MA4402 Computer Mathematics - University of Limerick Computer Mathematics Natalia Kopteva Autumn...

MA4402 Computer Mathematics

http://www.staff.ul.ie/natalia/MA4402.html

Natalia Kopteva

Autumn 2012

MA4402 (Autumn 2012) Computer Mathematics 0 / 12

Part 3 Sequences, Series, Power Series §3.0Motivation

One MA4402 Learning Outcome:

Employ sequences and series for an efficient representation ofmathematical functions in algorithms.

Part 3 Sequences, Series, Power Series 1 / 19

MOTIVATION:

Computers, Calculators, Phones are all capable of performingbasic arithmetic operations cheaply and accurately.By basic arithmetic I mean +, −, × and /.Most of these devices can do much more than this. They cancalculate sin x and cos x , log x and ex .But HOW???NOTE: Evaluating these functions is a non-trivial task.Since sin x and cos x ... are not basic arithmetic operations, thecomputer must approximate them somehow.Moreover, whatever method is used to approximate them, it mustrely entirely on the four basic operations.

NOTE: There are a number of approaches to approximate thesefunctions, but we shall consider only one of them:

Power Series Representationof functions such as sin x and cos x

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .To prepare for this, we shall first consider Sequences and Series.


§3.1 Sequences

DefinitionA sequence is an finite/infinite list of terms (or numbers) arranged in adefinite order, that is, there is a rule by which each term after the firstmay be found.

EXAMPLES:

1 {1,2,3,4,5, . . .}NOTE: the . . . at the end of a sequence indicates that we have aninfinite number of terms.

2 {1,4,9,16,25, . . .}NOTE: Denote the nth term by an.

Then a1 = 11 = 1, a2 = 22 = 4, a3 = 32 = 9...So this sequence can be described by the formula an = n2.


3 Let a sequence be defined by the formula of its nth term:

an =

(1 +

1n

)n

.

Then we get a1 =

(1 +

11

)1

= 2, a2 =

(1 +

12

)2

=(3

2

)2,...

So we get

{2,

(32

)2

,

(43

)3

,

(54

)4

, . . .

}.

4 The Fibonacci sequence is defined as followsa1 = a2 = 1, an+2 = an + an+1.

From this definition we get: {1,1,2,3,5,8,13,21,34, . . .}.


NOTATION: A sequence is frequently written as {a1,a2,a3,a4, . . .}.

Shorthand for such a sequence is {an}∞n=1

Another EXAMPLE: List the first 4 terms of the following sequence:an = n2 + 1.

Solution: The first four terms are a1, a2, a3 and a4.Therefore using the rule we get

a1 = (1)2 + 1 = 2; a2 = (2)2 + 1 = 5; a3 = (3)2 + 1 = 10;

a4 = (4)2 + 1 = 17.


NOTE: one can use computer languages/packages to evaluate + plotgraphs of sequences. For example, the package Maple for an = 1 + 1

n :


Same example an = 1 + 1n , but now with a graph:


Bounded Sequences

DefinitionThe sequence {an} has a lower bound L if an > L for all n.

Examples:(i) The Fibonacci sequence has lower bound 1 as an > 1 for all n > 1.

(ii)The sequence {n2} has lower bound 1 as n2 > 1 for all n > 1.

DefinitionThe sequence {an} has an upper bound M if an 6 M for all n.

Example:The sequence {1 − 1

n } has upper bound 1 as 1 − 1n 6 1 for all n > 1.


DefinitionThe sequence {an} is boundedif it has a lower bound L and an upper bound M

so L 6 an 6 M for all n.

Examples:

(i) an = sin(

n +1n

).

The sequence {an} is bounded as for all n > 1:

−1 6 sin(

n +1n

)6 1.

(ii) The sequence {2−n}∞n=1 is the sequence {12 ,

14 ,

18 , . . .}.

So it is bounded: 0 6 2−n 6 1.


Another EXAMPLE:

an =

{1 + 1

2n , if n is odd,

2n, if n is even.

