1 Algebra - Partial Fractions

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1 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) a n x n + a n 1 x n a 1 x + a 0. a n 0 and n is the degree of the polynomial, n deg(p ). a n, a n 1,..., a 1, a 0 are given real numbers called coefficients. a n x n is the leading term, a n the leading coefficient, and a 0 is the constant term. Note: n is a non-negative integer and is the highest power of x occurring. Example 1. The following are polynomials in x: x 3 + 3x (4x + 1)(x ) 1 x6 17x Example 1.3 The above polynomials have the following degrees: 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 x + 5x + 6 x x x 3 + x + 1 x + x 1 x 5 1 x 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. N D Q + M D where deg(m) < deg(d) and deg(q) deg(n) deg(d)

2 Example 1.7 (1) () x 6 x 4 + x x x x x x + 1 x 4 + x 3 + x + 3x x + 10 x x 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, where M is now a number. N(x) x a Q(x) + M x a 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 then R(x) N(x) D(x) x4 + x 3 x x + 4, x 3 + 4x + 5x + x x 3 + 4x + 5x + x 4 + x 3 x x + 4 x 4 + 4x 3 + 5x + x x 3 7x 3x + 4 x 3 8x 10x 4 x + 7x + 8 Q(x) quotient M(x) remainder Hence x 4 + x 3 x x + 4 x 3 + 4x + 5x + x + x + 7x + 8 x 3 + 4x + 5x + x + 7x + 8 and is proper, as required. x 3 + 4x + 5x + We may combine two rational functions into one by cross-multiplying, e.g. 1 x 1 + x + 1(x + ) + (x 1) (x 1)(x + ) 3x x + x

3 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) (II) (III) (IV) (V) Use long division to obtain a proper rational function. Factorise the denominator. Select the summand types. Cross multiply to get rid of the denominator. 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 ax + bx + c where b < 4ac) Summand Types If the denominator has Type 1: a unique factor ax + b, then there will be a summand of the form where A is a constant to be determined. A ax + b, Type : precisely n factors ax + b, then there will be summands of the form A 1 ax + b, A (ax + b), A 3 (ax + b),..., A n 3 (ax + b), n where the n constants A 1, A, A 3,..., A n are to be determined. Type 3: a unique factor ax + bx + c, then there will be a summand of the form Ax + B ax + bx + c, where A and B are a constants to be determined. Type 4: precisely n factors ax + bx + c, then there will be summands of the form A 1 x + B 1 ax + bx + c, A x + B (ax + bx + c), A 3 x + B 3 (ax + bx + c),..., A n x + B n 3 (ax + bx + c), n where the n constants A 1, A, A 3,..., A n and B 1, B, B 3,..., B n are to be determined.

4 Example 1.1 Let f(x) be any polynomial such that each of the given rational functions is proper: Summands: f(x) (x + 1)(x ) f(x) (x + 1) 3 (x ) f(x) (x + 1)(x ) f(x) (x + 1) (x ) A x B x A x B (x + 1) + C (x + 1) + D 3 x Ax + B x C x Ax + B x Cx + D (x + 1) + E x 1.0. 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 into partial fractions. 5x 1 5x 1 x x 5x 1 x x 5x 1 (x + 1)(x ) A x B x A(x + 1)(x ) (x + 1) + A(x ) + B(x + 1) B(x + 1)(x ) (x ) Let x 1 then 6 3A and A Let x then 9 3B and B 3. Hence 5x 1 x x x x.

