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The real line R consists of all the real numbers x Î (-¥ , ¥ ) and is synonimous with the collection of all decimals x = ± t_mt_m-1…t₁t₀× d₁d₂d₃ … d_n …, with the usual arithmetic operations of addition and multiplication, subtraction, division etc. Here ‘± ’ represents the sign of the decimal, ‘t_mt_m-1…t₁t₀^’ the integer part and ‘d₁d₂d₃…d_n…’ the decimal or the fractional part. The left most digit t_m is the most significant digit, t_m-1 is the next most significant digit, and so on till t₀, and then the decimal digits d₁, d₂, … in that order are the most significant ones. Note that the decimal part continues ad infinitum and that a terminating decimal stands for the number with the rest of the decimal digits assumed as zero. If at least one of the digits at the unit (t₀), tenth (t₁), hundredth (t₂), … place, or at any of the first (d₁), second (d₂), … , n-th (d_n) decimal places is non-zero a decimal with a ‘+’ sign represents a positive real number, and one with a ‘-’ sign a negative real number. If all of t_i’s and d_i’s are zero the decimal represents the number zero. Given two decimals we also know which is larger. Let a, b Î R = (-¥ , ¥ ) and b > a. Sub-intervals of the real line R are of the form: open (-¥ , b), semi-open (-¥ , b], open (a, ¥ ), semi-closed [a, ¥ ), open (-¥ , ¥ ), open (a, b), semi-open (a, b], semi-closed [a, b), closed [a, b]. Of these the last four are finite intervals and the first five infinite intervals. We would use Z to denote the set of integers, and N for the set of non-negative integers, or the natural numbers. The real line R could be picturized as an infinite straight line with 0, and the units ± 1, ± 2, … marked on it. Given a point x on the real line one could find its associated decimal representation ± t_m…t₁× d₁…d_n …, and conversely. The absolute value |x| of the decimal x = ± t_m…t₁× d₁ …d_n … is the number t_m…t₁× d₁…d_n… . The integral part of the positive number t_m…t₁× d₁…d_n… is the decimal t_m…t₁× 0, or the number t₀ + 10 t₁ + 100 t₂ + 10^mt_m.

The real numbers satisfy: a + (b + c) = (a + b) + c (associativity of addition), a(bc) = (ab)c (associativity of multiplication), a + b = b + a (commutativity of addition), a b = ba (commutativity of multiplication), a(b + c) = ab + ac (distributivity of multiplication over addition), and the triangle inequality: |a+b| £ |a| + |b|, a, b Î R .

Exercise: Prove that: (i) |a-b| ³ |a| - |b|, in which the equality holds iff ab ³ 0 and |a| ³ |b|; (ii) max {a, b} = [(a+b) + |a-b|]/2; (iii) min {a, b} = [(a+b) - |a-b|]/2. (The results (ii) and (iii) imply that picking up the largest and the smallest numbers in a list could be done through arithmetic operations!).

A subest S of R is called upper bounded, or, bounded above if for some M Î R , x £ M for all x Î S. Such an M is called an upper bound for S. The least upper bound (lub) of a set S is the smallest upper bound of S, i.e., an M such that x £ M for all x Î S and such that for any e > 0 there is an x Î S such that x > M-e . A constructive procedure to show the existence of lub of an upper bounded set is as follows: Without loss of generality (i.e., by adding to each element a fixed positive number and then dividing by a high power of 10, if necessary) we could assume that all elements of S are less than 1 and that at least one element is greater than zero. Consider the subset of S of numbers having the largest first decimal digit. Call this digit as d₁. Next consider the subset of this set having the largest second decimal digit. Call this digit as d₂. Continue this procedure to get d₃, d₄, … , d_n, … ad infinitum. It then is clear that M = 0× d₁d₂d₃ … d_n … is the least upper bound of S.

S is called lower bounded, or, bounded below if for some m Î R , x ³ m for all x Î S. Such an m is a lower bound for S. The greatest lower bound (glb) of a set S is the largest lower bound of S, i.e., an m such that x ³ m for all x Î S and such that for any e > 0 there is an x Î S such that x < m+e . Similar to the construction of lub we can constructively show that each lower bounded set has a glb.

S is called bounded if there is an M Î R such that |x| £ M for all x Î S. Thus, a bounded set is both upper and lower bounded, and conversely.

If S is a subset of R , sup S stands for lub of S if it is upper bounded and ¥ , otherwise. Similarly, inf S denotes glb of S if it is lower bounded and -¥ otherwise. Thus sup S and inf S are meaningful for all subsets of R , whereas lub S and glb S make sense only if S is, respectively, upper and lower bounded.

For a subset S of R , max S denotes the largest, and, min S denotes the smallest element of S. Such elements need not exist. If S is an upper bounded subset of R , sup S denotes the smallest M such that for all x Î S, x £ M. If S is not bounded above we define sup S to be equal to +¥ . If S is a lower bounded subset of R , inf S denotes the largest m such that for all x Î S, x ³ m. If S is not bounded below we define inf S to be equal to -¥ . Note that given any e > 0, howsoever small, there is an element x in S such that x > sup S - e . Similarly, there is an element x in S such that x < inf S + e .

Theorem. If S is a subset of R , then: (i) S has a least upper bound iff S is bounded above; (ii) S has a greatest lower bound iff S is bounded below; and (iii) S has a least upper bound as well as a greatest lower bound iff S is bounded.

Let x_n, n ³ 1 be real numbers. The notation {x_n}º {x_n}_n³ 1 is an abbreviation for a sequence {x₁, x₂, x₃, x₄, … }. (Note that the x_n’s in a sequence {x_n} need not be distinct, whereas the x_n’s in a countable set S = {x₁, x₂, x₃, x₄, …} are supposed to be distinct). If n₁ < n₂ < n₃ < … are natural numbers, the sequence {x_nk}_k³ 1 is called a subsequence of the sequence {x_n}_n³ 1.

{x_n} is called bounded if there exists a constant M such that |x_n| £ M for all n. It is called upper bounded if for some M, x_n £ M for all n, and lower bounded if for some M, x_n ³ M for all n.

A sequence {x_n}_nÎ N is called convergent, or is said to converge if there is a real number x such that given an arbitrary e > 0, there is an integer N such that |x_n – x| < e , for all n > N. The number x is said to be the limit of the sequence and we write: lim_n® ¥ x_n = x, or that x_n ® x.

It is immediate from the definiton of the convergence of a sequence that {x_n}_n³ 1 is convergent iff {x_n}_n³ m is convergent where m ³ 1 is any natural number. What it means is that a finite number of terms of a sequence can be dropped from it without affecting the convergence or divergence of the sequence. It is also clear that every convergent sequence is bounded.

We call x a limit point of a sequence {x_n} if given an arbitrary d > 0, x_n Î (x-d , x+d ) for an infinity of n’s. Similarly, a is called a limit point of a set S if for each d > 0, the set SÇ (x-d , x+d ) has an infinity of points in it.

If {x_n} is an upper bounded sequence lim sup_n® ¥ x_n denotes the largest of the limit points of the sequence. If the sequence is not bounded above lim sup_n® ¥ x_n is taken to be +¥ . If {x_n} is a lower bounded sequence lim inf_n® ¥ x_n denotes the smallest of the limit points of the sequence. If the sequence is not bounded below lim sup_n® ¥ x_n is taken to be -¥ .

