The WZ method and zeta function identities
|
|
- Edward Osborne
- 6 years ago
- Views:
Transcription
1 The WZ method and zeta function identities Ira M. Gessel Department of Mathematics Brandeis University American Mathematical Society 2010 Fall Eastern Sectional Meeting Syracuse University October 3, 2010
2 There are several interesting formulas that give quickly converging series for generating functions for the zeta function, due to Almkvist, Borwein, Bradley, Granville, Koecher, Leshchiner, and Rivoal.
3 There are several interesting formulas that give quickly converging series for generating functions for the zeta function, due to Almkvist, Borwein, Bradley, Granville, Koecher, Leshchiner, and Rivoal. For example, Koecher s formula is ζ(2k+3)x 2k = k=0 ( 1) n+1 n 3( ) 2n n n=1 ( x 2 /n 2 ) n 1 k=1 (1 x 2 k 2 ).
4 Khodabakhsh and Tatiana Hessami Pilehrood used the WZ method of Wilf and Zeilberger to prove some of these formulas.
5 Khodabakhsh and Tatiana Hessami Pilehrood used the WZ method of Wilf and Zeilberger to prove some of these formulas. I will give a brief explanation of the WZ method and I will explain how the WZ proofs of these zeta function identities are closely related to classical hypergeometric series identities, and how this connection allows us to generalize them.
6 The WZ method We consider a grid graph with a weight function defined on every directed edge with the following property: For all vertices A and B, all paths from A to B have the same weight. (Path-invariance)
7 This property is equivalent to the existence of a potential function defined on the vertices: the weight of an edge is the difference in the potential of its endpoints:
8 Some observations To check that a weighted grid graph has the path invariance property, it is sufficient to check it on each rectangle: b a d a + b = c + d c
9 Some observations The weights of a directed edge and its reversal are negatives of each other.
10 Some observations We can add arbitrary other edges to the graph, and they will have uniquely determined weights that satisfy the pathinvariance property
11 In particular, from one path-invariant weighted grid graph, we can get many more by taking a different grid on the same set of points:
12 In particular, from one path-invariant weighted grid graph, we can get many more by taking a different grid on the same set of points:
13 In particular, from one path-invariant weighted grid graph, we can get many more by taking a different grid on the same set of points:
14 In particular, from one path-invariant weighted grid graph, we can get many more by taking a different grid on the same set of points:
15 The WZ method Suppose that we have a a path-invariant weighted grid graph, where the vertices are in Z Z. Let f (i, j) be the weight on the edge from (i, j) to (i + 1, j) and let g(i, j) be the weight on the edge from (i, j) to (i, j + 1): f(i, j + 1) g(i, j) g(i + 1, j) (i, j) f(i, j)
16 The WZ method Suppose that we have a a path-invariant weighted grid graph, where the vertices are in Z Z. Let f (i, j) be the weight on the edge from (i, j) to (i + 1, j) and let g(i, j) be the weight on the edge from (i, j) to (i, j + 1): f(i, j + 1) g(i, j) g(i + 1, j) (i, j) f(i, j) The path invariance property is f (i, j) + g(i + 1, j) = g(i, j) + f (i, j + 1)
17 The WZ method Suppose that we have a a path-invariant weighted grid graph, where the vertices are in Z Z. Let f (i, j) be the weight on the edge from (i, j) to (i + 1, j) and let g(i, j) be the weight on the edge from (i, j) to (i, j + 1): f(i, j + 1) g(i, j) g(i + 1, j) (i, j) f(i, j) The path invariance property is f (i, j) + g(i + 1, j) = g(i, j) + f (i, j + 1) Note: Traditionally the order of the parameters is switched.
18 A pair of functions (f, g) satisfying this identity is called a WZ-pair. For our purposes, we want to assume that f and g have a particular form: we want each to be of the form Γ(a 1 i + b 1 j + u 1 ) Γ(a k i + b k j + u k ) Γ(c 1 i + d 1 j + v 1 ) Γ(c l i + d l j + v l ) zi w j where the a n, b n, c n, and d n are integers, and the u n, v n, z, and w are complex numbers.
19 A pair of functions (f, g) satisfying this identity is called a WZ-pair. For our purposes, we want to assume that f and g have a particular form: we want each to be of the form Γ(a 1 i + b 1 j + u 1 ) Γ(a k i + b k j + u k ) Γ(c 1 i + d 1 j + v 1 ) Γ(c l i + d l j + v l ) zi w j where the a n, b n, c n, and d n are integers, and the u n, v n, z, and w are complex numbers. This implies that f (i + 1, j)/f (i, j), f (i, j + 1)/f (i, j), and f (i, j)/g(i, j) are rational functions.
20 A pair of functions (f, g) satisfying this identity is called a WZ-pair. For our purposes, we want to assume that f and g have a particular form: we want each to be of the form Γ(a 1 i + b 1 j + u 1 ) Γ(a k i + b k j + u k ) Γ(c 1 i + d 1 j + v 1 ) Γ(c l i + d l j + v l ) zi w j where the a n, b n, c n, and d n are integers, and the u n, v n, z, and w are complex numbers. This implies that f (i + 1, j)/f (i, j), f (i, j + 1)/f (i, j), and f (i, j)/g(i, j) are rational functions. Note: We will want to allow i and j to be complex numbers.
