MTL100 Calculus Tutorial 3 PDF

MTL100 - Calculus Tutorial 3 Department of Mathematics Indian Institute of Technology Delhi (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 1 / 83 Question 1 D...

MTL100 - Calculus Tutorial 3 Department of Mathematics Indian Institute of Technology Delhi (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 1 / 83 Question 1 Determine if the following limits exist: (a) lim [x] x→0 (b) lim sgn(x) x→0 (c) lim sin( x1 ) x→0 √ (d) lim x sin( x1 ) x→0 (e) lim x cos( x1 ). x→0 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 2 / 83 Question 1 (a) Determine if the following limit exist: lim [x]. x→0 Solution: Recall the following theorem, Theorem lim f (x) = L exists ⇐⇒ lim f (x) = lim f (x) = L. x→a x→a− x→a+ Let us write the function f (x) = [x] near x = 0 as ( −1 if −1 ≤ x < 0 f (x) = 0 if 0 ≤ x < 1. Clearly, lim [x] = −1 and lim+ [x] = 0. x→0− x→0 Note that lim [x] 6= lim+ [x]. x→0− x→0 Hence lim [x] does not exist. x→0 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 3 / 83 Question 1 (b) Determine if the following limit exist: lim sgn(x). x→0 Solution: Recall the following: We say that lim f (x) = L if, for any sequence {xn } with xn 6= x0 and x→x0 xn → x0 , we have f (xn ) → L as n → ∞. Let us write the function f (x) = sgn(x) as  1  if x > 0 sgn(x) = −1 if x < 0  0 if x = 0.  (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 4 / 83 1 (b) Contd..     1 1 Consider the two sequences {xn } = and {yn } = −. n n Note that lim xn = lim yn = 0. n→∞ n→∞ However, lim f (xn ) = 1 and lim f (yn ) = −1. n→∞ n→∞ Hence, lim sgn(x) does not exist. x→0 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 5 / 83 Question 1 (c) Determine if the following limit exist: lim sin( x1 ). x→0 Solution:     1 1 Consider the sequences {xn } = and {yn } =. nπ 2nπ + π/2 Note that lim xn = lim yn = 0. n→∞ n→∞   1 Let f (x) = sin. Then we have, x lim f (xn ) = 0 and lim f (yn ) = 1. n→∞ n→∞ 1
Hence, lim sin x does not exist. x→0 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 6 / 83 Question 1 (d) Determine if the following limit exist: √ lim x sin( x1 ). x→0 Solution: Recall the following result: Suppose f (x) is bounded in an interval containing c and lim g (x) = 0. x→c Then lim f (x)g (x) = 0. x→c √   1 Let f (x) = sin and g (x) = x. x Then lim g (x) = 0 and |f (x)| ≤ 1 for x ∈ [−1, 1]. x→0 Hence, by the above result, √   1 lim x sin = 0. x→0 x (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 7 / 83 Question 1 (e) Determine if the following limit exist: lim x cos( x1 ). x→0 Solution:   1 Let f (x) = cos and g (x) = x. x Then lim g (x) = 0 and |f (x)| ≤ 1 for x ∈ [−1, 1]. x→0 Hence,   1 lim x cos = 0. x→0 x (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 8 / 83 Question 2 Determine if the following limits exist: x−|x| lim x and lim x 1+sin x. x→0 x→∞ x−|x| Solution: First we consider lim x. x→0 Note that ( x if x ≥ 0 |x| = −x if x < 0. Now we have x − |x| x −x lim = lim+ = 0, x→0+ x x→0 x and x − |x| x +x lim = lim = 2. x→0− x x→0 − x (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 9 / 83 2 Contd.. Note that x − |x| x − |x| lim+ 6= lim. x→0 x x→0− x Hence, the limit does not exist. Now, we consider lim x 1+sin x. x→∞ Recall the following: We say that lim f (x) = L if, for any sequence {xn } with xn → ∞, we x→∞ have f (xn ) → L as n → ∞. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 10 / 83 2 Contd.. Let f (x) = x 1+sin(x).   3π Consider the two sequences {xn } = 2nπ + and {yn } = {nπ}. 2 So we have, lim xn = lim yn = ∞. n→∞ n→∞ Now note that sin(xn ) = −1 and sin(yn ) = 0 for all n ≥ 1. Therefore, lim f (xn ) = 1 and lim f (yn ) = ∞. n→∞ n→∞ Hence, the limit does not exist. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 11 / 83 Question 3 Show that the function f is continuous only at x = 1/2. ( x, if x is rational f (x) = 1 − x, if x is irrational. Solution: We have, |f (x) − f (1/2)| = |x − 1/2|. Let  > 0. Then |f (x) − 1/2| <  whenever |x − 1/2| < δ = . Hence, by the definition of continuity, f is continuous at x = 1/2. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 12 / 83 3 Contd.. Let us recall the following result: For any r ∈ R there exists a sequence {xn } of rational (irrational) numbers such that lim xn = r. n→∞ 1 Let c ∈ R be an arbitrary real number and c 6=. 2 Then by the above result there exist a sequence {xn } of rational numbers and a sequence {yn } of irrational numbers such that lim xn = lim yn = c. n→∞ n→∞ Note that lim f (xn ) = c and lim f (yn ) = 1 − c. n→∞ n→∞ 1 Since lim f (xn ) 6= lim f (yn ) for c 6= , f is discontinuous at c. n→∞ n→∞ 2 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 13 / 83 Question 4 Determine which of the following functions are uniformly continuous in the interval mentioned: 2 (a) e x sin(x 2 ) on (0, 1) (b) | sin x| on [0, ∞) √ (c) x sin x on (0, ∞) (d) sin(x 2 ) on R. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 14 / 83 Question 4 (a) Determine if the following function is uniformly continuous: 2 e x sin(x 2 ) on (0, 1). Solution: 2 The functions e x and sin(x 2 ) both are continuous on [0, 1]. 2 Hence, e x sin(x 2 ) is continuous on [0, 1]. Recall the following theorem: Theorem A continuous function f (x) on a closed and bounded interval [a, b] is uniformly continuous. 2 By the above theorem, e x sin(x 2 ) is uniformly continuous on [0, 1] and hence uniformly continuous on the given interval (0, 1) ⊂ [0, 1]. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 15 / 83 Question 4 (b) Determine if the following function is uniformly continuous: | sin x| on [0, ∞). Solution: Let f (x) = | sin x|. Let  > 0 be arbitrary. Now, for x, y ∈ [0, ∞), |f (x) − f (y )| = || sin x| − | sin y || ≤ | sin x − sin y |     x +y x −y = 2 cos sin 2 2 x −y ≤2 = |x − y |. 2 Choose δ = . Then we have |x − y | < δ ⇒ |f (x) − f (y )| < . Hence, by the definition, | sin x| is uniformly continuous on [0, ∞). (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 16 / 83 Question 4 (c) Determine if the following function is uniformly continuous: √ x sin x on (0, ∞). Recall the following proposition: Proposition Let f be a real valued function on S ⊆ R. Then f is uniformly continuous function on S if and only if for any two sequences {xn }, {yn } in S such that |xn − yn | → 0, we have |f (xn ) − f (yn )| → 0 as n → ∞. Solution: 1 For n ∈ N, let xn = 2n2 π + and yn = 2n2 π. n 1 Then, |xn − yn | = → 0 as n → ∞. n (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 17 / 83 4 (c) Contd.. √ Let f (x) = x sin x. Now, √ r 1 |f (xn ) − f (yn )| = sin(2n2 π + 1/n) − 2n2 π sin(2n2 π) 2n2 π + n √ r   1 1 = 2π + 3 n sin → 2π as n → ∞. n n √ Hence, x sin x is not uniformly continuous on (0, ∞). (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 18 / 83 Question 4 (d) Determine if the following function is uniformly continuous: sin(x 2 ) on R. Solution: √ r π Let xn = nπ and yn = nπ +. 2 π √ r π 2 Now, |xn − yn | = nπ − nπ + = → 0. √ r 2 π nπ + nπ + 2 But | sin(xn2 ) − sin(yn2 )| = 1 6→ 0. Hence, sin(x 2 ) is not uniformly continuous on R. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 19 / 83 Question 5 Determine if the following functions are differentiable at 0. Also find f 0 (0), if it exists: ( 1 e − x2 if x 6= 0 (a) f (x) = 0 if x = 0. (b) f (x) = e −|x| , x ∈ R. ( x cos x1 if x 6= 0 (c) f (x) = 0 if x = 0. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 20 / 83 Question 5 (a) Determine if the following function is differentiable at 0. Find f 0 (0), if it exists. ( 1 e − x2 if x 6= 0 f (x) = 0 if x = 0. Solution: A real valued function f (x) is said to be differentiable at x0 if the limit f (x) − f (x0 ) lim x→x0 x − x0 exists. This limit is called the derivative of f at x0 , denoted by f 0 (x0 ). (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 21 / 83 5 (a) Contd.. Here, 1 f (x) − f (0) e − x2 lim = lim x→0 x −0 x→0 x 1 x = lim 1 x→0 e x2 1 1 Since e x 2 ≥ x2 , we have 1 1 x x 0≤ 1 ≤ 1 = |x|. e x2 x2 1 1 f (x)−f (0) Thus, lim x 1 = 0, i.e., lim x 1 = 0. This implies, lim x−0 = 0. x→0 x2 x→0 x2 x→0 Hence f is differentiable at 0 and f 0 (0) = 0. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 22 / 83 Question 5 (b) Determine if the following function is differentiable at 0. Find f 0 (0), if it exists. f (x) = e −|x| , x ∈ R. Solution: We have ( e −x if x ≥ 0 e −|x| = ex if x ≤ 0. Now, f (x) − f (0) e −x − 1 lim+ = lim+ x→0 x −0 x→0 x ex − 1 = − lim ( by replacing x by − x) x→0− x = −1. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 23 / 83 5 (b) Contd.. Also, we have f (x) − f (0) ex − 1 lim = lim x→0− x −0 x→0− x = 1. f (x)−f (0) Therefore lim x−0 does not exist. x→0 Hence f is not differentiable at x = 0. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 24 / 83 Question 5 (c) Determine if the following function is differentiable at 0. Find f 0 (0), if it exists. ( x cos x1 if x 6= 0 f (x) = 0 if x = 0. Solution: Note that f (x) − f (0) x cos x1 lim = lim x→0 x −0 x→0 x 1 = lim cos. x→0 x Recall that: The limit of a function g (x) at some point x0 is L if for any sequence {xn } with xn 6= x0 and xn → x0 as n → ∞, we have g (xn ) → L as n → ∞. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 25 / 83 5 (c) Contd.. 1 If we take the sequence xn = then clearly xn → 0 but 1 nπ cos = cos(nπ) = (−1)n. xn 1 Hence the limit lim cos does not exist and so the function is not x→0 x differentiable at 0. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 26 / 83 Question 6 Determine if f 0 (x) is continuous at 0 for the following functions: ( x 3 sin x1 if x 6= 0 (a) f (x) = 0 if x = 0. ( x 2 cos x1 if x 6= 0 (b) f (x) = 0 if x = 0. ( 1 x 2 ln |x| if x 6= 0 (c) f (x) = 0 if x = 0. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 27 / 83 Question 6 (a) Determine if f 0 (x) is continuous at 0 for the following function. ( x 3 sin x1 if x 6= 0 f (x) = 0 if x = 0. Solution: We shall use the following: 1 1 lim h2 sin = 0, lim h cos = 0. (See tutorial 2) h→0 h h→0 h h3 sin h1 − 0 When x = 0, we have f 0 (0) = lim = lim h2 sin h1 = 0. h→0 h h→0 0 2 1 1 Now for x 6= 0, f (x) = 3x sin x − x cos x. Also note that lim f 0 (x) = f 0 (0) = 0. x→0 Hence f 0 is continuous at 0. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 28 / 83 Question 6 (b) Determine if f 0 (x) is continuous at 0 for the following function. ( x 2 cos x1 if x 6= 0 f (x) = 0 if x = 0. Solution: h2 cos h1 − 0 1 We have f 0 (0) = lim = lim h cos = 0. h→0 h h→0 h For x 6= 0, 1 1 f 0 (x) = 2x cos − sin. x x 1 Now consider xn =. Then lim xn = 0. But 2nπ + π2 n→∞ lim f (xn ) = lim (0 − 1) = −1 6= 0 = f 0 (0). 0 n→∞ n→∞ 0 Therefore, f is not continuous at 0. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 29 / 83 Question 6 (c) Determine if f 0 (x) is continuous at 0 for the following function. ( 1 x 2 ln |x| if x 6= 0 f (x) = 0 if x = 0. 1 Solution: For x 6= 0, f (x) = x 2 ln = −x 2 ln |x|. |x| f (h) − f (0) −h2 ln h lim+ = lim+ = 0, by L’Hospital’s rule. h→0 h h→0 h f (h) − f (0) −h2 ln(−h) lim = lim = 0, by L’Hospital’s rule. h→0− h h→0− h f (h) − f (0) f (h) − f (0) Since lim+ = lim , we get f 0 (0) exists and h→0 h h→0− h equal to zero. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 30 / 83 6 (c) Contd.. For x > 0, f (x) = −x 2 ln x, and f 0 (x) = −2x ln x − x. So lim+ f 0 (x) = lim+ (−2x ln x − x) = 0. x→0 x→0 For x < 0, f (x) = −x 2 ln(−x), and f 0 (x) = −2x ln (−x) − x. So, lim f 0 (x) = lim+ (2x ln x + x) = 0. x→0− x→0 Since lim f (x) = lim+ f 0 (x) = 0, lim f 0 (x) exists and equal to 0. 