Infinite series-Calculus and Analytical Geometry
Download as PPTX, PDF2 likes559 views
The document provides a detailed overview of infinite sequences and series, including concepts such as convergence, divergence, and various convergence tests. It explains the characteristics of sequences (bounded, unbounded, non-decreasing) and the criteria for determining the convergence of series using methods like the ratio test and the comparison test. Additionally, it covers Taylor and Maclaurin series, demonstrating how to derive them and their applications.
Infinite series-Calculus and Analytical Geometry
- 1. 1Presented by: Group-E (The Anonymous) Bikash Dhakal Gagan Puri Bikram Bhurtel Bimal Pradhan Rabin B.K
- 2. 2 Infinite sequence and series of convergence and divergence Different types of test Absolute and conditional convergence Taylor and Maclaurin series Contents
- 3. Infinite sequence and series of convergence and divergence Infinite sequence: an infinite sequence of a numbers whose domain is the natural numbers than or equal to some fixed integer n. For instance , {1,2,3,4,…..}; {√n}; {(-1n)/n} are sequences. Normally, the general term of an infinite sequence is noted by u ( also known as nth term of the sequence. 3
- 4. Sequence General term {1,2,3,4,…..} N {√n} √n 2,4,6,8,…… 2n 1,8,27,64,…… n3 Constant Sequence : A sequence with every term is same fixed value is called a constant sequence. For instance, the sequence {2,2,2,2….. } is a constant sequence. 4
- 5. Bounded and Unbounded Sequence: • A sequence {un} is bounded if there are two fixed value k, KϵR such that k≤ un ≤ K, for all n. • A sequence {un} is bounded above if there is a real value k ϵ R such that un ≤ k, for all n. In such case, the sequence is called unbounded below. • A sequence{un} is bounded below if there is a real value k ϵ R such that k ≤ un for all n. In such case, the sequence is called unbounded above. • A sequence{un} is called unbounded if it is no bounded above and bounded below. 5
- 6. Convergent and Divergent of Infinite sequence: Convergent sequence: sequence approach to a single value as n tends to infinity. definition: Let {an} be infinite sequence. The sequence {an} converses to a number L if for every positive number ԑ > 0 there corresponds an integer N such that for all n, |an –L| < ԑ for n>N. mathematically, the sequence {a} converges to a number L if ԑ > 0 | an -L|< ԑ n>N. In such condition we observe =L Divergent sequence: sequence has no fixed value as n tends to infinity. Definition: An infinite sequence {an} of real numbers is called divergent if has no fixed finite value. 𝑙𝑖𝑚 𝑛→∞ 𝑎 𝑛 𝑙𝑖𝑚 𝑛→∞ 𝑎 𝑛 6
- 7. Non-Decreasing Sequence: A sequence {an} with the property an < an +1 for all n, is called a non-decreasing sequence. For instance the following sequences are non-decreasing. (i)1,2,3,……,n,…. (ii)1/2,2/3,3/4,…..,n/(n+1),…… Both are bounded below but not bounded above. Least Upper Bound: Let {an} be an infinite sequence. If there is a value M such that an ≤M for all n ……..(i) and if there is no value less than M satisfies the relation (i), then M is called least upper bound of the sequence {an} 7
- 8. Greater Lower Bound: Let {an} be an infinite sequence. If there is a value Q such that an ≥Q for all n ……..(i) and if there is no value greater than Q satisfies the relation (i), then Q is called greatest lower bound of the sequence {an} Eg: a) The sequence 1,2,3,….,n,…. has no upper bound but it has 1 as a greatest lower bound b) The sequence 1,1/2,1/3,…,1/n,…has upper bound any value among 1,2,3,…. But the least upper bound of the sequence is 1. 8
