MATH19832/19542 Mathematics 0C2/1C2. General information for the academic year 2012-2013: Lecturer: Dr Theodore Voronov, Room 2.109, School of Mathematics, Alan Turing Building, email: [email protected] . Lectures: Tuesdays 11am in Renold/D7 and Wednesdays 9am in Sackville/C14. Tutorials: Thursdays 11am : George Begg/C3 (optometry only) and Fridays 9am : rooms announced on the webpage (foundation). Assessment: coursework 1 (deadline in week 4): 10%, coursework 2 (deadline in week 10): 10%, 2 hour end of semester 2 exam: 80%. Course webpage: http://www.maths.manchester.ac.uk/service/MATH19832/ .

§3

More Algebra: Progressions and Binomial Theorem

3.1

Sequences. Arithmetic progressions

Definition 1. A sequence is an ordered set of numbers {ak }, each being called a term. A sequence can be finite or infinite. Example 1. 2, 5, 10, 17, 26, ..., k 2 + 1, ..., 10001 is a finite sequence of 100 numbers, the kth term being obtained by squaring k and adding 1. The term ak = k 2 + 1 is known as the general term. Definition 2. A series is the sum of the terms in a sequence: a1 + a2 + . . . + ak + . . . Example 2. The series corresponding to the example above is 2 + 5 + 10 + 17 + 26 + ... + (k 2 + 1) + ... + 10001 The sum is 338450. (This follows from a general formula that one may deduce for the case of n terms.) Often, the sigma notation is used with series. The series in the example would be written as

100 X (k 2 + 1) . k=1

1

Note that it is possible to have sequences and series with infinitely many terms. The sum of an infinite series is defined as the limit:   a1 + a2 + a3 + . . . = lim a1 + a2 + a3 + · · · + an n→∞

(at the left-hand side there is an infinite series and at the right-hand side, under the limit sign, there stands the sum of the first n terms). The limit may or may not exist. We shall not develop here general theory of infinite series (which deals with various “convergence tests”, i.e., tests for whether or not the limit and hence the infinite sum make sense), but concentrate on one particular class of infinite series. Example 3. Consider the sum ∞ X 3 = 0.3 + 0.03 + 0.003 + 0.0003 + ... = 0.3333... . 10k k=1

We can say that the sum of this series is 13 . (This is precisely the expression that we get when we divide 1 by 3 using a calculator.) We shall come back to that type of infinite series in the next subsection. Now we shall concentrate on a special types of sequences, called the arithmetic progressions, and their associated (finite) series. Given a sequence a1 , a2 , a3 , a4 , ..., ak , ak+1 , ..., (finite or infinite), we can associate with it the corresponding sequence of differences: d1 = a2 − a1 , d2 = a3 − a2 , d3 = a4 − a3 , . . . , dk = ak+1 − ak , . . . . If we view the terms ak as analogous to the values of a function f (x), so k is regarded as analogous to the argument x (unlike x, the number k takes only discrete values such as 1, 2, 3, . . . ), then introducing the differences dk = ak+1 − ak is somewhat similar to taking the derivative1 f 0 (x) of a function f (x). The sequence {dk } describes the “growth” or “rate of change” of the original sequence {ak }. Example 4. If for all k, dk = 0, that means that ak+1 − ak = 0, i.e., all the terms ak are the same, ak = a. This is an analog of a constant function. (Compare with a function f (x) having a zero derivative, which implies that it is a constant.) 1

The derivative is actually defined via the differences f (x + h) − f (h), as the limit of the difference ratio
f (x + h) − f (h) /h when h → 0. Unlike that, for sequences, the “step” h of the argument k is fixed, it is always 1, so there is no way of taking any limits, and we have to deal with the difference ak+1 − ak as such.

