Download 1 Algebra - Partial Fractions

1
Algebra - Partial Fractions
Definition 1.1 A polynomial P (x) in x is a function that may be written in the form
P (x) = an xn + an−1 xn−1 + ... + a1 x + a0 .
an 6= 0 and n is the degree of the polynomial, n = deg(P ).
an , an−1 , ..., a1 , a0 are given real numbers called coefficients.
an xn is the leading term, an the leading coefficient, and a0 is the constant term.
Note: n is a non-negative integer and is the highest power of x occurring.
Example 1.2 The following are polynomials in x:
x3 + 3x2 − 2
(4x + 1)(x − 2)
1 6
x
2
3
2
− 17x + 1
Example 1.3 The above polynomials have the following degrees:
3
2
6
0
Definition 1.4 A rational function R(x) is the ratio of two polynomials N (x), D(x)
R(x) =
N (x)
D(x)
R(x) is called proper if deg(N ) < deg(D) and improper if deg(N ) ≥ deg(D).
Example 1.5 The rational functions on the left are proper, while those on the right are improper:
x+1
x2 + 5x + 6
x3 + x2 + 1
x2 + 2x − 1
x2
x3 + 1
x5 − 1
x5 + 1
Theorem 1.6 An improper rational function, R(x), may be expressed as the sum of a polynomial Q(x) and a proper rational function.
M
N
=Q+
D
D
where deg(M ) < deg(D) and deg(Q) = deg(N ) − deg(D)
Example 1.7 (1)
1
x
=1−
x+1
x+1
(2)
x6 − 2x4 + x2 − 2
12x + 10
= x4 + x3 + x2 + 3x + 6 + 2
2
x −x−2
x −x−2
Corollary 1.8 If M = 0 then N = QD and so D is a factor of N (i.e. N factorises into QD).
An important special case occurs when the denominator is linear D(x) = x − a,
N (x)
M
= Q(x) +
x−a
x−a
where M is now a number.
Corollary 1.9 The Remainder Theorem: M = N (a)
N (x) = (x − a)Q(x) + N (a)
Corollary 1.10 x − a is a factor of the polynomial N (x) if and only if N (a) = 0.
The process of finding Q(x), M (x) is called polynomial long division.
Example 1.11 Let
R(x) =
N (x)
x4 + 2x3 − 2x2 − x + 4
=
,
D(x)
x3 + 4x2 + 5x + 2
then
x− 2
x3 + 4x2 + 5x + 2
Hence
x4 + 2x3 − 2x2 − x + 4
x4 + 4x3 + 5x2 + 2x
−2x3 − 7x2 − 3x + 4
−2x3 − 8x2 − 10x − 4
x2 + 7x + 8
= Q(x) quotient
= M (x) remainder
x4 + 2x3 − 2x2 − x + 4
x2 + 7x + 8
=
x
−
2
+
x3 + 4x2 + 5x + 2
x3 + 4x2 + 5x + 2
x2 + 7x + 8
is proper, as required.
and 3
x + 4x2 + 5x + 2
We may combine two rational functions into one by cross-multiplying, e.g.
2
1(x + 2) + 2(x − 1)
3x
1
+
=
= 2
x−1 x+2
(x − 1)(x + 2)
x +x−2
It is sometimes necessary (e.g. integration) to carry out the reverse procedure, i.e. to split a
rational function into the sum of two (or more) simpler ones, known as summands. There are
five steps.
(I)
Use long division to obtain a proper rational function.
(II)
Factorise the denominator.
(III) Select the summand types.
(IV) Cross–multiply to get rid of the denominator.
(V)
Determine the constants.
Note: In step (II) it is always possible to factorise a polynomial so that all the factors are linear
(of the form ax + b) or irreducible quadratics (of the form ax2 + bx + c where b2 < 4ac).
1.0.1
Summand Types
If the denominator has
Type 1: a unique factor ax + b, then there will be a summand of the form
A
,
ax + b
where A is a constant to be determined.