- 9. Non-Decreasing Sequence Theorem: A non-decreasing sequence of real numbers converges if and only if it is bounded from above. If a non decreasing sequence converges, it converges to its least upper bound. 9
- 10. Infinite Series: Given a sequence of numbers {an} an expression a1+a2+a3+……+ an +…..= an is called infinite series. The number a is known as nth term ( or general term) of the series. If we take only the finite number of terms from the infinite series, then series of such finite terms is called partial sum of the series. Partial sum: Given a series with finite k-terms in the form ∑an is called kth partial sum of the series ∑an and denoted by Sk i.e. Sk = ∑ai 10
- 11. Convergence and Divergence of an Infinite Series: Let an be an infinite series. If there is a finite value L such that 𝑙𝑖𝑚 𝑛→∞ 𝑖=1 𝑛 𝑎𝑖 =L i.e. 𝑙𝑖𝑚 𝑛→∞ Sn=L then we say the given series is convergent to L. Otherwise, the series is divergent. Telescoping series: A series in which on expansion of nth partial sum, every term except first and last term is cancelled out is called telescoping series. 11
- 12. Theorem: The necessary condition for the convergence of an infinite series an, is 𝑙𝑖𝑚 𝑛→∞ an =0. But, the condition is not sufficient. Proof: Let ∑an be an infinite series. Let, Sn = a1+a2+…..+an. Suppose that the series ∑an is convergent to a finite value S. So, 𝑙𝑖𝑚 𝑛→∞ Sn =S. And also, 𝑙𝑖𝑚 𝑛→∞ Sn-1 =S. Now, an = Sn – Sn-1 Therefore, 𝑙𝑖𝑚 𝑛→∞ an = 𝑙𝑖𝑚 𝑛→∞ (Sn-Sn-1) = 𝑙𝑖𝑚 𝑛→∞ Sn – 𝑙𝑖𝑚 𝑛→∞ Sn-1 Thus the series converses then necessarily 𝑙𝑖𝑚 𝑛→∞ an = 0. 12
- 13. But, the condition is not sufficient. That is = 0 may not imply that the series converges. Take the series ∑(1/n). Here, an=1/n Then, = =0. But, ∑( 1 𝑛 )= 1+ 1 2 + 1 3 +….=1+ 1 2 +( 1 3 + 1 4 )+( 1 5 + 1 6 + 1 7 + 1 8 )+…=1+ 1 2 + 1 2 + 𝟏 𝟐 + 𝟏 𝟐 +… =∞ This means the given series is divergent. Thus, the condition 𝑙𝑖𝑚 𝑛→∞ 𝑎 𝑛= 0 is only the necessary condition for convergency of an infinite series but not a sufficient. Note: The second part of above theorem can be solved by using p-series with p=1. So, by p-test ∑(1/n) is divergent. 𝑙𝑖𝑚 𝑛→∞ 𝑎 𝑛 𝑙𝑖𝑚 𝑛→∞ 𝑎 𝑛 13 𝑙𝑖𝑚 𝑛→∞ 1 𝑛
- 14. nth term test for divergence: if lim an fails to exist or different from zero then the series ∑an is divergent. Geometric series: A series of the form a+ar+ar²+….+arⁿ+…. = arⁿ where a is the non-zero first term r is fixed ratio. Such series is known as geometric series. Geometric Ratio Test Theorem: The geometric series a+ar+ar²+….. converses to a/(1-r) if |r|<1 and diverges if |r|>1. 14
- 15. Combining series : If the two series are convergent, then the basic mathematical operation between the series does not destroy the behavior. Theorem: If ∑an =A and ∑bn=B are two convergent series. Then, i) Sum Rule: ∑(an+bn) = ∑ an+ ∑ bn =A+B ii) Difference rule: ∑(an-bn)= ∑ an- ∑ bn =A-B iii) Constant multiple rule: ∑(kan)= k ∑ an = kA for any constant number k. 15
- 16. • Integral test:- A series of positive terms f(1)+f(2)+....f(x)+…. Where ,f(x) decreases as n increases, the series f(x) converses if the is finite. The series f(x) diverges , if the improper integral is an infinite. 𝑛=1 ∞ 𝑢 𝑛 1 ∞ 𝑓 𝑥 𝑑𝑥 1 ∞ 𝑓 𝑥 𝑑𝑥 Theorem (statement only):- A series of non-negative term converges , if and only if its partial sum are bounded above. 16
