THE GAMMA FUNCTION AND THE ZETA FUNCTION
|
|
- Julia King
- 5 years ago
- Views:
Transcription
1 THE GAMMA FUNCTION AND THE ZETA FUNCTION PAUL DUNCAN Abstract. The Gamma Function and the Riemann Zeta Function are two special functions that are critical to the study of many different fields of mathematics. In this paper we will discuss characterizations and properties of each, and how the two are connected. Contents. The Gamma Function.. Convex Functions.. The Gamma Function 4. Values of the Riemann Zeta Function 8 3. Characterization of the Zeta Function 9 Acknowledgments References. The Gamma Function In this paper we will start by characterizing the Gamma function. Its development is motivated by the desire for a smooth extension of the factorial function to R. Recall the recursive definition of the factorial function, namely that (n + )! = (n + )n! and! =. Since it is only defined on the integers, there are many possible continuous extensions to R. However, since we know that e n < n! < n n, we can imagine narrowing our options using some notion of fast growth. Fortunately, it turns out that we can find a unique extension using the concept of convexity... Convex Functions. Recall from calculus that a twice differentiable function is called convex if f (x) for all x. We would like an extension of this definition to functions that are not differentiable. Definition.. For a function f(x) define φ f (x, x ) = f(x ) f(x ) x x which can be thought of as the slope of the secant line between two points. We say that f(x) is convex on an interval if for every x 3 on the interval, φ(x, x 3 ) is a monotonically increasing function of x. Date: August 9, 3.
2 PAUL DUNCAN Alternatively, f(x) is convex if, for all x, x, x 3, Ψ f, where Ψ f is defined as Ψ f (x, x, x 3 ) = φ f (x, x 3 ) φ f (x, x 3 ) (.) x x = (x 3 x )f(x ) + (x x 3 )f(x ) + (x x )f(x 3 ) (.3). (x x )(x x 3 )(x 3 x ) Remark.4. The values of φ f (x, x ) and Ψ f (x, x, x 3 ) are invariant under permutation of their arguments. First we will need some tools to determine whether a given function is convex. Theorem.5. The sum of convex functions, limit function of a convergent sequence of convex functions, and sum of a convergent series of convex functions are all convex. Proof. From the second statement of the definition of convexity we can see that for two functions f(x) and g(x) if Ψ f (x, x, x 3 ) and Ψ g (x, x, x 3 ) then Ψ f+g (x, x, x 3 ) = Ψ f (x, x, x 3 ) + Ψ g (x, x, x 3 ). If we look at the same inequality for a sequence of convex functions f (x), f (x)... that converges to f(x) we see that Ψ f (x, x, x 3 ) = lim Ψ f n (x, x, x 3 ), n so f(x) is also convex. The third part of the theorem then follows from the first two since a series is a sequence of partial sums. Definition.6. We say that a function is weakly convex on an interval if for every x, x,..., x n on that interval f( x+x+...+xn n ) n (f(x ) + f(x ) f(x n )). Lemma.7. If a function is weakly convex and continuous, then it is convex. Proof. Suppose f(x) is weakly convex. Choose x < x numbers in the interval and p n two integers. Apply the definition of weak convexity in the case where p numbers have the value x and the other n p have the value x. We get f ( p n x + ( p n ) ) x p n f(x ) + ( p n ) f(x ). Assume f(x) is continuous and let t be a real number such that t [, ]. Choose a sequence of rational numbers converging to t. We can write each number as p n, so we can apply the above equation. But f(x) is continuous, so we can take the limit of the sequence to obtain f(t(x ) + ( t)x ) tf(x ) + ( t)f(x ). We want to show that for any x, x, x 3, we have Ψ f (x, x, x 3 ). We can assume x < x 3 < x since we can permute the arguments of Ψ f, so the denominator for expression (.3) is positive. Set t = x3 x x x. We know that t (, ) and t = x x3 x x. So tx + ( t)x = (x 3 x )x + (x x 3 )x x x = x 3. Plugging this into the previous equation we get f(x 3 ) x 3 x x x f(x ) + x x 3 x x f(x ).
3 THE GAMMA FUNCTION AND THE ZETA FUNCTION 3 It is now clear that the numerator of (.3) is also positive, so f(x) must be convex. Remark.8. The continuous condition is necessary, but in order to construct an example of a function that is weakly convex, but not convex, one must use the Axiom of Choice. The converse of the lemma is also true. Definition.9. We say f(x) is log convex on an interval if log(f(x)) is defined and convex on that interval. Remark.. If a function is log convex, then it is also convex. From Theorem.5 we know that the product of log convex functions is log convex. More surprisingly, the same turns out to be true for their sum. Theorem.. Let f(x) and g(x) be log convex functions defined on a common interval. Then f(x) + g(x) is also log convex. Proof. It is enough to prove the statement for weakly convex functions, since we can add continuity to then obtain the theorem. Assume f(x) and g(x) are weakly log convex. Then both are positive and for every x, x or Similarly, We need to show (f( x + x log f( x + x ) (log f(x ) + log f(x )), (f( x + x )) f(x )f(x ). (g( x + x )) g(x )g(x ). ) + g( x + x )) (f(x ) + g(x ))(f(x ) + g(x )). Let S = {(a, b, c) R 3 ac b }. We have (f(x ), f( x + x ), f(x)), (g(x ), g( x + x ), g(x )) S, and we want to show their sum is in S. We will show this by showing that S is closed under addition. By the quadratic formula we know that ac b ax + bx + c x R, so S = {(a, b, c) R 3 ax + bx + c addition. x R}, which is clearly closed under Theorem.. If φ(x) is a positive continuous function defined on the interior of the integration interval, then b a φ(t)t x dt is a log convex function of x for every interval on which the proper or improper integral exists.
