Classnotes - MA Series and Matrices

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1 Classnotes - MA-2 Series and Matrices Department of Mathematics Indian Institute of Technology Madras This classnote is only meant for academic use. It is not to be used for commercial purposes. For suggestions and improvements, contact Arindama Singh.

2 Contents I Series 4 Series of Numbers 5. Preliminaries Sequences Series Some results on convergence Comparison tests Improper integrals Convergence tests for improper integrals Tests of convergence for series Alternating series Series Representation of Functions 3 2. Power series Determining radius of convergence Taylor s formulas Taylor series Fourier series II Matrices 5 3 Matrix Operations 5 3. Examples of linear equations Basic matrix operations Transpose and adjoint Elementary row operations Row reduced echelon form Determinant Computing inverse of a matrix Rank and Linear Equations Matrices as linear maps Linear independence

3 4.3 Gram-Schmidt orthogonalization Determining linear independence Rank Solvability of linear equations Gauss-Jordan elimination Matrix Eigenvalue Problem Eigenvalues and eigenvectors Characteristic polynomial Special types of matrices Diagonalization Bibliography 97 Index 98 3

4 Part I Series 4

5 Chapter Series of Numbers. Preliminaries We use the following notation: = the empty set. N = {, 2, 3,...}, the set of natural numbers. Z = {..., 2,,,, 2,...}, the set of integers. Q = { p q : p Z, q N}, the set of rational numbers. R = the set of real numbers. R + = the set of all positive real numbers. N Z Q R. The numbers in R Q is the set of irrational numbers. Examples are 2, 3. etc. Along with the usual laws of +,, <, R satisfies the Archimedian property: If a > and b >, then there exists an n N such that na b. Also R satisfies the completeness property: Every nonempty subset of R having an upper bound has a least upper bound (lub) in R. Explanation: Let A be a nonempty subset of R. A real number u is called an upper bound of A if each element of A is less than or equal to u. An upper bound l of A is called a least upper bound if all upper bounds of A are greater than or equal to l. Notice that Q does not satisfy the completeness property. For example, the nonempty set A = {x Q : x 2 < 2} has an upper bound, say, 2. But its least upper bound is 2, which is not in Q. Similar to lub, we have the notion of glb, the greatest lower bound of a subset of R. Let A be a nonempty subset of R. A real number v is called a lower bound of A if each element of A is greater than or equal to v. A lower bound m of A is called a greatest lower bound if all lower bounds of A are less than or equal to m. The completeness property of R implies that Every nonempty subset of R having a lower bound has a greatest lower bound (glb) in R. 5

6 The lub acts as a maximum of a nonempty set and the glb acts as a minimum of the set. In fact, when the lub(a) A, this lub is defined as the maximum of A and is denoted as max(a). Similarly, if the glb(a) A, this glb is defined as the minimum of A and is denoted by min(a). Moreover, both Q and R Q are dense in R. That is, if x < y are real numbers then there exist a rational number a and an irrational number b such that x < a < y and x < b < y. We may not explicitly use these properties of R but some theorems, whose proofs we will omit, can be proved using these properties. These properties allow R to be visualized as a number line: From the Archemedian property it follows that the greatest integer function is well defined. That is, for each x R, there corresponds, the number [x], which is the greatest integer less than or equal to x. Moreover, the correspondence x [x] is a function. We visualize R as a straight line made of expansible rubber of no thickness! Let a, b R, a < b. [a, b] = {x R : a x b}, the closed interval [a, b]. (a, b] = {x R : a < x b}, the semi-open interval (a, b]. [a, b) = {x R : a x < b}, the semi-open interval [a, b). (a, b) = {x R : a < x < b}, the open interval (a, b). (, b] = {x R : x b}, the closed infinite interval (, b]. (, b) = {x R : x < b}, the open infinite interval (, b). [a, ) = {x R : x a}, the closed infinite interval [a, ). (a, ) = {x R : x b}, the open infinite interval (a, ). (, ) = R, both open and closed infinite interval. We also write R + for (, ) and R for (, ). These are, respectively, the set of all positive real numbers, and the set of all negative real numbers. A neighborhood of a point c is an open interval (c δ, c + δ) for some δ >. { x if x The absolute value of x R is defined as x = x if x < Thus x = x 2. And a = a or a ; x y is the distance between real numbers x and y. Moreover, if a, b R, then a = a, ab = a b, a = a if b, a + b a + b, a b a b. b b Let x R and let a >. The following are true: 6

7 . x = a iff x = ±a. 2. x < a iff a < x < a iff x ( a, a). 3. x a iff a x a iff x [ a, a]. 4. x > a iff a < x or x > a iff x (, a) (a, ) iff x R \ [ a, a]. 5. x a iff a x or x a iff x (, a] [a, ) iff x R \ ( a, a). Therefore, for a R, δ >, x a < δ iff a δ < x < a + δ. The following statements are useful in proving equalities from inequalities: Let a, b R.. If for each ɛ >, a < ɛ, then a =. 2. If for each ɛ >, a < b + ɛ, then a b. Exercises for.. Solve the following inequalities and show their solution sets in R. (a) 2x < x + 3 (b) x 3 < 2x + (c) 6 x Solve the following and show the solution sets in R. (a) 2x 3 = 7 (b) 5 2 < (c) 2x 3 (d) 2x 3. x 3. Show the solution set of x 2 x 2 in R. 4. Graph the inequality: x + y..2 Sequences Since the fox runs times faster than the rabbit, the fox starts km behind. By the time he reaches the point where from the rabbit has started, the rabbit has moved ahead m. By the time the fox reaches the second point, the rabbit has moved ahead m. This way to surpass the rabbit, the fox must touch upon infinite number of points. Hence the fox can never surpass the rabbit! The question is whether / + / + is a number. We rather take the partial sums, +, + +, /,... which are numbers and ask whether the sequence of these numbers approximates certain real number? 7

