Appendix A.1.1 Introduction As discussed in the Chapter 9 on Sequences and Series, a sequence a1, a2, ..., an, ... having infinite number of terms is called infinite sequence and its indicated sum, i.e., a1 + a2 + a3 + ... + an + ... is called an infinte series associated with infinite sequence. This series can also be expressed in abbreviated form using the sigma notation, i.e., ∞ aa1 + a2 + a3 + . . . + an + . . . = ∑ k k =1 In this Chapter, we shall study about some special types of series which may be required in different problem situations. A.1.2 Binomial Theorem for any Index In Chapter 8, we discussed the Binomial Theorem in which the index was a positive integer. In this Section, we state a more general form of the theorem in which the index is not necessarily a whole number. It gives us a particular type of infinite series, called Binomial Series. We illustrate few applications, by examples. We know the formula (1 + x)n = n C0 + n C1 x + . . . + n Cn xn Here, n is non-negative integer. Observe that if we replace index n by negative integer or a fraction, then the combinations n Cr do not make any sense. We now state (without proof), the Binomial Theorem, giving an infinite series in which the index is negative or a fraction and not a whole number. Theorem The formula mm −1 mm −1 m − 2m () 2 ()( ) 3(1+ x)=1+ mx + x + x + ... 12. 123 .. holds whenever x <1. Remark 1. Note carefully the condition | x | < 1, i.e., – 1< x < 1 is necessary when m is negative integer or a fraction. For example, if we take x = – 2 and m = – 2, we obtain (−2)(−3212 −2 =1+−2 −2 +−2 + ... (−) ()() )()1.2 or 1= 1 + 4 + 12 + . . . This is not possible 2. Note that there are infinite number of terms in the expansion of (1+ x)m, when m is a negative integer or a fraction ⎡⎛ b ⎞⎤mm ⎛ b ⎞m Consider (ab)m = ⎢a⎜1+⎟ = a ⎜1++⎥⎟⎣⎝ a ⎠⎦ ⎝ a ⎠ ⎡ b mm −1 ⎛⎞ ⎤() b 2 am ⎢1+ m ++ ...⎥= ⎜⎟a 1.2 ⎝⎠ ⎥a⎢⎣⎦ mm −1mm−1 () m−22= a + ma b + ab + ... 1.2 b This expansion is valid when <1 or equivalently when | b | < | a |.aThe general term in the expansion of (a + b)m is mr r( −)(m − 2 ... m − r +) −mm 1 )( 1 ab 1.2.3...r We give below certain particular cases of Binomial Theorem, when we assume x <1, these are left to students as exercises: 1. (1 + x) – 1 = 1 – x + x2 – x3 + . . . 2. (1 – x) – 1 = 1 + x + x2 + x3 + . . . 3. (1 + x) – 2 = 1 –2 x + 3x2 – 4x3 + . . . 4. (1 – x) – 2 = 1 +2x + 3x2 + 4x3 + . . . 1 ⎛ x ⎞− 2Example 1 Expand 1 , when | x | < 2.⎜− ⎟⎝ 2 ⎠ Solution We have 1 ⎛ 1 ⎞⎛ 1 ⎞⎛ 3 ⎞ ⎛ x ⎞− 2 ⎜−⎟ ⎜−⎟⎜ −⎟ 2⎝ 2 ⎠⎛−x ⎞⎝ 2 ⎠⎝ 2 ⎠⎛ x ⎞⎜1−⎟ =1+ ⎜ ⎟+ ⎜−⎟+ ... ⎝ 2 ⎠1 ⎝ 2 ⎠ 12 . ⎝ 2 ⎠ x 3x2 =1++ + ... 