4 4 PAUL DUNCAN Proof. Suppose f(t, x) is defined and continuous for t [a, b] and arbitrary x, and also log convex as a function of x. For n Z and h = (b a)/n, construct F n (x) = h(f(a, x) + f(a + h, x) f(a + (n )h, x)). Since F n (x) is a sum of log convex functions, it is log convex. If we take the limit as n goes to infinity, we get b a f(t, x)dt. If the integral is improper, then it is the limit of a sequence of log convex functions, so it is also log convex. The theorem is the case in which f(t, x) = φ(t)t x. With log convexity, we have enough to define a unique Γ(x)... The Gamma Function. Definition.3. We define Γ(x) = e t t x dt for x >. Note that the integral converges for positive real x. Theorem.4. Suppose a function f(x) satisfies the following three conditions: () f(x + ) = xf(x) () The domain of f(x) contains all x >, and f(x) is log convex for these x (3) f() =. Then n x n! f(x) = lim n x(x + )...(x + n). Moreover, Γ(x) satisfies (), (), (3) for x >. In particular, (), (), (3) characterize Γ(x) uniquely. Proof. First, we will show that Γ(x) satisfies the three conditions. Integration by parts shows that Γ(x) satisfies the first condition. Γ(x + ) = lim δ,ɛ δ ɛ e t t x dt = lim δ,ɛ ( e t t x δ ɛ + x = lim δ,ɛ ( e δ δ x + e ɛ ɛ x + x δ ɛ δ e t t x dt) = xγ(x) ɛ e t t x dt) The second condition comes from Theorem.8 and the third is easily checked. Since the gamma function satisfies the requirements, we know that such a function exists. Let f(x) be a function that satisfies these conditions. Then it is enough to show that f(x) = Γ(x) for x (, ], since the first condition then implies that they agree on the full domain. For x (, ] and integer n, we can use the second condition to obtain log f( + n) log f(n) log f(x + n) log f(n) ( + n) n (x + n) n log f( + n) log f(n). ( + n) n By combining the first and third conditions we know that f(n) = (n )!, so we can write log f(x + n) log(n )! log(n ) log n, x which can be rearranged to get log(n ) x (n )! f(x + n) log n x (n )!.
5 THE GAMMA FUNCTION AND THE ZETA FUNCTION 5 Now, since log is a monotonically increasing function, this implies (n ) x (n )! f(x + n) n x (n )!. From the first condition it is clear that f(x + n) = x(x + )...(x + n)f(x), so (n ) x (n )! x(x + )...(x + n ) f(x) (n) x (n )! x(x + )...(x + n). Since this is true for all n, we can write n x n! x(x + )...(x + n) f(x) n x n! x + n x(x + )...(x + n) n Now we can take the limit as n goes to infinity and get the formula n x n! f(x) = lim n x(x + )...(x + n). But since Γ(x) also satisfies the three conditions, it is equal to f(x).. We use the limit from Theorem.4 to define Γ(x) for x <. Note that Γ(x) is still not defined for non-positive integers. With some manipulation, we can use the formula to find a useful alternative definition for the gamma function. Corollary.5. The gamma function can be expressed as the product Γ(x) = e Cx e x j x + x j for x R \ Z, where C is Euler s constant. Proof. n x n! Γ(x) = lim n x(x + )...(x + n) = lim n ex log n e x/ x/+x/ x/+...+x/n x/n x x + x +... n x + n = lim n ex(log n / /... /n) e x/ e x/ x x/ + x/ +... e x/n x/n +. j= Recall C = lim n n log n exists and is called Euler s constant. So Γ(x) = e Cx n x lim e x j n + x = e Cx j x j= j= e x j + x j Although differentiability was not one of our criteria for the uniqueness of the gamma function, we would hope that it would be infinitely differentiable. In fact it is. Theorem.6. The function Γ(x) is infinitely differentiable on its domain of definition.
6 6 PAUL DUNCAN Proof. It suffice to show that log Γ(x) is infinitely differentiable, since the chain rule then tells us that the same is true for Γ(x). Moreover, we only need to consider the case of x > because of the relation Γ(x + ) = xγ(x). We know that log Γ(x) is defined since Γ(x) > for all x >. From the product equation in Corollary. we can write log Γ(x) = Cx log x + ( x j log( + x j )). Now we can differentiate termwise if the series thus obtained is uniformly convergent. The series in question is C x + ( j x + j ) = C x + x j(x + j). j= For x (, r] for arbitrary r j= j= x j(x + j) r j, which is independent of x. Since the right side converges, we can apply the Weierstrass M-test to show that our series converges uniformly, and log Γ(x) is differentiable on that interval. But r is arbitrary, so the same is true for R >. If we differentiate again, we get x j= j= (x + j). Since the jth summand is less than /j, we can again apply the Weierstrass M- test. Repeating the process shows that log Γ(x), and by extension Γ(x) is infinitely differentiable. j= Lemma.7. The gamma function satisfies the relation ( x ) ( ) x + Γ Γ = c x Γ(x) for some constant c. Proof. Consider the function ( x ) ( ) x + f(x) = x Γ Γ. Since f(x) is a product of log convex functions, it is log convex. Moreover, ( ) x + ( x ) f(x + ) = x+ Γ Γ + = xf(x). Since f(x) satisfies the first two conditions of Theorem., we know that f(x) = cγ(x) for some constant c and with slight rearrangement we have the lemma. The product formula for the gamma function is already useful in deriving an alternate formula for sin(x). Theorem.8. The sine function can be expressed as the infinite product sin(πx) = πx ( x j ). j=