21 We illustrate our approach with the simplest interesting WZ pair, which is associated with the binomial theorem. One form of this WZ pair is ( Γ(i + j) i + j 1 f (i, j) = Γ(i + 1)Γ(j) x i (1 x) j = i ( Γ(i + j) i + j 1 g(i, j) = Γ(i)Γ(j + 1) x i (1 x) j = i 1 ) x i (1 x) j ) x i (1 x) j Let s first see the connection between this WZ pair and the binomial theorem.
22 We sum along the two paths P 1 and P 2 : (0, n) (m, n) P 2 P 1 (0, 0) (m, 0) Summing along P 1 gives m 1 i=0 n 1 n 1 ( m + j 1 f (i, 0) + g(m, j) = 1 i 1 j=0 and summing along P 2 gives n 1 m 1 g(0, j) + j=0 i=0 f (i, n) = 0 + j=0 m 1 i=0 ( i + n 1 i ) x m (1 x) j. ) x i (1 x) n.
23 So we have the identity n 1 ( m + j 1 1 i 1 j=0 ) x m (1 x) j = m 1 i=0 ( i + n 1 i ) x i (1 x) n. Taking the limit as m we get the binomial theorem in the form ( ) i + n 1 1 = x i (1 x) n. i i=0
24 So we have the identity n 1 ( m + j 1 1 i 1 j=0 ) x m (1 x) j = m 1 i=0 ( i + n 1 i ) x i (1 x) n. Taking the limit as m we get the binomial theorem in the form ( ) i + n 1 1 = x i (1 x) n. i i=0 Conversely, starting with the binomial theorem in this form, it s easy to find f and g, using Gosper s algorithm.
25 We can use this WZ pair to obtain identities for the sum L(x, α) = i=0 α α + i x i, which can be expressed in terms of the Lerch transcendent as α Φ(x, 1, α).
26 We can use this WZ pair to obtain identities for the sum L(x, α) = i=0 α α + i x i, which can be expressed in terms of the Lerch transcendent as α Φ(x, 1, α). Note that and L(x, 1) = x 1 log(1 x) L(x, 1/2) = x 1/2 tanh 1 x.
27 The sum L(x, α) is a generating function for the polygorithm Li s (x) = i=1 x i i s which reduces to the zeta function for x = 1.
28 The sum L(x, α) is a generating function for the polygorithm Li s (x) = i=1 x i i s which reduces to the zeta function for x = 1. We have L(x, α) = i=0 α α + i x i = 1 + α = 1 + = x i=1 m=0 1 i x i 1 + α/i ( 1) m α m+1 x i i m+1 i=1 ( 1) m 1 α m Li m (x). m=1
29 Let s look at f (i, j) = Γ(i + j) Γ(i + 1)Γ(j) x i (1 x) j. We want to modify f and g so that something like i=0 αx i /(α + i). i=0 f (i, 0) becomes
30 Let s look at f (i, j) = Γ(i + j) Γ(i + 1)Γ(j) x i (1 x) j. We want to modify f and g so that i=0 f (i, 0) becomes something like i=0 αx i /(α + i). If we replace i with α + i in f (i, j), then we get f (α + i, j) = Γ(α + i + j) Γ(α + i + 1) x i x α (1 x) j Γ(j)
31 Let s look at f (i, j) = Γ(i + j) Γ(i + 1)Γ(j) x i (1 x) j. We want to modify f and g so that i=0 f (i, 0) becomes something like i=0 αx i /(α + i). If we replace i with α + i in f (i, j), then we get f (α + i, j) = Now if we set j = 0, we get f (α + i, 0) = Γ(α + i + j) Γ(α + i + 1) x i x α (1 x) j Γ(j) α α + i x i x α α Γ(0) which is just what we want except for a constant factor
32 Let s look at f (i, j) = Γ(i + j) Γ(i + 1)Γ(j) x i (1 x) j. We want to modify f and g so that i=0 f (i, 0) becomes something like i=0 αx i /(α + i). If we replace i with α + i in f (i, j), then we get f (α + i, j) = Now if we set j = 0, we get f (α + i, 0) = Γ(α + i + j) Γ(α + i + 1) x i x α (1 x) j Γ(j) α α + i x i x α α Γ(0) which is just what we want except for a constant factor which is unfortunately 0.
33 Clearly we can multiply a WZ pair by a constant.
34 Clearly we can multiply a WZ pair by a constant. Can we replace i with α + i? What about the Γ(0)?
35 Clearly we can multiply a WZ pair by a constant. Can we replace i with α + i? What about the Γ(0)? The first question is easy: Lemma. If ( f (i, j), g(i, j) ) is a WZ pair then so is ( f (α + i, β + j), g(α + i, β + j) ) for any α and β.
36 To deal with the Γ(0) we need to do a little more. We observe that ( f (i, j), g(i, j) ) is a WZ pair, then so is ( p(j)f (i, j), p(j)g(i, j) ), where p(j) is a function that is periodic with period 1. We take p(j) to be ( 1) j Γ(j)Γ(1 j) = ( 1) j π sin πj.
37 To deal with the Γ(0) we need to do a little more. We observe that ( f (i, j), g(i, j) ) is a WZ pair, then so is ( p(j)f (i, j), p(j)g(i, j) ), where p(j) is a function that is periodic with period 1. We take p(j) to be ( 1) j Γ(j)Γ(1 j) = ( 1) j π sin πj. This is what Wilf and Zeilberger call shadowing.