5 Always check your answer by combining back into a single rational function: x x (x ) + 3(x + 1) (x + 1)(x ) 5x 1 (x + 1)(x ). () Method of Comparing Coefficients Example 1.14 Split into partial fractions x + 1 x 3 + 7x x + 1 x 3 + 7x x + 1 x(x + 7) A x + Bx + C x + 7 x + 1 Ax(x + 7) x + (Bx + C)x(x + 7) (x + 7) x + 1 A(x + 7) + (Bx + C)x Let x 0 then 1 7A and A 1 7. Comparing coefficients of x gives 1 A + B. Hence B 6 7 Comparing coefficients of x gives 0 C i.e. C 0. So x x 3 + 7x 7 x + x x 1 7x + 6x 7(x + 7) Example 1.15 Split the improper rational function into partial fractions. x 4 + x 3 x x + 4 x 3 + 4x + 5x + The first step is to use polynomial long division, x 4 + x 3 x x + 4 x 3 + 4x + 5x + x + x + 7x + 8 x 3 + 4x + 5x + The second step is to factorise the cubic denominator f(x) x 3 + 4x + 5x +. Now f(0), f(1)1, f(-1)0. By the factor theorem x + 1 is a factor. Hence x 3 + 4x + 5x + (x + 1)(x + ax + b) (x + 1)(x + 3x + ) Now take the proper rational function and split into partial fractions x + 7x + 8 x 3 + 4x + 5x + x + 7x + 8 (x + 1) (x + ) A x B (x + 1) + C x +

6 Cross multiplying: x + 7x + 8 A(x + 1) (x + ) (x + 1) Let x 1 then B. Let x then C. + B(x + 1) (x + ) + C(x + 1) (x + ) (x + 1) (x + ) A(x + 1)(x + ) + B(x + ) + C(x + 1) Comparing coefficients of x gives 1 A + C and so A 3. Hence Finally x + 7x + 8 x 3 + 4x + 5x + 3 x (x + 1) x + x 4 + x 3 x x + 4 x 3 + 4x + 5x + Algebra - Sequences and Series x + 3 x (x + 1) x + Definition.1 A sequence is an ordered set of numbers, each being called a term. Example., 5, 10, 17, 6,..., k + 1,..., is a sequence of 100 numbers, the k th term being obtained by squaring k and adding 1. The term k + 1 is known as the general term. Definition.3 A series is the sum of the terms in a sequence. Example.4 The series corresponding to the example above is The sum is (k + 1) Often, the sigma notation is used with series. The series in the example would be written as 100 k1 (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 k k1 we can say that the sum of this series is 1 3. In this section, we will concentrate on two special types of sequence, the arithmetic and geometric progressions, and their associated series.

7 Definition.5 An arithmetic progression is a sequence of the form: a, a + d, a + d, 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.6 5, 7, 9, 11, 13, 15,..., 77 is an arithmetic progression with a 5 and d. Theorem.7 The sum S n of the first n terms of an arithmetic progression is S n 1 n (a + (n 1)d) Proof: We write down the sum twice: once forwards, then backwards, and add them. S n a + (a + d) (a + (n )d) + (a + (n 1)d) S n (a + (n 1)d) + (a + (n )d) (a + d) + a S n (a + (n 1)d) + (a + (n 1)d) (a + (n 1)d) + (a + (n 1)d) Note that all of the terms in S n are the same, and there are n of them. Hence, we see that S n 1 n (a + (n 1)d) as required. Example.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, so S n 1 37 (10 + (37 1)) Note that since the last term is a + (n 1)d, we can rewrite the formula S n 1 n(a + z), where z is the last term. Example.9 Find the sum of the first n natural numbers: (n 1) + n Solution: We have a 1 and d 1, so S n 1n ( + (n 1)) 1 n(n + 1). Example.10 The 8 th term of an arithmetic progression is 1, 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:

8 13 th term a + 1d 8 8 th term a + 7d 1 subtracting 5d 15 which gives d 3. Substituting, we get a 10. Now, the sum S ( 0 + (14). 3 ) 15. Definition.11 A geometric progression is a sequence of the following form a, ar, ar, ar 3,..., ar k 1,..., ar n 1 where a and r are constants: a is the first term and r is the common ratio. Example.1 3, 6, 1, 4, 48, 96,... is a geometric progression with a 3 and r. 1, 1, 1 4, 1 8, 1 16,... is a geometric progression with a 1 and r 1. Theorem.13 The sum S n of the first n terms of a geometric progression is if r 1. Otherwise if r 1 then S n an. S n a (1 rn ) 1 r Proof: The case when r 1 is obvious, so we assume r 1. Now multiplying by r: and subtracting S n a + ar + ar ar n + ar n 1 rs n ar + ar + ar ar n 1 + ar n (1 r)s n a ar n from which we deduce the result. Example.14 Calculate the sum of the first six terms of a geometric progression with a 1 and r 10. Solution: Using the formula: S 6 1. (1 106 )