8 For example, we may approximate 2 by the usual division procedure. We get the sequence,.4,.4,.44,.442,.442,.4423,.44235, ,... Does it approximate 2? In general, we define a sequence of real numbers as a function f : N R. The values of the function are f(), f(2), f(3),... These are called the terms of the sequence. With f(n) = x n, the nth term of the sequence, we write the sequence in many ways such as (x n ) = (x n ) n= = {x n } n= = {x n } = (x, x 2, x 3,...) showing explicitly its terms. For example, x n = n defines the sequence f : N R with f(n) = n, that is, the sequence is (, 2, 3, 4,...), the sequence of natural numbers. Informally, we say the sequence x n = n. The sequence x n = /n is the sequence (,,,,...); formally, {/n} or (/n) The sequence x n = /n 2 is the sequence (/n 2 ), or {/n 2 }, or (, 4, 9, 6,...). The constant sequence {c} for a given real number c is the constant function f : N R, where f(n) = c for each n N. It is (c, c, c,...). A sequence is an infinite list of real numbers; it is ordered like natural numbers, and unlike a set of numbers. There are sequences which approximate a real number and there are sequences which do not approximate any real number. For example, {/n} approximates the real number, whereas {n} approximates no real number. Also the sequence (,,,,,,...), which may be written as {( ) n }, approximates no real number. We would say that the sequence {/n} converges to and the other two sequences diverge. The sequence {n} diverges to and the sequence {( ) n } diverges. Look at the sequence {/n} closely. We feel that eventually, it will approximate, meaning that whatever tolerance I fix there is a term in the sequence, after which every term is away from within that tolerance. What does it mean? Suppose I am satisfied with an approximation to within the tolerance 5. Then, I see that the terms of the sequence, starting with and then /2, /3,..., all of them are within 5 units away from. In fact, /n < 5 for all n. Now, you see, bigger the tolerance, it is easier to fix a tail of the sequence satisfying the tolerance condition. Suppose I fix my tolerance as /5. Then I see that the sixth term onwards, all the terms of the sequence are within /5 distance away from. That is, /n < /5 for all n 6. If I fix my tolerance as. Then we see that /n < for all n +. This leads to the formal definition of convergence of a sequence. Let {x n } be a sequence. Let a R. We say that {x n } converges to a iff for each ɛ >, there exists an m N such that if n m is any natural number, then x n a < ɛ. 8

9 Example.. Show that the sequence {/n} converges to. Let ɛ >. Take m = +. That is, m is the natural number such that m < m. Then ɛ ɛ m < ɛ. Moreover, if n > m, then n < < ɛ. That is, for any such given ɛ >, there exists an m m, (we have defined it here) such that for every n m, we see that /n < ɛ. Therefore, {/n} converges to. Notice that in Example., we could have resorted to the Archimedian principle and chosen any natural number m > /ɛ. Now that {/n} converges to, the sequence whose first terms are like (n) and st term onward, it is like (/n) also converges to. Because, for any given ɛ >, we choose our m as /ɛ +. Moreover, the sequence whose first terms are like {n} and then onwards it is, /2, /3,... converges to for the same reason. That is, convergence behavior of a sequence does not change if first finite number of terms are changed. For a constant sequence x n = c, suppose ɛ > is given. We see that for each n N, x n c = < ɛ. Therefore, the constant sequence {c} converges to c. Sometimes, it is easier to use the condition x n a < ɛ as a ɛ < x n < a + ɛ. Such an open interval (a ɛ, a + ɛ) for some ɛ > is called a neighborhood of a. A sequence thus converges to a implies the following:. Each neighborhood of a contains a tail of the sequence. 2. Every tail of the sequence contains numbers arbitrarily close to a. We say that a sequence {x n } converges iff it converges to some a. A sequence diverges iff it does not converge to any real number. There are two special cases of divergence. Let {x n } be a sequence. We say that {x n } diverges to iff for every r >, there exists an m N such that if n > m is any natural number, then x n > r. We call an open interval (r, ) a neighborhood of. A sequence thus diverges to implies the following:. Each neighborhood of contains a tail of the sequence. 2. Every tail of the sequence contains arbitrarily large positive numbers. In this case, we write lim n x n = ; we also write it as x n as n or as x n. We say that {x n } diverges to iff for every r >, there exists an m N such that if n > m is any natural number, then x n < r. Calling an open interval (, s) a neighborhood of, we see that a sequence diverges to implies the following: 9

10 . Each neighborhood of contains a tail of the sequence. 2. Every tail of the sequence contains arbitrarily small negative numbers. In this case, we write lim n x n = ; we also write it as x n as n or as x n. We use a unified notation for convergence to a real number and divergence to ±. For l R {, }, the notations lim x n = l, lim x n = l, x n l as n, x n l n all stand for the phrase limit of {x n } is l. When l R, the limit of {x n } is l means that {x n } converges to l; and when l = ±, the limit of {x n } is l means that {x n } diverges to ±. Example.2. Show that (a) lim n = ; (b) lim ln(/n) =. (a) Let r >. Choose an m > r 2. Let n > m. Then n > m > r. Therefore, lim n =. (b) Let r >. Choose a natural number m > e r. Let n > m. Then /n < /m < e r. Consequently, ln(/n) < ln e r = r. Therefore, ln(/n). We state a result connecting the limit notion of a function and limit of a sequence. We use the idea of a constant sequence. A sequence {a n } is called a constant sequence if a n = α for each n, where α is a fixed real number. We state some results about sequences which will be helpful to us later. Theorem.. (Sandwich Theorem): Let {x}, {y n }, and {z n } be sequences such that x n y n z n holds for all n greater than some m. If x n l and z n l, then y n l. Theorem.2. Limits of sequences to Limits of functions: Let a < c < b. Let f : D R be a function where D contains (a, c) (c, b). Let l R. Then lim f(x) = l iff for each non-constant x c sequence {x n } converging to c, the sequence of functional values {f(x n )} converges to l. The same way, limit of a sequence {a n } as n is related to the limit of a function f(x) as x provided some conditions are satisfied. Theorem.3. Let k N. Let f(x) be a function defined for all x k. Let {a n } be a sequence of real numbers such that a n = f(n) for all n k. If lim f(x) = l, then lim a n = l. x n As an application, consider the function ln x. We know that it is defined on [, ). Using L Hospital s rule, we have ln x lim x x = lim x x =. ln n Therefore, lim n n = lim ln x x x =. Our main goal is to study when an infinite sum can represent a number.