38 So (after multiplying f and g by αx α ) we get a modified WZ pair Γ(α + i + j)γ(1 j) f (i, j) = α x i (x 1) j Γ(α + i + 1) Γ(α + i + j)γ( j) g(i, j) = α x i (x 1) j Γ(α + i) and we have f (i, 0) = i=0 i=0 α α + i x i.
39 To transform this sum, we consider three paths: (0, 0) (m, 0) The sum along the black path is equal to the sum along the red path plus the sum along the green path. We take the limit as m and we hope that the green path goes to 0 and the red path gives us something interesting.
40 More precisely, we pick positive integers s and t and define h(k) to be the weight of a path from (sk, tk) to (s(k + 1), t(k + 1)). (0, 0) h(0) (s, t) h(1) (2s, 2t) h(2) (3s, 3t)
41 More precisely, we pick positive integers s and t and define h(k) to be the weight of a path from (sk, tk) to (s(k + 1), t(k + 1)). (0, 0) h(0) (s, t) h(1) (2s, 2t) h(2) (3s, 3t) It s not hard to see that as m the sum along the green path goes to 0, so we have the identity i=0 α α + i x i = h(k) k=0
42 What is h(k)? It s the weight of any path from (sk, tk) to (s(k + 1), t(k + 1)). So if we take the path that goes down first and then to the right (sk, tk) (sk, t(k + 1)) (s(k + 1), t(k + 1)) we get h(k) = tk 1 j= t(k+1) s(k+1) 1 g(sk, j) + f (i, t(k + 1)) i=sk
43 It s not hard to see that h(k) will be a rational function of α, x, and k times ( (tk)! (α) (s t)k ( 1) t x s ) k (α + 1) sk (1 x) t, where (u) m is the rising factorial, u(u + 1) (u + m 1) if m 0 (u) m = ( 1) (1 u) m if m < 0.
44 The simplest cases are i=0 α α + i x i = = = k=0 k! ( x) k (α + 1) k (1 x) k+1 (s = 1, t = 1) (1 + k α + xk + xα) k=0 (2k)! ( x) k (α + 1) k (1 α) k+1 (1 x) 2k+2 (s = 1, t = 2) (1 + 2k + α x xk) k=0 k! (α) k (α + 1) 2k+1 ( x 2 ) k (1 x) k+1 (s = 2, t = 1)
45 The simplest cases are i=0 α α + i x i = = = k=0 k! ( x) k (α + 1) k (1 x) k+1 (s = 1, t = 1) (1 + k α + xk + xα) k=0 (2k)! ( x) k (α + 1) k (1 α) k+1 (1 x) 2k+2 (s = 1, t = 2) (1 + 2k + α x xk) k=0 k! (α) k (α + 1) 2k+1 ( x 2 ) k (1 x) k+1 (s = 2, t = 1) Note: The first formula is a 2 F 1 linear transformation.
46 For α = 1 the first formula gives log(1 x) = k=1 ( ) 1 x k ( = log 1 + x ) k 1 x 1 x
47 For α = 1 the first formula gives log(1 x) = k=1 ( ) 1 x k ( = log 1 + x ) k 1 x 1 x In the second formula, we can t directly set α = 1 because of the 1 α in the denominator. But by taking an appropriate limit, we can get log(1 x) = ( k 2 x + k 2 (2k 2)! 3k + 1) k! 2 k=1 ( ) x k (1 x) 2.
48 The third formula gives log(1 x) = (x 2) k=1 (k 1)! k! (2k)! x 2k 1 ( 1) k (1 x) k.
49 The third formula gives log(1 x) = (x 2) k=1 (k 1)! k! (2k)! This is equivalent to the known formula sin 1 x = x k! 2 1 x 2 (2k + 1)! (2x)2k. k=0 x 2k 1 ( 1) k (1 x) k.
50 For α = 1/2 the first formula gives tanh 1 x = 2x 1 x which is also equivalent to k=0 k! (k + 1)! (2k + 2)! ( 4x 2 1 x 2 sin 1 x = x k! 2 1 x 2 (2k + 1)! (2x)2k. k=0 ) k,
51 For α = 1/2 the first formula gives tanh 1 x = 2x 1 x which is also equivalent to k=0 k! (k + 1)! (2k + 2)! ( 4x 2 1 x 2 sin 1 x = x k! 2 1 x 2 (2k + 1)! (2x)2k. k=0 ) k, The second formula gives another formula equivalent to this.
52 The third formula gives tanh 1 x = 4x 1 x 2 (3 + 4k 2(1 + k)x 2 ) k=0 (2k)! (2k + 2)! (4k + 4)! ( 4x 4 1 x 2 ) k.
53 The simplest interesting polylogarithm formula is the dilogarithm formula obtained from the second transformation formula Li 2 (x) = + i=1 x i i 2 = x 1 x ( 1) k( (1 + 2k (k + 1)x)(H 2k+1 H k 1 ) 1 ) k=1 (k 1)! k! (2k + 1)! where H n is the harmonic number 1 + 1/ /n. x 2k (1 x) k+1
54 The simplest interesting polylogarithm formula is the dilogarithm formula obtained from the second transformation formula Li 2 (x) = + i=1 x i i 2 = x 1 x ( 1) k( (1 + 2k (k + 1)x)(H 2k+1 H k 1 ) 1 ) k=1 (k 1)! k! (2k + 1)! where H n is the harmonic number 1 + 1/ /n. x 2k (1 x) k+1 We can t set x = 1, but we can set x = 1, to get a formula for ζ(2).
