EDEXCEL ANALYTICAL METHODS FOR ENGINEERS H1

UNIT 2 - NQF LEVEL 4

OUTCOME 1

TUTORIAL 3 – SERIES

Algebraic methods: polynomial division; quotients and remainders; use of factor and remainder
theorem; rules of order for partial fractions (including linear, repeated and quadratic factors);
reduction of algebraic fractions to partial fractions

Exponential, trigonometric and hyperbolic functions: the nature of algebraic functions; relationship
between exponential and logarithmic functions; reduction of exponential laws to linear form;
solution of equations involving exponential and logarithmic expressions; relationship between
trigonometric and hyperbolic identities; solution of equations involving hyperbolic functions

Arithmetic and geometric: notation for sequences; arithmetic and geometric progressions;
the limit of a sequence; sigma notation; the sum of a series; arithmetic and geometric series;
Pascal’s triangle and the binomial theorem

Power series: expressing variables as power series functions and use series to find approximate
values (eg exponential series, Maclaurin’s series, binomial series)

You should judge your progress by completing the self assessment exercises.

© D.J.Dunn www.freestudy.co.uk 1
FACTORIALS

You should already know that a factorial number is denoted with ! so 4! means factorial 4.

4! = (4)(3)(2)(1) = 24 n!=(n)(n-1)(n-2)(n-3).....1 Without proof, 0! Is always taken as 1

SUMMATION of a SERIES

Many expressions can be represented by a series of numbers added together.

1 1 1 1 1
This might be a series of numbers like x = 1 + + + + + ..........
2 3 4 5 n
It might be a power series of x like y = x + x + x + x + ..........x n
0 1 2 3

In general a series may be written as x = u1 + u2 + .......un where u is the term in the series.
1 1 1 1 1
Consider the following calculation. x = 1 + + + + + ..........
2 3 4 5 n
The value of x is the sum of a series of fractions with the numerators forming a descending series of
n=n n
1 1
integers (whole numbers). We could write this more simply as: x = ∑ or ∑
n =1 n 1 n
The symbol ∑ is a capital letter Sigma and means the ‘sum of ’
The limits of the variable between the summation takes place is shown.

WORKED EXAMPLE No. 1

4n 2 x n
n
Write out the series represented by ∑ 2 for the first 3 terms.
1 2n + n − 2
SOLUTION
4n 2 4 4n 2 16 16
n=1 = = 4 n = 2 = =
2n + n − 2 2 + 1 − 2
2
2n + n − 2 8 + 4 − 2 10
2
2
4n 36 36
n=3 = =
2n + n − 2 18 + 3 − 2 19
2

n
4n 2 x n 16x 2 36x 3 4n 2 x n
hence ∑ 2n 2 + n − 2 = 4x +
10
+
19
+ .....
2n 2 + n − 2
+ .....
1

WORKED EXAMPLE No. 2

n
n
x
Write out the series represented by ∑ (− 1)n −1 n!
for the first 4 terms.
1
SOLUTION
xn x1 n
x2 x2
n = 1 (− 1)n −1 = (− 1)0 =x n=2 (− 1)n −1 x = (− 1)1
=−
n! 1 n! 2 2
n 3
n 1 x 2 x x3 n 4
x4
n = 3 (− 1) − = (− 1) = n=4 (− 1)n −1 x = (− 1)3 x =−
n! (3)(2) 6 n! (4)(3)(2) 24

n
n
x x 2 x3 x 4 n 1 x
n
Hence ∑ (− 1)n −1 n!
= x−
2
+ −
6 24
+ .....(− 1) −
n!
+ ....
1

©D.J.Dunn www.freestudy.co.uk 2
LIMITING VALUES
2n 2 + n − 4
Consider the following equation. x= Suppose we wish to know if this has a value
7n 2 + 6n − 20
when n = ∞. A simple way to find out is to make use of the fact that 1/∞ = 0

If we rearrange the equation by dividing through by the highest order of n (n2 in this case) we get
2 + 1/n − 4/n 2
x=
7 + 6/n − 20/n 2

2+0−0 2
Now put n = ∞ x= = so there is a limiting value when n = ∞
7+0−0 7
5
We write this as Lt x =
n →∞ 7