From this definition we have:

a1 = 1 + 12 = 3

2 (here n = 1, odd);

a2 = 22 = 4 (here n = 2, even);

a3 = 1 + 123 = 1 + 1

8 = 98 (here n = 3, odd);

a4 = 24 = 16 (here n = 4, even); etc.

So we get the sequence: {32 ,4,

98 ,16, . . .}

This sequence has a lower bound 1 (as 1 + 12n > 1 and 2n > 1); but no

upper bound,

so it is not bounded.


Increasing/Decreasing Sequences, etc

DefinitionThe sequence {an} is

positive if an > 0 for all n.

negative if an 6 0 for all n.

increasing if an+1 > an for all n.

decreasing if an+1 < an for all n.

monotonic if it is either increasing or decreasing.

alternating if anan+1 < 0 for all n. i.e., consecutive terms haveopposite signs.


NOTE: For a positive sequence, one can—replace the criterion an+1 > an by the equivalent an+1

an> 1;

—and the criterion an+1 < an by the equivalent an+1an

< 1.

EXAMPLES (Increasing/Decreasing Sequences)1 The sequence {an = n}∞n=1 is increasing

as an+1 − an = (n + 1) − n = 1 > 0.

2 The sequence {an = 1 − n}∞n=1 is decreasingas an+1 − an = (1 − (n + 1)) − (1 − n) = −1 < 0.

3 The positive sequence {an = 2n}∞n=1 is increasingas an+1

an= 2n+1

2n = 2 > 1.

4 The sequence {an}∞n=1 with an = 1+

(−1)n

n is neither increasing nordecreasingas we have {0, 3

2 ,23 ,

54 , . . .}.

(This sequence keeps bouncing backwards and forwards around1 and so neither increases nor decreases.)Part 3 Sequences, Series, Power Series 12 / 19

Limits of Sequences

DefinitionAs n→∞, the terms an may approach some number L.

Then we say that the sequence {an} converges to the limit L and writelim

n→∞an = L.

The series is said to be convergent.

Example: limn→∞ 1

n= 0; lim

n→∞(5 −

1n2

)= 5.


DefinitionIf the sequence {an} does not converge to a finite limit, it is said to bedivergent.

Example: If an = n, then {an} diverges (as it goes to +∞.)


Further EXAMPLES:

1 If an = −2n, then {an} diverges (as it goes to −∞).

2 If an = (−1)n, then {an} = {−1,+1,−1,+1 . . .} just diverges.


Elementary Properties of Convergent SequencesSuppose lim

n→∞an = a and limn→∞bn = b.

(i) If cn = an + bn, then limn→∞ cn = a + b.

(ii) If cn = an × bn, then limn→∞ cn = a× b.

(iii) If cn = anbn

and b 6= 0, then limn→∞ cn =

ab

.

Example:Set an = 1 + 1

n and bn = 5 − 2n and check the above properties (i)–(iii).

NOTE: these properties will be used in all remaining Examples of thissection §3.1.


NOTE: sequences such as 1n , 1

n2 , 1n3 , 1√

n converge to 0 as n→∞.

We shall use this observations in the next few examples.

Further EXAMPLES:

3 limn→∞ 7n − 5

2 + 3n= lim

n→∞ 7 − 5n

2n + 3

=7 − 00 + 3

=73

.

NOTE: we divided both the numerator and denominator by thestrongest term n

4 limn→∞ 3

√n − 1

8 − 5√

n=


Further EXAMPLES:

5 limn→∞ 2n2 − n − 1

5n2 + n − 3= lim

n→∞2 − 1

n − 1n2

5 + 1n − 3

n2

=2 − 0 − 05 + 0 + 0

=25

.

NOTE: we divided both the numerator and denominator by thestrongest term n2


Further EXAMPLES:

7 limn→∞ n2 − 5n4 + 10

3n2 + n − 7n4 = ...

8 limn→∞ n2 − 5n3 + 10

3n2 + n − 7n4 = ...

9 limn→∞ n2 − 5n4 + 10

3n2 + n − 7n3 = ...