7 THE GAMMA FUNCTION AND THE ZETA FUNCTION 7 Proof. Define a new function φ(x) = Γ(x)Γ( x) sin πx for non-integer arguments. Our first goal is to show that φ(x) is constant. First we want to extend φ(x) to R so that it is periodic with period, continuous, and infinitely differentiable. If we replace x by x + in each of the components, we get Γ(x + ) = xγ(x), Γ( x) = Γ( x), sin π(x + ) = sin πx. x Combining them, we see that φ(x + ) = φ(x). By Lemma.7 we have Γ( x )Γ(x + ) = c x Γ(x) for some constant c. Replace x by x to get Now So φ( x )φ(x + Γ( x )Γ( x ) = cx Γ( x). ) = Γ( x )Γ( x ) sin(πx = c Γ(x)Γ( x) sin πx. 4 )Γ(x + )Γ( x ) cos( πx ) (.9) φ( x )φ(x + ) = c 4 φ(x). We want φ(x) to be continuous and infinitely differentiable on R. It is on nonintegral arguments because Γ(x) and sin(x) are, and to continue those properties to the real line we extend φ(x) by writing φ(x) = Γ( + x)γ( x) sin πx x = Γ( + x)γ( x)(π π3 x 3! + π5 x 4 5!...). Using that expansion, we define φ() = π and since φ(x) is periodic with period, φ(n) = π for n Z. Now φ is continuous and infinitely differentiable. We also know that φ(x) is positive everywhere by combining the original definition, the periodicity, and the fact that φ(n) = π for n Z. Now we want to show that φ(x) is constant. Let g(x) = d dx log φ(x). From (.9) we can derive (.) 4 (g(x ) + g(x + )) = g(x). Since g(x) is continuous on [, ], it is bounded, e.g. g(x) M on that interval for some M. But since φ(x) is periodic, so must g(x) be, so it is bounded everywhere. Using (.) we can write g(x) 4 g(x ) + 4 g(x + ) M 4 + M 4 = M. This process can be repeated an arbitrary number of times, so g(x) =. By the definition of g(x), this means that log φ(x) is linear. However it is also periodic, so it must be constant, which means that φ(x) is also constant. Since we know
8 8 PAUL DUNCAN φ() = π, we know that φ(x) = π for all x. Now we can rewrite the definition of φ(x) π sin πx = Γ(x)Γ( x) = π xγ(x)γ( x). A straightforward substitution of the formula for the gamma function in Corollary. yields sin(πx) = πx ( x i ). i=. Values of the Riemann Zeta Function The Riemann zeta function is another extraordinarily useful special function. We will only talk about its real part here. Definition.. We define for s >. Definition.. We define for s,..., s k >. ζ(s,..., s k ) = ζ(s) = n= n s n >n >...>n k > n s...ns k k From calculus we know that the formula for ζ converges for s >, but we cannot easily compute its values. At certain integers, however, its values have a simple form. Theorem.3. We can obtain the values of the zeta function and multiple zeta function with arguments of s, namely ζ() = π π4, ζ(, ) = 6. Proof. We want to find some C m (N) such that N ( x N j ) = ( ) j C (N) j x j. j= We expand the product to form N ( x j ) = ( x )( x )...( x N ) j= j= = x( N ) + x ( N (N ) )... Now we can collect terms and see that C (N) = and for m N C m (N) = j. j...j m j >j >...>j m>
9 THE GAMMA FUNCTION AND THE ZETA FUNCTION 9 If for all m, lim N C m (N) = C m, then ( x j ) = ( ) j C (N) j x j. j= j= Since every term in the summation of C (N) is positive, it is clear that C m (N) < (C (N) ) m. We can see that C m (N) increases as N does and we know that C is finite from calculus, so each C m (N) must converge. We can also see that C m = ζ(,,..., ) with m arguments. Now let f(x) = ( x i ) = ζ()x + ζ(, )x.... i= Now, by Theorem.8, we can write sin πx = πxf(x ) = πx ζ()πx 3 + ζ(, )πx But from calculus we know that sin πx = πx π3 x 3 + π5 x ! 5! Equating coefficients, we have a formula for zeta values involving, in particular ζ() = π π4 6 and ζ(, ) =. Theorem.4. We can also compute ζ(4) = π4 9. Proof. Notice that ζ() = ( n ) = n 4 + n + n >n n n = ζ(4) + ζ(, ). n <n n n= Solving for ζ(4), we get n= ζ(4) = ζ() ζ(, ) = π4 36 π4 = π Characterization of the Zeta Function Soon we will be able to see that the gamma and zeta functions are related, but first we must extend the zeta function. From its definition, it is meaningless to evaluate in the interval (, ), but we can use a similar function to derive a useful continuation. Notation 3.. For r Z >, we define ζ r (s) := n> Theorem 3.. For s >, b n n b n = ζ(s) = Moreover, ζ r (s) is defined for s >. { : r n r : r n. rs r s ζ r(s).
10 PAUL DUNCAN Proof. Notice that r s ζ(s) = n= (rn) s = n= and that b n = ra n. Now we can write So, { a n : r n n s a n = : r n ζ r (s) = ζ(s) r( r s ζ(s)) = ζ(s)( r r s ) = rs r s ζ(s). ζ(s) = rs r s ζ r(s). Now we want to show that ζ r (s) is defined for s >. Let B N = N n= b n. It is easy to see that B N r for all N. Let s >. We want to show that the series that defines ζ r (s) is Cauchy. Using the fact that B k B k = b k, we can see that for any N > M N k=m b k k s = B N N s B N M M s + B k ( k s (k + ) s ). The first two terms go to zero, so it suffices to show that the sum converges. Using calculus we can see that k s (k + ) s = s k k+ k=m x s+ dx s (k + ) s+ and we can combine that with our bound on B N to yield N k=m B k ( k s (k + ) s ) rs (k + ) s+ = rs k which we know converges because s >. We therefore extend ζ(s) to R > by defining k (k + ) s+, ζ(s) = rs r s ζ r(s). Now we have enough for a functional equation relating the zeta and gamma functions. Theorem 3.3. Let then ξ(s) = ξ( s) for s R > \ {}. Proof. Let θ(u) = n= ξ(s) = π s/ Γ( s )ζ(s) e πnu = + (e πu + e 4πu + e 9πu...). We need the following lemma from Fourier analysis: Lemma 3.4. θ(u) satisfies the functional equation θ(/u) = u / θ(u).