0 x→0− x→0 x→0 0 0 0 As f (0) = 0 = lim f (x), f is continuous at 0. x→0 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 31 / 83 Question 7 Question Let f be differentiable on R and sup |f 0 (x)| < 1. Select s0 ∈ R and define x∈R sn = f (sn−1 ). Prove that {sn } is a convergent sequence. Solution: First let us recall the Mean Value Theorem: Mean Value Theorem If f is continuous on [a, b] and differentiable on (a, b), then there exists c ∈ (a, b) such that f (b) − f (a) = f 0 (c)(b − a). (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 32 / 83 7 Contd.. Let α = sup |f 0 (x)| < 1. Then by mean value theorem there exists x∈R some s between sn and sn−1 such that |sn − sn−1 | = |f 0 (s)(sn−1 − sn−2 )| ≤ α|sn−1 − sn−2 | ≤ αn−1 |s1 − s0 |. Now if n ≥ m, then |sn − sm | ≤ |sn − sn−1 | + |sn−1 − sn−2 | + · · · + |sm+1 − sm | ≤ (αm + αm+1 + · · · + αn−1 )|s1 − s0 | αm ≤ |s1 − s0 |. 1−α (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 33 / 83 7 Contd.. As 0 < α < 1, lim αm = 0. Hence for  > 0, there exists N ∈ N m→∞ such that 1−α |αm − 0| <  ∀ m ≥ N. |s1 − s0 | This implies that |sn − sm | <  for all n, m ≥ N. Hence {sn } is a Cauchy sequence in R, and hence, it is convergent. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 34 / 83 Question 8 Question Let f be differentiable on R and |f (x) − f (y )| ≤ (x − y )2. Then show that f is constant. Solution: First we recall the following result: Sandwich Theorem Let f (x), h(x) and g (x) be real-valued functions such that f (x) ≤ h(x) ≤ g (x) in an interval containing c, and lim f (x) = lim g (x) = L. x→c x→c Then lim h(x) = L. x→c (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 35 / 83 8 Contd.. Also, recall that Theorem If f is differentiable on (a, b) and f 0 ≡ 0, then f is constant. Let c ∈ R. Since |f (x) − f (y )| ≤ (x − y )2 , we have that −(x − c)2 ≤ f (x) − f (c) ≤ (x − c)2. For the right-handed derivative (RHD), we have f (x) − f (c) −(x − c) ≤ ≤ (x − c). x −c Therefore, by sandwich theorem, f (x) − f (c) f 0 (c + ) (i.e. RHD) = lim+ = 0. x→c x −c (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 36 / 83 8 Contd.. For the left-handed derivative (LHD), we have f (x) − f (c) −(x − c) ≥ ≥ (x − c). x −c Again, it follows from the Sandwich Theorem, that f (x) − f (c) f 0 (c − )(i.e. LHD) = lim = 0. x→c − x −c This implies that f 0 (c) = 0. Since c ∈ R is arbitrary, f 0 ≡ 0 on R and so f is a constant function. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 37 / 83 Question 9 Evaluate the following limits: e x − (1 + x) (a) lim x→0 x2 1 − cos t − t 2 /2 (b) lim t→0 t4 2 (c) lim x 2 (e −1/x − 1). x→∞ (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 38 / 83 Question 9 (a) Evaluate the following limit: e x − (1 + x) lim. x→0 x2 Solution: By L’Hospital’s Rule e x − (1 + x) ex − 1 lim = lim x→0 x2 x→0 2x ex = lim x→0 2 1 =. 2 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 39 / 83 Question 9 (b) Evaluate the following limit: 1 − cos t − t 2 /2 lim. t→0 t4 Solution: By L’Hospital’s Rule 1 − cos t − t 2 /2 sin t − t lim = lim t→0 t4 t→0 4t 3 cos t − 1 = lim t→0 12t 2 − sin t = lim t→0 24t − cos t = lim t→0 24 1 = −. 24 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 40 / 83 Question 9 (c) Evaluate the following limit: 2 lim x 2 (e −1/x − 1). x→∞ Solution: By L’Hospital’s Rule 2 2 e −1/x − 1 lim x 2 (e −1/x − 1) = lim 1 x→∞ x→∞ x2 2 e −1/x.(2/x 3 ) = lim x→∞ −2/x 3 2 = lim −e −1/x x→∞ = −1. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 41 / 83 Question 10 Question Find an approximation of sin x when error is of magnitude no greater than 5 × 10−4 and |x| < 3/10. Theorem Let I be an open interval around a. Let f be a function defined on I. Suppose that f (m+1) (x) exists on I. Then for each x 6= a in I , there exists c between x and a such that f (m) (a) f (x) = f (a) + f 0 (a)(x − a) + · · · + (x − a)m + Rm (x), m! f (m+1) (c) where Rm (x) = (x − a)m+1. (m + 1)! (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 42 / 83 10 Contd.. Solution: There exists c between 0 and x such that sin(m+1) (c) m+1 Rm (x) = x. (m + 1)! sin(m+1) (c) m+1 |x|m+1 3m+1 Thus, |Rm (x)| = x ≤ < m+1. (m + 1)! (m + 1)! 10 (m + 1)! As, 3m+1 = 3m−3 34 < 10m−3 · 5 · (m + 1)! if and only if m ≥ 3, we get |Rm (x)| < 5 × 10−4 if and only if m ≥ 3. Hence, an approximation of sin x when error magnitude is no greater f (2) (0) 2 f (3) (0) 3 x3 than 5 × 10−4 is f (0) + f 0 (0)x + x + x =x−. 2! 3! 3! (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 43 / 83 Question 11 Question x3 Estimate the error in the approximation of sinh x = x + when |x| < 0.5. 3! Theorem Let I be an open interval around a. Let f be a function defined on I. Suppose that f (m+1) (x) exists on I. Then for each x 6= a in I , there exists c between x and a such that f (m) (a) f (x) = f (a) + f 0 (a)(x − a) + · · · + (x − a)m + Rm (x), m! f (m+1) (c) where Rm (x) = (x − a)m+1. (m + 1)! (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 44 / 83 11 Contd.. Solution: There exists c between 0 and x such that sinh(4) (c) 4 sinh(c) 4 R3 (x) = x = x. 4! 4! As, c ∈ (−.5,.5), | sinh(c)| < 1. Thus, sinh(c) 4 |x|4 (0.5)4 1 |R3 (x)| = x < < = 4. 4! 4! 4! 2 4! x3 Hence, the error estimate in the approximation x + of sinh x when 3! 1 |x| < 0.5 is 384. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 45 / 83 Question 12 Find the radius of convergence of the following power series: ∞ X (a) (n + 1 + 2n )x n n=0 ∞ X 1 2n (b) x , a 6= 0 an n=0 ∞ X 1 n (c) x n!nn n=0 ∞ X n! n (d) x nn n=0 ∞ 2 X nn n 2 (x − 1). (e) (n + 1) n n=1 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 46 / 83 Question 12 Theorem ∞ X Let an (x − a)n be a power series with center a. Let R denote the n=0 radius of convergence of the power series. Then, 1 an+1 1 = lim = lim sup |an | n. R n→∞ an n→∞ (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 47 / 83 Question 12 (a) Find the radius of convergence of the power series: ∞ X (n + 1 + 2n )x n. n=0 Solution: Here, an = n + 1 + 2n and an+1 = n + 2 + 2n+1 for all n ∈ N. Therefore, n 2 1 an+1 n + 2 + 2n+1 2n + 2n +2 = lim = lim = lim n 1 = 2. R n→∞ an n→∞ n + 1 + 2n n→∞ 2n + 2n +1 1 Hence, R = is the radius of convergence of the given series. 2 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 48 / 83 Question 12 (b) Find the radius of convergence of the power series: ∞ X 1 2n x , where a 6= 0. an n=0 Solution: Let a be a non-zero real number. ∞ X 1 n Consider the power series y. an n=0 1 1 Here, an = and an+1 = n+1 for all n ∈ N. an a Therefore, 1 an+1 1 1 = lim = lim =. R n→∞ an n→∞ a a ∞ X 1 n Thus, R = |a| is the radius of convergence of y. an n=0 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 49 / 83 12 (b) Contd... ∞ X 1 n Since y converges for |y | < R and diverges for |y | > R, we an n=0 ∞ X 1 2n get that x converges for |x 2 | < R and diverges for |x 2 | > R. an n=0 ∞ 1 2n X √ p Therefore, x converges for |x| < R = |a| and diverges for an √ n=0 p |x| > R = |a|. ∞ X 1 2n Hence, the radius of convergence of the power series x is an p n=0 | a |. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 50 / 83 Question 12 (c) Find the radius of convergence of the power series: ∞ X 1 n x. n!nn n=0 Solution: 1 1 Here, an = n and an+1 = for all n ∈ N. n!n (n + 1)!(n + 1)n+1 Therefore, 1 an+1 n!nn = lim = lim R n→∞ an n→∞ (n + 1)!(n + 1)n+1 1 = lim. n→∞ (n + 1)2 (1 + 1 )n n (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 51 / 83 12 (c) Contd.. Recall that  n 1 lim 1+ =e n→∞ n and 1 lim = 0. n→∞ (n + 1)2 Therefore, 1 1 1 1 = lim · lim