The Bolzano-Weierstrass Principle. Every bounded sequence {x_n}_{n³ 1} of real numbers possesses a convergent subsequence {x_nk}_{k³ 1}.

Proof: After division by a power of 10 and a translation we could assume the sequence {x_n} to be contained in [0, 1). Consider the ten semi-closed subintervals I₁₀ = [0, 10^-1), I₁₁ = [10^-1, 2´ 10^-1) …, I₁₉ = [9´ 10^-1, 1), each of length 10^-1. An infinity of x_n’s must lie in one of these intervals, I_1d1, say. Pick one member x_n1 from I_1d1 and then divide I_1d1 into 10 semi-closed subintervals I₂₀, … , I₂₉, each of lengh 10^-2. One of these, I_2d2, say, contains an infinity of the points of the sequence. Pick one member x_n2, (n₂ > n₁), from I_2d2 and then divide I_2d2 into 10 semi-closed subintervals I₃₀, … , I₃₉, each of lengh 10^-3. Continuing like this we get the subsequence {x_nk} of {x_n} convergent to the decimal x = 0× d₁d₂d₃d₄d₅ … d_k… . #

{x_n} is called Cauchy if for each e > 0 there exists an integer N such that:|x_n–x_m| < e , for all m, n > N.

The Cauchy Convergence Criterion. A real sequence {x_n} is convergent iff it is Cauchy.

Proof: A sequnce satisfying the Cauchy criterion is bounded. Hence by the Bolzano-Weierstrass principle there exists a convegent subsequence {x_nk}, say, with limit x. Hence, there is an M such that |x_nk – x| < e /2, k > M. Also, given an e > 0, there is an N such that |x_n – x_m| < e /2, for all m, n > N. Choosing k > M large enough so that n_k > N, we have |x_n – x| < |x_n – x_nk| + | x_nk - x| < e /2 + e /2 = e , n > N. Conversely, let {x_n} be convergent, say, {x_n} ® x. Then given an e > 0, there is an N such that |x_n – x| < e /2, for all n > N. Hence |x_n – x_m| £ |x_n – x| + |x – x_m| < e /2 + e /2 = e , for all m, n > N, i.e., {x_n} is Cauchy. #

The Cauchy convergence criterion could be alternately stated as follows: {x_n} converges iff for arbitrary n > m, |x_n - x_m| ® 0, as m ® ¥ . The property that every Cauchy sequence in R converges to a point in R is termed as the completeness of R .

Exercise. Prove that: (i) lim_n® ¥ x_n exists iff lim inf_n® ¥ x_n = lim sup_n® ¥ x_n.; (ii) if x_n £ x, lim sup_n® ¥ x_n £ x.; (ii) if x_n ³ x, lim inf_n® ¥ x_n ³ x; (iii) if x_n £ y_n, lim sup_n® ¥ x_n £ lim inf_n® ¥ y_n; (iii) if x_n £ y_n £ z_n, and lim sup_n® ¥ x_n = lim inf_n® ¥ z_n = c, then lim_n® ¥ y_n = c (sandwich theorem).

A sequence {x_n} is called: (i) increasing if x_n+1 > x_n for all n, (ii) decreasing if if x_n+1 < x_n for all n, (iii) non-increasing if x_n+1 £ x_n for all n, (iii) non-decreasing if x_n+1 ³ x_n for all n, (iv) monotone if {x_n} is either non-increasing or non-decreasing, and (v) strictly monotone if it is increasing or decreasing. {x_n}¯ L means {x_n} is non-increasing and lim x_n = L. {x_n}­ L means {x_n} is non-decreasing and lim x_n = L.

Theorem. Every bounded monotone sequence {x_n} is convergent.

Proof: Using the Bolzano-Weierstrass principle, onsider a convergent subsequence {x_nk} ® L. Then given an arbitrary e > 0, there is a K such that | x_nk - L| < e , k > K. Taking N = n_K+1, irrespectively of whether {x_nk} is non-decreasing or non-increasing, |x_n - L| < e , n > N. Hece {x_n} is Cauchy and so convergent. #

Alternately, if {x_n} is monotone but not Cauchy, for some e > 0, there exist pairs (m_k, n_k), k ³ 1, such that m_k < n_k < m_k+1 < n_k+1, k ³ 1 and |x_mk - x_nk| ³ e . Then |x_mk – x_mk+p| ³ pe ® ¥ , as p ® ¥ . Hence {x_n} is un bounded. #

Let lim_n® ¥ a_n = a, lim_n® ¥ b_n = b, and lim_n® ¥ c_n = c ¹ 0. Then, (i) lim_n® ¥ (a_n+b_n) = a+b, (ii) lim_n® ¥ (a_n-b_n) = a-b, (iii) lim_n® ¥ (a_nb_n) = ab, (iv) lim_n® ¥ (1/c_n) = 1/c, and (v) lim_n® ¥ (a_n/c_n) = a/c.

Proof: For (i) and (ii), given an e > 0, choose N such that |a_n-a|, |a_n-a| < e /2, n > N. Then |a_n+b_n –(a+b)| £ |a_n-a| + |b_n-b| < e /2 + e /2 = e , and |a_n-b_n –(a-b)| £ |a_n-a| + |b_n-b| < e /2 + e /2 = e , n > N. # For (iii) given an 1 > h > 0 choose N such that |a_n-a| < h , |b_n-b| < h , n > N. Then, for n > N, |a_nb_n – ab| £ |a_n–a||b_n| + |a||b_n – b| £ h (|b|+h ) + |a|h < h (|b|+1) + |a|h . Taking h = e /(|a|+|b|+1), we have |a_nb_n – ab| < e , n > N. # For (iv) given an |c|/2 > h > 0 choose N such that |c_n-c| < h , n > N. Note that this implies |c_n| ³ |c|-h > |c|/2. Then, for n > N, |1/c_n – 1/c| £ |c_n–c|/(|c||c_n|) < h /(|c||c|/2) = 2h /c². Taking h = e c²/2, we have |1/c_n – 1/c| < e , n > N. # Finally (v) follows from (iii) and (iv). #

Let a_n be real numbers. Just like a sequence {a_n} is an infinite extension of an n-tuple {a₁, a₂, …, a_n}, an infinite sum or a series S _n³ 1 a_n º a₁ + a₂ + … + a_n + … , is an infinite extension of the notion of a finite sum a₁+a₂+…+a_n of n terms. The sum s_n = a₁+a₂+…+a_n is called the n-th partial sum of the series S _n³ 1 a_n. To associate a possible meaning to a series, we call S _n³ 1 a_n convergent if the sequence {s_n}_n³ 1 of its partial sums is convergent. The limit s = lim_n® ¥ s_n, if it exists, is called the sum of the series. If lim_n® ¥ s_n, does not exist the series is said to be divergent, i.e., if a series is not convergent it is divergent and then the notion of its sum remains undefined. If {s_n} ® ¥ , one says that S _n³ 1 a_n diverges to +¥ . Similarly, if {s_n} ® -¥ , one says that S _n³ 1 a_n diverges to -¥ . A series which is divergent, but neither diverges to +¥ nor to -¥ , is called oscillatory.