11 Exercises for.2. Show the following: ln n (a) lim n n (d) lim n x n = for x < = (b) lim n n/n = x n (e) lim n n! (c) lim x /n = for x > n ( x) n = (f) lim + = e x. n n 2. Find the limit of the sequence {a n } or show that it diverges. (a) a n = 2n + 3 (b) a n = n + n 4 (c) sin ( π n 2n n 3/2 2 + ) n.3 Series A series is an infinite sum of numbers. As it is, two numbers can be added; so by induction, a finite of them can also be added. For an infinite sum to be meaningful, we look at the sequence of partial sums. Let {x n } be a sequence. The series x + x x n + is meaningful when another sequence, namely, n x, x + x 2, x + x 2 + x 3,..., x k,... is convergent. The infinite sum itself is denoted by k= x n and also by x n. We say that the series x n is convergent iff the sequence {s n } is convergent, where the nth partial sum s n is given by s n = n k= x n. Thus we may define convergence of a series as follows: We say that the series x n converges to l R iff for each ɛ >, there exists an m N such that for each n m, n k= x k l < ɛ. In this case, we write x n = l. Further, we say that the series x n converges iff the series converges to some l R. The series is said to be divergent iff it is not convergent. Similar to convergence, if the sequence of partial sums {s n } diverges to ±, we say that the series xn diverges to ±. That is, the series x n diverges to iff for each r >, there exists m N such that for each n m, n k= x k > r. We write it as x n =. Similarly, the series x n diverges to iff for each r >, there exists m N such that for each n m, n k= x k < r. We write it as x n =. Notice that converges to a real number and diverges to ± both are written the same way. In the unified notation, we say that a series a n sums to l R {, }, when either the series converges to the real number l or it diverges to ±. In all these cases we write a n = l. Moreover, if a series converges to some real number l, then the partial sums can be thought of as approximations to the number l. k=

12 There can be series which diverge but neither to nor to. For example, the series ( ) n = n= neither diverges to nor to. But it is a divergent series. Can you see why? Example.3. (a) The series n= 2 n converges to. Because, if {s n} is the sequence of partial sums, then s n = n k= 2 = k 2 (/2)n /2 = 2 n. (b) The series diverges to. To see this, let s n = n k= be the partial sum k up to n terms. Let m be the natural number such that 2 m n < 2 m+. Then s n = n k= k m ( = ( + 3) ) ( > ( + 4) ) = = + m. 2 ( 2 m k=2 m k ( 2 m ) k=2 m 2 m ) As n, we see that m. Consequently, s n. That is, the series diverges to. This is called the harmonic series. (c) The series n diverges to. (d) The series + + diverges. It neither diverges to nor to. Because, the sequence of partial sums here is,,,,,,,.... Example.4. Let a. Consider the geometric series ar n = a + ar + ar 2 + ar 3 +. n= The nth partial sum of the geometric series is s n = a + ar + ar 2 + ar 3 + ar n = a( rn ). r (a) If r <, then r n. The geometric series converges to lim s n = n Therefore, ar n = ar n = a r. n= n= 2 a r.

13 (b) If r, then r n diverges. The geometric series ar n diverges. Exercises for.3. You drop a ball from a meters above a flat surface. Each time the ball hits the surface after falling a distance h, it rebounds a distance rh, where r is positive but less than. Find the total distance the ball travels up and down. 2. Express in the form m/n for m, n N..4 Some results on convergence If a series sums to l, then it cannot sum to s where s l. Theorem.4. If a series a n sums to l R {, }, then l is unique. Proof. Suppose the series a n sums to l and also to s, where both l, s R {, }. Suppose that l s. We consider the following exhaustive cases. Case : l, s R. Then s l >. Choose ɛ = s l /3. We have natural numbers k and m such that for every n k and n m, n a j l ɛ j= and n a j s ɛ. Fix one such n, say M > max{k, m}. Both the above inequalities hold for n = M. Then s l = s This is a contradiction. M a j + j= M a j l j= j= M a j s + j= M a j l 2ɛ < s l. Case 2: l R and s =. Then there exists a natural number k such that for every n k, we have n a j l <. Since the series sums to, we have m N such that for every n m, j= n a j > l +. j= Now, fix an M > max{k, m}. Then both of the above hold for this n = M. Therefore, j= M a j < l + j= and M a j > l +. j= This is a contradiction. 3

14 Case 3: l R and s =. It is similar to Case 2. Choose less than l. Case 4: l =, s =. Again choose an M so that n j= a n is both greater than and also less than leading to a contradiction. The results in the following theorem are sometimes helpful in ascertaining the convergence of a series without knowing what the sum of the series is. Theorem.5. () (Cauchy Criterion) A series a n converges iff for each ɛ >, there exists a k N such that n j=m a j < ɛ for all n m k. (2) (Weirstrass Criterion) Let a n be a series of non-negative terms. Suppose there exists c R such that each partial sum of the series is less than c, i.e., for each n, n j= a j < c. Then a n is convergent. The following result sometimes helps in ascertaining that a given series diverges. Theorem.6. If a series a n converges, then the sequence {a n } converges to. Proof: Let s n denote the partial sum n k= a k. Then a n = s n s n. If the series converges, say, to l, then lim s n = l = lim s n. It follows that lim a n =. It says that if lim a n does not exist, or if lim a n exists but is not equal to, then the series a n diverges. n n The series diverges because lim 3n + n 3n + = 3. n= The series ( ) n diverges because lim( ) n does not exist. Notice what Theorem.6 does not say. The harmonic series diverges even though lim n =. Recall that for a real number l our notation says that l+ =, l = and l (± ) = ±. Similarly, we accepted the convention that + = and =. We use these conventions in the statement of our next result. Theorem.7. () If a n converges to a and b n sums to b, then the series (a n + b n ) sums to a + b; (a n b n ) sums to a b; and kb n sums to kb; where k is any real number. (2) If a n converges and b n diverges, then (a n + b n ) and (a n b n ) diverge. (3) If a n diverges and k, then ka n diverges. Proofs of the statements in Theorem.7 are left as exercises. However, write the first statement in the above theorem as separate statements taking b as a real number, as, and as. Notice that sum of two divergent series can converge. For example, both (/n) and ( /n) diverge but their sum converges. Since deleting a finite number of terms of a sequence does not alter its convergence, omitting a finite number of terms or adding a finite number of terms to a convergent (divergent) series implies 4