Type 2: precisely n factors ax + b, then there will be summands of the form
A1
A2
A3
An
,
,
, ...,
,
2
3
ax + b (ax + b) (ax + b)
(ax + b)n
where the n constants A1 , A2 , A3 , ..., An are to be determined.
Type 3: a unique factor ax2 + bx + c, then there will be a summand of the form
Ax + B
,
ax2 + bx + c
where A and B are a constants to be determined.
Type 4: precisely n factors ax2 + bx + c, then there will be summands of the form
A2 x + B2
A 3 x + B3
An x + Bn
A1 x + B1
,
,
,
...,
,
ax2 + bx + c (ax2 + bx + c)2 (ax2 + bx + c)3
(ax2 + bx + c)n
where the 2n constants A1 , A2 , A3 , ..., An and B1 , B2 , B3 , ..., Bn are to be determined.
Example 1.12 Let f (x) be any polynomial such that each of the given rational functions is
proper:
Summands:
1.0.2
f (x)
(x + 1)(x − 2)
=
A
B
+
x+1 x−2
f (x)
(x + 1)3 (x − 2)
=
C
D
A
B
+
+
+
2
3
x + 1 (x + 1)
(x + 1)
x−2
(x2
f (x)
+ 1)(x − 2)
=
Ax + B
C
+
2
x +1
x−2
(x2
f (x)
+ 1)2 (x − 2)
=
Ax + B
Cx + D
E
+ 2
+
2
2
x +1
(x + 1)
x−2
Finding the Constants
Having chosen the summands, the next step is to find the values of the ‘constants to be determined’. First, multiply both sides by the left–hand denominator.
(1) Method of Convenient Values
Example 1.13 Split
5x − 1
−x−2
x2
into partial fractions.
5x − 1
5x − 1
=
−x−2
(x + 1)(x − 2)
x2
=
5x − 1 =
B
A
+
x+1 x−2
A(x + 1)(x − 2) B(x + 1)(x − 2)
+
(x + 1)
(x − 2)
= A(x − 2) + B(x + 1)
Let x = −1 then −6 = −3A and A = 2
Let x = 2 then 9 = 3B and B = 3.
Hence
2
3
5x − 1
=
+
.
−x−2
x+1 x−2
x2
Always check your answer by combining back into a single rational function:
2
3
2(x − 2) + 3(x + 1)
5x − 1
+
=
=
.
x+1 x−2
(x + 1)(x − 2)
(x + 1)(x − 2)
(2) Method of Comparing Coefficients
Example 1.14 Split
x2 + 1
x3 + 7x
into partial fractions
x2 + 1
x2 + 1
A Bx + C
=
=
+ 2
x3 + 7x
x(x2 + 7)
x
x +7
x2 + 1 =
Ax(x2 + 7) (Bx + C)x(x2 + 7)
+
x
(x2 + 7)
x2 + 1 = A(x2 + 7) + (Bx + C)x
Let x = 0 then 1 = 7A and A = 17 .
Comparing coefficients of x2 gives 1 = A + B. Hence B =
6
7
Comparing coefficients of x gives 0 = C i.e. C = 0. So
1
6
x
x2 + 1
7
= + 27
3
x + 7x
x x +7
=
6x
1
+
7x 7(x2 + 7)
Example 1.15 Split the improper rational function
x4 + 2x3 − 2x2 − x + 4
x3 + 4x2 + 5x + 2
into partial fractions.
The first step is to use polynomial long division,
x2 + 7x + 8
x4 + 2x3 − 2x2 − x + 4
=
x
−
2
+
x3 + 4x2 + 5x + 2
x3 + 4x2 + 5x + 2
The second step is to factorise the cubic denominator f (x) = x3 + 4x2 + 5x + 2.