n = 0 · = 0. R n→∞ (n + 1)2 n→∞ 1 + 1 e n Hence, the radius of convergence of the power series is R = ∞. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 52 / 83 Question 12 (d) Find the radius of convergence of the power series: ∞ X n! n x. nn n=0 Solution: n! (n + 1)! Here, an = n and an+1 = for all n ∈ N. n (n + 1)n+1 Therefore, 1 an+1 (n + 1)!nn = lim = lim R n→∞ an n→∞ n!(n + 1)n+1 nn 1 = lim = lim
n. n→∞ (n + 1)n n→∞ 1 + n1 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 53 / 83 12 (d) Contd.. Recall that 1 n   lim 1 + = e. n→∞ n Thus, 1 1 1 = lim
=. 1 n R n→∞ 1+ n e Hence, R = e is the radius of convergence of the given series. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 54 / 83 Question 12 (e) Find the radius of convergence of the power series: ∞ 2 X nn n2 (x − 1)n. n=1 (n + 1) Solution: 2 nn Here, an = for all n ∈ N. (n + 1)n2 Therefore, 1 2 n 1 1 nn = lim sup |an | n = lim sup 2 R n→∞ n→∞ (n + 1)n nn 1 = lim sup = lim sup
n. n→∞ (n + 1)n n→∞ 1 + n1 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 55 / 83 12 (e) Contd.. Recall that 1 n   lim 1 + = e. n→∞ n Thus, 1 1 1 1 = lim sup |an | n = lim sup
=. 1 n R n→∞ n→∞ 1+ n e Hence R = e is the radius of the convergence of the given series. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 56 / 83 Question 13 Write the Taylor’s series around 0 and find the radius of convergence. (a) 1 1+x (b) sinh x (c) e x sinh x (d) x sin x. Taylor’s series Let f (x) be an infinitely differentiable function at 0. Then, the Taylor ∞ X f (n) (0) n series of f (x) around 0 is the power series x , where f (n) (0) n! n=0 denotes the nth derivative of f evaluated at the point 0. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 57 / 83 Question 13 (a) Write the Taylor’s series around 0 and find the radius of convergence. 1 f (x) =. 1+x Solution: −1 (−1)(−2) (3) (−1)(−2)(−3) f (1) (x) = 2 , f (2) (x) = 3 , f (x) =. (1 + x) (1 + x) (1 + x)4 Procedding in this way, we get (−1)(−2) · · · (−n) (−1)n n! f (n) (x) = =. (1 + x)(n+1) (1 + x)(n+1) Therefore, f (0) = 1, and f (n) (0) = (−1)n n! for all n ∈ N. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 58 / 83 13 (a) Contd.. The Taylor series of f (x) around 0 is given by ∞ ∞ ∞ X f (n) (0) n X (−1)n n! n X f (x) = x = x = (−1)n x n. n! n! n=0 n=0 n=0 Here, an = (−1)n for all n ∈ N. Therefore, 1 an+1 (−1)n+1 = lim = lim = 1. R n→∞ an n→∞ (−1)n Hence, the radius of convergence of the Taylor series of the given function is R = 1. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 59 / 83 Question 13 (b) Write the Taylor’s series around 0 and find the radius of convergence. f (x) = sinh(x). Solution: Note that f (1) (x) = cosh(x), f (2) (x) = sinh(x), f (3) (x) = cosh(x) and f (4) (x) = sinh(x). Procedding in this way, we get f (2n−1) (x) = cosh(x) and f (2n) (x) = sinh(x) for all n ∈ N. Therefore, f (2n−1) (0) = 1 and f (2n) (0) = 0 for all n ∈ N. The Taylor series of f (x) around 0 is given by ∞ ∞ X f (n) (0) X 1 f (x) = xn = x 2n+1. n! (2n + 1)! n=0 n=0 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 60 / 83 13 (b) Contd.. 1 Let an = x 2n+1 for all n ∈ N. (2n + 1)! an+1 (2n + 1)!x 2n+3 lim = lim n→∞ an n→∞ (2n + 3)!x 2n+1 x2 = lim = 0. n→∞ (2n + 2)(2n + 3) an+1 Thus, for all x ∈ R, lim < 1. n→∞ an By ratio test, the Taylor series converges for all x ∈ R. Hence, the radius of convergence of the Taylor series is R = ∞. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 61 / 83 Question 13 (c) Write the Taylor’s series around 0 and find the radius of convergence. f (x) = e x sinh x. Solution: e 2x − 1 Note that f (x) = e x sinh x =. 