2

The next in complexity case is the one where the differences dk are not necessarily zero, but are the same for all k. This means “uniform growth” and is analogous to the linear functions. There is a special name for that. Definition 3. An arithmetic progression is a sequence {ak } such that the corresponding differences dk = ak+1 − ak do not depend on k, d1 = d2 = · · · = d, so that the sequence {ak } is 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 = a1 is the first term and d is called the common difference (or simply difference). Each term is obtained by adding d to the previous term. Note the formula for the kth term: ak = a + (k − 1)d . For example, a1 = a. Example 5. The sequence 5, 7, 9, 11, 13, 15, ..., 77 is an arithmetic progression with a = 5 and d = 2. One can see that here we have 37 terms. Indeed, for the last term we have 77 = 5 + 2 · (37 − 1) = 5 + 2 · 36 = 5 + 72 = 77. Theorem 1. The sum Sn of the first n terms of an arithmetic progression with the first term a and difference d is Sn =

 1  n 2a + (n − 1)d . 2

Proof. We write down the sum twice: once forwards, then backwards, and add them. + (a + d)

+...+

(a + (n − 2)d) + (a + (n − 1)d)

Sn

= a

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 claimed. Remark 1. Note that since the last term is an = a + (n − 1)d, we can rewrite the formula for the sum of an arithmetic progression in the form (easy to remember) Sn =

1 n(a1 + an ) . 2

3

Example 6. Find the sum: 5 + 7 + 9 + 11 + 13 + 15 + ... + 77 . Solution: We have an arithmetic progression with 37 terms. Here a = 5 and d = 2, so Sn = 1 37 (10 2

+ (37 − 1)2) = 1517.

Example 7. 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 = 12 n (2 + (n − 1)) = 12 n(n + 1). Example 8. The 8th term of an arithmetic progression is − 21 , 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 th

8

term

subtracting

= a + 12d

= −8

= a + 7d

= − 12

5d

= − 15 2


which gives d = − 32 . Substituting, we get a = 10. Now, the sum S15 = 21 15 20 + 14 · (− 32 ) = − 15 . 2

4

3.2

Geometric progressions

Now we shall consider another type of ‘progression’, called ‘geometric progression’. Consider a sequence a1 , a2 , a3 , a4 , ..., ak , ak+1 , ..., (finite or infinite), such that all the terms are non-zero: ak 6= 0 for all k. Then, instead of considering the differences dk = ak+1 − ak , as we did in the previous subsection, we may consider the ratios: r1 = a2 /a1 , r2 = a3 /a2 , r3 = a4 /a3 , . . . , rk = ak+1 /ak , . . . (we need indeed that all terms be non-zero to be able to divide). If all the ratios are 1, then a1 = a2 = a3 = . . ., so we again arrive at a constant sequence. Now, the next interesting case is when the ratios rk do not depend on k. The special name for that is as follows. Definition 4. 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 called the common ratio (or simply, ratio). Here the common term is ak = ark−1 . (This is similar to an arithmetic progression considered before, but now we measure the “growth” of a sequence in a multiplicative way, not additive. So instead of ak = a1 + (k − 1)d, as it is for an arithmetic progression, we now have ak = a1 rk−1 .) Example 9. 3, 6, 12, 24, 48, 96, ... is a geometric progression with a = 3 and r = 2. Example 10. 1 1 1 1 1, , , , , ... 2 4 8 16 is a geometric progression with a = 1 and r = 12 . Theorem 2. 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.

5

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 11. Calculate the sum of the first six terms of a geometric progression with a = 1 and r = 10. Solution: Using the formula: S6 =

1 · (1 − 106 ) −999999 = = 111111 . 1 − 10 −9

Alternatively, we may notice that this is the series 1 + 10 + 100 + 1000 + 10000 + 100000 = 111111 . Example 12. 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 . This is the condition for n. We can use logarithms to solve this: 2n > 3334 ⇔ ln(2n ) > ln(3334) ⇔ n ln 2 > ln 3334 ⇔ n >

ln 3334 . ln 2

Using a calculator we may find that, approximately, ln 3334 ≈ 8.112 , ln 2 ≈ 0.693, and ln 3334 8.112 ≈ ≈ 11.703 . ln 2 0.693 We can see that 211 = 2048 < 3334 (because 210 = 25 · 25 = 32 · 32 = 1024) and 212 = 4096 > 3334. So n = 12 is indeed the solution. After exactly 12 terms, the sum will be greater than 9999 (in fact, the sum of 12 terms is 12285).