WORKED EXAMPLE No. 3

Find the limiting value of the following expressions when n →∞

4n 2 n n 3 + 5n 2 + 5
i) 2 ii) iii)
2n + n − 2 (3n − 2)(5n + 7) n2

SOLUTION
4n 2 4 n 1/n 0
i) = =2 ii) = = =0
2n + n − 2 2 + 1/n − 2/n 2
2
(3n − 2)(5n + 7) (3 − 2/n )(5 + 7/n ) 15

n 3 + 5n 2 + 5 1 + 5/n + 5/n 3 1 + 0 + 0 1
iii) = = = =∞
n2 1/n 0 0

SELF ASSESSMENT EXERCISE No.1

x 2n -1
1. Write out the series represented by (− 1)
n −1
for the first 3 terms.
(2n - 1)!

x3 x5 x7 n −1 x
2n -1
Answer x− + − + .......(− 1)
6 120 5040 (2n - 1)!
2. Find the limiting value of the following expressions when n →∞
2
i. 5 + 2 ii.
5n - 2
iii.
(n + 2)(n + 3)
n 2n + 1 n (n + 1)(n + 4 )
Answers 5, 2.5 and 0

©D.J.Dunn www.freestudy.co.uk 3
CONVERGENCE and DIVERGENCE

1 1 1 1 1
In the series x = 1 + + + + + .......... we might think that since each term is smaller than the
2 3 4 5 n
one before it, then the value of x would tend to converge on some figure as we add more and more
terms. In fact the value of x will go on getting bigger so in the limit as n→∞ the value of x will also
tend to infinity. This series has no limiting value. We must be very careful dealing with series
because the value of each term may get bigger and bigger (divergence) or it might get smaller and
smaller (convergence) and if it converges there is a limiting value.

In order to find the value of a series when n = ∞ we write the series in the form:
n =∞
x = u1 + u2 + u3 + ..... un or x = ∑ un
n =1
If the term at n = ∞ is not zero then it seems likely that series has no limiting value and is
divergent. If Lt u n = 0 the series might converge to a limiting value but this is not certain.
n→∞
n =∞
⎛1⎞ 1 1 1
For example consider again the following series i. x = ∑ ⎜⎝ n ⎟⎠ The series is 1 + + + .....
2 3 n
n =1
⎛1⎞
Lt ⎜ ⎟ = 0 It was shown earlier that this series in divergent so having a zero value does not prove
n → ∞⎝ n ⎠
the series is convergent.

WORKED EXAMPLE No. 4

n =∞
⎛ 3n + 2 ⎞
Determine if the following series is divergent. x = ∑ ⎜⎝ 10n + 1 ⎟⎠
n =1
SOLUTION

5 8 11 3n + 2
The series is + + + .....
11 21 31 10n + 1

⎛ 3n + 2 ⎞ ⎛ 3 + 2/n ⎞ ⎛ 3 ⎞
Lt ⎜ ⎟=⎜ ⎟ = ⎜ ⎟ and since this is not zero the series is divergent.
n → ∞⎝ 10n + 1 ⎠ ⎝ 10 + 1/n ⎠ ⎝ 10 ⎠

D'ALEMBERT'S RATIO
n =∞
For any series x = ∑ un
n =1

⎛u ⎞
Lt ⎜⎜ n +1 ⎟⎟ < 1 the series converges.
n → ∞⎝ u n ⎠

⎛u ⎞
Lt ⎜⎜ n +1 ⎟⎟ > 1 the series diverges.
n → ∞⎝ u n ⎠

This does not tell us what happens if the result is unity.

©D.J.Dunn www.freestudy.co.uk 4
 WORKED EXAMPLE No. 5

n=∞
⎛ 2n − 1 ⎞
Determine if the following series is divergent. x = ∑ ⎜ n −1 ⎟
n =1 ⎝ 2 ⎠
SOLUTION
3 5 7 2n − 1
The series is x = 1 + + 2 + 3 + .............. n −1
2 2 2 2
2n − 1 2(n + 1) − 1 2n +
We note that u n = and u n +1 = = n
2 n −1
2 n + 1 −1 2

=
( )
u n +1 (2n + 1) 2n -1
= =
( )
2 n (2n + 1) 2n 2-1 (2n + 1)
=
(2 + 1/n )
u n
2 (2n − 1)
n
2 (2n − 1)
n
2(2n − 1) 2(2 − 1/n )