Now f(0)=2, f(1)=12, f(-1)=0. By the factor theorem x + 1 is a factor. Hence
x3 + 4x2 + 5x + 2 = (x + 1)(x2 + ax + b) = (x + 1)(x2 + 3x + 2)
Now take the proper rational function and split into partial fractions
x2 + 7x + 8
A
B
C
x2 + 7x + 8
=
=
+
+
3
2
2
2
x + 4x + 5x + 2
(x + 1) (x + 2)
x + 1 (x + 1)
x+2
Cross–multiplying:
A(x + 1)2 (x + 2) B(x + 1)2 (x + 2) C(x + 1)2 (x + 2)
x + 7x + 8 =
+
+
(x + 1)
(x + 1)2
(x + 2)
2
= A(x + 1)(x + 2) + B(x + 2) + C(x + 1)2
Let x = −1 then 2 = B.
Let x = −2 then −2 = C.
Comparing coefficients of x2 gives 1 = A + C and so A = 3. Hence
x2 + 7x + 8
2
3
2
−
=
+
3
2
2
x + 4x + 5x + 2
x + 1 (x + 1)
x+2
Finally
2
x4 + 2x3 − 2x2 − x + 4
3
2
2
=x−2 +
+
−
3
2
2
x + 4x + 5x + 2
x + 1 (x + 1)
x+2
Algebra - Sequences and Series
Definition 2.1 A sequence is an ordered set of numbers, each being called a term.
Example 2.2
2, 5, 10, 17, 26, ..., k 2 + 1, ..., 10001
is a sequence of 100 numbers, the k th term being obtained by squaring k and adding 1. The
term k 2 + 1 is known as the general term.
Definition 2.3 A series is the sum of the terms in a sequence.
Example 2.4 The series corresponding to the example above is
2 + 5 + 10 + 17 + 26 + ... + (k 2 + 1) + ... + 10001
The sum is 338450.
Often, the sigma notation is used with series. The series in the example would be written as
100
X
(k 2 + 1)
k=1
It is also possible to have sequences and series with infinitely many terms. The sum of an
infinite series is defined as a limit, which may or may not exist. For example, in
∞
X
3
= 0.3 + 0.03 + 0.003 + 0.0003 + ... = 0.3333...
k
10
k=1
we can say that the sum of this series is 13 .
In this section, we will concentrate on two special types of sequence, the arithmetic and
geometric progressions, and their associated series.
Definition 2.5 An arithmetic progression is a sequence of the form:
a, a + d, a + 2d, a + 3d, ..., a + (k − 1)d, ..., a + (n − 1)d
where a and d are constants: a is the first term and d the common difference.
Each term is obtained by adding d to the previous term. Note that the k th term is a + (k − 1)d.
Example 2.6
5, 7, 9, 11, 13, 15, ..., 77
is an arithmetic progression with a = 5 and d = 2.
Theorem 2.7 The sum Sn of the first n terms of an arithmetic progression is
1
Sn = n (2a + (n − 1)d)
2
Proof: We write down the sum twice: once forwards, then backwards, and add them.
Sn
= a
+ (a + d)
+...+
(a + (n − 2)d) + (a + (n − 1)d)
Sn
= (a + (n − 1)d) + (a + (n − 2)d) +...+
(a + d)
2Sn
= (2a + (n − 1)d) + (2a + (n − 1)d) +...+
(2a + (n − 1)d) + (2a + (n − 1)d)
+ a
Note that all of the terms in 2Sn are the same, and there are n of them. Hence, we see that
Sn = 21 n (2a + (n − 1)d) as required.
Example 2.8 Find the sum of the 37 terms of the sequence
5, 7, 9, 11, 13, 15, ..., 77
Solution: We have a = 5 and d = 2, so Sn = 12 37 (10 + (37 − 1)2) = 1517. Note that since the
last term is a + (n − 1)d, we can rewrite the formula Sn = 12 n(a + z), where z is the last term.
Example 2.9 Find the sum of the first n natural numbers:
1 + 2 + 3 + 4 + ... + (n − 1) + n
Solution: We have a = 1 and d = 1, so Sn = 21 n (2 + (n − 1)) = 12 n(n + 1).
Example 2.10 The 8th term of an arithmetic progression is − 12 , and the 13th term is −8. Find
the first term, the common difference, and the sum of the first 15 terms.