- 17. 𝑛=𝑁 ∞ 𝑓 𝑥 𝑑𝑥 Theorem (statement only):- Let {uₙ}be a sequence of positive terms suppose that u =f(x) , where f is a continuous , positive , decreasing function of x for all x ≥ N (N is a positive integer ) then the series both converses or both diverges. P-series test: The series: 1 1 𝑝 + 1 2 𝑝 + 1 3 𝑝 + …….. + 1 𝑛 𝑝 + … . Converges if p>1 and diverges if p≤1 17
- 18. 𝑠 𝑛 = 𝑟=1 𝑛 𝑢 𝑟 ≤ 𝑟=1 𝑛 𝑣𝑟 ≤ 𝑡 𝑛 Direct comparison test :- (statement):- if ∑uₙ and ∑vₙ are two series of positive termms and uₙ and vₙ then, 1. If ∑vₙ converges ,∑uₙ also converges, 2. If ∑uₙ diverges ,∑vₙ also diverges. •Proof let sₙ and tₙ denote the nth partial sums of the series ∑uₙ and ∑vₙ respectively , clearly {sₙ} and {tₙ} are both monotonic increases . Since ∑vₙ converses , {tₙ} is bounded above ,but since uₙ ≤ vₙ i.e sₙ ≤ tₙ so , {sₙ} is bounded above for all n Hence ,{sₙ} being monotonic increasing and bounded above so sₙ ≤ tₙ shows that the sequence {tₙ} is above not bounded above and is divergent and hence the series ∑vₙ is also divergent. 18
- 19. .,,,)(lim divergesanthendivergesbnif bn an n Suppose for all n > N (N is an integer). obnandan 0 divergesbothorconvergesbothbnandanthenL bn an n ,,,0)(lim .,,,,0)(lim convergesanthenconvergesbnifand bn an n 19 Limit Comparison Test
- 20. Let ∑ be a series with positive terms and suppose that Then, If L<1 then the series converges. If L>1 then the series diverges. If L=1 then the series inconclusive and further test needed. an Lnan n 1 )(lim 20 The ratio Test
- 21. The ratio Test Let ∑ be a series with ≥0 for n ≤ N and suppose that Then, If L < 1 then the series converges. If L > 1 then the series diverges. If L = 1 then the test is inconclusive. thn an an Lnan n 1 )(lim 21
- 22. 1 ∞ 𝑢 𝑛 = 𝑢1 − 𝑢2 + 𝑢3 − 𝑢4+. . . . . . 𝑛=1 ∞ −1 𝑛+1 𝑢 𝑛 = 𝑢1 − 𝑢2 + 𝑢3 − 𝑢4+. . . . 𝑢 𝑛=1 ≤ 𝑢 𝑛 𝑛 ≥ 𝑁 lim 𝑛→∞ 𝑢 𝑛 = 0 Alternating series :- Let {un} be a sequence of positive numbers. A series of the form Example of alternative harmonic series Leibnitz’s theorem (alternative series test): The series converges if all thee of the following conditions are satisfied. ●The un’s are all positive ● For all for some integer N Absolute and conditional convergence 1 − 1 2 + 1 3 − 1 4 +. . . + −1 1+𝑛 𝑛 +. . . 22
- 23. Let the series converges. Then converges. By direct comparison test, the series ● A series of convergence absolutely if the corresponding series converges Theorem (the absolute convergence test) If converges then also converses Proof: 𝑛=1 ∞ 𝑢 𝑛 𝑛=1 ∞ 𝑢 𝑛 𝑛=1 ∞ 𝑢 𝑛 𝑛=1 ∞ 𝑢 𝑛 − 𝑢 𝑛 ≤ 𝑢 𝑛 ≤ 𝑎 𝑛 For all n 0 ≤ 𝑢 𝑛 + 𝑢 𝑛 ≤ 2 𝑢 𝑛 2 𝑛=1 ∞ 𝑎 𝑛 Absolute convergence 23
- 24. 𝑛=1 ∞ 𝑢 𝑛 + 𝑢 𝑛 𝑢 𝑛 = 𝑢 𝑛 + 𝑢 𝑛 − 𝑢 𝑛 𝑛=1 ∞ 𝑢 𝑛 = 𝑛=1 ∞ 𝑢 𝑛 + 𝑢 𝑛 − 𝑛=1 ∞ 𝑢 𝑛 Converges Converges Hence, it also Converges and then Also Converges 24
- 25. If converges absolutely and be any arrangement of the sequence then converges absolutely and Conditional convergence:- If a series converges but doesn’t converge, then the series is called convergent conditionally 𝑛=1 ∞ 𝑢1 𝑛 𝑛=1 ∞ 𝑢2 𝑛 𝑢1 𝑛 𝑛=1 ∞ 𝑢2 𝑛 𝑛=1 ∞ 𝑢1 𝑛 = 𝑛=1 ∞ 𝑢2 𝑛 𝑛=1 ∞ 𝑢 𝑛 𝑛=1 ∞ 𝑢 𝑛 25
- 26. Let f be a function with derivatives of all orders throughout same interval containing a as an interior point then the Taylor series generated by f at x = a, 𝑘=0 ∞ 𝑓 𝑘(𝑎)(𝑥−𝑎) 𝑘 𝑘! is given by: 26 Taylor series 𝑘=0 ∞ 𝑓 𝑘(𝑎)(𝑥−𝑎) 𝑘 𝑘! = f(a) + 𝑓′ 𝑎 𝑥−𝑎 1! + 𝑓′′(𝑎)(𝑥−𝑎)2 2! - - - - - + 𝑓 𝑘(𝑎) (𝑥−𝑎) 𝑘 𝑘!