If the associated series S _n³ 1 |a_n| is convergent, the series S _n³ 1 a_n is called absolutely convergent. If {b_n}_n³ 1 is a permutation of the sequence {a_n}_n³ 1, S _n³ 1 b_n is called a re-ordering or a re-arrangement of the series S _n³ 1 a_n. Each subsequence {s_nk}_k³ 1 of {s_n}_n³ 1 defines a re-grouping S _k³ 1 (a_nk-1+1 + … + a_nk), (n₀ = 1), of the series S _n³ 1 a_n. In testing the convergence of a series, the corresponding result on the sequences implies that a finite number of terms of a series can be dropped without affecting its convergence or divergence.

The Cauchy Convergence Criterion. Let s_n denote the n-th partial sum of a series S _{n³ 1} a_n. The series conveges iff for each e > 0, there exists a N = N(e ) such that |a_m+1+…+a_n| = |s_n-s_m| < e for all n > m > N.

The Cauchy convergence criterion for a series is equivalent to the following statement: The series S _n³ 1 a_n conveges iff for arbitrary n > m > N, |s_n - s_m| ® 0, N ® ¥ . In particular: if S _n³ 1 a_n then a_n ® 0, n ® ¥ .

The Weierstrass’ M-Test. A series S a_n of positive terms is convergent if there is a constant M such that s_n = a₁+…+a_n < M, for all n ³ 1.

Comparison Test. If S _{n³ 1} b_n is a convergent series of positive terms and |a_n| £ b_n, n ³ 1, then S _{n³ 1} a_n is convergent.

Proof: The result follows as |a_m+1 + … + a_n| £ |b_m+1 + … + b_n|. #

Corollary. If a_n, b_n > 0, and a_n+1/a_n £ b_n+1/n_n, n ³ 1, then S a_n converges if S b_n converges.

Proof: We have a_n = (a_n/a_n-1) (a_n-1/a_n-2)… (a₂/a₁)a₁ £ (b_n/b_n-1) (b_n-1/b_n-2)… (b₂/b₁)a₁ = (a₁/b₁)b_n, n ³ 1. #

Root Test. Let q = lim_{n® ¥} |a_n|^1/n exist. Then, the series S _{n³ 1} a_n converges if q < 1, and diverges is q > 1. (If q = 1, the test is inconclusive).

Proof: If q < 1, we can choose e > 0 such that q+e < 1. Now, we can find a N such that |a_n|^1/n < q+e , n > N. Then, for n > m > N, |s_n – s_m| £ |a_m+1| + … + |a_n| < (q+e )^m+1 + … + (q+e )ⁿ < (q+e )^N/[1-(q+e )] ® 0, N ® ¥ . Consequently S _n³ 1 a_n converges. Next, if q > 1, we can choose e > 0 such that q-e > 1. Next, we can find a N such that |a_n|^1/n > q-e , n > N. Hence, for n > N, |a_n| ³ (q-e )ⁿ ® ¥ , as n ® ¥ . Consequently, as a_n does not tend to zero as n ® ¥ , the series S _n³ 1 a_n diverges. (For both the series S [n(n+1)]^-1 (s_n = 1-1/(n+1) ® 1, n ® ¥ , convergent), and S 1 (s_n = n ® ¥ , divergent), q = lim_n® ¥ |a_n|^1/n = 1, illustrating the inconclusive cases!).

Ratio Test. If q = lim_{n® ¥} |a_n+1/a_n| exists, the series S _{n³ 1} a_n converges if q < 1, and diverges if q > 1. (If q = 1, the test is inconclusive).

For the series S 1/n and S 1/n², both ratio and the root tests give q = 1. Hence, even though q = 1 for both the series, one of them diverges while the other converges.

Proof: If q < 1, we can choose e > 0 such that q+e < 1. Now, we can find a N such that |a_n+1/a_n| < q+e , n > N. Then, |a_N+n| < |a_N+1|(q+e )^n-1, n ³ 1. Hence the series S _n³ 1 a_N+n obtained by dropping the first N terms, is dominated by the convergent geometric series S _n³ 1 |a_N+1|(q+e )^n-1 implying the convergence of S _n³ 1 a_n. If q > 1, we can choose e > 0 such that q-e > 1. Now, we can find a N such that |a_n+1/a_n| > q-e , n > N. Then, |a_N+n| > |a_N+1|(q+e )^n-1 ® ¥ , as n ® ¥ . Thus as |a_n| does not tend to zero as n ® ¥ , S _n³ 1 a_n diverges. (The inconclusiveness of the test in the case q = 1 is illustrated by the series S [n(n+1)]^-1 and S 1). #

Integral Test. Let f(x) > 0 be a function defined on [1, ¥ ) such that f(x) £ f(y) if x > y and let a_n = f(n), n ³ 1. Then the series S a_n converges iff ò _{(1,¥ )} f(x)dx < ¥ .

Proof: Since a_n ³ ò _(n,n+1) f(x)dx ³ a_n+1, n ³ 1, s_n ³ ò _(1,n+1) f(x)dx ³ s_n+1 – a₁. Hence if {s_n} is bounded ò _(1,¥ ) f(x)dx < ¥ , and if the integral is bounded {s_n+1} is bounded. The test follows from the Weierstrass’ M-test. #

Examples. Taking f(x) = 1/x, the associated series and the integral are S 1/n, and ò _(1,¥ ) (1/x)dx = ¥ , hence S 1/n diverges. Taking g(x) = 1/x², the series and the integrals are S 1/n², and ò _(1,¥ ) (1/x²)dx = [-1/x]_(1,¥ ) = 1 < ¥ , hence S 1/n² converges. More generally, defining ln^[0](x) = x, ln^[1](x) = ln (ln^[0](x), ln^[2](x) = ln (ln^[1](x), ln^[3](x) = ln (ln^[2](x), and so on, so that ln^[q+1](x) = ln (ln^[q](x), q ³ 0, and taking h(x) = 1/(ln^[0](x) ln^[1](x) ln^[2](x)… ln^[q-1](x) [ln^[q](x)]^p), the associated series is S _n³ m (n ln (n) ln (ln n)… ln^[q-1](n) [ln^[q](n)]^p)^-1, and the associated integral is I = ò _(m,¥ ) (x ln x ln(ln x)… ln^[q-1](x) [ln^[q](x)]^p)^-1 dx. Putting t = ln^[q](x), dt = d(ln (ln ( … (ln x) … ))) = [1/ln^[q-1](x)] [1/ln^[q-2](x)] … [1/ln(x)] [1/x] dx. Hence, I = ò _(ln[q](m),¥ ) t^-pdt < ¥ iff p > 1. Hence the series converges if p > 1 and diverges for p £ 1.

Cauchy’s Condensation Test. If a_n+1 ³ a_n ³ 0, n ³ 1, S a_n converges iff S 2ⁿa_2n is converges.

Proof: As s_2n+1 - s_2n ³ 2ⁿa_2n ³ (s_2n+2 - s_2n+1)/2 summing for 0 £ n £ m-1, s_2m – a₁ ³ a₁ + S ₁£ n£ m-1 2ⁿa_2n ³ (s_2m+1 – s₂)/2, showing that the sequence of partial sums of S a_n is bounded iff the sequence of partial sums of S 2ⁿa_2n is so. The result follows. #

Leibnitz Test (Alternating series). If a_n ¯ 0, n ® ¥ , the series a₁ – a₂ + a₃ – a₄ + … converges.

Proof: For n > m, |s_n-s_m| = |a_m+1 - a_m+2 + … +(-1)^n-ma_n| Î [a_m+1-a_m+2, a_m+1]. Hence |s_n-s_m| ® 0, m ® ¥ . #

Theorem (Regrouping). If a series S a_n is convergent, every regrouped series is convergent with the same limit.

