math guide - 34.14

• We can solve the previous problem using the algebra technique.

			5				A			+	B	+			C	+		D					+			E
s-----------------------2( s + 1) 3				= ----							--s				------------------		------------------							(---------------s + 1)
							s2								( s + 1) 3		( s + 1)					2
=			A( s + 1) 3			+ Bs( s + 1) 3 + Cs2 + Ds2( s + 1) + Es2( s + 1) 2
			-----------------------------------------------------------------------------------------------------------------------------------------												s2( s + 1) 3
=			s4( B + E) + s3( A + 3B + D + 2E) + s2( 3A + 3B + C + D + E) + s( 3A + B) + ( A)
			--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------																	s2( s + 1) 3
																												–1
	0 1 0 0 1					A						0						A							0 1 0 0 1				0				5
	1 3 0 1 2					B						0						B							1 3 0 1 2				0				–15
	3 3 1 1 1					C				=		0						C				=			3 3 1 1 1				0			=	5
	3 1 0 0 0					D						0						D							3 1 0 0 0				0				10
	1 0 0 0 0					E						5						E							1 0 0 0 0				5				15

5	=	5	+	–15	+	5	+	10	+	15
s-----------------------2( s + 1) 3		----		--------s		------------------		------------------		(---------------s + 1)
		s2				( s + 1) 3		( s + 1) 2

Figure 34.17 An algebra solution to partial fractions

34.2.4 Summation and Series

b

• The notation ∑ xi is equivalent to xa + xa + 1 + xa + 2 + … + xb assuming a

i = a

and b are integers and b ≥ a . The index variable i is a placeholder whose name does not matter.

• Operations on summations:

b	a
∑ xi	= ∑ xi
i = a	i = b

math guide - 34.15

b	b
∑ α xi	= α ∑ xi
i = a	i = a	b
b	b

∑ xi + ∑ yj = ∑ ( xi + yi)

i = a	j = a	i = a
b	c	c

∑ xi + ∑ xi = ∑ xi

i = a			i = b + 1
	b			d
	∑ xi			∑ yj =
i = a			j = c

i = a

b d

∑∑ xiyj

i= a j = c

• Some common summations:

N

∑i

i= 1

N – 1

∑α

i= 0

∞

=	1
	--N( N + 1)
	2

1 – α N

i---------------, α

=1 – α

, α

N

≠ 1	for both real and complex α .
= 1

∑ α	i	1		α		< 1 for both real and complex α . For α		≥ 1 , the summation
		= ------------,
		1 –	α

i = 0

does not converge.

math guide - 34.16

34.2.5Practice Problems

1.Convert the following polynomials to multiplied terms as shown in the example.

e.g.,	x2 + 2x + 1 = ( x + 1) ( x + 1)
a)	x2 – 2x + 1	d)	x2 + x + 10
b)	x2 – 1	e)
c)	x2 + 1	f)

2. Solve the following equation to find ‘x’.

2x2 + 8x = –8

(ans.	2x2 + 8x = –8
	x2 + 4x + 4 = 0
	( x + 2) 2 = 0
	x = –2, –2

3. Reduce the following expression to partial fraction form.

x + 4

----------------

x2 + 8x

(ans.

x + 4

=

x + 4

=

A

+

B

=

Ax + 8A + Bx

x-------------------( x + 8)

--

--------------------------------

x

2

+ 8x

x x + 8

x

2

+ 8x

( 1) x + 4 = ( A + B) x + 8A

8A = 4

A = 0.5

1 = A + B

B = 0.5

----------------

x + 4

=

0.5------ +

-----------

0.5

x

2

+ 8x

x

x + 8