55 Li 2 ( 1) = 1 π2 2ζ(2) = 12 = ( 1) k( (2 + 3k)(H 2k+1 H k 1 ) 1 ) k=1 (k 1)! k! (2k + 1)! 1 2 k+1
56 What do we do with a more complicated identity? Let s look at the Bailey-Borwein-Bradley formula ζ(2n + 1)x 2n = n=0 k=1 1 k 2 x 2
57 What do we do with a more complicated identity? Let s look at the Bailey-Borwein-Bradley formula ζ(2n + 1)x 2n = n=0 = 3 k=1 ( 2k k k=1 1 ) (k 2 x 2 ) 1 k 2 x 2 k 1 m=1 4x 2 m 2 x 2 m 2
58 What do we do with a more complicated identity? Let s look at the Bailey-Borwein-Bradley formula ζ(2n + 1)x 2n = n=0 = 3 ( 2k k=1 k = 3/2 1 x 2 k=1 1 k 2 x 2 k 1 1 4x 2 m 2 ) (k 2 x 2 ) x 2 m 2 m=1 (2) k (1 2x) k (1 + 2x) k k=0 ( 3 2 ) k(2 x) k (2 + x) k ( ) 1 k. 4
59 What do we do with a more complicated identity? Let s look at the Bailey-Borwein-Bradley formula ζ(2n + 1)x 2n = n=0 = 3 ( 2k k=1 k = 3/2 1 x 2 k=1 1 k 2 x 2 k 1 1 4x 2 m 2 ) (k 2 x 2 ) x 2 m 2 m=1 (2) k (1 2x) k (1 + 2x) k k=0 ( 3 2 ) k(2 x) k (2 + x) k ( ) 1 k. 4 This identity is associated with a WZ pair that corresponds to Gauss s theorem (the nonterminating form of Vandermonde s theorem).
60 Gauss s theorem is k=0 (a) k (b) k Γ(c)Γ(c a b) = k! (c) k Γ(c a)γ(c b). We want to transform the summand of Gauss s theorem into a constant times 1 (k + 1) 2 x 2 = (1 + x) k (1 x) k (1 x 2 )(2 + x) k (2 x) k. If we replace k by 1 + x + k in (a) k (b) k /k! (c) k and then set a = 0, b = 2x, and c = 1 2x, we get 2x Γ(0)(1 x 2 ) (1 + x) k(1 x) k (2 + x) k (2 x) k
61 Gauss s theorem is k=0 (a) k (b) k Γ(c)Γ(c a b) = k! (c) k Γ(c a)γ(c b). We want to transform the summand of Gauss s theorem into a constant times 1 (k + 1) 2 x 2 = (1 + x) k (1 x) k (1 x 2 )(2 + x) k (2 x) k. If we replace k by 1 + x + k in (a) k (b) k /k! (c) k and then set a = 0, b = 2x, and c = 1 2x, we get 2x Γ(0)(1 x 2 ) (1 + x) k(1 x) k (2 + x) k (2 x) k By shadowing as before we can get rid of the Γ(0) in the denominator.
62 But in trying to find a WZ pair we have a problem: Gauss s theorem has four parameters (the summation index and three others) but to get a WZ pair all we need is the summation index and one parameter. How do we know which one to use?
63 But in trying to find a WZ pair we have a problem: Gauss s theorem has four parameters (the summation index and three others) but to get a WZ pair all we need is the summation index and one parameter. How do we know which one to use? We use all of them, and instead of a WZ pair we get a WZ 4-tuple (or WZ form). This gives a path-invariant grid graph in Z 4 rather than Z 2. Everything works as before, except that we have more choices.
64 The WZ 4-tuple f = ( f 1 (i, j, k, l),..., f 4 (i, j, k, l) ) corresponding to Gauss s theorem is ( G f = i, G j, G k, G ), l where k Γ(i + k) Γ(i + l) Γ(j + k) Γ(j + l) Γ(1 k) G = ( 1). Γ(i) Γ(j) Γ(l) Γ(i + j + k + l) What this means is that if we put a weight of f t (i, j, k, l) on the edge from (i, j, k, l) to the adjacent point (i, j, k, l) + e t, where e t = (0,..., 1 t,..., 0), then we get a path-invariant grid graph.
65 The WZ 4-tuple f = ( f 1 (i, j, k, l),..., f 4 (i, j, k, l) ) corresponding to Gauss s theorem is ( G f = i, G j, G k, G ), l where k Γ(i + k) Γ(i + l) Γ(j + k) Γ(j + l) Γ(1 k) G = ( 1). Γ(i) Γ(j) Γ(l) Γ(i + j + k + l) What this means is that if we put a weight of f t (i, j, k, l) on the edge from (i, j, k, l) to the adjacent point (i, j, k, l) + e t, where e t = (0,..., 1 t,..., 0), then we get a path-invariant grid graph. Note that i, j, k, and l don t need to be integers.