§3.2 Sequences in Root Algorithms

How are square roots computed on a computer??One possible way: define a sequence by a recursive rule.

Square Root√

p

a1 = 1, an+1 = 12

(an + p

an

)NOTE: this is in fact the Newton-Raphson Method applied to theequation x2 = p (whose positive root is

√p); see §4 for further details.

Particular Case√

2

a1 = 1, an+1 = 12

(an + 2

an

)Part 3 Sequences, Series, Power Series 1 / 6

EXAMPLE: compute the first 3 terms in the√

2 algorithm.

a1 = 1,

a2 = 12

(a1 +

2a1

)= 1

2

(1 + 2

1

)= 3

2 ,

a3 = 12

(a2 +

2a2

)= 1

2

(32 + 2

3/2

)≈ 1.4166.

A SIMPLE COMPUTER CODE will produce as accurate anapproximation as required (see next page).


QUESTION: how we can deduce from the recursive rule

an+1 = 12

(an + 2

an

)defining our sequence that it converges to

√2??

DISCUSSION:Assuming that the sequence {an} converges to some number L > 0,

i.e. limn→∞an = L ,

let n→∞ in the recursive rule:

limn→∞an+1 = lim

n→∞ 12

(an + 2

an

)L = 1

2

(L + 2

L

)so we have

2L = L + 2L ⇒ L = 2

L ⇒ L2 = 2 ⇒ L =√

2.

CONCLUSION: if the sequence {an} converges to some number L > 0,

then L =√

2


EXERCISE: evaluate the first 4 terms in the√

3 sequence.

Solution:an+1 = 1

2

(an + 3

an

)So we get:

a1 = 1,

a2 =

a3 =

a4 =


EXAMPLE: the sequence is defined by the recursive rulea1 = 1, an+1 = 1

4

(3an + 33

a3n

)Assuming that the sequence {an} converges to some number L > 0,find L.

Solution:

Answer: L = 4√

33


§3.3 Series

A series is formed when the terms of a sequence are added together.

For example,2,5,8,11,14, · · · is an infinite sequence,but 2 + 5 + 8 + 11 + 14 + · · · is an infinite series.


The Sigma Notation

The Sigma Notation:The Greek letter

∑(pronounced sigma), which means the sum of, is

generally used to express a series in a concise way.

Example: 2 + 4 + 8 + 16 + 32 = 2 + 22 + 23 + 24 + 25 =

5∑n=1

2n.

Note: 2n is the nth term in the sequence.Since we are summing the terms from 1 to 5 inclusive, the least valueof n is placed below the

∑sign and the greatest value of n is placed

above.


NOTE: A finite series always ends with the last term even if severalmiddle terms are omitted:

Example:20∑

n=1

12n + 2

=14+

16+

18+ · · ·+ 1

42.

NOTE: An infinite series may also be written in sigma notation, where∞ is used to indicate that there is no upper limit for n:

Example: 2 + 4 + 6 + 8 + · · · =∞∑

n=1

2n.


EXERCISE: Write the following series using∑

notation:

(i) 1 + 4 + 9 + 16 + · · ·+ 81

(ii) 1 − x + x2 − x3 + x4 − · · ·

(ii) −1 + 12x − 1

3x2 + 14x3 − 1

5x4 + · · ·

NOTE: In a series where the sign alternates from positive to negative,(−1)n may be used for the sign.

(−1)n results in even terms being positive and odd terms beingnegative.

(−1)n+1 results in even terms being negative and odd terms beingpositive.


Infinite Series

Definition (Infinite Series)Given a sequence {an}, the sum of all infinitely many terms in thissequence ∞∑

n=1

an = a1 + a2 + a3 + · · ·

is called an infinite series.

NOTE: it may be counterintuitive that we add infinitely many terms andstill can get a finite number, but this may be the case.


EXAMPLE:

consider∞∑

n=1

1n2 = 1 +

122 +

132 +

142 +

152 + · · · .