6

Example 13. Calculate the sum of the first eight terms of a geometric progression with a = 1 and r = 12 . Solution: Using the formula:  1· 1− S8 = 1−


1 8 2 1 2

 =

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 3. The sum of an infinite geometric progression S with |r| < 1 is S=

a , 1−r

where a is the first term and d is the common ratio. 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 = lim Sn = n→∞

a (1 − 0) a = 1−r 1−r

which completes the proof. Example 14. We check the aforementioned series: 0.3 + 0.03 + 0.003 + 0.0003 + 0.00003 + ... = 0.3333... which has first term a = 0.3 and common ratio r = 0.1. The sum is S = 3 9

=

1 3

as we would expect.

7

a 0.3 = = 1−r 1 − 0.1

3.3

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. We can obtain directly the following: (a + b)1

=

a+b

2

=

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

(a + b)

.......................................................................................... In each term of the expansions, the power of a plus the power of b is equal to n. That is, each term is of the form an−r br , where r takes the values 0, 1, ..., n successively, multiplied by a number called a binomial coefficient. As we shall shortly see, the coefficients in the expansions may be obtained as follows. Consider the so called Pascal’s Triangle: Pascal’s Triangle 1 1 1 1 1 1 1 1 1 ...

8 ...

1 4

10 20

35 56

...

3

10

21

1

6

15

28 ...

3

5

7

2

4

6

1

5 15

35 70

...

1 6 21

56 ...

1 1 7 28

...

1 8

...

1 ...

...

The rule for forming these numbers should be clear. A number in each row is obtained as the sum of the two nearest neighbors in the previous row. (Sometimes Pascal’s Triangle is written slightly differently, so that the numbers are arranged in the shape of a right triangle like the coefficients in the expansions above. Then the rule is that to obtain a number in a row, one adds two numbers in the previous row, the one strictly above and one to the left.)

8

If we introduce the notation Cnr for the rth number in the nth row of Pascal’s triangle, then the rule defining these numbers reads as follows: C00 = 1 ,

C0r = 0 for r > 0 ,

Cn0 = 1 ,

r−1 r Cnr = Cn−1 + Cn−1

Alternative notations: Cnr

for r > 0 .

  n = Cr = . r n

Pronounced “n choose r”, because these numbers can be related with combinatorial choice: the
number Cnr = nr is exactly the number of combinations of r elements of n elements. Theorem 4 (The Binomial Theorem). If n is a positive integer, and a and b are any numbers then n

(a + b) =

n X

Cnr an−r br ,

r=0

where the coefficients Cnr = n Cr =


n r

are the numbers in Pascal’s triangle defined above.

Remark 2. Where is the ‘binomial’ here? A binomial is a polynomial with two terms. We may consider a binomial ax + b instead of a + b above. Then the theorem gives a formula for the nth power of such a binomial. Proof of the Binomial Theorem. The statement is clearly true for small values of n such as n = 1 and n = 2, where it can be checked directly. Since (a + b)n+1 = (a + b)n (a + b), it is possible to deduce the statement for all natural n by assuming that the formula holds for some n and deduce it for n + 1 by multiplying through by a + b. Here is where the defining property of the numbers in Pascal’s triangle arises. So assume the validity of the formula (a + b)n = Cn0 an + Cn1 an−1 b + Cn2 an−2 b2 + . . . + Cnr−1 an−r+1 br−1 + Cnr an−r br + . . . , for some n > 0. (Note that Cn0 = 1 for all n.) Multiply both sides by a + b. We obtain   (a + b)n+1 = Cn0 an+1 + Cn1 an b + Cn2 an−1 b2 + . . . + Cnr an−r+1 br + . . . +   0 n 1 n−1 2 r−1 n−r+1 r b + ... = Cn a b + Cn a b + . . . + Cn a


Cn0 an+1 + Cn1 + Cn0 an b + Cn2 + Cn1 an−1 b2 + . . . + Cnr + Cnr−1 an−r+1 br + . . . = r 0 1 2 an+1−r br + . . . , Cn+1 an+1 + Cn+1 an b + Cn+1 an−1 b2 + . . . + Cn+1

which is the desired formula for n+1. Therefore we can prove the formula for all n = 1, 2, 3, 4, . . . starting from the known cases and increasing the power by one at each step.