66 Now if we sum f in the direction (1, 0, 0, 0) (i.e., parallel to the x 1 -axis) starting at the point (1 + x, 1, 0, 2x) we get 2x k=0 1 (1 + k) 2 x 2 = 2x 1 k 2 x 2. k=1
67 Now if we sum f in the direction (1, 0, 0, 0) (i.e., parallel to the x 1 -axis) starting at the point (1 + x, 1, 0, 2x) we get 2x k=0 1 (1 + k) 2 x 2 = 2x 1 k 2 x 2. k=1 To transform the sum, we compute the value of the kth step starting at the same point and going in another direction. We look for something interesting and we have to check that the sum along the connecting path" goes to 0.
68 Now if we sum f in the direction (1, 0, 0, 0) (i.e., parallel to the x 1 -axis) starting at the point (1 + x, 1, 0, 2x) we get 2x k=0 1 (1 + k) 2 x 2 = 2x 1 k 2 x 2. k=1 To transform the sum, we compute the value of the kth step starting at the same point and going in another direction. We look for something interesting and we have to check that the sum along the connecting path" goes to 0. We find that the direction (1, 2, 1, 1) gives us what we want: the sum starting at (1 + x, 1, 0, 2x) gives 3x 1 x 2 (2) n (1 2x) n (1 + 2x) n n=0 ( 3 2 ) n(2 x) n (2 + x) n ( ) 1 n. 4
69 An advantage of this approach is that it gives us something more; if we start at the point (x, 1, 0, y x) we get the more general formula n=0 1 (n + x)(n + y) = 1 2xy (1 + x + y + 3n) n=0 n! (1 + x y) n(1 + y x) n ( 3 2 ) n(1 + x) n (1 + y) n ( ) 1 n. 4
Applications of Computer Algebra to the Theory of Hypergeometric Series
Applications of Computer Algebra to the Theory of Hypergeometric Series A Dissertation Presented to The Faculty of the Graduate School of Arts and Sciences Brandeis University Department of Mathematics
More informationExperimental Determination of Apéry-like Identities for ζ(2n +2)
Experimental Determination of Apéry-lie Identities for ζ(n + David H. Bailey, Jonathan M. Borwein, and David M. Bradley CONTENTS. Introduction. Discovering Theorem. 3. Proof of Theorem.. An Identity for
More informationHankel determinants, continued fractions, orthgonal polynomials, and hypergeometric series
Hankel determinants, continued fractions, orthgonal polynomials, and hypergeometric series Ira M. Gessel with Jiang Zeng and Guoce Xin LaBRI June 8, 2007 Continued fractions and Hankel determinants There
More informationarxiv: v1 [math.ca] 7 Mar 2013
A SIMPLE PROOF OF ANDREWS S 5 F 4 EVALUATION IRA M. GESSEL arxiv:1303.1757v1 [math.ca] 7 Mar 2013 Department of Mathematics Brandeis University Waltham, MA 02453 gessel@brandeis.edu Abstract. We give a
More informationSEARCHING FOR STRANGE HYPERGEOMETRIC IDENTITIES BY SHEER BRUTE FORCE. Moa APAGODU. Doron ZEILBERGER 1
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (2008), #A36 1 SEARCHING FOR STRANGE HYPERGEOMETRIC IDENTITIES BY SHEER BRUTE FORCE Moa APAGODU Department of Mathematics, Virginia Commonwealth
More informationQuadratic Transformations of Hypergeometric Function and Series with Harmonic Numbers
Quadratic Transformations of Hypergeometric Function and Series with Harmonic Numbers Martin Nicholson In this brief note, we show how to apply Kummer s and other quadratic transformation formulas for
More informationExperimental mathematics and integration
Experimental mathematics and integration David H. Bailey http://www.davidhbailey.com Lawrence Berkeley National Laboratory (retired) Computer Science Department, University of California, Davis October
More informationNew Multiple Harmonic Sum Identities
New Multiple Harmonic Sum Identities Helmut Prodinger Department of Mathematics University of Stellenbosch 760 Stellenbosch, South Africa hproding@sun.ac.za Roberto Tauraso Dipartimento di Matematica Università
More informationSumming Series: Solutions
Sequences and Series Misha Lavrov Summing Series: Solutions Western PA ARML Practice December, 0 Switching the order of summation Prove useful identity 9 x x x x x x Riemann s zeta function ζ is defined
More informatione x = 1 + x + x2 2! + x3 If the function f(x) can be written as a power series on an interval I, then the power series is of the form
Taylor Series Given a function f(x), we would like to be able to find a power series that represents the function. For example, in the last section we noted that we can represent e x by the power series
More informationHypergeometric series and the Riemann zeta function
ACTA ARITHMETICA LXXXII.2 (997) Hypergeometric series and the Riemann zeta function by Wenchang Chu (Roma) For infinite series related to the Riemann zeta function, De Doelder [4] established numerous
More informationSums and Products. a i = a 1. i=1. a i = a i a n. n 1
Sums and Products -27-209 In this section, I ll review the notation for sums and products Addition and multiplication are binary operations: They operate on two numbers at a time If you want to add or
More informationMath Lecture 4 Limit Laws
Math 1060 Lecture 4 Limit Laws Outline Summary of last lecture Limit laws Motivation Limits of constants and the identity function Limits of sums and differences Limits of products Limits of polynomials
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 information2014 Summer Review for Students Entering Algebra 2. TI-84 Plus Graphing Calculator is required for this course.