15 the convergence (divergence) of the new series. Of course, the sum of the convergent series will be affected. For example, ( ) ( ) = 2 n 2 n 2 4. However, n=3 n=3 n= ( 2 n 2 ) = n= ( 2 n ). This is called re-indexing the series. As long as we preserve the order of the terms of the series, we can re-index without affecting its convergence and sum. Exercises for.4. Find the n-th partial sum of the series and then sum the series Find the limits of the following series: ( ) (a) 2 + ( )n 4n (b) n 5 n (2n ) 2 (2n + ) 2 n= n= 3. Find the values of x for which the series converges. And then find the limit for those values of x. ( ) n ( ) n ( ) n (a) (b) sin x ((x 3)/2) n n=.5 Comparison tests n= There are various ways to determine whether a series converges or not, occasionally, some information on its sum is also obtained. Theorem.8. (Comparison Test) Let a n and b n be series of non-negative terms. Suppose there exists k > such that a n kb n for each n greater than some natural number m.. If b n converges, then a n converges. 2. If a n diverges to, then b n diverges to. Proof: () Consider all partial sums of the series having more than m terms. We see that a + + a m + a m+ + + a n a + + a m + k n j=m+ Since b n converges, so does n j=m+ b j. By Weirstrass criterion, a n converges. (2) Similar to (). Caution: The comparison test holds for series of non-negative terms. b j. 5

16 Theorem.9. (Ratio Comparison Test) Let a n and b n be series of non-negative terms. Suppose there exists m N such that for each n > m, a n >, b n >, and a n+ a n b n+ b n.. If b n converges, then a n converges. 2. If a n diverges to, then b n diverges to. Proof: For n > m, a n = a n a n a n a n 2 am+ a m a m b n b n b n b n 2 bm+ b m b m = a m b m b n. By the Comparison test, if b n converges, then a n converges. This proves (). And, (2) follows from () by contradiction. Theorem.. (Limit Comparison Test) Let a n and b n be series of non-negative terms. a n Suppose that there exists m N such that for each n > m, a n >, b n >, and lim = k. n b n. If k > then b n and a n converge or diverge to, together. 2. If k = and b n converges, then a n converges. 3. If k = and b n diverges to then a n diverges to. Proof: () Let ɛ = k/2 >. The limit condition implies that there exists m N such that k 2 < a n b n < 3k 2 By the Comparison test, the conclusion is obtained. for each n > m. (2) Let ɛ =. The limit condition implies that there exists m N such that < a n b n < for each n > m. Using the right hand inequality and the Comparison test we conclude that convergence of b n implies the convergence of a n. (3) If k >, lim(b n /a n ) = /k. Use (). If k =, lim(b n /a n ) =. Use (2). Example.5. For each n N, n! 2 n. That is, Since n= n! 2 n. 2 is convergent, is convergent. Therefore, adding to it, the series n n! n= + + 2! + 3! + + n! + is convergent. In fact, this series converges to e. To see this, consider s n = + + 2! + + ( n!, t n = + n) n. 6

17 By the Binomial theorem, t n = + + ( ) + + 2! n n! Thus taking limit as n, we have [ ( n )( 2 n e = lim n t n lim n s n. ) ( n ) ] s n. n Also, for n > m, where m is any fixed natural number, t n + + ( ) + + [ ( /n)( 2/n) ( m ) ] 2! n m! n Taking limit as n we have e = lim t n s m. n Since m is arbitrary, taking the limit as m, we have Therefore, lim m s m = e. That is, the series Example.6. Determine when the series Let a n = Since e lim m s m. n= n= n! = e. n + 7 n(n + 3) n + 5 and b n =. Then n3/2 a n n(n + 7) = b n (n + 3) n + 5 n + 7 n(n + 3) n + 5 converges. as n. is convergent, Limit comparison test says that the given series is convergent. n3/2 Exercises for.5. Determine whether the following series converge or diverge: (a) 2 + cos n n + 3 (b) (c) n n 2 (d) (g) (h) (i) n= n=3 ln ln n (e) n= n= ( 3 n) n (f) a n, where a = /2, a n+ = n + ln n n + a n. n= a n, where a = 3, a n+ = n n + a n. n= n= sin n n= a n, where a n = n/2 n if n is prime; else, a n = /2 n. n= ( ) n (5/4) n

18 .6 Improper integrals There is a nice connection between integrals and series. To see this connection, we consider the so called improper integrals. In the definite integral b f(x)dx we required that both a, b are finite and also the range of f(x) a is a subset of some finite interval. However, there are functions which violate one or both of these requirements, and yet, the area under the curves and above the x-axis remain bounded. Such integrals are called Improper Integrals. Suppose f(x) is continuous on [, ). It makes sense to write f(x)dx = lim b b f(x)dx provided that the limit exists. In such a case, we say that the improper integral f(x)dx converges and its value is given by the limit. We say that the improper integral diverges iff it is not convergent. Obviously, we are interested in computing the value of an improper integral, in which case, the integral is required to converge. Integrals of the type b f(x) dx can become improper when f(x) a is not continuous at a point in the interval [a, b]. Here are the possible types of improper integrals.. If f(x) is continuous on [a, ), then 2. If f(x) is continuous on (, b], then 3. If f(x) is continuous on (, ), then f(x) dx = c f(x) dx + c a b f(x) dx = lim f(x) dx = b b a b lim a a f(x) dx, for any c R. 4. If f(x) is continuous on (a, b] and discontinuous at x = a, then b a f(x) dx = lim t a+ b t f(x) dx. 5. If f(x) is continuous on [a, b) and discontinuous at x = b, then b a f(x) dx = lim t b t a f(x) dx. f(x) dx. f(x) dx. 6. If f(x) is continuous on [a, c) (c, b] and discontinuous at x = c, then b a f(x) dx = c a f(x) dx + b c f(x) dx. 8