Solution: We can solve for a and d:
13th term
= a + 12d
= −8
8th term
= a + 7d
= − 12
subtracting
= − 15
2
5d
¡
¢
which gives d = − 32 . Substituting, we get a = 10. Now, the sum S15 = 21 15 20 + (14).− 32 =
− 15
.
2
Definition 2.11 A geometric progression is a sequence of the following form
a, ar, ar2 , ar3 , ..., ark−1 , ..., arn−1
where a and r are constants: a is the first term and r is the common ratio.
Example 2.12
3, 6, 12, 24, 48, 96, ...
is a geometric progression with a = 3 and r = 2.
1 1 1 1
1, , , , , ...
2 4 8 16
is a geometric progression with a = 1 and r = 21 .
Theorem 2.13 The sum Sn of the first n terms of a geometric progression is
Sn =
a (1 − rn )
1−r
if r 6= 1. Otherwise if r = 1 then Sn = an.
Proof: The case when r = 1 is obvious, so we assume r 6= 1. Now
Sn
=
a
+
ar
+
ar2
+...+ arn−2 +
arn−1
rSn
=
ar
+
ar2
+
ar3
+...+ arn−1 +
arn
multiplying by r:
and subtracting
(1 − r)Sn = a − arn
from which we deduce the result.
Example 2.14 Calculate the sum of the first six terms of a geometric progression with a = 1
and r = 10.
Solution: Using the formula:
S6 =
−999999
1. (1 − 106 )
=
= 111111
1 − 10
−9
which is as expected as the series is
1 + 10 + 100 + 1000 + 10000 + 100000
Example 2.15 In the following geometric progression
3, 6, 12, 24, 48, 96, ...
after how many terms will the sum be greater than 9999 ?
Solution: We have a = 3 and r = 2. The formula for Sn is
Sn =
3. (1 − 2n )
= 3 (2n − 1)
1−2
so we must find n for which 3 (2n − 1) > 9999. This means that 2n − 1 > 3333 or 2n > 3334.
We can use logarithms to solve this:
2n > 3334 =⇒ ln 2n > ln 3334
ln 3334
=⇒ n ln 2 > ln 3334 =⇒ n >
ln 2
and using a calculator
8.112
= 11.703
0.693
and so after 12 terms, the sum will be greater than 9999 (in fact it is 12285).
n>
Example 2.16 Calculate the sum of the first eight terms of a geometric progression with a = 1
and r = 21 .
Solution: Using the formula:
³
´
8
1. 1 − 12
S8 =
=
1 − 21
28 −1
28
1
2
=
2. (28 − 1)
510
=
≈ 1.992
8
2
256
In fact, if we continue adding terms of the above progression, we get values closer and closer
to 2. This is an example of a convergent series. Provided |r| < 1 a geometric series always
converges, and the sum can be computed by the following formula.
Theorem 2.17 The sum to infinity S∞ of a geometric progression with |r| < 1 is
a
S∞ =
1−r
a (1 − rn )
, and we note that as n gets larger, the value of rn becomes
1−r
smaller, since |r| < 1. In fact, we can say that as n → ∞, rn → 0. Thus, letting n → ∞ in Sn ,
Proof: We have Sn =
we get
S∞ =
which completes the proof.
a
a (1 − 0)
=
1−r
1−r
Example 2.18 We check the aforementioned series:
0.3, 0.03, 0.003, 0.0003, 0.00003, ...
a
which has first term a = 0.3 and common ratio r = 0.1. The sum to infinity is S∞ =
=
1−r
0.3
= 13 as we would expect.
1 − 0.1
3
Algebra - Binomial Theorem
The binomial theorem is used when we want to multiply out expressions of the form (a + b)n .