11 THE GAMMA FUNCTION AND THE ZETA FUNCTION We will not prove this here, but a proof can be found on page 5 of Elkies notes[]. Notice for every integer n we have e πnu u s/ du u = = e t π s/ n t s/ dt πn t = π s/ Γ( s ) n. e t s/ dt t t By summing over n we get (θ(u) )u s/ du u = ξ(s). Now we can write ξ(s) = We can substitute /u for u to get (θ(u) )u s/ du u + (θ(u) )u s/ du u = s + θ(u)u s/ du u + (θ(u) )u s/ du u. θ(u)u s/ du u = θ(u )u s/ du u = θ(u)u ( s)/ du u = s + (θ(u) )u ( s)/ du u using the functional equation. Combining these we find ξ(s) = s s + (θ(u) )(u s/ + u ( s)/ ) du u. Since the right side is symmetric in s and s, we can see that ξ(s) = ξ( s). Acknowledgments. It is a pleasure to thank my mentors, Preston Wake and Yun Cheng, for their unceasing help and considerable patience. References [] Emil Artin. The Gamma Function. Holt, Reinhart, and Winston [] Noam Elkies. The Riemann zeta function and its functional equation. elkies/m59./zeta.pdf. [3] Tobias Oekiter, Hubert Partl, Irene Hyna and Elisabeth Schlegl. The Not So Short Introduction to LATEX ε.
Mathematics 324 Riemann Zeta Function August 5, 2005
Mathematics 324 Riemann Zeta Function August 5, 25 In this note we give an introduction to the Riemann zeta function, which connects the ideas of real analysis with the arithmetic of the integers. Define
More informationWe start with a simple result from Fourier analysis. Given a function f : [0, 1] C, we define the Fourier coefficients of f by
Chapter 9 The functional equation for the Riemann zeta function We will eventually deduce a functional equation, relating ζ(s to ζ( s. There are various methods to derive this functional equation, see
More informationEconomics 204 Summer/Fall 2011 Lecture 5 Friday July 29, 2011
Economics 204 Summer/Fall 2011 Lecture 5 Friday July 29, 2011 Section 2.6 (cont.) Properties of Real Functions Here we first study properties of functions from R to R, making use of the additional structure
More information1 + lim. n n+1. f(x) = x + 1, x 1. and we check that f is increasing, instead. Using the quotient rule, we easily find that. 1 (x + 1) 1 x (x + 1) 2 =
Chapter 5 Sequences and series 5. Sequences Definition 5. (Sequence). A sequence is a function which is defined on the set N of natural numbers. Since such a function is uniquely determined by its values
More informationChapter 8: Taylor s theorem and L Hospital s rule
Chapter 8: Taylor s theorem and L Hospital s rule Theorem: [Inverse Mapping Theorem] Suppose that a < b and f : [a, b] R. Given that f (x) > 0 for all x (a, b) then f 1 is differentiable on (f(a), f(b))
More information7: FOURIER SERIES STEVEN HEILMAN
7: FOURIER SERIES STEVE HEILMA Contents 1. Review 1 2. Introduction 1 3. Periodic Functions 2 4. Inner Products on Periodic Functions 3 5. Trigonometric Polynomials 5 6. Periodic Convolutions 7 7. Fourier
More informationbe the set of complex valued 2π-periodic functions f on R such that
. Fourier series. Definition.. Given a real number P, we say a complex valued function f on R is P -periodic if f(x + P ) f(x) for all x R. We let be the set of complex valued -periodic functions f on
More informationMeasure and Integration: Solutions of CW2
Measure and Integration: s of CW2 Fall 206 [G. Holzegel] December 9, 206 Problem of Sheet 5 a) Left (f n ) and (g n ) be sequences of integrable functions with f n (x) f (x) and g n (x) g (x) for almost
More informationTHE INVERSE FUNCTION THEOREM
THE INVERSE FUNCTION THEOREM W. PATRICK HOOPER The implicit function theorem is the following result: Theorem 1. Let f be a C 1 function from a neighborhood of a point a R n into R n. Suppose A = Df(a)
More informationChapter 11 - Sequences and Series
Calculus and Analytic Geometry II Chapter - Sequences and Series. Sequences Definition. A sequence is a list of numbers written in a definite order, We call a n the general term of the sequence. {a, a
More informationMathematical Methods for Physics and Engineering
Mathematical Methods for Physics and Engineering Lecture notes for PDEs Sergei V. Shabanov Department of Mathematics, University of Florida, Gainesville, FL 32611 USA CHAPTER 1 The integration theory
More informationMathematic 108, Fall 2015: Solutions to assignment #7
Mathematic 08, Fall 05: Solutions to assignment #7 Problem # Suppose f is a function with f continuous on the open interval I and so that f has a local maximum at both x = a and x = b for a, b I with a
More informationRiemann integral and volume are generalized to unbounded functions and sets. is an admissible set, and its volume is a Riemann integral, 1l E,
Tel Aviv University, 26 Analysis-III 9 9 Improper integral 9a Introduction....................... 9 9b Positive integrands................... 9c Special functions gamma and beta......... 4 9d Change of
More informationTHE GAMMA FUNCTION THU NGỌC DƯƠNG
THE GAMMA FUNCTION THU NGỌC DƯƠNG The Gamma unction was discovered during the search or a actorial analog deined on real numbers. This paper will explore the properties o the actorial unction and use them
More informationMetric Spaces and Topology
Chapter 2 Metric Spaces and Topology From an engineering perspective, the most important way to construct a topology on a set is to define the topology in terms of a metric on the set. This approach underlies
More informationDRAFT - Math 101 Lecture Note - Dr. Said Algarni
2 Limits 2.1 The Tangent Problems The word tangent is derived from the Latin word tangens, which means touching. A tangent line to a curve is a line that touches the curve and a secant line is a line that
More informationSome approximation theorems in Math 522. I. Approximations of continuous functions by smooth functions
Some approximation theorems in Math 522 I. Approximations of continuous functions by smooth functions The goal is to approximate continuous functions vanishing at ± by smooth functions. In a previous homewor
More informationAn idea how to solve some of the problems. diverges the same must hold for the original series. T 1 p T 1 p + 1 p 1 = 1. dt = lim
An idea how to solve some of the problems 5.2-2. (a) Does not converge: By multiplying across we get Hence 2k 2k 2 /2 k 2k2 k 2 /2 k 2 /2 2k 2k 2 /2 k. As the series diverges the same must hold for the
More informationIowa State University. Instructor: Alex Roitershtein Summer Homework #5. Solutions
Math 50 Iowa State University Introduction to Real Analysis Department of Mathematics Instructor: Alex Roitershtein Summer 205 Homework #5 Solutions. Let α and c be real numbers, c > 0, and f is defined
More informationNotes on uniform convergence
Notes on uniform convergence Erik Wahlén erik.wahlen@math.lu.se January 17, 2012 1 Numerical sequences We begin by recalling some properties of numerical sequences. By a numerical sequence we simply mean
More informationContinuity. Chapter 4
Chapter 4 Continuity Throughout this chapter D is a nonempty subset of the real numbers. We recall the definition of a function. Definition 4.1. A function from D into R, denoted f : D R, is a subset of
More informationAnalysis Finite and Infinite Sets The Real Numbers The Cantor Set
Analysis Finite and Infinite Sets Definition. An initial segment is {n N n n 0 }. Definition. A finite set can be put into one-to-one correspondence with an initial segment. The empty set is also considered
More informationSequences. Chapter 3. n + 1 3n + 2 sin n n. 3. lim (ln(n + 1) ln n) 1. lim. 2. lim. 4. lim (1 + n)1/n. Answers: 1. 1/3; 2. 0; 3. 0; 4. 1.