19 In each case, if the limit of the concerned integral is finite, then we say that the improper integral (on the left) converges, else, the improper integral diverges; the finite value as obtained from the limit is the value of the improper integral. A convergent improper integral converges to its value. Two important sub-cases of divergent improper integrals are when the limit of the concerned integral is or. In these cases, we say that the improper integral diverges to or to as is the case. Example.7. For what values of p R, the improper integral value, when it converges? Case : p =. b dx b x = dx = ln b ln = ln b. p x Since lim ln b =, the improper integral diverges to. b Case 2: p <. b dx x = x p+ p p + b = p (b p ). Since lim b p =, the improper integral diverges to. b Case 3: p >. b dx x = p p (b p ) = ( ) p b. p Since lim b =, we have bp Hence, the improper integral dx b x = lim dx p b x p = lim b Example.8. For what values of p R, the improper integral dx x p ( ) p b = p p. converges? What is its dx x converges to for p > and diverges to for p. p p Case : p =. dx x = lim dx p a + a x = lim [ln ln a] =. a + Therefore, the improper integral diverges to. Case 2: p <. dx x = lim dx p a + a x = lim a p p a + p Therefore, the improper integral converges to /( p). dx x p converges? = p. 9

20 Case 3: p >. dx x = lim a p p a + p Hence the improper integral diverges to. The improper integral Exercises for.6 ( ) = lim a + p a =. p dx x converges to for p < and diverges to for p. p p. Evaluate the following improper integrals: 2 (a) x 2 x dx (b) x + x2 + 2x dx (d) 2 2e x sin x dx (e) 4 dx x (f) (c) dx ( + x 2 )( + tan x) dx x 2.7 Convergence tests for improper integrals Sometimes it is helpful to be sure that an improper integral converges, even if we are unable to evaluate it. Theorem.. (Comparison Test) Let f(x) and g(x) be continuous functions on [a, ). Suppose that f(x) g(x) for all x a.. If a 2. If a g(x) dx converges, then f(x) dx converges. a f(x) dx diverges to, then g(x) dx diverges to. a Proof: Since f(x) g(x) for all x a, As lim b b a b a f(x)dx g(x)dx = l for some l R, lim b b a b to l. This proves (). Proof of (2) is similar to that of (). We also use a similar result stated below, without proof. a g(x)dx. f(x)dx exists and the limit is less than or equal Theorem.2. (Limit Comparison Test) Let f(x) and g(x) be positive continuous functions f(x) on [a, ). If lim x g(x) = L, where < L <, then f(x)dx and g(x)dx either both a a converge, or both diverge. Theorems. and.2 talk about non-negative functions. The reason is the following result, which we will not prove: Theorem.3. Let f(x) be a continuous function on [a, b), for b R or b =. If the improper integral b f(x) dx converges, then the improper integral b f(x) dx also converges. a a 2

21 Example.9. (a) (b) 2 sin 2 x x 2 dx x2 dx converges because sin2 x for all x and x 2 x2 diverges to because x2 x for all x 2 and 2 dx x diverges to. dx x 2 converges. dx (c) converges or diverges? + x2 [ / ] x 2 Since lim = lim =, the limit comparison test says that the given improper integral and both converge or diverge together. The latter converges, so does x + x 2 x 2 x + x2 dx the x 2 former. However, they may converge to different values. (d) Does the improper integral dx + x 2 = lim b [tan b tan ] = π 2 π 4 = π 4. lim x ( dx x = lim 2 b b ) =. dx e x + converge? / e x + e = lim e x x x e x + =. Also, e 2 implies that for all x, e x x 2. So, e x x 2. Since dx e x dx x 2 converges, also converges. By limit comparison test, the given improper integral converges. Example.. Show that Γ(x) = e t t x dt converges for each x >. Fix x >. Since lim t e t t x+ =, there exists t such that < e t t x+ < for t > t. That is, < e t t x < t 2 for t > t. Since t 2 dt is convergent, t 2 dt is also convergent. By the comparison test, t t e t t x dt is convergent. The integral t e t t x dt exists and is not an improper integral. Next, we consider the improper integral e t t x dt. Let < a <. 2

22 For a t, we have < e t t x < t x. So, a e t t x dt < Taking the limit as a +, we see that the a t x dt = ax x < x. e t t x dt is convergent, and its value is less than or equal to /x. Therefore, is convergent. e t t x dt = e t t x dt + t e t t x dt + t e t t x dt The function Γ(x) is defined on (, ). For x >, using integration by parts, Γ(x + ) = [ ] t x e t dt = t x e t xt x ( e t ) dt = xγ(x). It thus follows that Γ(n + ) = n! for any non-negative integer n. We take! =. Example.. Test the convergence of e t2 dt. Since e t2 is continuous on [, ], e t2 dt exists. For t >, we have t < t 2. So, < e t2 test, e t2 dt is convergent. < e t. Since e t dt is convergent, by the comparison Now, a e t2 dt = a e t2 d( t) = a e t2 dt. Taking limit as a, we see that e t2 dt is convergent and its value is equal to e t2 dt. Combining the three integrals above, we conclude that e t2 dt converges. The Gamma function takes other forms by substitution of the variable of integration. Substituting t by rt we have Γ(x) = r x e rt t x dt for < r, < x. Substituting t by t 2, we have Γ(x) = 2 e t2 t 2x dt for < x. Using multiple integrals it can be shown that Γ(/2) = 2 e t2 dt = π. Example.2. Show that Γ(/2) = 2 e t2 dt = π. ( Γ = 2) e x x /2 dx = 2 e t2 dt (x = t 2 ) 22

23 To evaluate this integral, consider the double integral of e x2 y 2 over two circular sectors D and D 2, and the square S as indicated below. Since the integrand is positive, we have D < S < D 2. Now, evaluate these integrals by converting them to iterated integrals as follows: e r2 r dr π/2 dθ < R R e x2 dx e y2 dy < R 2 e r2 r dr π ( R ) 2 π ( e R2) < e x2 dx < ( e 2R2) 4 4 Take the limit as R to obtain ( ) 2 π e x2 dx = 4 From this, the result follows. π/2 Example.3. Prove: B(x, y) = tx ( t) y dt converges for x >, y >. We write the integral as a sum of two integrals: B(x, y) = /2 t x ( t) y dt + Setting u = t, the second integral looks like /2 t x ( t) y dt = /2 /2 t x ( t) y dt u y ( u) x dt Therefore, it is enough to show that the first integral converges. Notice that here, < t /2. Case : x. For < t < /2, t >. Therefore, for all y >, the function ( t) y is well defined, continuous, and bounded on (, /2]. So is the function t x. Therefore, the integral /2 t x ( t) y dt exists and is not an improper integral. Case 2: < x <. Here, the function t x is well defined and continuous on (, /2]. By Example.8, the integral /2 t x dt converges. Since t x ( t) y t x for < t /2, we conclude that 23 dθ