- 27. Step 1: Writing Taylor’s series 𝑘=0 ∞ 𝑓 𝑘(𝑎)(𝑥−𝑎) 𝑘 𝑘! = f(a) + 𝑓′ 𝑎 𝑥 − 𝑎 + 𝑓′(𝑎)(𝑥−𝑎)2 2! - - - - - + 𝑓 𝑘(𝑎) (𝑥−𝑎) 𝑘 𝑘! Step 2: Fin respective derivative and plug into the series according to the series (till fifth or sixth derivative is better and recommended) f(a=1) = ln(1)=0 𝑓′ 𝑎 = 1 = 1 𝑥 = 1 1 = 1 𝑓′′ 𝑎 = 1 = −1 𝑥2 = −1 (1)2 = -1 𝑓′′′ 𝑎 = 1 = 2 𝑥3 = 2 (1)3 = 2 𝑓′𝑣 𝑎 = 1 = −6 𝑥4 = −6 (1)4 = -6 𝑓 𝑣 𝑎 = 1 = 24 𝑥4 = 24 (1)4 = 24 𝑓′𝑣 𝑎 = 1 = −100 𝑥5 = −100 (1)5 = -100 27 Example: Taylor series expansion of ln(x) with center a= 1.
- 28. Step 3: Plugging in the respective values in the series 𝑘=0 ∞ 𝑓 𝑘(𝑎)(𝑥−𝑎) 𝑘 𝑘! = f(a) + 𝑓′ 𝑎 𝑥 − 𝑎 + 𝑓′(𝑎)(𝑥−𝑎)2 2! + - - - - + 𝑓 𝑘(𝑎) (𝑥−𝑎) 𝑘 𝑘! = f(1) + 𝑓′ 1 𝑥−1 1! + 𝑓′′(1)(𝑥−1)2 2! + 𝑓′′′(1)(𝑥−1)3 3! + - - - - + 𝑓 𝑘(𝑎) (𝑥−𝑎) 𝑘 𝑘! Step 4: Replacing the values 𝑘=0 ∞ 𝑓 𝑘(𝑎)(𝑥−𝑎) 𝑘 𝑘! = 0 + 1. 𝑥−1 1! + (−1)(𝑥−1)2 2! + (2)(𝑥−1)3 3! + ……… = 1. 𝑥−1 1! − (𝑥−1)2 2! + 2(𝑥−1)3 3! - ………… 28
- 29. Let f be a function with derivatives of order k for k=1,2,3,…….,n in same interval containing a as an interior point. Then for any integer n from 0 through N, the Taylor polynomial of order n generated by f at x = a is the polynomial, 29 Taylors polynomial of order n 𝑃𝑛 𝑥 = 𝑓 𝑎 + 𝑓′ 𝑎 𝑥 − 𝑎 1! + 𝑓′′ (𝑎)(𝑥 − 𝑎)2 2! + ⋯ + 𝑓 𝑘 𝑎 𝑥 − 𝑎 𝑘 𝑘! + … + 𝑓 𝑎 𝑛 𝑥 − 𝑎 𝑛 𝑛!
- 30. Taylor series centered at 0 is called a Maclaurin series Let f be a function with derivatives of all orders throughout same interval containing 0 as an interior point then the Maclaurin series generated by f at x = 0 is, 30 Maclaurin series 𝑘=0 ∞ 𝑓 𝑘(0)(𝑥) 𝑘 𝑘! = f(0) + 𝑓′ 0 𝑥 + 𝑓′′(0)(𝑥)2 2! - - - - - + 𝑓 𝑘(0) (𝑥) 𝑘 𝑘!
- 31. Step 1: Write Maclaurin’s series 𝑘=0 ∞ 𝑓 𝑘(0)(𝑥) 𝑘 𝑘! = f(0) + 𝑓′ 0 𝑥 1! + 𝑓′(0)(𝑥)2 2! - - - - - + 𝑓 𝑘(0) (𝑥) 𝑘 𝑘! Step 2: Find respective derivative and plug into the series according to the series (till fifth or sixth derivative is better and recommended) 𝑓′ 𝑥 = 0 = cosx= cos(0)= 1 𝑓′′ 𝑥 = 0 = -sinx= sin(0)= 0 𝑓′′′ 𝑥 = 0 = -cosx= -cos(0)= -1 𝑓′𝑣 𝑥 = 0 = sinx= sin(0) = 0 𝑓 𝑣 𝑥 = 0 = cosx= cos(0)= 1 𝑓 𝑣′ 𝑥 = 0 = -sinx= -sin(0)=0 … …. 𝑓2𝑛+1 𝑥 = 0 = (−1) 𝑛 cosx 𝑓2𝑛 𝑥 = 0 = (−1) 𝑛 sinx 31 Example: sinx converges to sinx for all x.