Let us do the summation gradually: one term, then 2 terms, then 3terms, etc.:

S1 = 1 —this it the first term;S2 = 1 + 1

22 —this it the sum of first 2 terms;S3 = 1 + 1

22 + 132 —this it the sum of first 3 terms;

S4 = 1 + 122 + 1


In general, Sn is the sum of the first n terms in our sequence:Sn = 1 + 1

22 + 132 + 1

42 + 152 + · · ·+ 1

n2


For the series∞∑

n=1

1n2 = 1 +

122 +

132 +

142 +

152 + · · · ,

here are a few values of Sn obtained using a computer package:



n=1

1n2 = 1 +

122 +

132 +

142 +

152 + · · · ,

plot a few values of Sn obtained using a computer package:

We see that as n→∞, the values Sn approach π2

6 ≈ 1.644934068.This implies that if we add all infinitely many terms of the sequence 1

n2 ,

we get∞∑

n=1

1n2 =

π2

6≈ 1.644934068


Definition (Convergent Infinite Series)Given a sequence {an}, let the sum of its first n terms

Sn =

n∑k=1

ak = a1 + a2 + a3 + · · ·+ an

If limn→∞Sn = S, where S is some real number, then we write∞∑

n=1

an = S and say that the series∞∑

n=1

an converges to the sum S.

NOTE: if limn→∞Sn is ±∞ or does not exist,

then we say that the series∞∑

n=1

an diverges.


EXAMPLE of a DIVERGENT series: Harmonic Series

consider∞∑

n=1

1n= 1 +

12+

13+

14+

15+ · · · .

Let us do the summation gradually: one term, then 2 terms, then 3terms, etc.:

S1 = 1 —this it the first term;S2 = 1 + 1

2 —this it the sum of first 2 terms;S3 = 1 + 1


S4 = 1 + 12 + 1


In general, Sn is the sum of the first n terms in our sequence:Sn = 1 + 1

2 + 13 + 1

4 + 15 + · · ·+ 1

n



n=1

1n= 1 +

12+

13+

14+

15+ · · · ,

here are a few values of Sn obtained using a computer package:



n=1

1n= 1 +

12+

13+

14+

15+ · · · ,

plot a few values of Sn obtained using a computer package:

We see that as n→∞, the values Sn do not approach any finitenumber (no matter how large n we consider), but go to +∞.This implies that if we add all infinitely many terms of the sequence 1

n ,

we get +∞ so∞∑

n=1

1n

diverges


Further EXAMPLES:

1 Telescoping Series∞∑

n=1

1n(n + 1)

=1

1 · 2+

12 · 3

+1

3 · 4+ · · ·

2 Series∞∑

n=1

1 = 1 + 1 + 1 + 1 + · · ·

3 Geometric Series∞∑

n=1

12n−1 = 1 +

12+

14+

18+ · · ·

4 More general Geometric Series∞∑

n=1

xn−1 = 1+ x + x2 + x3 + · · ·


Geometric Series

Geometric series:∞∑

n=1

xn−1 = 1 + x + x2 + x3 + · · ·

Convergence/Divergence of Geometric Series

∞∑n=1

xn−1 = 1 + x + x2 + x3 + · · · ={ 1

1−x , if x ∈ (−1,1),divergent, otherwise

Proof:

Case (i): If x = 1, then∞∑

n=1

xn−1 = 1+x +x2 +x3 + · · · = 1+1+1+ · · ·

is divergent (see previous page).


Case (ii): Let x 6= 1. Then one can show that

Sn =

n∑k=1

xk−1 = 1 + x + x2 + · · · xn−1 =1 − xn

1 − x(Exercise!)

Now, if |x | < 1, then limn→+∞ xn = 0, so

limn→+∞ sn =

11 − x

.

Otherwise, limn→+∞ xn does NOT exist, so the series is divergent.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Example The rational number 0.323232323232 . . . can be representedas

0.32 + 0.0032 + 0.000032 + · · · = 32100

+32

1002 +32

1003 + · · ·

=32100

(1 +1

100+

11002 +

11003 + · · · ) = 32

100· 1

1 − 1100

=3299

.