4 32 A.1.3 Infinite Geometric Series From Chapter 9, Section 9.5, a sequence a1, a2, a3, ..., an is called G.P., if ak +1 = r (constant) for k = 1, 2, 3, ..., n–1. Particularly, if we take a1 = a, then theak resulting sequence a, ar, ar2, ..., arn–1 is taken as the standard form of G.P., where a is first term and r, the common ratio of G.P. Earlier, we have discussed the formula to find the sum of finite series a + ar + ar2 + ... + arn – 1 which is given by a(1− rn )Sn = . 1− r In this section, we state the formula to find the sum of infinite geometric series a + ar + ar2 + ... + arn – 1 + ... and illustrate the same by examples. 24Let us consider the G.P. 1, , ,... 39 2Here a = 1, r = . We have 3 2 n 1−⎛⎞⎜⎟ n⎡3 ⎛⎞2 ⎤⎝⎠ n == 31S ⎢−2 ⎜⎟⎥ ... (1)3⎢ ⎝⎠⎥1−⎣ ⎦ 3 2 n⎛⎞Let us study the behaviour of ⎜⎟as n becomes larger and larger. 3⎝⎠ n 1 5 10 20 2 3 n⎛⎞ ⎜⎟⎝⎠ 0.6667 0.1316872428 0.01734152992 0.00030072866 2 n⎛⎞We observe that as n becomes larger and larger, ⎜⎟becomes closer and closer to3⎝⎠ 2 n⎛⎞zero. Mathematically, we say that as n becomes sufficiently large, ⎜⎟becomes3⎝⎠ →∞⎛⎞sufficiently small. In other words, as n , 23 n → 0 . Consequently, we find that ⎜⎟⎝⎠ the sum of infinitely many terms is given by S = 3. Thus, for infinite geometric progression a, ar, ar2, ..., if numerical value of common ratio r is less than 1, then a 1− rn n() a ar Sn = = −1−r 1− r 1− r arn In this case, rn →0 as n →∞ since || 1r < and then → 0 . Therefore,1 − r Sn → a as n →∞ .1−r Symbolically, sum to infinity of infinite geometric series is denoted by S. Thus, we have S = a 1− r For example 111 1(i) 1++ ++ ... == 2 22223 11− 2 111 112(ii) 1−+ −+ ... = == 23 122 2 ⎛ 1 ⎞ 31−−1+⎜⎟⎝ 2 ⎠ 2 Example 2 Find the sum to infinity of the G.P. ; −55 −5 , , ,.... 416 64 −51Solution Here a = and r =− r <. Also || 1.44 −5 −5 44Hence, the sum to infinity is = =− 1.151+ 44 A.1.4 Exponential Series Leonhard Euler (1707 – 1783), the great Swiss mathematician introduced the number e in his calculus text in 1748. The number e is useful in calculus as π in the study of the circle. Consider the following infinite series of numbers 11111++ ++ + ... ... (1)1! 2!3! 4! The sum of the series given in (1) is denoted by the number e Let us estimate the value of the number e. Since every term of the series (1) is positive, it is clear that its sum is also positive. Consider the two sums 111 1+++ ... ++ ... ... (2)3! 4! 5! n! 111 1 and +++ .... ++ ... ... (3)234 2n−1222 Observe that 1111 11 = and 2 = , which gives < 23!6 24 3!2 1111 11 = and 3 , which gives < 3 = 4!24 28 4!2 1111 11 = and 4 = , which gives < 4.5!120 2 16 5!2 Therefore, by analogy, we can say that 11<−1 , when n > 2 n!2n We observe that each term in (2) is less than the corresponding term in (3), ⎛ 111 1 ⎞⎛ 111 1 ⎞+++ ... + < +++ ... ++ ... Therefore ⎜ ⎟⎜ 234 2n−1 ⎟ ... (4)⎝ 3! 4! 5! n!⎠⎝ 222 ⎠ ⎛ 11 ⎞Adding ⎜1++ ⎟ on both sides of (4), we get,⎝ 1! 2! ⎠⎛ 11 ⎞⎛ 111 1 ⎞1++ ++++ ... ++ ... ⎜⎟⎜ ⎟⎝ 1! 2!⎠⎝ 3! 4!5! n! ⎠ ⎧⎛ 11 ⎞⎛ 111 1 ⎞⎫ ... (5)< ⎨⎜1+ + ⎟+⎜ 2 + 3 + 4 + ... + 2n−1 + ...⎟⎬ ⎩⎝ 1! 