Chapter 3 Sequences Both the main elements of calculus (differentiation and integration) require the notion of a limit. Sequences will play a central role when we work with limits. Definition 3.. A Sequence
More informationChapter 7: Techniques of Integration
Chapter 7: Techniques of Integration MATH 206-01: Calculus II Department of Mathematics University of Louisville last corrected September 14, 2013 1 / 43 Chapter 7: Techniques of Integration 7.1. Integration
More informationAdvanced Calculus Math 127B, Winter 2005 Solutions: Final. nx2 1 + n 2 x, g n(x) = n2 x
. Define f n, g n : [, ] R by f n (x) = Advanced Calculus Math 27B, Winter 25 Solutions: Final nx2 + n 2 x, g n(x) = n2 x 2 + n 2 x. 2 Show that the sequences (f n ), (g n ) converge pointwise on [, ],
More informationSome Interesting Properties of the Riemann Zeta Function
Some Interesting Properties of the Riemann Zeta Function arxiv:822574v [mathho] 2 Dec 28 Contents Johar M Ashfaque Introduction 2 The Euler Product Formula for ζ(s) 2 3 The Bernoulli Numbers 4 3 Relationship
More informationMath 0230 Calculus 2 Lectures
Math 00 Calculus Lectures Chapter 8 Series Numeration of sections corresponds to the text James Stewart, Essential Calculus, Early Transcendentals, Second edition. Section 8. Sequences A sequence is a
More informationd(x n, x) d(x n, x nk ) + d(x nk, x) where we chose any fixed k > N
Problem 1. Let f : A R R have the property that for every x A, there exists ɛ > 0 such that f(t) > ɛ if t (x ɛ, x + ɛ) A. If the set A is compact, prove there exists c > 0 such that f(x) > c for all x
More information3 (Due ). Let A X consist of points (x, y) such that either x or y is a rational number. Is A measurable? What is its Lebesgue measure?
MA 645-4A (Real Analysis), Dr. Chernov Homework assignment 1 (Due ). Show that the open disk x 2 + y 2 < 1 is a countable union of planar elementary sets. Show that the closed disk x 2 + y 2 1 is a countable
More informationINTRODUCTORY LECTURES COURSE NOTES, One method, which in practice is quite effective is due to Abel. We start by taking S(x) = a n
INTRODUCTORY LECTURES COURSE NOTES, 205 STEVE LESTER AND ZEÉV RUDNICK. Partial summation Often we will evaluate sums of the form a n fn) a n C f : Z C. One method, which in ractice is quite effective is
More informationReal Analysis Math 131AH Rudin, Chapter #1. Dominique Abdi
Real Analysis Math 3AH Rudin, Chapter # Dominique Abdi.. If r is rational (r 0) and x is irrational, prove that r + x and rx are irrational. Solution. Assume the contrary, that r+x and rx are rational.
More informationExamples 2: Composite Functions, Piecewise Functions, Partial Fractions
Examples 2: Composite Functions, Piecewise Functions, Partial Fractions September 26, 206 The following are a set of examples to designed to complement a first-year calculus course. objectives are listed
More informationMath 141: Lecture 11
Math 141: Lecture 11 The Fundamental Theorem of Calculus and integration methods Bob Hough October 12, 2016 Bob Hough Math 141: Lecture 11 October 12, 2016 1 / 36 First Fundamental Theorem of Calculus
More information1.4 Techniques of Integration
.4 Techniques of Integration Recall the following strategy for evaluating definite integrals, which arose from the Fundamental Theorem of Calculus (see Section.3). To calculate b a f(x) dx. Find a function
More information4. We accept without proofs that the following functions are differentiable: (e x ) = e x, sin x = cos x, cos x = sin x, log (x) = 1 sin x
4 We accept without proofs that the following functions are differentiable: (e x ) = e x, sin x = cos x, cos x = sin x, log (x) = 1 sin x x, x > 0 Since tan x = cos x, from the quotient rule, tan x = sin
More information(x 3)(x + 5) = (x 3)(x 1) = x + 5. sin 2 x e ax bx 1 = 1 2. lim
SMT Calculus Test Solutions February, x + x 5 Compute x x x + Answer: Solution: Note that x + x 5 x x + x )x + 5) = x )x ) = x + 5 x x + 5 Then x x = + 5 = Compute all real values of b such that, for fx)
More information2 Sequences, Continuity, and Limits
2 Sequences, Continuity, and Limits In this chapter, we introduce the fundamental notions of continuity and limit of a real-valued function of two variables. As in ACICARA, the definitions as well as proofs
More informationMATH 409 Advanced Calculus I Lecture 25: Review for the final exam.