⎛u ⎞ 2+0 2 1
Lt ⎜⎜ n +1 ⎟⎟ = = = This is less than 1 so the series is convergent
n → ∞⎝ u n ⎠ 2(2 + 0) 4 2

SELF ASSESSMENT EXERCISE No. 2

Find the limiting value of the following expressions when n →∞

1. Determine if the following are divergent or convergent.

n = ∞⎛
2 n −1 ⎞ n =∞
⎛ n ⎞
i. x = ∑ ⎜ n + 9 ⎟⎟ ii)
⎜ x= ∑ ⎜⎝ n + 1 ⎟⎠
n =1 ⎝ ⎠ n =1

Answers
⎛u ⎞
i. Lt ⎜⎜ n +1 ⎟⎟ = 2 hence divergent.
n → ∞⎝ u n ⎠

⎛u ⎞
ii. Lt ⎜⎜ n +1 ⎟⎟ = 1 hence indeterminate
n → ∞⎝ u n ⎠

but Lt u n = 1 and since this is not zero it must be divergent.

n →∞

©D.J.Dunn www.freestudy.co.uk 5
ABSOLUTE CONVERGENCE

Without proof it can be shown that if we determine the sum of the modulli of each term in a series
such that S = u1 + u 2 + u 3 + ........ u n = ∑ u n and if this is convergent, then ∑ u n is also
convergent and has a definite value hence the use of the words absolute convergence.

It follows that if all the terms are positive anyway, then if the series is convergent it is absolutely
convergent

WORKED EXAMPLE No. 6

3 5 7
Test the following series to see if it has absolute convergence. 1 − + 2 − 3 + .....
2 2 2

SOLUTION

n =∞
3 5 7 ⎛ 2n − 1 ⎞
∑ un = 1 + + 2 + 3 + ..... ∑ ⎜ n −1 ⎟
2 2 2 n =1 ⎝ 2 ⎠
This is the same as example whence
⎛u ⎞ 2+0 2 1
Lt ⎜⎜ n +1 ⎟⎟ = = =
n → ∞⎝ u n ⎠ 2(2 + 0) 4 2
3 5 7
This is less than 1 so the series is convergent and the series 1 − + 2 − 3 + ..... is absolutely
2 2 2
convergent.

WORKED EXAMPLE No. 7

x 2 x3 x n
Determine if the following power series is convergent or divergent. 1 + x + + + .....
2! 3! n!
SOLUTION

xn x n +1
un = u n +1 =
n! (n + 1)!
u n +1 n!x n +1 x
= n =
un x (n + 1)! n
⎛u ⎞ x
Lt ⎜⎜ n +1 ⎟⎟ = = 0 for all values of x
n → ∞⎝ u n ⎠ ∞
This is less than 1 so the series is absolutely convergent for all values of x.

©D.J.Dunn www.freestudy.co.uk 6
 SELF ASSESSMENT EXERCISE No. 3

Test the following series for convergence.

x 2 x3 x 4 xn
1. x − + − + .....(− 1)n −1
2 3! 4! n!
3 5 7 2n -1
x x x n −1 x
2. x − + − + .......(− 1)
3! 5! 7! (2n - 1)!
Answers

u n +1
1. Lt = x so if |x|<1 the series is absolutely convergent but if |x|>1it is divergent.
n→∞ u n

u n +1
2. Lt = 0 hence the series is absolutely convergent for all x.
n →∞ u n

PASCAL'S TRIANGLE

n n(n - 1)(n - 2)(n - 3) ......(n - {r − 1})

An important factorial expression is Cr =
r!
The top line is the first r factors of n and the bottom line is factorial r

If we evaluated all the values of nCr from r = 0 to r = n we would find the values are symmetrical.
For example take the case n = 5
5 1 5 (5)(4) (5)(4)(3) (5)(4)(3)(2)
C0 = = 1 5 C1 = = 5 5 C 2 = = 10 5 C3 = = 10 5 C 4 = =5
1 1 (2)(1) (3)(2)(1) (4)(3)(2)(1)
5 (5)(4)(3)(2)(1)
C5 = =1
(5)(4)(3)(2)(1)

Pascal's Triangle is made of rows as shown.