24 /2 t x ( t) y dt converges. By setting t as t, we see that B(x, y) = B(y, x). By substituting t with sin 2 t, the Beta function can be written as B(x, y) = 2 π/2 (sin t) 2x (cos t) 2y dt, for x >, y >. Changing the variable t to t/( + t), the Beta function can be written as B(x, y) = Again, using multiple integrals it can be shown that t x+ dt for x >, y >. ( + t) x+y B(x, y) = Γ(x)Γ(y) Γ(x + y) for x >, y >. Exercises for.7. Test for convergence, the following improper integrals: π/2 cos x dx (a) dx (b) (π 2x) /3 x sin x (d) (g) π/2 2 2 dx x dx x(ln x) p (e) (h) dx ex x.8 Tests of convergence for series 2 (f) (c) dx e x + e x ln x dx dx x(ln x) p Hint: p <, p =, p >. Put x = e t. Theorem.4. (Integral Test) Let a n be a series of positive terms. Let f : [, ) R be a continuous, positive and non-increasing function such that a n = f(n) for each n N.. If f(t)dt is convergent, then a n is convergent. 2. If f(t)dt diverges to, then a n diverges to. Proof: Since f is a positive and non-increasing, the integrals and the partial sums have a certain relation. 24

25 n+ f(t) dt a + a a n a + n f(t) dt. If n f(t) dt is finite, then the right hand inequality shows that a n is convergent. If n f(t) dt =, then the left hand inequality shows that a n diverges to. Notice that when the series converges, the value of the integral can be different from the sum of the series. Moreover, Integral test assumes implicitly that {a n } is a monotonically decreasing sequence. Further, the integral test is also applicable when the interval of integration is [m, ) instead of [, ). Example.4. Show that n= converges for p > and diverges for p. np For p =, the series is the harmonic series; and it diverges. Suppose p. Consider the function f(t) = /t p from [, ) to R. This is a continuous, positive and decreasing function. dt = lim tp b t p+ p + b { = ( ) p lim b b if p > p = p if p <. Then the Integral test proves the statement. We note that for p >, the sum of the series n p need not be equal to (p ). There are simple tests which are applicable to series of positive terms, whether the terms are decreasing or not. We discuss those next. Theorem.5. (D Alembert Ratio Test) Let a n+ a n be a series of positive terms. Suppose lim = l. n a n. If l <, then a n converges. 2. If l > or l =, then a n diverges to. 3. If l =, then no conclusion is obtained. Proof: () Given that lim(a n+ /a n ) = l <. Choose δ such that l < δ <. There exists m N such that for each n > m, a n+ /a n < δ. Then a n a m+ = a n a n a n a n 2 am+2 a m+ < δ n m. Thus, a n < δ n m a m+. Consequently, a m+ + a m a n < a m+ ( + δ + δ 2 + δ n m ). Since δ <, this approaches a limit as n. Therefore, the series a m+ + a m+2 + a n + 25

26 converges. In that case, the series a n = (a + + a m ) + a m+ + a m+2 + converges. (2) Given that lim(a n+ /a n ) = l >. Then there exists m N such that for each n > m, a n+ > a n. Then a m+ + a m a n > a m+ (n m). Since a m+ >, this approaches as n. Therefore, the series a m+ + a m+2 + a n + diverges to. In that case, the series a n = (a + + a m ) + a m+ + a m+2 + diverges to. The other case of l = is similar. (3) For the series (/n), lim(a n+ /a n ) = lim(n/(n + )) =. This series, as we know is divergent to. But the series (/n 2 ) is convergent although lim(a n+ /a n ) =. Example.5. Does the series Write a n = n!/(n n ). Then a n+ a n = n= n! n n converge? (n + ( )!nn n ) n (n + ) n+ (n! = n + e By D Alembert s ratio test, the series converges. Then it follows that the sequence { n! n n } converges to. Theorem.6. (Cauchy Root Test) Let a n be a series of positive terms. Suppose lim n (a n ) /n = l.. If l <, then a n converges. 2. If l > or l =, then a n diverges to. 3. If l =, then no conclusion is obtained. < as n. Proof: () Suppose l <. Choose δ such that l < δ <. Due to the limit condition, there exists an m N such that for each n > m, (a n ) /n < δ. That is, a n < δ n. Since < δ <, δ n converges. By Comparison test, a n converges. (2) Given that l > or l =, we see that (a n ) /n > for infinitely many values of n. That is, the sequence {(a n ) /n } does not converge to. Therefore, a n is divergent. It diverges to since it is a series of positive terms. (3) Once again, for both the series (/n) and (/n 2 ), we see that (a n ) /n has the limit. But one is divergent, the other is convergent. a n+ Remark: In fact, for a sequence {a n } of positive terms if lim n a n and the two limits are equal. exists, then lim n (a n ) /n exists 26