Theorem (Rearrangement). If S a_n is absolutely convergent, every rearranged series is convergent with the same limit.

Proof: Let S denote the sum of the series. By the absolute convergence, given any e > 0, there exists an N₀ such that |s_m – S| = |a₁ + … + a_m – S| < e /2, and |a_m+1| + … + |a_n| < e /2, for all n > m > N₀. Now consider a rearranged series S b_n. Since each term of S a_n is represented precisely once in S b_n, fixing m as before, we can find an N such that for l > N all of a₁, … , a_m occur amongst b₁, … , b_l. Also we can find n > m such that all of b₁, … , b_l occur amongst a₁, … , a_n. Now, |b₁ + … + b_l - S| £ |b₁ + … + b_l – (a₁ + … + a_m)| +| a₁ + … + a_m – S| < |a_m+1| + … + |a_n| + e /2 < e /2 + e /2 = e . It follows that lim_l® ¥ b₁ + … + b_l = S, i.e., S b_n converges with sum S. #

Theorem. If S a_n is conditionally convergent and s is any given number, there exists a rearrangement S b_n of the series S a_n such that the series S b_n is convergent and its sum is s .

Proof: Let {p _n} and {n _n} denote the subsequences of {a_n} of respectively positive and negative terms. To begin with the construction of S b_n, in the first step, if s is positive keep on adding terms from {p _n} till the sum is greater than s , and if s is negative keep on adding terms from {n _n} till the sum is less than s . Subsequently, in the second step, if the sum is greater than s keep adding terms from {n _n} till the sum is less than s , and if the sum is less than s keep adding terms from {p _n} till the sum becomes greater than s . Continue this process ad infinitum by performing third, fourth, etc. steps. Let the last term added at the n-th step be called b _n and the sum after the addition of b _n as s _n. If b _n is positive 0 < s _n - s £ b _n, and if b _n is negative b _n £ s _n - s < 0. Hence |s _n - s | £ |b _n|. Since b _n ® 0, it follows that s _n ® s , as n ® ¥ . Also if t _n is any other sum during the n-th step |t _n - s | < |s _n-1 - s |, so that also t _n ® 0 as n ® ¥ . Hence the sequence of the partial sums of S b_n converges to s . #

Convergence of a Power Series. The complex power series S _{n³ 0} a_n(z-a)ⁿ converges if |z-a| < R, and diverges if |z-a| > R, with the radius of convergence R given by R = liminf_{n® ¥} |a_n|^-1/n.

Proof: If R < ¥ and |z-a| < R, for some e > 0, |z-a| < R-2e . Now for some N, n > N implies |a_n|^-1/n > R-e , i.e., |a_n| < 1/(R-e )ⁿ. Hence, |a_n(z-a)ⁿ| < [(R-2e )/(R-e )]ⁿ, n > N so that, after a certain number of terms, being dominated by a geometric series with common ratio (R-2e )/(R-e ) < 1, the series S _n³ 0 a_n(z-a)ⁿ converges. Next, if R = ¥ , given any complex z there exists an N such that |a_n|^-1/n > 2|z-a|, i.e., |a_n(z-a)ⁿ| < 2^-n for all n > N. Hence, after a certain number of terms, being dominated by a geometric series with common ratio ½, the series

On the other hand, if |z-a| > R, for some e > 0, |z-a| > R+2e . Also, for an infinity of n’s |a_n|^-1/n < R+e , i.e., |a_n| > 1/(R+e )ⁿ, so that |a_n(z-a)ⁿ| > [(R+2e )/(R+e )]ⁿ ® ¥ , n ® ¥ . Hence the series S _n³ 0 a_n(z-a)ⁿ can not converge. #

Examples on Convergence on the Circle of Convergence (a) The power series S _n³ 1 zⁿ/n² converges uniformly on the circle of convergence |z| = 1. (b) The geometric series S _n³ 0 zⁿ converges nowhere on the circle of convergence |z| = 1. (c) The power series

A real valued function f(x) defined in a neighborhood of a point x = a is said to be continuous at x = a if given an arbitrary e > 0, howsoever small, there exists a d > 0 such that |f(x)–f(a)| < e , for all {x : |x–a| < d }. Note that |f(x)–f(a)| < e , and |x–a| < d are, resp., equivalent to f(a) - e < f(x) < f(a) + e , and x Î (a-d , a+d ). Also note that if f is defined only on the left of x = a the condition 0 < |x–a| < d is assumed to be applicable only to the points x that satisfy x < a; similarly, if f is defined only on the right of x = a, the condition 0 < |x –a| < d applies only to x > a. If f(x) is continuous at each point of an interval I, f(x) is said to be continuous on the interval I. In the sequel a function would mean a real valued function, unless otherwise stated.

The definition of continuity of f(x) at x = a implies that if f(a) > 0 there is a neighborhood (a-d , a+d ), (d > 0) of the point x = a, in which f(x) is positive. For, with e = f(a)/2, the d obtained satisfies f(x) > f(a) - f(a)/2 = f(a)/2, x Î (a-d , a+d ). Similarly if f(a) is negative, choosing e = -f(a)/2 we have f(x) < f(a) + (-f(a)/2) =f(a)/2 < 0, x Î (a-d , a+d ), i.e., f(x) is negative in the entire neighborhood (a-d , a+d ).

• f(x) is continuous at x = a iff whenever x_n ® a, f(x_n) ® f(a).

• If f(x) and g(x) are continuous at x = a, then: (a) f(x) + g(x), (b) f(x) – g(x), and, (c) f(x)g(x) are continuous at x = a. Also, (d) f(x)/g(x) is continuous at x = a, if g(a) ¹ 0.

• If f(x) is continuous at x = g(a) and g(x) is continuous at x = a, then f(g(x)) is continuous at x = a.

Theorem. Let f(x) be a continuous real valued function defined on a closed interval [a, b]. Then: (i) f(x) is bounded, i.e., the set {f(x) : x Î [a, b]} is a bounded subset of R ; (ii) f(x) attains its maxima, i.e., there is a point y Î [a, b] such that f(y) ³ f(x) for all x Î [a, b]; (iii) f(x) attains its minima, i.e., there is a point z Î [a, b] such that f(z) £ f(x) for all x Î [a, b]; (iv) f(x) is uniformly continuous, i.e., given an arbitrary e > 0, there exists a d > 0, such that |f(y)–f(z)| < e , for all y, z Î [a, b] satisfying |y–z| < d .