1. Solving Linear Equations 2. Solving Linear Systems of Equations 3. Multiplying Polynomials and Solving Quadratics 4. Writing the Equation of a Line 5. Laws of Exponents and Scientific Notation 6. Solving
More informationAnalytic Aspects of the Riemann Zeta and Multiple Zeta Values
Analytic Aspects of the Riemann Zeta and Multiple Zeta Values Cezar Lupu Department of Mathematics University of Pittsburgh Pittsburgh, PA, USA PhD Thesis Overview, October 26th, 207, Pittsburgh, PA Outline
More informationA polynomial expression is the addition or subtraction of many algebraic terms with positive integer powers.
LEAVING CERT Honours Maths notes on Algebra. A polynomial expression is the addition or subtraction of many algebraic terms with positive integer powers. The degree is the highest power of x. 3x 2 + 2x
More informationMULTIPLE HARMONIC SUMS AND MULTIPLE HARMONIC STAR SUMS ARE (NEARLY) NEVER INTEGERS
#A0 INTEGERS 7 (207) MULTIPLE HARMONIC SUMS AND MULTIPLE HARMONIC STAR SUMS ARE (NEARLY) NEVER INTEGERS Khodabakhsh Hessami Pilehrood The Fields Institute for Research in Mathematical Sciences, Toronto,
More informationSolving Linear and Rational Inequalities Algebraically. Definition 22.1 Two inequalities are equivalent if they have the same solution set.
Inequalities Concepts: Equivalent Inequalities Solving Linear and Rational Inequalities Algebraically Approximating Solutions to Inequalities Graphically (Section 4.4).1 Equivalent Inequalities Definition.1
More informationInterpolation of Rational Functions on a Geometric Mesh
CONSTRUCTIVE THEORY OF FUNCTIONS, Varna 5 additional information (to be provided by the publisher) Interpolation of Rational Functions on a Geometric Mesh Dimitar K. Dimitrov We discuss the Newton-Gregory
More informationSect Least Common Denominator
4 Sect.3 - Least Common Denominator Concept #1 Writing Equivalent Rational Expressions Two fractions are equivalent if they are equal. In other words, they are equivalent if they both reduce to the same
More informationIntroduction to standard and non-standard Numerical Methods
Introduction to standard and non-standard Numerical Methods Dr. Mountaga LAM AMS : African Mathematic School 2018 May 23, 2018 One-step methods Runge-Kutta Methods Nonstandard Finite Difference Scheme
More informationIntegration. 2. The Area Problem
Integration Professor Richard Blecksmith richard@math.niu.edu Dept. of Mathematical Sciences Northern Illinois University http://math.niu.edu/ richard/math2. Two Fundamental Problems of Calculus First
More informationClosed-form Second Solution to the Confluent Hypergeometric Difference Equation in the Degenerate Case
International Journal of Difference Equations ISS 973-669, Volume 11, umber 2, pp. 23 214 (216) http://campus.mst.edu/ijde Closed-form Second Solution to the Confluent Hypergeometric Difference Equation
More informationCalculus with business applications, Lehigh U, Lecture 03 notes Summer
Calculus with business applications, Lehigh U, Lecture 03 notes Summer 01 1 Lines and quadratics 1. Polynomials, functions in which each term is a positive integer power of the independent variable include
More informationPLC Papers. Created For:
PLC Papers Created For: Algebra and proof 2 Grade 8 Objective: Use algebra to construct proofs Question 1 a) If n is a positive integer explain why the expression 2n + 1 is always an odd number. b) Use
More informationINTRODUCTION TO SIGMA NOTATION
INTRODUCTION TO SIGMA NOTATION The notation itself Sigma notation is a way of writing a sum of many terms, in a concise form A sum in sigma notation looks something like this: 3k The Σ sigma) indicates
More informationOn A Central Binomial Series Related to ζ(4).