2 f (1) (x) = e 2x , f (2) (x) = 2e 2x , f (3) (x) = 22 e 2x and f (4) (x) = 23 e 2x. Proceeding in this way, we get f (n) (x) = 2n−1 e 2x for all n ∈ N. Therefore, f (n) (0) = 2n−1 for all n ∈ N. The Taylor series of f (x) around 0 is given by ∞ ∞ X f (n) (0) X 2n−1 f (x) = xn = x n. n! n! n=0 n=1 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 62 / 83 13 (c) Contd.. 2n−1 Here, an = for all n ∈ N. n! Therefore, 1 an+1 2 = lim = lim = 0. R n→∞ an n→∞ n + 1 Hence, the radius of convergence of the Taylor series of the given function is R = ∞. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 63 / 83 Question 13 (d) Write the Taylor’s series around 0 and find the radius of convergence. f (x) = x sin x. Solution: f (1) (x) = x cos x + sin x. f (2) (x) = −x sin x + 2 cos x. f (3) (x) = −x cos x − 3 sin x. f (4) (x) = x sin x − 4 cos x. Proceeding in this way, we get   (n)  nπ  (n − 1)π f (x) = x sin + x + n sin +x. 2 2 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 64 / 83 13 (d) Contd..   (n) (n − 1)π Therefore, f (0) = n sin for all n ∈ N. 2 ( (n) 0 if n = 2m − 1, Note that f (0) = (−1)m−1 2m if n = 2m. The Taylor series of f (x) around 0 is given by ∞ ∞ ∞ X f (n) (0) X (−1)m−1 2m 2m X (−1)m−1 2m f (x) = xn = x = x. n! (2m)! (2m − 1)! n=0 m=1 m=1 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 65 / 83 13 (d) Contd.. (−1)m−1 2m Let am = x for all m ∈ N. (2m − 1)! Thus, for all x, am+1 (2m − 1)!x 2m+2 lim = lim m→∞ am m→∞ (2m + 1)!x 2m x2 = lim = 0 < 1. m→∞ 4m2 + 2m By ratio test, the Taylor series converges for all x ∈ R. Hence, the radius of convergence of the Taylor series is R = ∞. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 66 / 83 Question 14 Question Obtain the Taylor’s series around 0 for the following series using term by term differentiation/integration and calculate the radius of convergence. Is this the maximal interval of validity of the series? (a) tan−1 (x). (b) sin−1 (x). (c) sinh−1 (x). 1 (d). (1 + x 2 )2 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 67 / 83 Question 14 Theorem (Term by term differentiation and integration) ∞ X Suppose that f (x) = an x n converges for |x| < R. Then, n=0 ∞ X (a) nan x n−1 converges in |x| < R and is equal to f 0 (x). n=0 ∞ Z X 1 (b) an x n+1 converges in |x| < R and is equal to f (x)dx. n+1 n=0 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 68 / 83 Question 14 (a) tan−1 (x). Solution: Note that d 1 (tan−1 x) = dx 1 + x2 and ∞ 1 X = (−1)n x 2n for x ∈ (−1, 1). 1 + x2 n=0 By termwise integration, we get ∞ ! Z x Z x X 1 tan−1 (x) = 2 dt = (−1)n t 2n dt 0 1 + t 0 n=0 ∞ x ∞ (−1)n 2n+1 X Z X = (−1)n t 2n dt = x. 0 = 2n + 1 n=0 n 0 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 69 / 83 14 (a) Contd.. (−1)n 2n+1 Fix x ∈ R, and let bn = x. 2n + 1 (−1)n+1 x 2n+3 bn+1 2n+3 2n + 1 2 Then, lim = lim (−1)n x 2n+1 = lim x = x 2. n→∞ bn n→∞ n→∞ 2n + 3 2n+1 ∞ X (−1)n 2n+1 So, if we want the power series x to converge, we 2n + 1 n=0 need x 2 < 1, by ratio test. ∞ X (−1)n 2n+1 Thus, the radius of convergence of the series x is 1. = 2n + 1 n 0 ∞ X (−1)n x 2n+1 Hence, converges for |x| < 1. = 2n + 1 n 0 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 70 / 83 14 (a) Contd.. Leibnitz test: Suppose {an }∞ n=1 is a sequence of positive numbers such that an ≥ an+1 for all n ∈ N lim an = 0. n→∞ ∞ X Then, the alternating series (−1)n+1 an converges. n=1 ∞ ∞ X (−1)n x 2n+1 X (−1)n At x = 1, = converges by Leibnitz test. = 2n + 1 = 2n + 1 n 0 n 0 ∞ ∞ X (−1)n x 2n+1 X (−1)n+1 At x = −1, = converges by Leibnitz 2n + 1 2n + 1 n=0 n=0 test. Hence, the maximum interval of convergence is [−1, 1]. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 71 / 83 Question 14 (b) sin−1 (x). Solution: Note that d 1 −1 (sin−1 x) = √ = (1 − x 2 ) 2. dx 1 − x2 Hence, for −1 < x < 1, 1 1 1·3 4 1·3·5 6 √ = 1 + x2 + x + x + ··· 1−x 2 2 2·4 2·4·6 ∞   X 2n 1 2n = x. n 22n n=0 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 72 / 83 14 (b) Contd.. By termwise integration, we get ∞   Z x Z x X ! −1 1 2n 1 2n sin (x) = √ dt = t dt 0 1 − t2 0 n 22n n=0 ∞ Z x  ∞   X 2n 1 2n X 2n 1 x 2n+1 = t dt =. 0 n 22n n 22n 2n + 1 n=0 n=0   2n 1 Fix x ∈ R, and let bn = 2n x 2n+1. n 2 (2n + 1) Then, 2n+2 1 2n+3