The nth row is made of all the numbers nCr for r = 0 to
r=n
The zero row is 0C0 = 1
The 1st row is 1C0 = 1 and 1C1 =1
The 2nd row is 2C0 = 1 2C1 = 2 2C2 = 1
The 3rd row is 3C0 = 1 3C1 = 3 3C2 = 3 3C3 = 1

©D.J.Dunn www.freestudy.co.uk 7
BINOMIAL THEOREM

Consider a polynomial made up of n repeated factors of (y + x)

If n = 2 then (x+y)n = y2 +2xy + x2

If n = 3 then (x+y)n = y3 + 3xy2 +3x2y + x3
If n = 4 then (x+y)n = y4 + 4xy3 + 6x2y2 + 4x3y + x4
If n = 5 then (x+y)n = y5 + 5xy4 + 10 x2y3 + 10x3y2 + 5x4y + x5

Let y = 1.

If n = 2 then (x + 1)n = 1 +2x + x2

If n = 3 then (x + 1)n = 1 + 3x +3x2 + x3
If n = 4 then (x + 1)n = 1 + 4x + 6x2 + 4x3 + x4
If n = 5 then (x + 1)n = 1+ 5x + 10 x2 + 10x3 + 5x4 + x5

This is a series with each term containing x to a power in ascending order from 0 to n.
Examine this and you will see that the coefficients match the rows of Pascal's Triangle. It follows
that the coefficients are nCr.

Since xo = 1 and we normally write x1 as simply x the equation is more commonly written as:
(x + 1)n = 1 + 4x + 6x2 + 4x3 + x4 + .......xn

Noting that nC0 = nCn = 1 (1 + x)n = nC0 + nC1 x + nC2 x2 + nC3 x3 + nC4 x4 + ....... + nCn xn

(1 + x)n = 1 + nC1 x + nC2 x2 + nC3 x3 + nC4 x4 + ....... + xn

( )
(1 + x) n = ∑ r = 0 n C r x r
r=n

Returning to (x + y)n we can see that this is a series such that :

(x + y)n = nCnyn + nCn-1xyn-1 + nCn-2 x2yn-2 + nCn-3x3yn-3 + ... nC0xn
(x + y)n = yn + nCn-1xyn-1 + nCn-2 x2yn-2 + nCn-3x3yn-3 + ... xn

This result can be obtained expanding (1 + x)n as shown in the next example.

WORKED EXAMPLE No. 8

Using the binomial theorem expand S = (z + a)n

SOLUTION

First we must rearrange the expression into a form that can be expanded.
n
n⎛ a⎞
(z + a) = z ⎜1 + ⎟ let a/z = x (z + a) n = z n (1 + x )n
n

⎝ z⎠
Expanding we get
S = zn(1 + x)n = zn[1 + nC1 x + nC2 x2 + nC3 x3 + nC4 x4 + ....... + xn]
S = zn(1 + x)n = zn[1 + nC1 (a/z) + nC2 (a/z)2 + nC3 (a/z)3 + nC4 (a/z)4 + ....... + (a/z)n]
S = zn(1 + x)n = zn[1 + nC1 z-1a + nC2 z-2a2 + nC3 z-3a3 + nC4 z-4a4 + ....... + z-nan]
S = zn(1 + x)n = zn + nC1 zn-1a + nC2 zn-2a2 + nC3 zn-3a3 + nC4 zn-4a4 + ....... + an
Note if we change z to y and a to x we get the reult for expanding (x + y)n
yn + nC1 yn-1x + nC2 yn-2x2 + nC3 yn-3x3 + nC4 yn-4x4 + ....... + xn

©D.J.Dunn www.freestudy.co.uk 8
 WORKED EXAMPLE No. 9

Expand (1 + x)6 with the binomial theorem.