On A Central Binomial Series Related to ζ. Vivek Kaushik November 8 arxiv:8.67v [math.ca] 5 Nov 8 Abstract n n n 7π 3 by We prove a classical binomial coefficient series identity n evaluating a double
More information2015 SUMMER MATH PACKET
Name: Date: 05 SUMMER MATH PACKET College Algebra Trig. - I understand that the purpose of the summer packet is for my child to review the topics they have already mastered in previous math classes and
More informationPade Approximations and the Transcendence
Pade Approximations and the Transcendence of π Ernie Croot March 9, 27 1 Introduction Lindemann proved the following theorem, which implies that π is transcendental: Theorem 1 Suppose that α 1,..., α k
More informationBeukers integrals and Apéry s recurrences
2 3 47 6 23 Journal of Integer Sequences, Vol. 8 (25), Article 5.. Beukers integrals and Apéry s recurrences Lalit Jain Faculty of Mathematics University of Waterloo Waterloo, Ontario N2L 3G CANADA lkjain@uwaterloo.ca
More informationODD MAGNE OGREID AND PER OSLAND
DESY 97-45 hep-th/9868 December 997 SUMMING ONE- AND TWO-DIMENSIONAL SERIES RELATED TO THE EULER SERIES arxiv:hep-th/9868v 6 Jan 998 ODD MAGNE OGREID AND PER OSLAND Abstract We present results for some
More information4 Holonomic Recurrence Equations
44 4 Holonomic Recurrence Equations The main algorithmic idea to find hypergeometric term representations of hypergeometric series goes bac to Celine Fasenmyer (who is often called Sister Celine: Her idea
More informationArea. A(2) = sin(0) π 2 + sin(π/2)π 2 = π For 3 subintervals we will find
Area In order to quantify the size of a -dimensional object, we use area. Since we measure area in square units, we can think of the area of an object as the number of such squares it fills up. Using this
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 informationA2 HW Imaginary Numbers
Name: A2 HW Imaginary Numbers Rewrite the following in terms of i and in simplest form: 1) 100 2) 289 3) 15 4) 4 81 5) 5 12 6) -8 72 Rewrite the following as a radical: 7) 12i 8) 20i Solve for x in simplest
More informationSeries Solutions of ODEs. Special Functions
C05.tex 6/4/0 3: 5 Page 65 Chap. 5 Series Solutions of ODEs. Special Functions We continue our studies of ODEs with Legendre s, Bessel s, and the hypergeometric equations. These ODEs have variable coefficients
More informationRamanujan-Type Series Related to Clausen s Product
Ramanujan-Type Series Related to Clausen s Product arxiv:.9v [math.nt] 7 Jan John M. Campbell York University Toronto, ON Canada maxwell8@yorku.ca Abstract Infinite series are evaluated through the manipulation
More information8.7 MacLaurin Polynomials
8.7 maclaurin polynomials 67 8.7 MacLaurin Polynomials In this chapter you have learned to find antiderivatives of a wide variety of elementary functions, but many more such functions fail to have an antiderivative
More informationTABLE OF CONTENTS. Introduction to Finish Line Indiana Math 10. UNIT 1: Number Sense, Expressions, and Computation. Real Numbers
TABLE OF CONTENTS Introduction to Finish Line Indiana Math 10 UNIT 1: Number Sense, Expressions, and Computation LESSON 1 8.NS.1, 8.NS.2, A1.RNE.1, A1.RNE.2 LESSON 2 8.NS.3, 8.NS.4 LESSON 3 A1.RNE.3 LESSON
More informationCh 4 Differentiation
Ch 1 Partial fractions Ch 6 Integration Ch 2 Coordinate geometry C4 Ch 5 Vectors Ch 3 The binomial expansion Ch 4 Differentiation Chapter 1 Partial fractions We can add (or take away) two fractions only
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 informationSection September 6, If n = 3, 4, 5,..., the polynomial is called a cubic, quartic, quintic, etc.
Section 2.1-2.2 September 6, 2017 1 Polynomials Definition. A polynomial is an expression of the form a n x n + a n 1 x n 1 + + a 1 x + a 0 where each a 0, a 1,, a n are real numbers, a n 0, and n is a
More informationMatrices, Row Reduction of Matrices
Matrices, Row Reduction of Matrices October 9, 014 1 Row Reduction and Echelon Forms In the previous section, we saw a procedure for solving systems of equations It is simple in that it consists of only
More informationMSM120 1M1 First year mathematics for civil engineers Revision notes 3
MSM0 M First year mathematics for civil engineers Revision notes Professor Robert. Wilson utumn 00 Functions Definition of a function: it is a rule which, given a value of the independent variable (often
More informationCh1 Algebra and functions. Ch 2 Sine and Cosine rule. Ch 10 Integration. Ch 9. Ch 3 Exponentials and Logarithms. Trigonometric.
Ch1 Algebra and functions Ch 10 Integration Ch 2 Sine and Cosine rule Ch 9 Trigonometric Identities Ch 3 Exponentials and Logarithms C2 Ch 8 Differentiation Ch 4 Coordinate geometry Ch 7 Trigonometric
More informationMath 651 Introduction to Numerical Analysis I Fall SOLUTIONS: Homework Set 1
ath 651 Introduction to Numerical Analysis I Fall 2010 SOLUTIONS: Homework Set 1 1. Consider the polynomial f(x) = x 2 x 2. (a) Find P 1 (x), P 2 (x) and P 3 (x) for f(x) about x 0 = 0. What is the relation
More informationA Combinatorial Interpretation of the Numbers 6 (2n)! /n! (n + 2)!
