Explanation If { a n } is a sequence of real numbers. Then the expression S n = a 1 + a 2 + ... + a n = ∑ k = 1 n a k is termed as sequence of n th partial sum, i.e., the sum of first n terms of a sequence and S = lim n → ∞ S n = a 1 + a 2 + ... + a n + ... = ∑ n = 1 ∞ a n is known as sum of infinite terms of sequence termed as infinite series. The given series is termed as positive term series if in the series S = a 1 + a 2 + ... + a n + ... = ∑ n = 1 ∞ a n , a n > 0 ∀ n ∈ N . An infinite series S = lim n → ∞ S n = a 1 + a 2 + ... + a n + ... = ∑ n = 1 ∞ a n is said to be convergent or divergent if its n th partial sum S n = a 1 + a 2 + ... + a n = ∑ k = 1 n a k is convergent to finite value or divergent to infinity. Necessary and sufficient condition for convergence Necessary and sufficient condition that infinite series S = ∑ n = 1 ∞ a n is said to be convergent if lim n → ∞ a n = 0 . Suppose a n ≥ 0 ∀ n ∈ N , then S = ∑ n = 1 ∞ a n is said to be convergent if and only if S n = a 1 + a 2 + ... + a n = ∑ k = 1 n a k is bounded above. Tests for Convergence Let us determine the convergence or the divergence of a series by comparing it to one whose behavior is already known. Comparison test: Suppose 0 ≤ a n ≤ b n ∀ n ≥ k for some k. Then (1) If ∑ n = 1 ∞ b n is convergent then ∑ n = 1 ∞ a n is also convergent (2) If ∑ n = 1 ∞ a n is divergent then ∑ n = 1 ∞ b n is also divergent Limit Comparison Test: Suppose a n , b n ≥ 0 ∀ n ∈ N and lim n → ∞ a n b n = L (1) If L ∈ R , L > 0 , then both ∑ n = 1 ∞ a n and ∑ n = 1 ∞ b n converge or diverge together. (2) If L ∈ R , L = 0 and ∑ n = 1 ∞ b n converges , then ∑ n = 1 ∞ a n also converges (3) If L → ∞ and ∑ n = 1 ∞ b n diverges , then ∑ n = 1 ∞ a n also diverges Cauchy Test or Cauchy condensation Test : Suppose a n ≥ 0 ∀ n ∈ N and a n + 1 ≤ a n , then ∑ n = 1 ∞ a n converges if and only if ∑ k = 0 ∞ 2 k a 2 k converges. Geometric Series Test : The geometric series S = a + a r + ... + a r n − 1 + ... = ∑ n = 1 ∞ a r n − 1 , a > 0 is Convergent if | r | < 1 Divergent if | r | ≥ 1 p- Series Test : (1) ∑ n = 1 ∞ 1 n p converges if p > 1 and diverges if p ≤ 1 (2) ∑ n = 2 ∞ 1 n ( log n ) p converges if p > 1 and diverges if p ≤ 1 Ratio Test: D’ Alembert Test: Suppose a n > 0 and lim n → ∞ a n a n + 1 = L . Then (1) If L ∈ R , L > 1 , then ∑ n = 1 ∞ a n converges. (2) If L ∈ R , L < 1 , then ∑ n = 1 ∞ a n diverges. (3) If L ∈ R , L = 1 , then test gets fail Raabe’s Test: Suppose a n > 0 and lim n → ∞ n ( a n a n + 1 − 1 ) = L . Then (1) If L ∈ R , L > 1 , then ∑ n = 1 ∞ a n converges. (2) If L ∈ R , L < 1 , then ∑ n = 1 ∞ a n diverges...

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition)

3rd Edition

ISBN: 9781319019846

Author: Jon Rogawski, Colin Adams

Publisher: W. H. Freeman

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Question

Chapter 10.4, Problem 25E

To determine

Whether the infinite series is convergent or divergent.

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Chapter 10.4, Problem 24E

Chapter 10.4, Problem 26E

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The correct answer is D Could you explain and show the steps please

Taylor Series Approximation Example- H.W More terms used implies better approximation f(x) 4 f(x) Zero order f(x + 1) = f(x;) First order f(x; + 1) = f(x;) + f'(x;)h 1.0 Second order 0.5 True f(x + 1) = f(x) + f'(x)h + ƒ"(x;) h2 2! f(x+1) 0 x; = 0 x+1 = 1 x h f(x)=0.1x4-0.15x³- 0.5x2 -0.25x + 1.2 51 Taylor Series Approximation H.w: Smaller step size implies smaller error Errors f(x) + f(x,) Zero order f(x,+ 1) = f(x) First order 1.0 0.5 Reduced step size Second order True f(x + 1) = f(x) + f'(x)h f(x; + 1) = f(x) + f'(x)h + "(xi) h2 f(x,+1) O x₁ = 0 x+1=1 Using Taylor Series Expansion estimate f(1.35) with x0 =0.75 with 5 iterations (or & s= 5%) for f(x)=0.1x 0.15x³-0.5x²- 0.25x + 1.2 52

Could you explain this using the formula I attached and polar coorindates

Chapter 10 Solutions

Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition)

Ch. 10.1 - Prob. 1PQ Ch. 10.1 - Prob. 2PQ Ch. 10.1 - Prob. 3PQ Ch. 10.1 - Prob. 4PQ Ch. 10.1 - Prob. 5PQ Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E

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