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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.

Prepared by Tekleyohannes Negussie128



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 .




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.



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.



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



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.



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

= .



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 + + . . . + + . . .

= + + + . . . =



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



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.



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



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.



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 ,



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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