§3.4 Tests for Divergent/Convergent Series

Theorem If the series∑∞

n=0 an is convergent, then

limn→+∞an = 0.

From this we have

Divergence TestIf lim

n→+∞an 6= 0 then the series∑∞

n=0 an is divergent.

EXAMPLES:

(i) The series∞∑

n=1

(−1)n+1n = 1 − 2 + 3 − 4 + · · · is divergent as

limn→∞(−1)n+1n 6= 0;

(ii) The series∞∑

n=0

(−1)n = 1 − 1 + 1 − 1 + 1 + · · · is divergent as

limn→∞(−1)n 6= 0;


EXAMPLES:

(iii) The series∞∑

n=0

n2n − 1

= 0 + 1 +23+

35+

47+ · · · is divergent as

limn→∞ n

2n − 1=

126= 0.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .NOTE: if limn→+∞ an = 0, the series may be divergent (see 3 aboveexamples), or convergent, such as the divergent series

∞∑n=1

1n= +∞.

(was considered earlier).

In other words, limn→+∞an = 0 is a necessary condition for the

convergence of∞∑

n=0

an, but NOT sufficient.


Ratio test

Ratio testLet∑

n an a series such that

limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = p

If p < 1 the series is convergent.If p > 1 (or infinite) the series is divergent.If p = 1 the test is inconclusive.

NOTE: The ratio test is generally used to test for convergence ordivergence a series in which(i) the variable (generally n) appears in factorial form, e.g.,

∞∑n=0

3n!n + 1

(ii) the variable appears as a power, e.g.,∞∑

n=0

2n

n2


EXAMPLES:

1

∞∑n=1

1n!

is convergent (p = 0 < 1)

2

∞∑n=1

2n

n2 is divergent (p = 2 > 1)

3

∞∑n=1

n5

3n is convergent (p = 13 < 1)

4

∞∑n=1

(2n)!(n!)2 is divergent (p = 4 > 1)

5

∞∑n=1

1nq (for some positive q) —test is inconclusive (p = 1);

Recall:∞∑

n=1

1n

is divergent, while∞∑

n=1

1n2 is convergent.


6

∞∑n=0

10n

n!is convergent (as p = 0 < 1)

NOTE: this series converges to e10.

7

∞∑n=0

xn

n!is convergent for each x (as p = 0 < 1 for each fixed x)

(when we investigate convergence of the series, treat x as a fixedconstant)NOTE: this series converges to ex for each x .


8

∞∑n=0

(2x)n

(n2 + 1)3n is convergent for |x | < 32 (as p =

|2x |3 < 1)

9

∞∑n=0

n!xn

is divergent for each x 6= 0 (as p =∞ > 1 for each fixed x 6= 0)

10 x −x2

2+

x3

3+ · · · =

∞∑n=1

(−1)n+1xn

nis convergent for each |x | < 1

(as p = |x | < 1 for each fixed |x | < 1)NOTE: this series converges to ln(1 + x) for each |x | < 1.


§3.5 Functions as Infinite Power Series

Power SeriesA power series (about x = 0) is a series of the type:∞∑n=0

cnxn = c0 + c1x + c2x2 + · · ·,

where x is a variable, and c0, c1, · · · are given constants.

Exponential function: the power series representation is

ex =

∞∑n=0

xn

n!

NOTE: the Ratio Test shows (§3.4) that this series is convergent forany real x .

APPLICATION: one can use the sum of sufficiently many first terms

SN(x) =N∑

n=0

xn

n!to get an approximation of ex .

Then the evaluation of ex is reduced to basic operations (+, −, ×, /).Part 3 Sequences, Series, Power Series 1 / 12

NOTE: one can use computer languages/packages to evaluate the

sum of sufficiently many first terms SN(x) = 1 + x +x2

2!+ · · ·+ xN

N!:


Consider the power series representation of ex : ex =

∞∑n=0

xn

n!