MATH 49 Advanced Calculus I Lecture 25: Review for the final exam. Topics for the final Part I: Axiomatic model of the real numbers Axioms of an ordered field Completeness axiom Archimedean principle Principle
More informationMATH 311: COMPLEX ANALYSIS CONTOUR INTEGRALS LECTURE
MATH 3: COMPLEX ANALYSIS CONTOUR INTEGRALS LECTURE Recall the Residue Theorem: Let be a simple closed loop, traversed counterclockwise. Let f be a function that is analytic on and meromorphic inside. Then
More informationBernoulli Polynomials
Chapter 4 Bernoulli Polynomials 4. Bernoulli Numbers The generating function for the Bernoulli numbers is x e x = n= B n n! xn. (4.) That is, we are to expand the left-hand side of this equation in powers
More information1 The functional equation for ζ
18.785: Analytic Number Theory, MIT, spring 27 (K.S. Kedlaya) The functional equation for the Riemann zeta function In this unit, we establish the functional equation property for the Riemann zeta function,
More informationCalculus II Lecture Notes
Calculus II Lecture Notes David M. McClendon Department of Mathematics Ferris State University 206 edition Contents Contents 2 Review of Calculus I 5. Limits..................................... 7.2 Derivatives...................................3
More informationAP Calculus Chapter 9: Infinite Series
AP Calculus Chapter 9: Infinite Series 9. Sequences a, a 2, a 3, a 4, a 5,... Sequence: A function whose domain is the set of positive integers n = 2 3 4 a n = a a 2 a 3 a 4 terms of the sequence Begin
More informationSome Fun with Divergent Series
Some Fun with Divergent Series 1. Preliminary Results We begin by examining the (divergent) infinite series S 1 = 1 + 2 + 3 + 4 + 5 + 6 + = k=1 k S 2 = 1 2 + 2 2 + 3 2 + 4 2 + 5 2 + 6 2 + = k=1 k 2 (i)
More informationContinuity. Chapter 4
Chapter 4 Continuity Throughout this chapter D is a nonempty subset of the real numbers. We recall the definition of a function. Definition 4.1. A function from D into R, denoted f : D R, is a subset of
More informationMath 321 Final Examination April 1995 Notation used in this exam: N. (1) S N (f,x) = f(t)e int dt e inx.
Math 321 Final Examination April 1995 Notation used in this exam: N 1 π (1) S N (f,x) = f(t)e int dt e inx. 2π n= N π (2) C(X, R) is the space of bounded real-valued functions on the metric space X, equipped
More informationMATH Solutions to Probability Exercises
MATH 5 9 MATH 5 9 Problem. Suppose we flip a fair coin once and observe either T for tails or H for heads. Let X denote the random variable that equals when we observe tails and equals when we observe
More informationx s 1 e x dx, for σ > 1. If we replace x by nx in the integral then we obtain x s 1 e nx dx. x s 1
Recall 9. The Riemann Zeta function II Γ(s) = x s e x dx, for σ >. If we replace x by nx in the integral then we obtain Now sum over n to get n s Γ(s) = x s e nx dx. x s ζ(s)γ(s) = e x dx. Note that as
More informationSome expressions of double and triple sine functions
SUT Journal of Mathematics Vol 43, No 7, 51 61 Some expressions of double triple sine functions Hidekazu Tanaka Received March, 7 Abstract We show some expressions of double triple sine functions Then
More information1 Question related to polynomials
07-08 MATH00J Lecture 6: Taylor Series Charles Li Warning: Skip the material involving the estimation of error term Reference: APEX Calculus This lecture introduced Taylor Polynomial and Taylor Series
More informationDirichlet s Theorem. Martin Orr. August 21, The aim of this article is to prove Dirichlet s theorem on primes in arithmetic progressions:
Dirichlet s Theorem Martin Orr August 1, 009 1 Introduction The aim of this article is to prove Dirichlet s theorem on primes in arithmetic progressions: Theorem 1.1. If m, a N are coprime, then there
More informationMath 141: Lecture 19
Math 141: Lecture 19 Convergence of infinite series Bob Hough November 16, 2016 Bob Hough Math 141: Lecture 19 November 16, 2016 1 / 44 Series of positive terms Recall that, given a sequence {a n } n=1,
More informationSynopsis of Complex Analysis. Ryan D. Reece
Synopsis of Complex Analysis Ryan D. Reece December 7, 2006 Chapter Complex Numbers. The Parts of a Complex Number A complex number, z, is an ordered pair of real numbers similar to the points in the real
More informationCH 2: Limits and Derivatives
2 The tangent and velocity problems CH 2: Limits and Derivatives the tangent line to a curve at a point P, is the line that has the same slope as the curve at that point P, ie the slope of the tangent
More informationSection Taylor and Maclaurin Series
Section.0 Taylor and Maclaurin Series Ruipeng Shen Feb 5 Taylor and Maclaurin Series Main Goal: How to find a power series representation for a smooth function us assume that a smooth function has a power
More informationSUMMATION TECHNIQUES
SUMMATION TECHNIQUES MATH 53, SECTION 55 (VIPUL NAIK) Corresponding material in the book: Scattered around, but the most cutting-edge parts are in Sections 2.8 and 2.9. What students should definitely
More informationMATH 104 : Final Exam
MATH 104 : Final Exam 10 May, 2017 Name: You have 3 hours to answer the questions. You are allowed one page (front and back) worth of notes. The page should not be larger than a standard US letter size.