2! ⎠⎝ 222 ⎠⎭ ⎧⎛ 1111 1 ⎞⎫ = ⎨1+⎜1++ ++ + ... ++ ... 2342n−1 ⎟⎬ ⎩⎝ 22 22 ⎠⎭ 1 = 1+=1+ 2 = 311− 2 Left hand side of (5) represents the series (1). Therefore e < 3 and also e > 2 and hence 2 < e < 3. Remark The exponential series involving variable x can be expressed as 23 n x xxx x e 1 +++... =+ ++... 1!2! 3! n! e2x+3Example 3 Find the coefficient of x2 in the expansion of as a series in powers of x. Solution In the exponential series 23xx x ex =1++ ++ ... 1!2! 3! replacing x by (2x + 3), we get (2x + 3)(2x + 3)2 2x+3e =1++ + ...1! 2! (2x +3)nx(3+2 ) n Here, the general term is = . This can be expanded by then! n! Binomial Theorem as 1 ⎡ nn n−1 nn−22 n ⎤ ⎣3 + 1 ()+ C3 ()+ ... + 2x ⎦.C3 2x 22x ()n! nn−22 22C3Here, the coefficient of x2 is . Therefore, the coefficient of x2 in the whole n! series is ∞ nn−22 ∞ n−2(−13C3 2 nn )22∑ = ∑n! n!n=2 n=2 ∞ n–23 = ∑ [using n! = n (n – 1) (n – 2)!]2 (n − 2!)n=2 ⎡ 23 ⎤33 321++ ++ ... = ⎢⎥1!2! 3! ⎣⎦ =2e3 . Thus 2e3 is the coefficient of x2 in the expansion of e2x+3. Alternatively e2x+3 = e3 . e2x 3 ⎡ 2xx 2 (2 ) 3 ⎤(2 ) x e 1++ + +...⎢⎥= 1!2! 3! ⎣⎦ 2 32 3e . = 2eThus, the coefficient of x2 in the expansion of e2x+3 is 2! Example 4 Find the value of e2, rounded off to one decimal place. Solution Using the formula of exponential series involving x, we have 23 n x xxx x e 1 + ++...=+ ++... 1! 2! 3! n! Putting x = 2, we get 2345622 22 2221 ++++++...e =+ 1! 2! 3! 4! 5! 6! 4 2 44 = 12 2 ++++ + ...++ 3315 45 ≥ the sum of first seven terms ≥ 7.355. On the other hand, we have 2345 23 2 ⎛ 22 22 ⎞ 2 ⎛ 22 2 ⎞ e <⎜1++ ++⎟+⎜1++ 2 + 3 + ...⎟1! 2!3! 4! 5! 66 6⎝ ⎠⎝⎠⎛⎞ 4 ⎛ 1 ⎛⎞2 ⎞⎜ 1 ⎟ 214 = 7 +⎜1+ +⎜⎟+ ...⎟ = 7 +⎜ ⎟ = 7 += 7.4 .15 ⎜ 3 ⎝⎠ ⎟ 153 15⎝ ⎠⎜ 1−⎟ ⎝ 3 ⎠Thus, e2 lies between 7.355 and 7.4. Therefore, the value of e2, rounded off to one decimal place, is 7.4. A.1.5 Logarithmic Series Another very important series is logarithmic series which is also in the form of infinite series. We state the following result without proof and illustrate its application with an example. Theorem If | x | < 1, then 23xxloge (1 + x)= x −+− ... 23 The series on the right hand side of the above is called the logarithmic series. The expansion of loge (1+x) is valid for x = 1. Substituting x = 1 in the expansion of loge (1+x), we get 111log 2 =1– + – +...e 234 αβ are the roots of the equation x2 +Example 5 If ,− px q = 0 , prove that 22 33 2 23α+β α+β log e (1 + px + qx ) =(α +β ) x − x + x − ... 23 22 33 2233⎡α x α x ⎤⎡ β x β x ⎤α− + ... +β x −+ ... Solution Right hand side = ⎢ x −⎥⎢ −⎥23 23⎣ ⎦⎣⎦ = log (1+α x)+ log 1 ( +β x)e = log e (1 +α+β ( ) x +αβ x2 ) 2= log e (1+ px +qx ) = Left hand side. Here, we have used the facts α+β = p and αβ = q . We know this from the given roots of the quadratic equation. We have also assumed that both | α x |< 1 and | β x |< 1. — —

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