Proof: (i) If f(x) is not bounded there is no M such that |f(x)| < M for all x Î [a, b]. Hence there is a sequence of points x_n Î [a, b] such that |f(x_n)| ® ¥ . By the Bolzano-Weierstrass principle there is a subsequnce {x_nk} converging to a point c Î [a, b]. But then, since f(x_nk) diverges to infinity in magnitude it can not converge to the finite value f(c), f(x) cannot be continuous at x = c. #

(ii) By (i) S = {f(x) : a £ x £ b} is bounded and so has a lub M. If the lub M does not belong to S there is a sequence {x_n} of points in [a, b] such that f(x_n) ® M. As {x_n} is a bounded sequence, it has a convergent subsequence {z_k} ® z Î [a, b], say. Then, by the continuity of f(x) at x = z, M ¬ f(z_k) ® f(z), k ® ¥ , so that f(z) = M. Hence, after all, M is attained at the point x = z. #

(iii) The proof could be given analogously to (ii): Let f(x_n) ® inf {f(x) : x Î [a, b]. Since {x_n} is a bounded sequence, it has a convergent subsequence {z_n} ® z Î [a, b]. Then f(z_n) ® f(z) and so the miniima is attained at x = z. Alternatively, it follows from (ii) by considering the function –f(x) instead of f(x), as the negative of the maxima of –f(x) is the minima of f(x). #

(iv) If f(x) is not uniformly continuous, there is an e > 0 for which no d > 0 works. Hence, there exists a sequence of pairs (x_n, y_n) of points in [a, b] such that |x_n-y_n| < 1/n and |f(x_n)-f(y_n)| ³ e . Since {x_n} is a bounded sequence it has a convergent subsequence {x _k} ® c, say. The associated subsequence {h _k}, say, of the corresponding y_n’s is also convergent and moreover converges to the same limit c. Letting k ® ¥ in |f(x _k)-f(h _k)| ³ e , we have 0 = |f(c)- f(c)| ³ e , a contradiction. #

Intermediate Value Theorem. Let f(x) be continuous on [a, b]. If f(a) ¹ f(b) and g is a value in between them, there exists a point c Î (a, b) such that f(c) = g .

Proof: Considering the function ± [f(x)-g ] instead of f(x), if necessary, we could assume that f(a) < 0 = g < f(b). Let S = {x Î [a, b] : f(x) < 0}. S is non-empty and bounded. Let c = lub S. Then f(c) ³ 0, else there would be a point on right to c belonging to S. However, for any d > 0, there exists a point zd such that c-d < zd Î S, i.e.,f(zd ) < 0. Letting d ® 0, by continuity of f(x) at x = c there follows f(c) = limd ® ₀₊ f(zd ) £ 0. It follows that f(c) = 0. #

df(x)/dx º (d/dx)f(x) º f¢ (x) = lim_{h® 0} [f(x+h) – f(x)]/h.

If the limit does not exist we say that the derivative does not exist at the point x, or that f is not differentiable at the point x. The second derivative f² (x) is the derivative of the first derivative at a point x, i.e., f² (x) = (f¢ )¢ (x). The existence of f² (x) implies that f¢ (t) exists in a neighborhood of the point x, and that f¢ (x+h) = f¢ (x) + hf² (x) + o(h), h ® 0. In general, if g(x) = f⁽ⁿ⁾(x), the n-th derivative of f at x, then f⁽ⁿ⁺¹⁾(x) = g¢ (x). The zeroth order derivative f⁽⁰⁾(x), by convention, is taken to be f(x) itself.

The right and the left derivatives f¢ ₊(x) and f¢ _-(x) at a point x = a are defined by f¢ ₊(x) = lim_h® 0+ [f(x+h)–f(x)]/h and f¢ _-(x) = lim_h® 0- [f(x+h)–f(x)]/h, respectively. It is clear that f¢ (x) exists iff both f¢ ₊(x) and f¢ (x) exist and are equal.

By a quantity q(h) being o(h), h ® 0, means that q(h)/h ® 0, as h ® 0. We say Q(h) = O(h), h ® 0, if there exists a d > 0 and a constant M such that |Q(h)/h| < M, |h| < d .

f(x+h) = f(x) + f¢ (x)h + o(h), h ® 0.

Similarly, the existence of f¢ ₊(x) is equivalent to the relation

f(x+h) = f(x) + f¢ ₊(x)h + o(h), h ® 0+,

and the existence of f¢ _-(x) is equivalent to the relation

f(x+h) = f(x) + f¢ _-(x)h + o(h), h ® 0-.

(f(x) ± g(x))¢ = f¢ (x) ± g¢ (x),

(cf(x))¢ = cf¢ (x).

Since g, being differentiable, is continuous at the point x, [f(x+h)g(x+h)-f(x)g(x)]/h = [(f(x+h)-f(x))/h]g(x+h) + f(x)[g(x+h)-g(x)]/h ® f¢ (x)g(x) + f(x)g¢ (x). it follows that the derivative of a product of two functions could be obtained by

(f(x)g(x))¢ = f¢ (x)g(x) + f(x)g¢ (x).

(f₁(x)f₂(x) … f_n(x))¢ = f_1¢ (x)f₂(x) … f_n(x) + f₁(x)f_2¢ (x) … f_n(x) + … + f₁(x)f₂(x) … f_n¢ (x).

Proof: For, assuming the result for n-1, (f₁(x)f₂(x)…f_n(x))¢ = (f₁(x)f₂(x)…f_n-1(x))¢ f_n(x) + (f₁(x)f₂(x)…f_n-1(x))f_n¢ (x) = f₁¢ (x)f₂(x) … f_n(x) + f₁(x)f₂¢ (x) … f_n(x) + … + f₁(x)f₂(x) … f_n¢ (x). #

Choosing f₁(x) = f₂(x) = … = f_n(x) = x, and using (x)¢ = 1, we get

(d/dx)xⁿ º dxⁿ/dx º (xⁿ)¢ = nx^n-1, n = 1, 2, … .

If g(x) ¹ 0, using the previous result, f¢ (x) = [(f(x)/g(x))g(x)]¢ = (f(x)/g(x))¢ g(x) + (f(x)/g(x))g¢ (x), which on simplification gives the formula for the derivative of a quotient

(f(x)/g(x))¢ = [f¢ (x)g(x) – f(x)g¢ (x)]/[g(x)]², (g(x) ¹ 0).

f(g(x+h)) - f(g(x)) = f¢ (g(x))(g(x+h)-g(x)) + o(g(x+h)-g(x)).

Dividing by h and passing to the limit as h ® 0, we get

(f(g(x))¢ = f¢ (g(x))g¢ (x), (chain rule).

Leibnitz Formula. Let n be a natural number and let f⁽ⁿ⁾(x) and g⁽ⁿ⁾(x) exist. Then [f(x)g(x)]⁽ⁿ⁾ exists and is given by:

(d/dx)ⁿ[f(x)g(x)] = S _{0£ k£ n} ⁿC_kf^(n-k)(x)g^(k)(x).

[f(x)g(x)]⁽ⁿ⁺¹⁾ = S ₀£ k£ n ⁿC_k[f^(n+1-k)(x)g^(k)(x)+f^(n-k)(x)g^(k+1)(x)]

= f⁽ⁿ⁺¹⁾(x)g⁽⁰⁾(x) + S ₁£ k£ n [ⁿC_k+ⁿC_k-1]f^(n+1-k)(x) g^(k)(x) + f⁽⁰⁾(x)g⁽ⁿ⁺¹⁾(x)

= S ₀£ k£ n+1 ⁿ⁺¹C_kf^(n+1-k)(x)g^(k)(x), as ⁿC_k+ⁿC_k-1 = [ⁿC_k+ⁿC_k-1], 1 £ k £ n. #

L’Hospital’s Rule (I). If f(a), g(a) = 0, f¢ (a), g¢ (a) exist, and g¢ (a) ¹ 0, then

lim_{x® a} f(x)/g(x) = f¢ (a)/g¢ (a).

Proof: f(x)/g(x) = {f(a) + [f¢ (a)+o(1)](x-a)}/{g(a)+ [g¢ (a)+o(1)](x-a)} = [f¢ (a)+o(1)]/[g¢ (a)+o(1)] ® f¢ (a)/g¢ (a). #

Exercise 1. If f¢ (x) exists, prove that f is continuous at the point x.

