of Power Series
Theorem 7.4 (Integration theorem for Power Series)
Let be a power series with radius of convergence R 0. Then
has the same radius of convergence and
= = for
R.
Example 9 Show that
ℓn (1 + x) = = for 1.
Solution Note that:
= for 1.
Replacing x by t we get:
= for 1.
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Hence ℓn (1 + x) = = = for 1.
Therefore, ℓn (1 + x) = for 1.
Since converges for 1 x 1,
ℓn (1 + x) = = for 1 x 1
Therefore, ℓn 2 = .
Note that: The power series expansion of ℓn (1 + x) for 1 x 1 is known
as Mercator’s Series.
Example 10 Show that tan 1 x = for 1.
Solution If 1, then t 2 1.
Hence = for 1 and tan 1 x = =
= for 1.
Therefore, tan 1 x = for 1.
Since converges for = 1, tan 1 x = for
1. Therefore
= .
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Note that: The power series expansion of tan 1 x for 1
is known as Gregory’s Series.
Example 11 Find a power series representation of x .
Solution Since e x = x , = .
Hence = = =
Therefore, = x .
Example 12 Find a power series representation of and use it to verify that = 1.
Solution Since = + = +
= + 1 + x + + . . . + + . . .
= 1 + + + + . . . =
Hence = and = = 1.
Therefore, = and = 1.
Example 13 Find the power series expansion of and use it to evaluate
Solution = + = +
= + 1 + x + + . . . + + . . .
= + + + . . . =
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Hence = and = = .
Therefore, = and = .
Example 14 Evaluate using power series expansion of .
Solution = for x 1.
Hence = x + + . . .
= 1 + + . . . =
and = = 1
Therefore, = and = 1.
Example 15 Find the power series representation of for 1.
Solution Since = for 1 and = .
But = = = .
Therefore, = for 1.
Taylor Series
Suppose f is a function defined on an open interval I containing 0 by
f (x) = x I.
Then we say that we have a power series representation of f on I. The value of f at x in I can be
approximated by the partial sum of the convergent power series of f provided that
converges for each x in I.
Remark: 1. the partial sums of a power series are polynomials.
2. if f has a power series representation, then
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i) f must have derivatives of all order on I.
ii) this representation is unique and written as
f (x) = x I.
Definition 7.2 Suppose that f has derivatives of all orders at a, then the Taylor series
of f about a is the power series
If a = 0, then this series is called the Maclaurin Series.
Remark: Any polynomial in x is its own Taylor series.
Definition 7.3 The nth Taylor polynomial Pn and the nth Taylor remainder Rn of f about a
are defined by
Pn (x) = and Rn (x) =
where tk is strictly between a and x. Moreover Rn (x) = f (x) Pn (x).
Therefore, f (x) = if and only if Rn (x) = 0.
Example 16 Show that sin x = for all x .
Solution Let f (x) = sin x for all x .
Then f (2k) (x) = ( 1) k sin x and f (2k + 1) (x) = ( 1) k cos x for all k 0 and k Z.
Thus Pn (x) = , where m = [ (n 1)], greatest integer not
exceeding (n 1) and Rn (x) = , where strictly lies between 0 and x.
Now to show that f(x) = Pn (x) we need to show that Rn (x) = 0.
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Since 1,
and hence = = 0
Thus Rn (x) = 0.
Therefore, sin x = for all x .
Example 17 Show that cos x = for all x .
Solution Let f (x) = cos x for all x .
Then g (2k) (x) = ( 1) k cos x and g (2k + 1) (x) = ( 1) k + 1 sin x for all k 0 and k Z.
Thus Pn (x) = , where m = [ n], greatest integer not exceeding n .
Since Rn (x) = , where strictly lies between 0 and x, and
1, = = 0
and Pn (x) = = .
Therefore, cos x = for all x .
Example 18 Find the fifth Taylor polynomial P5 of
a) sin x b) cos x
Solutions a) P5 (x) = sin 0 + x cos 0 + + + +
= x +
Therefore, P5 (x) = x + .
b) P5 (x) = cos 0 + x ( sin 0) + + + +
= 1 +
Therefore, P5 (x) = 1 + .
Taylor Series about an Arbitrary Point
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Definition 7.4 If f has derivatives of all orders at a , then we call
the Taylor Series of f about the number a.
Example 19 Consider the function g(x) = ℓn x.
g is not defined on an open interval about 0. Hence g does not have a Taylor series of the form
. But g(x) = ℓn (1 + x 1), and hence ℓn x for
1 x 1 1.
Therefore, ℓn x = for 0 x 2.
Example 20 Express the polynomial g(x) = 2x3 9x2 + 11x 1 as a polynomial in x a, where a .
Solution Since g(x) = 2x3 9x2 + 11x 1,
g(a) = 2a3 9a2 + 11a 1, g (a) = 6a2 18a + 11,
g (a) = 12a 18, g (a) = 12, and g (n) (a) = 0 for all n 4.
Therefore, g(x) = .
In particular if a = 2, then g(x) = 1 (x 2) + 3 (x 2)2 + 2(x 2)3.
Example 21 Find the Taylor series of cosh x about a, where a .
Solution Let g(x) = cosh x.
Then g (2n) (a) = cosh a and g (2n + 1) (a) = sinh a for all n 0.
Therefore, cosh x = .
Example 22 Using the Taylor series of cos x, approximate cos ( ) with an error less than 0.004.
Solution cos x = x .
We need to find the smallest value of n for which
= 0.004.
Since 1, where f(x) = cos x and tk is strictly between and x, for all n 0.
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0.004 (n + 1) ! n 6.
Hence cos ( )
= 1 + 0.2586
Therefore, cos ( ) 0.2586 with an error less than 0.004.
Example 23 Approximate e with an error less than 0.000003.
Solution Let f(x) = e x , since e x = , e = .
We need to find the smallest value of n for which
= 0.000003.
3 0.000003, where tk (0, 1) (n + 1) ! 1,000,000 n 9.
Hence = e.
Therefore, e 2.718281526 with an error less than 0.000003.
Example 23 Find the Taylor series expansion of f(x) = 2 x about a = 1.
Solution f(1) = 2, f (x) = 2 x ℓn 2, f (1) = 2 ℓn 2, f (x) = 2 x (ℓn 2) 2, f (1) = 2 (ℓn 2) 2
f (x) = 2 x (ℓn 2) 3, f (1) = 2 (ℓn 2) 3
In general f (n) (x) = 2 x (ℓn 2) n and f (n) (1) = 2 (ℓn 2) n for n 0.
Therefore, 2 x = for every number a.
Note that: From the result of the above example we get:
2 x = for any x .
Binomial Series
The Taylor series about 0 of the function f given by
f (x) = (1 + x)s ,
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where s is any fixed number is called a binomial series.
Now consider the Maclaurin series of f(x) = (1 + x)s, where s is any fixed number.
Let c0 = 1, c1 = s and cn = s (s 1) (s 2) … (s (n 1)) for all n 2.
Then f(n) (x) = cn (1 + x)s – n and f(n) (0) = cn .
Hence the Maclaurin series of f (x) = (1 + x)s is given by:
f (x) = (1 + x)s = .
In particular if s = , the Maclaurin series of
f (x) = = 1 + x +
= 1 + x +
Therefore, = 1 + x + .
Definition 7.5 Let s be any number. Then we define the binomial coefficient by the formula:
= 1, = s and = for n
2.
In particular = s (s 1) and if s N, then = .
Now using this definition the above Maclaurin series of f(x) = (1 + x)s is given by:
(1 + x)s =
This series is called a binomial series.
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