9

Hence the binomial coefficients are contained in Pascal’s triangle. Using Pascal’s triangle is the quickest way of finding the binomial expansion for small n. However, it may be useful to have a closed expression for the binomial coefficients for arbitrary n and r. We need one more definition. Definition 5. The number denoted 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. Example 15. 1! = 1, 2! = 2, 3! = 3 · 2 = 6, 4! = 4 · 3 · 2 = 24, 5! = 5 · 4 · 3 · 2 · 1 = 120. Proposition 1. For every positive integer n and for all r = 0, 1, ..., n, Cnr =

n! . r!(n − r)!

Proof. We check that the RHS, i.e., the number

n! r!(n−r)!

satisfies the relations for the (non-zero)

numbers in Pascal’s Triangle. Denote temporarily C˜nr =

n! r!(n − r)!

(with the tilde). First note that when n = 0, we have C˜00 =

0! = 1 = C00 . 0!(0 − 0)!

Now, for all positive n and r = 1, 2, . . . , n, we have C˜nr + C˜nr−1 =

n! n!(n − r + 1) + n!r (n + 1)n! n! + = = = r!(n − r)! (r − 1)!(n − r + 1)! r!(n − r + 1)! r!(n − r + 1)! (n + 1)! r = C˜n+1 . r!(n + 1 − r)!

Therefore the numbers C˜nr are calculated by the same rule as the numbers in Pascal’s triangle, i.e., C˜ r = C r as claimed. n

n

Corollary 1.     n n = r n−r Example 16. In the case n = 5, we can compute the coefficients:

10

C50

=

5! 5!0!

=

5·4·3·2·1 5·4·3·2·1·1

=1

C51

=

5! 4!1!

=

5·4·3·2·1 4·3·2·1·1

=5

C52

=

5! 3!2!

=

5·4·3·2·1 3·2·1·2·1

= 10

C53

=

5! 2!3!

=

5·4·3·2·1 2·1·3·2·1

= 10

C54

=

5! 1!4!

=

5·4·3·2·1 1·4·3·2·1

=5

C55

=

5! 0!5!

=

5·4·3·2·1 1·5·4·3·2·1

=1

giving the fifth row of Pascal’s triangle. Example 17. (Here we have used an alternative notation for the coefficients.)             5 5 0 5 4 5 3 2 5 2 3 5 5 0 5 5 4 (a + b) = ab + a b+ ab + ab + ab + ab 0 1 2 3 4 5 = a5 + 5a4 b + 10a3 b2 + 10a2 b3 + 5ab4 + b5 . When expressing the binomial coefficients using factorials, it is convenient to get rid of common factors in the numerator and denominator of the fraction:

Cnr

  n n! n(n − 1)...(n − r + 1)(n − r)...3 · 2 · 1 n(n − 1) · ... · (n − r + 1) = = = = . r r!(n − r)! r!(n − r)...3 · 2 · 1 r!

Therefore the first few terms in the binomial expansion (for general n) are as follows: (a + b)n = an + n an−1 b +

n(n − 1) n−2 2 n(n − 1)(n − 2) n−3 3 a b + a b + ... . 2 6

Example 18. 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

11

Example 19. 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 20. 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! 8·7·6 = = 8 · 7 = 56 , 3!5! 3!
3 and so the required term is 56(3x)5 − x1 = −56 · 35 x2 = −56 · 243 = −13608x2 . C85 = C83 =

12