9 which is as expected as the series is Example.15 In the following geometric progression 3, 6, 1, 4, 48, 96,... after how many terms will the sum be greater than 9999? Solution: We have a 3 and r. The formula for S n is S n 3. (1 n ) 1 3 ( n 1) so we must find n for which 3 ( n 1) > This means that n 1 > 3333 or n > We can use logarithms to solve this: n > 3334 ln n > ln 3334 ln 3334 n ln > ln 3334 n > ln and using a calculator n > and so after 1 terms, the sum will be greater than 9999 (in fact it is 185). Example.16 Calculate the sum of the first eight terms of a geometric progression with a 1 and r 1. Solution: Using the formula: ( ) S (8 1) In fact, if we continue adding terms of the above progression, we get values closer and closer to. 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.17 The sum to infinity S of a geometric progression with r < 1 is S a 1 r Proof: We have S n a (1 rn ), and we note that as n gets larger, the value of r n becomes 1 r smaller, since r < 1. In fact, we can say that as n, r n 0. Thus, letting n in S n, we get which completes the proof. S a (1 0) 1 r a 1 r

10 Example.18 We check the aforementioned series: 0.3, 0.03, 0.003, , ,... which has first term a 0.3 and common ratio r 0.1. The sum to infinity is S as we would expect. 3 Algebra - Binomial Theorem a 1 r 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 (a + b) a + ab + b (a + b) 3 a 3 + 3a b + 3ab + b 3 (a + b) 4 a 4 + 4a 3 b + 6a b + 4ab 3 + b 4 (a + b) 5 a 5 + 5a 4 b + 10a 3 b + 10a b 3 + 5ab 4 + b 5 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 a n r b r, 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 Thus, the coefficients in the expansions may be obtained from the numbers in Pascal s Triangle. Denote ( ) this coefficient by n C r then each term is of the form n C r a n r b r. An alternative n notation is. An alternative way of writing n C r is r

11 Definition 3.1 k!, pronounced k factorial is the product of all integers from k down to 1, i.e. Also 0! 1. For example 5! k! k(k 1)(k ) Proposition 3. For any positive integer n, and r 0, 1,..., n n then n! C r (n r)!r! Example 3.3 In the case n 5, we compute the coefficients: 5 C 0 5! 5!0! C 1 5! 4!1! 5 C 5! 3!! 5 C 3 5!!3! 5 C 4 5! 1!4! 5 C 5 5! 0!5! 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 where n C r Example 3.5 n! (n r)!r!. (a + b) n n n C r a n r b r r0 (a + b) 5 5 C 0 a 5 b C 1 a 4 b + 5 C a 3 b + 5 C 3 a b C 4 ab C 5 a 0 b 5 a 5 + 5a 4 b + 10a 3 b + 10a b 3 + 5ab 4 + b 5.

12 An alternative way of writing is as before (a + b) 5 a a4 b a3 b a b ab4 + b 5. a 5 + 5a 4 b + 10a 3 b + 10a b 3 + 5ab 4 + b 5. Example 3.6 Expanding (1 + x) 5, we let a 1 and b x, which gives: (1 + x) 5 a 5 + 5a 4 b + 10a 3 b + 10a b 3 + 5ab 4 + b (x) + 10(x) + 10(x) 3 + 5(x) 4 + (x) x + 40x + 80x x x 5 Example 3.7 Approximate (0.98) 5 to 4 decimal places. Solution: We use the previous example, substituting x 0.01: (0.98) (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 x in the expansion of (3x 1 x )8. Solution: We need the term where x r ( 1 x )8 r x, which is clearly when r 5. Now we have 8 C 5 8! 3!5! and so the required term is 56(3x) 5 ( 1 x ) x 13608x.

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