bn+1 n+1 22n+2 (2n+3) x lim = lim 2n
1 n→∞ bn n→∞ n 22n (2n+1) x 2n+1 (2n + 1)2 = lim x 2 = x 2. n→∞ (2n + 2)(2n + 3) (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 73 / 83 14 (b) Contd.. ∞ 1 2n x 2n+1 X   So, if we want the power series to converge, we 22n n 2n + 1 n=0 need x 2 < 1, by ratio test. ∞ 1 2n x 2n+1 X   Thus, the radius of convergence of the series is 22n n 2n + 1 n=0 1. ∞ 1 2n x 2n+1 X   Hence, converges for |x| < 1. 22n n 2n + 1 n=0 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 74 / 83 14 (b) Contd.. ∞   1X 2n We claim that 2n is convergent. 2 (2n + 1) n n=0 4n   2n Use induction to prove that ≤√ for all n ∈ N. n 3n + 1 Therefore, for all n ∈ N, we get 42n   1 2n 1 ≤ √ 22n (2n + 1) n 22n (2n + 1) 3n + 1 1 = √ (2n + 1)( 3n + 1) 1 ≤ 3/2. n (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 75 / 83 14 (b) Contd.. ∞ X 1 Recall that converges if and only if p > 1. np n=0 ∞   X 1 2n By comparision test, the series 2n is convergent. 2 (2n + 1) n n=0 ∞ 1 2n x 2n+1 X   So, at x = 1 and x = −1, converges. 22n n 2n + 1 n=0 Hence, the maximum interval of convergence is [−1, 1]. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 76 / 83 Question 14 (c) sinh−1 (x). Solution: Note that d 1 −1/2 (sinh−1 x) = √ = (1 + x 2 ). dx 1+x 2 Hence, for −1 < x < 1, 1 1 1·3 4 1·3·5 6 √ = 1 − x2 + x − x + ···+ 1+x 2 2 2·4 2·4·6 ∞ (−1)n 2n 2n X   = x. = 22n n n 0 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 77 / 83 14 (c) Contd.. By termwise integration, we get Z x Z x X ∞   ! n 2n 1 (−1) sinh−1 (x) = √ dt = 2n t 2n dt 0 1 + t2 0 2 n n=0 ∞ Z x n ∞ (−1)n 2n x 2n+1     X (−1) 2n 2n X = t dt =. = 0 22n n = 22n n 2n + 1 n 0 n 0 (−1)n   2n 2n+1 Fix x ∈ R, and let bn = 2n x. 2 (2n + 1) n Then, (−1)n+1 2n+2 2n+3
bn+1 22n+2 (2n+3) n+1 x lim = lim (−1) n 2n 2n+1 bn n→∞ n→∞
2n x 2 (2n+1) n (2n + 1)2 = lim x 2 = x 2. n→∞ (2n + 2)(2n + 3) (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 78 / 83 14 (c) Contd.. ∞ (−1)n 2n x 2n+1 X   So, if we want the power series to converge, = 22n n 2n + 1 n 0 we need x 2 < 1, by ratio test. ∞ (−1)n 2n x 2n+1 X   Thus, the radius of convergence of the series = 22n n 2n + 1 n 0 is 1. ∞ (−1)n 2n x 2n+1 X   Hence, converges for |x| < 1. = 22n n 2n + 1 n 0 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 79 / 83 14 (c) Contd.. ∞   X 1 2n We have proved in part (b) that the series is 22n (2n + 1) n n=0 convergent. ∞ (−1)n 2n x 2n+1 X   So, at x = 1 and x = −1, converges. = 22n n 2n + 1 n 0 Hence, the maximum interval of convergence is [−1, 1]. (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 80 / 83 Question 14 1 (d). (1+x 2 )2 Solution: Note that   d 1 −2x = dx 1 + x2 (1 + x 2 )2 and ∞ 1 X = (−1)n x 2n for x ∈ (−1, 1). 1 + x2 n=0 By termwise differentiation, we get ∞ −2x X = (−1)n (2n)x 2n−1. (1 + x 2 )2 n=0 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 81 / 83 14 (d) Contd.. For x 6= 0, we get ∞ 1 X = (−1)n−1 nx 2n−2. (1 + x 2 )2 n=1 Fix x ∈ R, let bn = (−1)n−1 nx 2n−2 for all n ∈ N. bn+1 n+1 2 lim = lim x = x 2. n→∞ bn n→∞ n ∞ X By ratio test, bn converges if x 2 < 1, and diverges if x 2 > 1. n=0 (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 82 / 83 14 (d) Contd.. Thus, the radius of convergence of the Taylor series of given function is 1. At x = ±1, the series is divergent, as {(−1)n n}∞ n=1 does not converge to zero. Hence, the maximum interval of convergence is (−1, 1). (Maths Dept., IIT Delhi) MTL100 - Calculus, Tutorial-3 Semester-I, 2020-21 83 / 83