- 32. 𝑘=0 ∞ 𝑓 𝑘(0)(𝑥) 𝑘 𝑘! = f(0) + 𝑓′ 0 𝑥 1! + 𝑓′′(0)(𝑥)2 2! + 𝑓′′′ 0 (𝑥)3 3! + - - - - - + 𝑓 𝑘(0) (𝑥) 𝑘 𝑘! = 0 + 1 − 0.(𝑥)2 2! − 1.(𝑥)3 3! + 0.(𝑥)4 4! + 1. 𝑥 5 5! − … … + −1 𝑛 𝑥 2𝑛+1 𝑛+1 ! + 𝑅 𝑛 𝑥 … … . . = 1 − 1.(𝑥)3 3! − 1. 𝑥 5 5! − ………… 32 Step 3: Plugging in the values into the series
- 33. Estimating remainder in Maclaurin series If there are positive constants m such that 𝑓(𝑡) 𝑛+1 < 𝑚 for all t lies between a and x, inclusive then the remainder term, 𝑅 𝑛(𝑥) in Taylor’s theorem satisfies in inequality. If these conditions hold for every n and all other conditions of Taylor’s theorem are satisfied by f, then the series converges to f(x). 33 𝑅 𝑛(𝑥) ≤ 𝑀 𝑥 − 𝑎 𝑛+1 𝑛 + 1 !
- 34. Step 1: Write Maclaurin’s series 𝑘=0 ∞ 𝑓 𝑘(0)(𝑥) 𝑘 𝑘! = f(0) + 𝑓′ 0 𝑥 1! + 𝑓′(0)(𝑥)2 2! - - - - - + 𝑓 𝑘(0) (𝑥) 𝑘 𝑘! Step 2: Find respective derivative and plug into the series according to the series (till fifth or sixth derivative is better and recommended) 𝑓′ 𝑥 = 0 = cosx= cos(0)= 1 𝑓′′ 𝑥 = 0 = -sinx= sin(0)= 0 𝑓′′′ 𝑥 = 0 = -cosx= -cos(0)= -1 𝑓′𝑣 𝑥 = 0 = sinx= sin(0) = 0 𝑓 𝑣 𝑥 = 0 = cosx= cos(0)= 1 𝑓 𝑣′ 𝑥 = 0 = -sinx= -sin(0)=0 … …. 𝑓2𝑛+1 𝑥 = 0 = (−1) 𝑛 cosx 𝑓2𝑛 𝑥 = 0 = (−1) 𝑛 sinx 34 Example: sinx converges to sinx for all x.
- 35. 𝑘=0 ∞ 𝑓 𝑘(0)(𝑥) 𝑘 𝑘! = f(0) + 𝑓′ 0 𝑥 1! + 𝑓′′(0)(𝑥)2 2! + 𝑓′′′ 0 (𝑥)3 3! + - - - - - + 𝑓 𝑘(0) (𝑥) 𝑘 𝑘! = 0 + 1 − 0.(𝑥)2 2! − 1.(𝑥)3 3! + 0.(𝑥)4 4! + 1. 𝑥 5 5! − … … + −1 𝑛 𝑥 2𝑛+1 𝑛+1 ! + 𝑅 𝑛 𝑥 … … . . = 1 − 1.(𝑥)3 3! − 1. 𝑥 5 5! − ………… 35 Where , 𝑅 𝑛 𝑥 = −1 𝑛 𝑥 2𝑛+2 𝑛+2 !
- 36. For remainder, we have 𝑅 𝑛(𝑥) ≤ 1. 𝑥−0 𝑛+2 𝑛+2 ! = 𝑥 𝑛+2 𝑛+2 ! So when n→ ∞, 𝑅 𝑛(𝑥) → 0 36 𝑅 𝑛(𝑥) ≤ 𝑀 𝑥 − 𝑎 𝑛+1 𝑛 + 1 !
- 37. REFERENCE Calculus and Analytical Geometry (Binod Prasad Dhakal) Teachers note www.google.com www.wikipedia.com 37
- 38. Queries 38