More information2.1 The Tangent and Velocity Problems
2.1 The Tangent and Velocity Problems Tangents What is a tangent? Tangent lines and Secant lines Estimating slopes from discrete data: Example: 1. A tank holds 1000 gallons of water, which drains from
More informationAn Introduction to the Gamma Function
UNIVERSITY OF WARWICK Second Year Essay An Introduction to the Gamma Function Tutor: Dr. Stefan Adams April 4, 29 Contents 1 Introduction 2 2 Conve Functions 3 3 The Gamma Function 7 4 The Bohr-Möllerup
More informationswapneel/207
Partial differential equations Swapneel Mahajan www.math.iitb.ac.in/ swapneel/207 1 1 Power series For a real number x 0 and a sequence (a n ) of real numbers, consider the expression a n (x x 0 ) n =
More information17 The functional equation
18.785 Number theory I Fall 16 Lecture #17 11/8/16 17 The functional equation In the previous lecture we proved that the iemann zeta function ζ(s) has an Euler product and an analytic continuation to the
More informationTHEOREMS, ETC., FOR MATH 515
THEOREMS, ETC., FOR MATH 515 Proposition 1 (=comment on page 17). If A is an algebra, then any finite union or finite intersection of sets in A is also in A. Proposition 2 (=Proposition 1.1). For every
More informationProblem set 5, Real Analysis I, Spring, otherwise. (a) Verify that f is integrable. Solution: Compute since f is even, 1 x (log 1/ x ) 2 dx 1
Problem set 5, Real Analysis I, Spring, 25. (5) Consider the function on R defined by f(x) { x (log / x ) 2 if x /2, otherwise. (a) Verify that f is integrable. Solution: Compute since f is even, R f /2
More informationG: Uniform Convergence of Fourier Series
G: Uniform Convergence of Fourier Series From previous work on the prototypical problem (and other problems) u t = Du xx 0 < x < l, t > 0 u(0, t) = 0 = u(l, t) t > 0 u(x, 0) = f(x) 0 < x < l () we developed
More informationζ (s) = s 1 s {u} [u] ζ (s) = s 0 u 1+sdu, {u} Note how the integral runs from 0 and not 1.
Problem Sheet 3. From Theorem 3. we have ζ (s) = + s s {u} u+sdu, (45) valid for Res > 0. i) Deduce that for Res >. [u] ζ (s) = s u +sdu ote the integral contains [u] in place of {u}. ii) Deduce that for
More informationChapter 3a Topics in differentiation. Problems in differentiation. Problems in differentiation. LC Abueg: mathematical economics
Chapter 3a Topics in differentiation Lectures in Mathematical Economics L Cagandahan Abueg De La Salle University School of Economics Problems in differentiation Problems in differentiation Problem 1.
More informationMORE CONSEQUENCES OF CAUCHY S THEOREM
MOE CONSEQUENCES OF CAUCHY S THEOEM Contents. The Mean Value Property and the Maximum-Modulus Principle 2. Morera s Theorem and some applications 3 3. The Schwarz eflection Principle 6 We have stated Cauchy
More informationx y More precisely, this equation means that given any ε > 0, there exists some δ > 0 such that
Chapter 2 Limits and continuity 21 The definition of a it Definition 21 (ε-δ definition) Let f be a function and y R a fixed number Take x to be a point which approaches y without being equal to y If there
More informationSection 11.1 Sequences
Math 152 c Lynch 1 of 8 Section 11.1 Sequences A sequence is a list of numbers written in a definite order: a 1, a 2, a 3,..., a n,... Notation. The sequence {a 1, a 2, a 3,...} can also be written {a
More informationReview for the Final Exam
Math 171 Review for the Final Exam 1 Find the limits (4 points each) (a) lim 4x 2 3; x x (b) lim ( x 2 x x 1 )x ; (c) lim( 1 1 ); x 1 ln x x 1 sin (x 2) (d) lim x 2 x 2 4 Solutions (a) The limit lim 4x
More informationMathematics 104 Fall Term 2006 Solutions to Final Exam. sin(ln t) dt = e x sin(x) dx.
Mathematics 14 Fall Term 26 Solutions to Final Exam 1. Evaluate sin(ln t) dt. Solution. We first make the substitution t = e x, for which dt = e x. This gives sin(ln t) dt = e x sin(x). To evaluate the
More informationCALCULUS JIA-MING (FRANK) LIOU
CALCULUS JIA-MING (FRANK) LIOU Abstract. Contents. Power Series.. Polynomials and Formal Power Series.2. Radius of Convergence 2.3. Derivative and Antiderivative of Power Series 4.4. Power Series Expansion
More informationh(x) lim H(x) = lim Since h is nondecreasing then h(x) 0 for all x, and if h is discontinuous at a point x then H(x) > 0. Denote
Real Variables, Fall 4 Problem set 4 Solution suggestions Exercise. Let f be of bounded variation on [a, b]. Show that for each c (a, b), lim x c f(x) and lim x c f(x) exist. Prove that a monotone function
More informationSJÄLVSTÄNDIGA ARBETEN I MATEMATIK
SJÄLVSTÄNDIGA ARBETEN I MATEMATIK MATEMATISKA INSTITUTIONEN, STOCKHOLMS UNIVERSITET Gamma function related to Pic functions av Saad Abed 25 - No 4 MATEMATISKA INSTITUTIONEN, STOCKHOLMS UNIVERSITET, 6 9
More informationThis Week. Professor Christopher Hoffman Math 124
This Week Sections 2.1-2.3,2.5,2.6 First homework due Tuesday night at 11:30 p.m. Average and instantaneous velocity worksheet Tuesday available at http://www.math.washington.edu/ m124/ (under week 2)
More informationMath Camp II. Calculus. Yiqing Xu. August 27, 2014 MIT
Math Camp II Calculus Yiqing Xu MIT August 27, 2014 1 Sequence and Limit 2 Derivatives 3 OLS Asymptotics 4 Integrals Sequence Definition A sequence {y n } = {y 1, y 2, y 3,..., y n } is an ordered set
More informationg 2 (x) (1/3)M 1 = (1/3)(2/3)M.