27 a n+ To see this, suppose lim = l. Let ɛ >. Then we have an m N such that for all n > m, n a n l ɛ < a n+ < l + ɛ. Use the right side inequality first. For all such n, a n < (l + ɛ) n m a m. Then a n (a n ) /n < (l + ɛ)((l + ɛ) m a m ) /n l + ɛ as n. Therefore, lim(a n ) /n l + ɛ for every ɛ >. That is, lim(a n ) /n l. Similarly, the left side inequality gives lim(a n ) /n l. Notice that this gives an alternative proof of Theorem.6. Example.6. Does the series 2 ( )n n = converge? Let a n = 2 ( )n n. Then n= Clearly, its limit does not exist. But a n+ a n = (a n ) /n = { /8 if n even 2 if n odd. { 2 /n if n even 2 /n if n odd This has limit /2 <. Therefore, by Cauchy root test, the series converges. Exercises for.8. Test for convergence the following series: n 5 (a) (b) n + 2 n + n= n= n (d) (e) ln n n + (f) n n=2 n= n=3 (ln n) n 3 (2n ) (g) (h) n n / n(3 n + ) n= n= (c) n=2 ln n n /n (ln n) (ln n) 2 (i) n!e n 2. Show that neither the ratio test nor the root test determine convergence of the series.9 Alternating series n= n=2 (ln n) p. If the terms of a series have alternating signs, then these tests are not applicable. For example, the methods discussed so far fail on deciding whether the series ( ) n /n converges or not. Theorem.7. (Leibniz Alternating Series Test) Let {a n } be a sequence of positive terms decreasing to ; that is, for each n, a n a n+ >, and lim a n =. Then the series ( ) n a n converges, and its sum lies between a a 2 and a. n n= 27

28 Proof: The partial sum upto 2n terms is s 2n = (a a 2 ) + (a 3 a 4 ) + + (a 2n a 2n ) = a [ (a 2 a 3 ) + + (a 2n 2 a 2n ) a 2n. It is a sum of n positive terms bounded above by a and below by a a 2. Hence s 2n converges to some s such that a a 2 s a. The partial sum upto 2n + terms is s 2n+ = s 2n + a 2n+. It converges to s as lim a 2n+ =. Hence the series converges to some s with a a 2 s a. The bounds for s can be sharpened by taking s 2n s s 2n for each n >. Leibniz test now implies that the series is convergent to some s with /2 s. By taking more terms, we can have different bounds such as = 7 2 s = 2 In contrast, the harmonic series diverges to We say that the series a n is absolutely convergent iff the series a n is convergent. An alternating series a n is said to be conditionally convergent iff it is convergent but it is not absolutely convergent. Thus for a series of non-negative terms, convergence and absolute convergence coincide. As we just saw, an alternating series may be convergent but not absolutely convergent. The series is a conditionally convergent series. It shows that the converse of the following theorem is not true. Theorem.8. An absolutely convergent series is convergent. Proof: Let a n be an absolutely convergent series. Then a n is convergent. Let ɛ >. By Cauchy criterion, there exists an n N such that for all n > m > n, we have Now, a m + a m+ + + a n < ɛ. a m + a m+ + + a n a m + a m+ + + a n < ɛ. Again, by Cauchy criterion, the series a n is convergent. An absolutely convergent series can be rearranged in any way we like, but the sum remains the same. Whereas a rearrangement of the terms of a conditionally convergent series may lead to divergence or convergence to any other number. In fact, a conditionally convergent series can always be rearranged in a way so that the rearranged series converges to any desired number; we will not prove this fact. 28

29 Example.7. Do the series (a) ( ) n+ 2 n n= (b) n= cos n n 2 converge? (a) converges. Therefore, the given series converges absolutely; hence it converges. 2 n (b) cos n n 2 n ; and (n 2 ) converges. By comparison test, the given series converges absolutely; and hence it 2 converges. Example.8. Discuss the convergence of the series n= ( ) n+ n p. For p >, the series n p converges. Therefore, the given series converges absolutely for p >. For < p, by Leibniz test, the series converges. But n p does not converge. Therefore, the given series converges conditionally for < p. For p, lim ( )n+ n p Exercises for.9 n=. Therefore, the given series diverges in this case.. Which of the following series converge absolutely, converge conditionally, and diverge? ( ) n+ n+ n n+ ln n (a) (b) ( ) (c) ( ) n 3/2 n n n= n= n= ( ) n (d) + ( ) n (e) (f) ( 2/3) n n 2 n ln(n 3 ) n= n=2 n= (g) ( ) n tan n ( ) n (h) (i) ( ) n ( n + n + n 2 n ln n n= n= 29

30 Chapter 2 Series Representation of Functions 2. Power series A power series apparently is a generalization of a polynomial. A polynomial in x looks like p(x) = a + a x + a 2 x a n x n. A power series is an infinite sum of the same form. The question is though a polynomial defines a function when x R, when does a power series define a function? That is, for what values of x, a power series sums to a number? Let a R. A power series about x = a is a series of the form c n (x a) n = c + c (x a) + c 2 (x a) 2 + n= The point a is called the center of the power series and the real numbers c, c,, c n, are its co-efficients. If the power series converges to f(x) for all x D, for some subset D of R, then we say that the power series sums to the function f(x), whose domain is D. In such a case, we also say that the power series represents the function f(x). For example, the geometric series + x + x x n + is a power series about x = with each co-efficient as. We know that its sum is x for < x <. And we know that for x, the geometric series does not converge. That is, the series defines a function from (, ) to R and it is not meaningful for other values of x. Example 2.. Show that the following power series converges for < x < 4. 2 (x 2) + 4 (x 2)2 + + ( )n 2 n (x 2) n + 3

31 It is a geometric series with the ratio as r = ( /2)(x 2). Thus it converges for ( /2)(x 2) <. Simplifying we get the constraint as < x < 4. Notice that the power series sums to r = 2(x 2) = 2 x. Thus, the power series gives a series expansion of the function 2 for < x < 4. x Truncating the series to n terms give us polynomial approximations of the function 2. x A fundamental result for the power series is the following. It roughly says that if for some c >, a power series converges with x = c, then it converges for all x with x c. A similar result holds for the divergence of a power series. For this purpose, we consider power series about x =. Results on power series about any point a can be obtained from this particular case in a similar manner. Theorem 2.. (Convergence Theorem for Power Series) Suppose that the power series n= a nx n is convergent for x = c and divergent for x = d for some c >, d >. Then the power series converges absolutely for all x with x < c; and it diverges for all x with x > d. Proof: The power series converges for x = c means that a n c n converges. Thus lim a nc n =. Then we have an M N such that for all n > M, a n c n <. n Let x R be such that x < c. Write t = x. For each n > M, we have c a n x n = a n x n = a n c n x c n < x c n = t n. As t <, the geometric series n=m+ tn converges. By comparison test, for any x with x < c, the series n=m+ a nx n converges. However, M n= a nx n is finite. Therefore, the power series n= a nx n converges absolutely for all x with x < c. For the divergence part of the theorem, suppose, on the contrary that the power series converges for some α > d. By the convergence part, the series must converge for x = d, a contradiction. Notice that if the power series is about a point x = a, then we take t = x a and apply Theorem 2.. Also, for x =, the power series a n x n always converges. In view of Theorem 2., the following definition makes sense. Consider the power series n= a n(x a) n. The real number R = lub{c : the power series converges for all x with x a < c} 3