Exercise 2. If f¢ (x) exists, prove that f(x+h) = f(x) + O(h), h ® 0, but that the converse need not be true.

Note that if f¢ (x) and g¢ (x) exist

f(x+h)g(x+h) = [f(x) + f¢ (x)h + o(h)][ g(x) + g¢ (x)h + o(h)] = f(x)g(x) + [f¢ (x)g(x) + f(x)g¢ (x)] + o(h),

f(x+h)/g(x+h) – f(x)/g(x) = [f(x) + f¢ (x)h + o(h)]/[ g(x) + g¢ (x)h + o(h)] - f(x)/g(x)

= [f¢ (x)g(x) – f(x)g¢ (x)]/[g¢ (x)]² + o(h).

Rolle’s Theorem. If f(x) is continuous real valued on [a, b] and differentiable on (a, b) and f(a) = f(b) = 0, there exists a x on (a, b) such that f¢ (x ) = 0.

Proof: If f º 0, any point on (a, b) could be taken as x . If not without loss of generality we can assume that f takes a positive value. Then f assumes its maximum value at some point x Î (a, b), say. For h > 0, sufficiently small, we have f(x +h) – f(x ) £ 0, implying f¢ (x ) £ 0. But also we have f(x -h) – f(x ) £ 0, implying f¢ (x ) ³ 0. Hence f¢ (x ) = 0. #

Lagrange’s Mean Value Theorem of Differential Calculus. If f(x) is continuous on [a, b] and differentiable on (a, b), there exists a point x Î (a, b) such that f(b) - f(a) = (b - a)f¢ (x ).

Proof: Consider the function g(x) = f(x) – [f(a) + (x – a){f(b) – f(a)}/(b-a)]. It is continuous on [a, b], differentiable on (a, b), and it vanishes at x = a, b. Hence by Rolle’s theorem, there exists x Î (a, b) such that g¢ (x ) = 0, i.e., f¢ (x ) – [f(b) – f(a)]/(b-a) = 0. #

Cauchy’s form of the Mean Value Theorem. If f(x) and g(x) are continuous on [a, b], differentiable on (a, b), g¢ does not vanish on (a, b), and g(b) ¹ g(a), then there exists a point x Î (a, b) such that: [f(b)-f(a)]/[g(b)–g(a)] = f¢(x)/ g¢(x).

Proof: Consider the function h(x) = [f(x)–f(a)] [g(b)–g(a)] - [f(b)–f(a)] [g(x)–g(a)]. We have h(a) = h(b), h(x) is continuous on [a, b] and differentiable on (a, b). By Rolle’s theorem for some x Î (a, b), h¢ (x) = 0, i.e., f¢ (x )[g(b) – g(a)] = g¢ (x)[f(b) – f(a)]. #

L’Hospital’s Rule (II). Let f(x) and g(x) be continuous in a neighborhood of x = a, and differentiable in a deleated neighborhood U = (a-d , a+d )\{a}. If f(a), g(a) = 0, then lim_{x® a} f(x)/g(x) = lim_{x® a} f¢ (x)/g¢ (x), provided the latter limit exists.

Proof: By the Lagrange’s form of the mean value theorem, f(x)/g(x) = [f(x)-f(a)]/[g(x)–g(a)] = f¢ (x )/g¢ (x ), x Î U, where x lies in between a and x and therefore tends to a as x tends to a. The result follows. #

Taylor Series Approximation Theorem. Let n ³ 1. If f⁽ⁿ⁾(a) exists,

f(x) = f(a) + (x-a)f ¢ (a) + (x-a)²f² (a)/2! + … + (x-a)ⁿf⁽ⁿ⁾(a)/n! + o((x-a)ⁿ), x ® a.

lim_x® a [f(x) – {f(a) + (x-a)f ¢ (a) + (x-a)²f² (a)/2! + … + (x-a)ⁿf⁽ⁿ⁾(a)/n!}]/[(x-a)ⁿ/n!]

= lim_x® a [f² (x) – {f² (a) + (x-a)f⁽³⁾(a) + (x-a)²f⁽⁴⁾(a)/2! + … + (x-a)^n-2f⁽ⁿ⁾(a)/(n-2)!}]/ [(x-a)^n-2/(n-2)!]

= lim_x® a [f^(n-2)(x) – {f^(n-2(a) + (x-a)f^(n-1)(a) + (x-a)²f⁽ⁿ⁾(a)/2!}]/ [(x-a)²/2!]

= lim_x® a [f^(n-1)(x) – {f^(n-1)(a)+ (x-a)f⁽ⁿ⁾(a)}]/ [(x-a)]

= lim_x® a [{f^(n-1)(x) – f^(n-1)(a)}/(x-a) - f⁽ⁿ⁾(a)] = 0. #

L’Hospital’s Rule (III). Let f(x) and g(x) be differentiable on (a, b) and |f(x)|, |g(x)| ® ¥ as x ® a+. Then, if lim_{x® a+} f¢ (x)/g¢ (x) exists, lim_{x® a+} f(x)/g(x) = lim_{x® a+} f¢ (x)/g¢ (x).

Proof: Let e > 0 and L = lim_x® a+ f¢ (x)/g¢ (x). There exists a d > 0 such that |f¢ (x)/g¢ (x)-L| < e /2, if x Î (a, a+d ). Let 0 < h < d . It is clear that lim_x® a+ f(x)/g(x) = lim_x® a+ [f(x)-f(a+h)]/[g(x)-g(a+h)]. Hence there exists an 0 < h < h such that for a < x < a+h , |f(x)/g(x) - [f(x)-f(a+h)]/[g(x)-g(a+h)]| <e /2. By the Lagrange’s form of the mean value theorem for a < x < a+h , there exists a x satisfying x < x < a+h such that [f(x)-f(a+h)]/[g(x)-g(a+h)] = f¢ (x )/g¢ (x ) Î (L-e /2, L+e /2). Hence, |f(x)/g(x)–L| £ |f(x)/g(x)-[f(x)-f(a+h)]/[g(x)-g(a+h)] + [f(x)-f(a+h)]/[g(x)-g(a+h)] – L| < e /2 + e /2 = e , for all a < x < h . It follows that lim_x® a+ f(x)/g(x) = L. #

Let I[x, x+h] denote the closed interval with endpoints x and x+h, and I(x, x+h) the corresponding open interval. The result of the mean value theorem of differential calculus may be re-written as: f(x+h) = f(x) + hf¢ (x ), where f is continuous on I[x, x+h] and differentiable on I(x, x+h), and x Î I(x, x+h). A generalization of the same is:

Taylor’s Expansion with Mean Value Form of Remainder. If f is continuous on I[x, x+h] and (n+1)-times differentiable on I(x, x+h), then

f(x+h) = f(x) + hf ¢ (x) + h²f² (x)/2! + … + hⁿf⁽ⁿ⁾(x)/n! + hⁿ⁺¹f⁽ⁿ⁺¹⁾(x )/(n+1)!, x Î I(x, x+h).