COMPACTNESS If C R n is closed and bounded, then by B-W it is sequentially compact: any sequence of points in C has a subsequence converging to a point in C Conversely, any sequentially compact C R n is
More information8.7 Taylor s Inequality Math 2300 Section 005 Calculus II. f(x) = ln(1 + x) f(0) = 0
8.7 Taylor s Inequality Math 00 Section 005 Calculus II Name: ANSWER KEY Taylor s Inequality: If f (n+) is continuous and f (n+) < M between the center a and some point x, then f(x) T n (x) M x a n+ (n
More information1. Let A R be a nonempty set that is bounded from above, and let a be the least upper bound of A. Show that there exists a sequence {a n } n N
Applied Analysis prelim July 15, 216, with solutions Solve 4 of the problems 1-5 and 2 of the problems 6-8. We will only grade the first 4 problems attempted from1-5 and the first 2 attempted from problems
More informationLecture 9: Taylor Series
Math 8 Instructor: Padraic Bartlett Lecture 9: Taylor Series Week 9 Caltech 212 1 Taylor Polynomials and Series When we first introduced the idea of the derivative, one of the motivations we offered was
More informationFINAL REVIEW FOR MATH The limit. a n. This definition is useful is when evaluating the limits; for instance, to show
FINAL REVIEW FOR MATH 500 SHUANGLIN SHAO. The it Define a n = A: For any ε > 0, there exists N N such that for any n N, a n A < ε. This definition is useful is when evaluating the its; for instance, to
More informationPrinciple of Mathematical Induction
Advanced Calculus I. Math 451, Fall 2016, Prof. Vershynin Principle of Mathematical Induction 1. Prove that 1 + 2 + + n = 1 n(n + 1) for all n N. 2 2. Prove that 1 2 + 2 2 + + n 2 = 1 n(n + 1)(2n + 1)
More informationThe Prime Number Theorem
The Prime Number Theorem We study the distribution of primes via the function π(x) = the number of primes x 6 5 4 3 2 2 3 4 5 6 7 8 9 0 2 3 4 5 2 It s easier to draw this way: π(x) = the number of primes
More informationREVIEW OF DIFFERENTIAL CALCULUS
REVIEW OF DIFFERENTIAL CALCULUS DONU ARAPURA 1. Limits and continuity To simplify the statements, we will often stick to two variables, but everything holds with any number of variables. Let f(x, y) be
More informationMath 259: Introduction to Analytic Number Theory More about the Gamma function
Math 59: Introduction to Analytic Number Theory More about the Gamma function We collect some more facts about Γs as a function of a complex variable that will figure in our treatment of ζs and Ls, χ.
More informationMath 1102: Calculus I (Math/Sci majors) MWF 3pm, Fulton Hall 230 Homework 4 Solutions
Math 0: Calculus I (Math/Sci majors) MWF 3pm, Fulton Hall 30 Homework 4 Solutions Please write neatly, and show all work. Caution: An answer with no work is wrong! Problem A. Use Weierstrass (ɛ,δ)-definition
More informationPart II. Number Theory. Year
Part II Year 2017 2016 2015 2014 2013 2012 2011 2010 2009 2008 2007 2006 2005 2017 Paper 3, Section I 1G 70 Explain what is meant by an Euler pseudoprime and a strong pseudoprime. Show that 65 is an Euler
More informationIntroduction to Series and Sequences Math 121 Calculus II Spring 2015
Introduction to Series and Sequences Math Calculus II Spring 05 The goal. The main purpose of our study of series and sequences is to understand power series. A power series is like a polynomial of infinite
More informationSequences and Series of Functions
Chapter 13 Sequences and Series of Functions These notes are based on the notes A Teacher s Guide to Calculus by Dr. Louis Talman. The treatment of power series that we find in most of today s elementary
More informationThe Riemann Zeta Function
The Riemann Zeta Function David Jekel June 6, 23 In 859, Bernhard Riemann published an eight-page paper, in which he estimated the number of prime numbers less than a given magnitude using a certain meromorphic
More informationarxiv: v22 [math.gm] 20 Sep 2018
O RIEMA HYPOTHESIS arxiv:96.464v [math.gm] Sep 8 RUIMIG ZHAG Abstract. In this work we present a proof to the celebrated Riemann hypothesis. In it we first apply the Plancherel theorem properties of the
More informationAnalysis Qualifying Exam
Analysis Qualifying Exam Spring 2017 Problem 1: Let f be differentiable on R. Suppose that there exists M > 0 such that f(k) M for each integer k, and f (x) M for all x R. Show that f is bounded, i.e.,
More informationX. Numerical Methods
X. Numerical Methods. Taylor Approximation Suppose that f is a function defined in a neighborhood of a point c, and suppose that f has derivatives of all orders near c. In section 5 of chapter 9 we introduced
More informationBernoulli Numbers and their Applications
Bernoulli Numbers and their Applications James B Silva Abstract The Bernoulli numbers are a set of numbers that were discovered by Jacob Bernoulli (654-75). This set of numbers holds a deep relationship
More informationMathematics 3 Differential Calculus
Differential Calculus 3-1a A derivative function defines the slope described by the original function. Example 1 (FEIM): Given: y(x) = 3x 3 2x 2 + 7. What is the slope of the function y(x) at x = 4? y
More informationMATH 131A: REAL ANALYSIS (BIG IDEAS)
MATH 131A: REAL ANALYSIS (BIG IDEAS) Theorem 1 (The Triangle Inequality). For all x, y R we have x + y x + y. Proposition 2 (The Archimedean property). For each x R there exists an n N such that n > x.
More informationMcGill University Math 354: Honors Analysis 3
Practice problems McGill University Math 354: Honors Analysis 3 not for credit Problem 1. Determine whether the family of F = {f n } functions f n (x) = x n is uniformly equicontinuous. 1st Solution: The
More informationPRIME NUMBER THEOREM
PRIME NUMBER THEOREM RYAN LIU Abstract. Prime numbers have always been seen as the building blocks of all integers, but their behavior and distribution are often puzzling. The prime number theorem gives
More informationPrime Number Theory and the Riemann Zeta-Function
5262589 - Recent Perspectives in Random Matrix Theory and Number Theory Prime Number Theory and the Riemann Zeta-Function D.R. Heath-Brown Primes An integer p N is said to be prime if p and there is no
More informationMATH 409 Advanced Calculus I Lecture 12: Uniform continuity. Exponential functions.
MATH 409 Advanced Calculus I Lecture 12: Uniform continuity. Exponential functions. Uniform continuity Definition. A function f : E R defined on a set E R is called uniformly continuous on E if for every
More information