We shall limit ourselves to the case where n is a positive integer, but there are analogues for
negative, and non–integer values of n
(a + b)1
=
a+b
2
(a + b)
=
a2 + 2ab + b2
(a + b)3
=
a3 + 3a2 b + 3ab2 + b3
(a + b)4
=
a4 + 4a3 b + 6a2 b2 + 4ab3 + b4
(a + b)5
=
a5 + 5a4 b + 10a3 b2 + 10a2 b3 + 5ab4 + b5
In each term of the expansions, the power of a plus the power of b is equal to n, i.e. each term
is of the form an−r br , where r takes the values 0, 1, ..., n successively, multiplied by a number
called a coefficient. The coefficients in the expansions may be written as follows.
Pascal’s Triangle
1
1
1
1
1
1
1
1
1
8
3
5
7
6
15
1
4
10
20
35
56
1
3
10
21
28
2
4
6
1
5
15
35
70
1
1
6
21
56
1
7
28
1
8
1
Thus, the coefficients in the expansions may be obtained from the numbers in Pascal’s
Triangle. Denote
this coefficient by n Cr then each term is of the form n Cr an−r br . An alternative
Ã
!
n
notation is
. An alternative way of writing n Cr is
r
Definition 3.1 k!, pronounced k factorial is the product of all integers from k down to 1, i.e.
k! = k(k − 1)(k − 2)...3.2.1
Also 0! = 1.
For example 5! = 5.4.3.2.1 = 120.
Proposition 3.2 For any positive integer n, and r = 0, 1, ..., n n then
n!
Cr =
(n − r)!r!
Example 3.3 In the case n = 5, we compute the coefficients:
5
C0
=
5!
5!0!
=
5.4.3.2.1
5.4.3.2.1.1
=1
5
C1
=
5!
4!1!
=
5.4.3.2.1
4.3.2.1.1
=5
5
C2
=
5!
3!2!
=
5.4.3.2.1
3.2.1.2.1
= 10
5
C3
=
5!
2!3!
=
5.4.3.2.1
2.1.3.2.1
= 10
5
C4
=
5!
1!4!
=
5.4.3.2.1
1.4.3.2.1
=5
5
C5
=
5!
0!5!
=
5.4.3.2.1
1.5.4.3.2.1
=1
giving the fifth row of Pascal’s triangle.
Theorem 3.4 The Binomial Theorem If n is a positive integer, and a and b are any
numbers then
n
(a + b) =
n
X
n
Cr an−r br
r=0
where n Cr =
n!
.
(n − r)!r!
Example 3.5
(a + b)5 = 5 C0 a5 b0 + 5 C1 a4 b + 5 C2 a3 b2 + 5 C3 a2 b3 + 5 C4 ab4 + 5 C5 a0 b5
= a5 + 5a4 b + 10a3 b2 + 10a2 b3 + 5ab4 + b5 .
An alternative way of writing is
5
5×4 3 2 5×4×3 2 3 5×4×3×2 4
(a + b)5 = a5 + a4 b +
ab +
ab +
ab + b5 .
1
2×1
3×2×1
4×3×2×1
= a5 + 5a4 b + 10a3 b2 + 10a2 b3 + 5ab4 + b5 .
as before
Example 3.6 Expanding (1 + 2x)5 , we let a = 1 and b = 2x, which gives:
(1 + 2x)5 = a5 + 5a4 b + 10a3 b2 + 10a2 b3 + 5ab4 + b5
= 1 + 5(2x) + 10(2x)2 + 10(2x)3 + 5(2x)4 + (2x)5
= 1 + 10x + 40x2 + 80x3 + 80x4 + 16x5
Example 3.7 Approximate (0.98)5 to 4 decimal places.
Solution: We use the previous example, substituting x = −0.01:
(0.98) = 1 − 0.1 + 0.004 − 0.00008 + ... ≈ 0.9039
(we don’t care about the last two terms as they are obviously insignificant to 4 decimal places).
Example 3.8 Find the coefficient of x2 in the expansion of (3x − x1 )8 .
Solution: We need the term where xr ( x1 )8−r = x2 , which is clearly when r = 5. Now we have
8
C5 =
8!
8.7.6.5.4.3.2.1
=
= 56
3!5!
3.2.1.5.4.3.2.1
and so the required term is 56(3x)5 (− x1 )3 = −56.35 x2 = −13608x2 .