Survey

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Survey

Document related concepts

no text concepts found

Transcript

Bernoulli’s inequality (for integer cases) (1 + x)n ≥ 1 + nx (1) with conditions: (a) (1) is true ∀ n∈N ∪ {0} (b) (1) is true ∀ n∈2N and ∀x∈ R, x ≥ -1. and ∀x∈ R. (1 + x)n > 1 + nx The strict inequality: is true for every integer n ≥ 2 (2) and every real number x ≥ -1 with x ≠ 0. (The strict inequality is not discussed in the following.) Proof Use Mathematical Induction 1 Condition (a) Let P(n) be the proposition: (1 + x)n ≥ 1 + nx 0 (1 + x) = 1 ≥ 1 + 0x For P(0), ∀n∈N ∪ {0} and ∀x∈ R, x ≥ -1. ∴ P(0) is true. k ≥ 0, Assume P(k) is true for some k∈Z, k that is, (1 + x) ≥ 1 + kx ∀ k∈N ∪ {0} For P(k+1), (1 + x)k+1 = (1 + x)k(1 + x) ≥ (1 + kx) (1 + x) , by (3) and ∀x∈ R, and also note that since x ≥ -1. (3) x ≥ -1, the factor (x + 1) ≥ 0. 2 = 1 + (k + 1) x + kx ≥ 1 + (k + 1) x ∴ P(k+1) is also true. By the Principle of Mathematical Induction, P(n) is true ∀n∈N ∪ {0} and ∀x∈ R, x ≥ -1. Condition (b) Let P(n) be the proposition: (1 + x)n ≥ 1 + nx (1 + x)0 = 1 ≥ 1 + 0x For P(0), 2 and ∀x∈ R. ∴ P(0) is true. 2 (1 + x) = 1 + 2x + x ≥ 1 + 2x, For P(2), ∴ P(2) ∀ n∈2N since x2 ≥ 0, ∀x∈ R. is true. Assume P(k) is true for some k∈Z, that is, k ≥ 0, k (1 + x) ≥ 1 + kx ∀ k∈N ∪ {0} and ∀x∈ R, x ≥ -1. (4) For P(k + 2), (1 + x)k+2 = (1 + x)k (1 + x)2 ≥ (1 + kx)(1 + 2x), by (4) and P(2). 2 = 1 + (k + 2) x + 2kx ≥ 1 + (k + 2) x, since k > 0 and x2 ≥ 0, ∀x∈ R. ∴ P(k + 2) is also true. By the Principle of Mathematical Induction, P(n) is true ∀ n∈2N and ∀x∈ R. Condition (a) is discussed only in the following. 1 Proof Use A.M. ≥ G.M. 2 Consider the A.M. and G.M. of n positive numbers (1 + nx), 1, 1, ….,1 [with (n-1) “1”s] (1 + nx ) + 1 + 1 + .... + 1 n n + nx n ≥ n (1 + nx ).1.1....1 ≥ n (1 + nx ) (1 + x)n ≥ 1 + nx ∴ (1) Numbers should be positive before applying A.M. – G.M. theorem. Note : In the numbers used in A.M.-G.M. above, 1 > 0 and However, if 1 + nx ≥ 0, i.e. x ≥ − since it is given that x ≥ -1, or 1 + nx < 0, 1 n . x + 1 ≥ 0, L.H.S. of (1) = (1 + x)n ≥ 0 R.H.S. of (1) = 1 + nx < 0 ∴ (1 + x)n ≥ 0 > 1 + nx, which is always true. ∴ x ≥ -1 and not x ≥ − Proof 1 n can ensure (1) is correct. Use Binomial Theorem 3 ⎛n⎞ ⎛n⎞ ⎛n⎞ ⎛n⎞ (1 + x ) n = 1 + nx + ⎜ ⎟ x 2 + ⎜ ⎟ x 3 + ⎜ ⎟ x 4 + .... + ⎜ ⎟ x n ≥ 1 + nx ⎝n⎠ ⎝2⎠ ⎝ 3⎠ ⎝4⎠ (a) For x > 0, (b) For x = 0, (c) For -1 < x < 0, Put y = -x, obviously (1 + x)n ≥ 1 + nx (5) is true. (The proof below is not very rigorous.) then 0 < y < 1, ⎛n⎞ ⎛n⎞ ⎛n⎞ ⎛n⎞ (1 − y ) n = 1 − ny + ⎜ ⎟ y 2 − ⎜ ⎟ y 3 + ⎜ ⎟ y 