Proof: Consider the function g(t) = f(x+t) – [f(x) + tf ¢ (x) + t²f² (x)/2! + … + tⁿf⁽ⁿ⁾ (x)/n! + tⁿ⁺¹K/(n+1)!], where K is the constant that makes g(h) = 0. We have g(0) = g¢ (0) = … = g⁽ⁿ⁾(0) = 0. Since g(h) = 0, for some x ₁ Î (0, h), g¢ (x ₁) = 0. Hence, for some x ₂ Î (0, x ₁), g² (x ₂) = 0. Continuing like this, there exists a x _n+1 Î (0, x _n), g⁽ⁿ⁺¹⁾(x _n+1) = 0. Renaming x = x + x _n+1, we have K = f⁽ⁿ⁺¹⁾(x ), completing the proof. #

Generalized Mean Value Theorem. If f(t), g(t) are continuous on I[x, x+h], h ¹ 0, and are (n+1)-times differentiable on I(x, x+h) and g⁽ⁿ⁺¹⁾(t) ¹ 0, t Î I(x, x+h), for some x Î I(x, x+h)

[f(x+h) - S _{0£ k£ n} f^(k)(x)h^k/k!]/[g(x+h) - S _{0£ k£ n} g^(k)(x)h^k/k!] = f⁽ⁿ⁺¹⁾(x )/g⁽ⁿ⁺¹⁾(x ).

Proof: Let F(t) = [f(x+t) - S ₀£ k£ n f^(k)(x)t^k/k!], and G(t) = [g(x+t) - S ₀£ k£ n g^(k)(x)t^k/k!]. Note that F⁽ⁱ⁾(t) = [f⁽ⁱ⁾(x+t) - S ₀£ k£ n-i f^(k+i)(x)t^k/k!] and so F⁽ⁱ⁾(0) = 0, 0 £ i £ n.Consider the function H(t) = F(t)G(h) – F(h)G(t). We have H(0) = H(h) = 0. By the mean value theorem H¢ (x ₁) = 0 for some x ₁ Î ((0, h), i.e., F¢ (x ₁)G(h) – F(h)G¢ (x ₁) = 0. Now, for the function H¢ (t) = F¢ (t)G(h) – F(h)G¢ (t). We have H¢ (0) = H¢ (x ₁) = 0. By the mean value theorem, for some x ₂ Î ((0, x ₁), H² (x ₂) = 0, i.e., F² (x ₂)G(h) – F(h)G² (x ₂) = 0. Continuing like this, for the function H⁽ⁱ⁾(t) = F⁽ⁱ⁾(t)G(x _i) – F(x _i)G⁽ⁱ⁾(t), as H⁽ⁱ⁾(0) = H⁽ⁱ⁾(x _i) = 0, by the mean value theorem for some x _i+1 Î ((0, x _i), H⁽ⁱ⁺¹⁾ (x _i+1) = 0, i.e., F⁽ⁱ⁺¹⁾(x _i+1)G(h) – F(h)G⁽ⁱ⁺¹⁾(x _i+1) = 0, 1 £ i £ n. Putting x = x+x _n+1, the result for i = n+1 implies f⁽ⁿ⁺¹⁾(x )G(h) – F(h)g⁽ⁿ⁺¹⁾(x ) = 0. Note that by the mean value theorem G(h) = hⁿ⁺¹g⁽ⁿ⁺¹⁾(h ) ¹ 0. The result follows after division by the non-zero quantity G(h)g⁽ⁿ⁺¹⁾(x ). #

The function ln x is defined by: ln x = ò _(1,x) (1/t)dt, x > 0. Since for 0 < x < 1, ò _(1,x) º - ò _(x,1), the above definition implies that: ln x = - ò _(x,1) (1/x)dx, 0 < x < 1. Note that the change of variable s = 1/t leads to

ln (1/x) = ò _(1,1/x) (1/t)dt = - ò _(1,x) (1/s)ds = - ln x, x > 0.

The additive formula: ln xy = ln x + ln y, (x, y > 0), could be established as follows: Putting s = t/y,

ln xy = ò _(1,xy) (1/t)dt = ò _(1/y,x) [1/(sy)]yds = [ò _(1,x) -ò _(1,1/y)](1/s)ds = ln x – ln (1/y) = ln x + ln y. #

The defintion of ln x and the fundamental theorem of integral calculus implies that: d(ln x)/dx = 1/x, x > 0.

Note that the function ln x has been defined on the domain (0, ¥ ), and has range (-¥ , ¥ ). Since (ln x)¢ is positive throughout the domain ln x is increasing and therefore has its inverse function defined on the domain (-¥ , ¥ ), with range (0, ¥ ). This inverse function is called the exponential function e^x, i.e., e^x = ln^-1x, - ¥ < x < ¥ . Differentiating the identity ln (e^x) º x, (1/e^x)d(e^x)/dx = 1, i.e., d(e^x)/dx = e^x, -¥ < x < ¥ . Put x = e^a, y = e^b in ln xy = ln x + ln y, to get ln e^ae^b = a + b. Taking the exponetial of both sides, we have e^ae^b = e^a+b, which is the same as the multiplicative formula for the exponetials: e^x+y = e^xe^y, -¥ < x, y < ¥ .

log (1+x) = x – x²/2 + x³/3 – x⁴/4 + … , -1 < x £ 1.

Here, f(x) = log (1+x), f¢ (x) = 1/(1+x), f² (x) = -1/(1+x)², f¢ ¢ ¢ (x) = 2!/(1+x)³, … , f⁽ⁿ⁾(x) = (-1)^n-1(n-1)! /(1+x)ⁿ, … . Hence, log (1+x) = S ₁£ k£ n [(-1)^k-1(k-1)!/(1+x)^k|_x=0]x^k/k! + R_n(x) = S ₁£ k£ n (-1)^k-1x^k/k + R_n(x), where R_n(x) is the remainder term in the Taylor’s expansion. For 0 < x £ 1, R_n(x) = [(-1)ⁿ[x/(1+x )]ⁿ⁺¹/(n+1). Since for 0 < x £ 1, 0 < x < x, and so x/(1+x ) < 1, the remainder term (-1)ⁿ [x/(1+x )]ⁿ/(n+1) ® 0, as n ® ¥ . Hence the expansion is valid on (0, 1]. For –1 < x £ 0, we use the relation

log (1+x) = ò _(0,x) [1/(1+t)]dt = ò _(0,x) [1 – t + t² – t³ + … + (-1)^n-1t^n-1 + (-1)ⁿtⁿ/(1+t)]dt = S ₁£ k£ n (-1)^k-1x^k/k + R_n(x),

from which |R_n(x)| = |ò _(0,x) [(-1)ⁿtⁿ/(1+t)]dt| = ò _(0,|x|) [tⁿ/(1-t)]dt £ [1/(1-|x|)]|x|ⁿ⁺¹/(n+1) ® 0, as n ® ¥ . #

e^x = 1 + x + x²/2! + x³/3! + … , -¥ < x < ¥ .

Here, f(x) = e^x, f¢ (x) = e^x, f² (x) = e^x, f¢ ¢ ¢ (x) = e^x, … , f⁽ⁿ⁾(x) = e^x, … . Hence, e^x = S ₁£ k£ n-1 [e^x|_x=0]x^k/k! + [ex ]xⁿ/n! = S ₁£ k£ n-1 x^k/k + ex xⁿ/n!. As, n! = Ö (2p ) n^n+1/2e^-n, for each -¥ < x < ¥ , the remainder goes to zero as n ® ¥ . Alternately, if 2|x| £ m, |x|ⁿ/n! £ (|x|^m/m!)2^-(n-m) ® 0, n ® ¥ . #

e^xe^y = (S _n³ 0 xⁿ/n!) (S _m³ 0 y^m/m!) = S _l³ 0 S ₀£ n£ l (xⁿ/n!)[y^l-n/(l-n)!] = S _l³ 0 [S ₀£ n£ l ^lC_nxⁿy^l-n]/l! = S _l³ 0 (x+y)^l/l! = e^x+y. #

sin x = x – x³/3! + x⁵/5! – x⁷/7! + … , -¥ < x < ¥ ;

cos x = 1 – x²/2! + x⁴/4! – x⁶/6! + … , -¥ < x < ¥ .