32 is called the radius of convergence of the power series. That is, R is such non-negative number that the power series converges for all x with x a < R and it diverges for all x with x a > R. If the radius of convergence of the power series a n (x a) n is R, then the interval of convergence of the power series is (a R, a + R) if it diverges at both x = a R and x = a + R. [a R, a + R) if it converges at x = a R and diverges at x = a + R. (a R, a + R] if it diverges at x = a R and converges at x = a + R. That is, the interval of convergence of the power series is the open interval (a R, a+r) along with the point(s) a R and/or a + R, wherever it is convergent. Theorem 2. guarantees that the power series converges everywhere inside the interval of convergence, it converges absolutely inside the open interval (a R, a + R), and it diverges everywhere beyond the interval of convergence. Also, see that when R =, the power series converges for all x R, and when R =, the power series converges only at the point x = a, whence its sum is a. To determine the interval of convergence, you must find the radius of convergence R, and then test for its convergence separately for the end-points x = a R and x = a + R. 2.2 Determining radius of convergence The radius of convergence can be found out by ratio test and/or root test, or any other test. Theorem 2.2. The radius of convergence of the power series a n (x a) n is given by lim a n /n n provided that this limit is either a real number or equal to. Proof: Let R be the radius of convergence of the power series a n (x a) n. Let lim a n /n = r. n We consider three cases and show that () r > R =, (2) r = R =, (3) r = R =. r () Let r >. By the root test, the series is absolutely convergent whenever lim a n(x a) n /n < i.e., x a lim a n /n < i.e., x a < n n r. It also follows from the root test that the series is divergent when x a > /r. Hence R = /r. (2) Let r =. Then for any x a, lim a n (x a) n = lim x a a n /n =. By the root test, an (x a) n diverges for each x a. Thus, R =. (3) Let r =. Then for any x R, lim a n (x a) n /n = x a lim a n /n =. By the root test, the series converges for each x R. So, R =. Instead of the Root test, if we apply the Ratio test, then we obtain the following theorem. 32 n= n=

33 Theorem 2.3. The radius of convergence of the power series provided that this limit is either a real number or equal to. Example 2.2. For what values of x, do the following power series converge? (a) n!x n n= (b) n= x n n! (c) n= n= a n (x a) n a n, is given by lim n a n+ ( ) n x2n+ 2n + (a) a n = n!. Thus lim a n /a n+ = lim /(n + ) =. Hence R =. That is, the series is only convergent for x =. (b) a n = /n!. Thus lim a n /a n+ = lim(n + ) =. Hence R =. That is, the series is convergent for all x R. (c) Here, the power series is not in the form b n x n. The series can be thought of as ) x ( x2 3 + x4 5 + = x t n ( ) n t n 2n + n= for t = x 2 Now, for the power series ( ) n 2n +, a n = ( ) n /(2n + ). Thus lim a n /a n+ = lim 2n+3 =. Hence R =. That is, for t = 2n+ x2 <, the series converges and for t = x 2 >, the series diverges. Alternatively, you can use the geometric series. That is, for any x R, consider the series ) x ( x2 3 + x By the ratio test, the series converges if lim u n 2n + 3 = lim n u n+ n 2n + x2 = x 2 <. That is, the power series converges for < x <. Also, by the ratio test, the series diverges for x >. What happens for x =? For x =, the original power series is an alternating series; it converges due to Liebniz. Similarly, for x =, the alternating series also converges. Hence the interval of convergence for the original power series (in x) is [, ]. If R is the radius of convergence of a power series a n (x a) n, then the series defines a function f(x) from the open interval (a R, a + R) to R by f(x) = a + a (x a) + a 2 (x a) 2 + = a n (x a) n n= for x (a R, a + R). This function can be differentiated and integrated term-by-term and it so happens that the new series obtained by such term-by-term differentiation or integration has the same radius of convergence and they define the derivative and the integral of f(x). We state it without proof. 33

34 Theorem 2.4. Let the power series n= a n(x a) n have radius of convergence R >. Then the power series defines a function f : (a R, a + R) R. Further, f (x) and f(x)dx exist as functions from (a R, a + R) to R and these are given by f(x) = a n (x a) n, f (x) = n= na n (x a) n, n= f(x)dx = where all the three power series converge for all x (a R, a + R). n= a n (x a) n+ n + Caution: Term by term differentiation may not work for series, which are not power series. sin(n!x) For example, is convergent for all x. The series obtained by term-by-term differentiation is ; it diverges for all x. n 2 n= n! cos(n!x) n 2 n= Further, power series about the same point can be multiplied by using a generalization of multiplication of polynomials. We write the multiplication of power series about x = for simplicity. Theorem 2.5. Let the power series a n x n and b n x n have the same radius of convergence R >. Then their multiplication has the same radius of convergence R. Moreover, the functions they define satisfy the following: + C, If f(x) = a n x n, g(x) = b n x n then f(x)g(x) = c n x n for a R < x < a + R where c n = n k= a kb n k = a b n + a b n + + a n b + a n b. Example 2.3. Determine power series expansions of the functions (a) (a) For < x <, x = + x + x2 + x 3 +. Differentiating term by term, we have Differentiating once more, we get 2 ( x) 3 = 2 + 6x + 2x2 + = ( x) 2 = + 2x + 3x2 + 4x (x ) 3 (b) tan x. n(n )x n 2 for < x <. n=2 (b) + x = 2 x2 + x 4 x 6 + x 8 for x 2 <. Integrating term by term we have tan x + C = x x3 3 + x5 5 x7 + for < x <. 7 Evaluating at x =, we see that C =. Hence the power series for tan x. 34