4 − .... + ( −1) n ⎜ ⎟ y n ⎝n⎠ ⎝ 2⎠ ⎝ 3⎠ ⎝4⎠ (6) Put y = 0, we have: ⎛n⎞ ⎛n⎞ ⎛n⎞ ⎛n⎞ 0 = (1 − 1) n = 1 − n + ⎜ ⎟ − ⎜ ⎟ + ⎜ ⎟ − .... + ( −1) n ⎜ ⎟ ⎝ 2⎠ ⎝ 3⎠ ⎝ 4⎠ ⎝n⎠ ⎛n⎞ ⎛n⎞ ⎛n⎞ ⎛n⎞ ∴ ⎜ ⎟ − ⎜ ⎟ + ⎜ ⎟ − .... + ( −1) n ⎜ ⎟ = n − 1 > 0 ⎝ 2⎠ ⎝ 3⎠ ⎝ 4⎠ ⎝n⎠ Now (7) 0 < y < 1 , ∴ y2 > y3 > …> yn Therefore in (7), each term is multiplied by a factor that is smaller than the term before, n ⎛n⎞ 2 ⎛n⎞ 3 ⎛n⎞ 4 n⎛ ⎞ n ⎜ ⎟ y − ⎜ ⎟ y + ⎜ ⎟ y − .... + ( −1) ⎜ ⎟ y ≥ 0 ⎝ 2⎠ ⎝ 3⎠ ⎝ 4⎠ ⎝n⎠ (8) From (6), (1 – y)n ≥ 1 – ny or n (1 + x) ≥ 1 + nx for 0 < y < 1. for -1 < x < 0. 2 Extension of Bernoulli’s inequality Given x > -1, then (a) (b) (1 + x)r ≤ 1 + rx r (1 + x) ≥ 1 + rx for 0<r<1 for r < 0 or (9) r>1 (10) Firstly we give the proof that r is a rational number first. Use A.M. ≥ G.M. Proof 4 Since r ∈ Q, (a) p r= q Let 0 < r < 1, ∴ p < q, q – p > 0. Also 1 + x > 0, (1 + x)r = (1 + x ) = (b) p q px + q q q = (1 + x ) p 1q − p ≤ =1+ p q p(1 + x ) + 1 ⋅ ( q − p ) q (G.M. ≤ A.M.) x = 1 + rx Let r > 1, (i) If 1 + rx ≤ 0, then (ii) If 1 + rx > 0, (1 + x)r > 0 ≥ 1 + rx. rx > -1. 1 Since r > 1, we have 0 < r < 1 . By (a) we get: 1r 1 (1 + rx ) ≤ 1 + rx = 1 + x r r ∴ (1 + x) ≥ 1 + rx. Let r < 0, then - r > 0. Choose a natural number n sufficiently large such that 0 < -r/n < 1 and 1 > rx/n > -1. By (a), Since 0 < (1 + x ) −r n ≤1+ r n x > 0. −1 2 r r r r r 1 ≥ 1 − ⎛⎜ x ⎞⎟ = ⎛⎜ 1 − x ⎞⎟⎛⎜ 1 + x ⎞⎟ ⇒ ⎛⎜ 1 − x ⎞⎟ ≥ 1 + x ⎝n ⎠ ⎝ n ⎠⎝ n ⎠ ⎝ n ⎠ n (11) Hence by (11), r n r (1 + x )r ≥ ⎛⎜ 1 + x ⎞⎟ ≥ 1 + n ⎛⎜ x ⎞⎟ = 1 + rx . ⎝ n ⎠ ⎝n ⎠ Again, equality holds if and only if x = 0. Note : If r is irrational, we choose an infinite sequence of rational numbers r1, r2, r3, …., such that rn tends to r as n tends to infinity. For part (a), we can extend to irrational r: (1 + x ) r = lim (1 + x ) rn ≤ lim (1 + rn x ) = 1 + rx . n→∞ n→∞ Similar argument for part (b) completes the proof for the case where r ∈ R. 3 Proof Use analysis 5 f(x) = (1 + x)r – 1 – rx Let where x > –1 and r ∈ R\ {0, 1} (12) Then f(x) is differentiable and its derivative is: f ’(x) = r(1 + x)r-1 – r = r [(1 + x)r-1 – 1] from (13) we can get (a) If 0 < r < 1, f ’(x) = 0 ⇔ x = 0. then f ’(x) > 0 ∀ x∈ (–1, 0) and ∴ x = 0 is a global maximum point of (b) ∴ f(x) < f(0) = 0. ∴ (1 + x)r ≤ 1 + rx If (13) for f. 0 < r < 1. r < 0 or r > 1, then f ’(x) < 0 ∀ x∈ (-1, 0) and ∴ x = 0 is a global minimum point of ∴ f(x) > f(0) = 0. ∴ (1 + x)r ≥ 1 + rx for f ’(x) < 0 ∀ x∈ (0, +∞). r < 0 or f ’(x) > 0 ∀ x∈ (0, +∞). f. r > 1. Finally, please check that the equality holds for x = 0 or for r ∈ {0, 1}. The proof is complete. 4

Related documents