The complex exponential function eⁱq :

The function eⁱq may be defined by: e^iq = cos q + i sin q , -¥ < q < ¥ ,

so that: e^-iq = cos q - i sin q , cos q = (e^iq + e^-iq )/2, sin q = (e^iq - e^-iq )/(2i), -¥ < q < ¥ . Using i² = -1, and the expansions for sin q and cos q , we have

e^iq = 1 + iq + (iq )²/2! + (iq )³/3! +(iq )⁴/4! + … + (iq )ⁿ/n! + … , -¥ < q < +¥ ,

a comparision of which with the expansion e^x = 1 + x + x²/2! + x³/3! + … justifies the notation eⁱq .

arctan x º tan^-1x = x - x³/3 + x⁵/5 – x⁷/7 + … , |x| £ 1.

Here, f(x) = tan^-1x, f¢ (x) = 1/(1+x²). In this case, instead of a direct approach, we integrate the geometric series of 1/(1+x²) with remainder: 1/(1+x²) =1 – x² + x⁴ – x⁶ + … + (-1)ⁿx²ⁿ + (-1)ⁿ⁺¹ [x²ⁿ⁺²/(1+x²)], integrating which on (0, x) gives tan^-1x = x – x³/3 + x⁵/5 – x⁷/7 + … + (-1)ⁿx²ⁿ⁺¹/(2n+1) + (-1)ⁿ⁺¹ ò _(0,x) [x²ⁿ⁺²/(1+x²)]dx. The magnitude of the remainder does not exceed ò _(0,|x|) x²ⁿ⁺²dx = |x|²ⁿ⁺³/(2n+3) ® 0, n ® ¥ , if |x| £ 1. #

A function f(x) is said to have a local maxima at x = a, if there is a d > 0 such that f(a) ³ f(x), |x-a| < d . The point x = a is said to be a strict local maxima if there is a d > 0 such that f(a) > f(x), |x-a| < d .

Now suppose f(x) has a local maxima at x = a. By definition then f¢ ₊(x) £ 0 and f¢ _-(x) ³ 0 so that if f¢ (x) exists at x = a, we must have f¢ (a) = 0. Hence if f(x) is differentiable at x = a, the condition f¢ (a) = 0 is necessary for f(x) to have a maxima at x = a.

One could also reason as follows: if f¢ (x) < 0, by f(x) = f(a) + (x-a)f¢ (a) + o(x-a), x ® a, one concludes that there exists a d > 0, such that for all x satisfying |x-a| < d , there holds f(x) < f(a) if x > a and f(x) > f(a) if x < a, contradicting that x = a is a local maxima. Similarly, if f¢ (x) > 0, one concludes that there exists a d > 0, such that for all x satisfying |x-a| < d , there holds f(x) > f(a) if x > a and f(x) < f(a) if x < a, contradicting that x = a is a local maxima.

A function f(x) is said to have a local minima at x = a, if there is a d > 0 such that f(a) £ f(x), |x-a| < d . The point x = a is said to be a strict local minima if there is a d > 0 such that f(a) < f(x), |x-a| < d . As in the case of local maxima, if f¢ (a) exists also a local minima at x = a implies that f¢ (a) = 0.

Hence, if f(x) has a local maxima or a local minima at x = a and if f¢ (a) exists then f¢ (x) = 0, i.e., if f(x) is differentiable at x = a, f¢ (a) = 0 is a necessary condition for f(x) to have a local maxima-minima at x = 0. Thus, the points of local maxima-minima of f(x) are to be found amongst those x’s that satisfy the equation f¢ (x) = 0.

Note that for f(x) = x³, f¢ (0) = 0 but x = 0 is not a local maxima-minima point of f(x).

Theorem. If for some m ³ 1, f^(2m)(a) exists and is not equal to zero and f¢ (a) = … = f^(2m-1)(a) = 0, then x = a is a local maxima of f(x) if f^(2m)(a) < 0, and, x = a is a local minima of f(x) if f^(2m)(a) > 0.

Proof: As, f(x)-f(a) = (x-a)f¢ (a)+(x-a)²f² (a)/2!+…+(x-a)^2mf^(2m)(a)/(2m)!+o((x-a)^2m) = [f^(2m)(a)+o(1)](x-a)^2m/(2m)!, x ® a, so that there exists a d > 0 such that for 0 < |x-a| < d , f(x) – f(a) < 0 if f^(2m)(a) < 0 and f(x) – f(a) > 0 if f^(2m)(a) > 0, proving the result. #

Theorem. If for some m ³ 1, f¢ (a) = … = f^(2m-1)(a) = 0, and f^(2m)(x) exists in a deleted neighborhood of x = a and is negative (positive), then x = a is a local maxima (minima) of f(x).

f(x) = f(a) + (x-a)f ¢ (a) + (x-a)²f² (a)/2! + … + (x-a)^2m-1f^(2m-1)(a)/(2m-1)! + (x-a)^2mf^(2m)(x )/(2m)!, x Î I(a, x). #

Theorem. If for some n ³ 1, f¢ (a) = … = f^(n-1)(a) = 0, and f⁽ⁿ⁾(x) exists in a deleted neighborhood of x = a and the quantity (x-a)ⁿ f⁽ⁿ⁾(x) is negative (positive) there, then x = a is a local maxima (minima) of f(x).

f(x) = f(a) + (x-a)f ¢ (a) + (x-a)²f² (a)/2! + … + (x-a)^n-1f^(n-1)(a)/(n-1)! + (x-a)ⁿf⁽ⁿ⁾(x )/(n)!, x Î I(a, x),

since x -a and x-a have the same sign. #

A minima or a global minima of f(x) is the largest number m such that f(x) ³ m for all x in the domain of definition of f. If there exists a point a such that f(a) = m, x = a is called a minima point of f and we say that f(x) attains its minima. The maxima or global maxima, similarly, is the smallest number M such that f(x) £ M. A maxima point a satisfies f(a) = M, and we say that f attains its maxima. The maxima minima points are indeed also local maxima and minima points, without the converse being necessarily true.

Intermediate Value Theorem for a Derivative. Let f(x) be differentiable on [a, b]. If f¢ ₊(a) ¹ f¢ _-(b) and g is a value in between them, there exists a point c Î (a, b) such that f¢ (c) = g .

Proof: Using the function ± [f(x)-g x] instead of f(x), if necessary, we could assume that f¢ ₊(a) < 0 = g < f¢ _-(b). Then f(x) decreases at x = a and increases at x = b so that neither of these points can be a minima point of f(x). Let the minima point be c. Then, a < c < b and f¢ (c) = 0. #

One could contrast the statement of intermediate value theorem for a function f(x) with that for a derivative f¢ (x): whereas the former requires the continuity of the function f(x), in the latter the continuity of the derivative f¢ (x) is not needed.

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