EXERCISE: Use the series and the following table to estimate thevalues of e0.5, e1.5 and e0.3 to 5 decimal places.

x = 0.5 x = 1.5 x = 0.3S0 = 1 1 1

S1 = 1 + x 1.5 2.5S2 = (1 + x) + x2

2! 1.625000000 3.625000000S3 = (1 + x + x2

2! ) +x3

3! 1.645833333 4.187500000S4 = S3 +

x4

4! 1.648437500 4.398437500S5 = S4 +

x5

5! 1.648697917 4.461718750S6 = S5 +

x6

6! 1.648719618 4.477539062S7 = S6 +

x7

7! 1.648721168 4.480929129S8 = S7 +

x8

8! — 4.481564767S9 = S8 +

x9

9! — 4.481670707S10 = S9 +

x10

10! — 4.481686598


How one can get a power series representation??

QUESTION:How one can get a power series representation for any givenfunction??

ANSWER:There is a recipe called the Taylor series formula, which yields a powerseries representation for any function (no further details in this course).

Then one can use the Ratio test, to check where this power series isconvergent (i.e. where the power series is equal to the function ofinterest).


Further examples of power series representations

Trigonometric functions:

cos x =

∞∑n=0

(−1)nx2n

(2n)!= 1 −

x2

2!+

x4

4!+ · · ·

sin x =

∞∑n=0

(−1)nx2n+1

(2n + 1)!= x −

x3

3!+

x5

5!+ · · ·

NOTE: the Ratio Test shows (§3.4) that these series are convergent forany real x .

Important NOTE: x must be measured in radians, not in degrees(otherwise it doesn’t work).

Example: use cos(π3 ) = 1 −(π/3)2

2! +(π/3)4

4! + . . . to evaluate cos(π3 ) to3 decimal places.


EXAMPLE: Evaluate sin(10o) to 5 decimal places.

SOLUTION:The angle should be represented in radians:

10o = 10o π1800 = π

18 ≈ 0.1745329252 radians.

Use the series representation: sin x = x − x3

3! +x5

5! + · · · withx = 0.1745329252.

Note: it is convenient to fill in a table of type:n an Sn0 · ·1 · ·2 · ·3 · ·4 · ·

Answer: 0.17364 (see next page)


Intermediate RESULTS for the previous example using a simple loop:


Hyperbolic Functions:

cosh x =

∞∑n=0

x2n

(2n)!= 1 +

x2

2!+

x4

4!+ · · ·

sinh x =

∞∑n=0

x2n+1

(2n + 1)!= x +

x3

3!+

x5

5!+ · · ·

NOTE: the Ratio Test shows (§3.4) that these series are convergent forany real x .


QUESTION (Final Exam 2011):The hyperbolic cosine cosh(x) is a function defined by the series:

cosh(x) =+∞∑n=0

x2n

(2n)!

1 Use the ratio test to show that this series is convergent for all x .2 Use the series to estimate cosh(0.5) correct to 3 decimal places,

writing the partial sums in the following table.SOLUTION: (1) Ratio Test; (2) The Table:n an Sn0 · ·1 · ·2 · ·3 · ·4 · ·

Answer: 1.12 (see next page)


Intermediate RESULTS for the previous example using a simple loop:


Log function in base e ' 2.7182818:

ln(1 + x) =∞∑

n=1

(−1)n+1xn

n= x −

x2

2+

x3

3−

x4

4+ · · ·

Inverse Tangent function:

tan−1 x =

∞∑n=0

(−1)nx2n+1

2n + 1= x −

x3

3+

x5

5−

x7

7+ · · ·

NOTE: the Ratio Test shows (§3.4) that these two series areconvergent for |x | < 1.


How many terms??

QUESTION: How many terms in the power series representation of afunction one needs, to get certain accuracy / a certain number ofdecimal places correct??

Theoretical Approach:There is a theoretical estimate of the accuracy...Reference: the Lagrange remainder of the Taylor series.... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .Heuristic Approach:If the addition of the next term in the series does NOT change thedecimal places of interest, one can STOP.

NOTE: how many terms one needs to compute an approximate valueof f (x) depends on the

(i) required accuracy; (ii) function f ; (iii) value x .


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