SOLUTION
(1 + x)6 = ∑1 (6C r ) x r we know that
6

(1 + x) n = 6 C 0 + 6 C1x + 6 C 2 x 2 + 6 C3 x 3 + 6 C 4 x 4 + 6 C5 x 5 + 6 C 6 x 6
6
6
C 0 = 1 6 C1 = = 6 .... as solved earlier
1
6 6
C2 = C4 = 15 6 C3 = 20 6
C5 = 6 C1 = 6 6
C6 = 1
(1 + x)6 = 1 + 6x + 15x 2 + 20x 3 + 15x 4 + 6x 5 + x 6
Check this out by putting in any value of x say x = 2 (1+2)6 = 36 = 729
(1 + 2)6 = 1 + 26 + (6)(2) + (15)22 + (20)23 + (15)24 + (6)25
729 = 1 + 12 + 60 + 160 + 240 + 192 + 64
729 = 729

WORKED EXAMPLE No. 10

y = (1 + x ) and show that if x is very small then the

−3
Using the binomial theorem expand
answer is 1 – 3x

SOLUTION

n = -3 (1 + x) -3 = -3 C 0 + -3 C1x + -3 C 2 x 2 + -3 C3 x 3 + ........
−3 (−3)(−4) 2 (−3)(−4)(−5) 3
(1 + x) -3 = 1 + x+ x + x + ........
1 (1)(2) (1)(2)(3)
(1 + x)-3 = 1 − 3x + 6x 2 − 10x 3 + ........
The result is an infinite series and it is only useful for evaluation when x is small such that
higher powers are negligible.
Check if x = 0.02 then y = 1.02-3 = 0.9423 1 – 3x = 1 – 0.06 = 0.94

WORKED EXAMPLE No. 11

1
Using the binomial theorem expand y = and evaluate when x = 0.02
1+ x
SOLUTION
1
= (1 + x ) putting n = -1 we can expand.
−1
y=
1+ x
1
y= = (1 + x )−1 = 1 + n C1x + n C 2 x 2 + n C3 x 3 ...
1+ x
−1 (−1)(−2) 2 (−1)(−2)(−3) 3
y = 1+ x+ x + x + ....
1 (1)(2) (1)(2)(3)
1
y= = 1 − x + x 2 − x 3 .....
1+ x
1
y= = 1 − 0.02 + 0.0004 − 0.000008..... = 0.9804
1 + 0.02
Note for small values of y is quite accurately given by 1-x

©D.J.Dunn www.freestudy.co.uk 9
 SELF ASSESSMENT EXERCISE No. 4

1. Expand y = (1 + x )4 using the binomial theorem.

2. Expand y = (1 – p)-1 to four terms using the binomial theorem.

3. Expand y = (1 – q)-2 to four terms using the binomial theorem.

1 x
4. Expand y = and show that for small numbers y = 1 −
1+ x 2

5. Expand y=
(1 + x )4
and show that for small numbers y =1+
9x
1− x 2

MACLAURIN'S SERIES

This is a series that demands special attention and the student might also look up the Taylor Series
on which it is based.

The series is based on successive differentials.

f' ' (0) 2 f' ' ' (0) 3 f n (0) n
f(x) = f(0) + f' (0)x + x ++ x + ..... + x + ......
2! 3! n!

WORKED EXAMPLE No. 12

1
Expand into a series using Maclaurin's method.
1− x

SOLUTION

1 1 2
f(x) = f(0) = 1 f ′(x) = f ′(0) = 1 f ′′(x) = f ′(0) = 2
1− x (1 − x )2
(1 − x )3
6
f ′′(x) = f ′(0) = 6
(1 − x )4
Hence
1 2x 2 6x 3
= 1+ x + + + ......
1− x 2! 3!
1
= 1 + x + x 2 + x 3 + ....x n + ... but clearly if x = 1 the result is infinity
1− x
n
1
= ∑ xn
1− x 0

©D.J.Dunn www.freestudy.co.uk 10
 SELF ASSESSMENT EXERCISE No. 5

Expand the following Maclaurin's method theorem.

i. cos(x) ii. sin(x) iii. ex iv. sinh(x)

Check your answers in the table below.

TABLE OF FUNCTIONS AND THEIR SERIES

n
1
1− x
∑ xn
0
∞
(−1) n
cos(x) ∑ (2n )! x 2n
0
∞
(−1) n
sin(x) ∑ (2n + 1)!x 2n +1
0
∞
1
cosh(x) ∑ (2n )!x 2n
0
∞
1
sinh(x) ∑ (2n + 1)!x 2n +1
0
∞
1
ex ∑ n!x n
0
∞
(− 1)n +1 x n
ln(1 + x) ∑ n
1

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