1 2 3 47 6 23 11 Journal of Integer Sequences, Vol. 8 (2005), Article 05.2.3 A Combinatorial Interpretation of the Numbers 6 (2n)! /n! (n + 2)! Ira M. Gessel 1 and Guoce Xin Department of Mathematics Brandeis
More informationAlgebra & Trig Review
Algebra & Trig Review 1 Algebra & Trig Review This review was originally written for my Calculus I class, but it should be accessible to anyone needing a review in some basic algebra and trig topics. The
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 informationConstruction Of Binomial Sums For π And Polylogarithmic Constants Inspired By BBP Formulas
Applied Mathematics E-Notes, 77, 7-46 c ISSN 67-5 Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/ Construction Of Binomial Sums For π And Polylogarithmic Constants Inspired By BBP
More informationMOMENTS OF PRODUCTS OF ELLIPTIC INTEGRALS
MOMENTS OF PRODUCTS OF ELLIPTIC INTEGRALS JAMES G. WAN Abstract. We consider the moments of products of complete elliptic integrals of the first and second kinds. In particular, we derive new results using
More informationSolving Polynomial and Rational Inequalities Algebraically. Approximating Solutions to Inequalities Graphically
10 Inequalities Concepts: Equivalent Inequalities Solving Polynomial and Rational Inequalities Algebraically Approximating Solutions to Inequalities Graphically (Section 4.6) 10.1 Equivalent Inequalities
More informationOn the de Branges and Weinstein Functions
On the de Branges and Weinstein Functions Prof. Dr. Wolfram Koepf University of Kassel koepf@mathematik.uni-kassel.de http://www.mathematik.uni-kassel.de/~koepf Tag der Funktionentheorie 5. Juni 2004 Universität
More informationChapter 1 Review of Equations and Inequalities
Chapter 1 Review of Equations and Inequalities Part I Review of Basic Equations Recall that an equation is an expression with an equal sign in the middle. Also recall that, if a question asks you to solve
More information56 CHAPTER 3. POLYNOMIAL FUNCTIONS
56 CHAPTER 3. POLYNOMIAL FUNCTIONS Chapter 4 Rational functions and inequalities 4.1 Rational functions Textbook section 4.7 4.1.1 Basic rational functions and asymptotes As a first step towards understanding
More informationHankel Operators plus Orthogonal Polynomials. Yield Combinatorial Identities
Hanel Operators plus Orthogonal Polynomials Yield Combinatorial Identities E. A. Herman, Grinnell College Abstract: A Hanel operator H [h i+j ] can be factored as H MM, where M maps a space of L functions
More informationNOTES. [Type the document subtitle] Math 0310
NOTES [Type the document subtitle] Math 010 Cartesian Coordinate System We use a rectangular coordinate system to help us map out relations. The coordinate grid has a horizontal axis and a vertical axis.
More informationNanjing Univ. J. Math. Biquarterly 32(2015), no. 2, NEW SERIES FOR SOME SPECIAL VALUES OF L-FUNCTIONS
Naning Univ. J. Math. Biquarterly 2205, no. 2, 89 28. NEW SERIES FOR SOME SPECIAL VALUES OF L-FUNCTIONS ZHI-WEI SUN Astract. Dirichlet s L-functions are natural extensions of the Riemann zeta function.
More informationSolved problems: (Power) series 1. Sum up the series (if it converges) 3 k+1 a) 2 2k+5 ; b) 1. k(k + 1).
Power series: Solved problems c phabala 00 3 Solved problems: Power series. Sum up the series if it converges 3 + a +5 ; b +.. Investigate convergence of the series a e ; c ; b 3 +! ; d a, where a 3. Investigate
More informationNotes: Pythagorean Triples
Math 5330 Spring 2018 Notes: Pythagorean Triples Many people know that 3 2 + 4 2 = 5 2. Less commonly known are 5 2 + 12 2 = 13 2 and 7 2 + 24 2 = 25 2. Such a set of integers is called a Pythagorean Triple.
More informationTwo finite forms of Watson s quintuple product identity and matrix inversion
Two finite forms of Watson s uintuple product identity and matrix inversion X. Ma Department of Mathematics SuZhou University, SuZhou 215006, P.R.China Submitted: Jan 24, 2006; Accepted: May 27, 2006;
More informationSection Example Determine the Maclaurin series of f (x) = e x and its the interval of convergence.
Example Determine the Maclaurin series of f (x) = e x and its the interval of convergence. Example Determine the Maclaurin series of f (x) = e x and its the interval of convergence. f n (0)x n Recall from
More informationIrrationality of the Sums of Zeta Values
Mathematical Notes, vol. 79, no. 4, 2006, pp. 561 571. Translated from Matematicheskie Zametki, vol. 79, no. 4, 2006, pp. 607 618. Original Russian Text Copyright c 2006 by T. Hessami Pilehrood, Kh. Hessami
More informationA Symbolic Operator Approach to Several Summation Formulas for Power Series
A Symbolic Operator Approach to Several Summation Formulas for Power Series T. X. He, L. C. Hsu 2, P. J.-S. Shiue 3, and D. C. Torney 4 Department of Mathematics and Computer Science Illinois Wesleyan
More informationMath 115 Spring 11 Written Homework 10 Solutions
Math 5 Spring Written Homework 0 Solutions. For following its, state what indeterminate form the its are in and evaluate the its. (a) 3x 4x 4 x x 8 Solution: This is in indeterminate form 0. Algebraically,
More information2 = = 0 Thus, the number which is largest in magnitude is equal to the number which is smallest in magnitude.
Limits at Infinity Two additional topics of interest with its are its as x ± and its where f(x) ±. Before we can properly discuss the notion of infinite its, we will need to begin with a discussion on
More informationTHE GAMMA FUNCTION AND THE ZETA FUNCTION
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.
More informationCombinatorial Analysis of the Geometric Series
Combinatorial Analysis of the Geometric Series David P. Little April 7, 205 www.math.psu.edu/dlittle Analytic Convergence of a Series The series converges analytically if and only if the sequence of partial
More informationPolyGamma Functions of Negative Order
Carnegie Mellon University Research Showcase @ CMU Computer Science Department School of Computer Science -998 PolyGamma Functions of Negative Order Victor S. Adamchik Carnegie Mellon University Follow
More informationAutomatic Proofs of Identities: Beyond A=B
Bruno.Salvy@inria.fr FPSAC, Linz, July 20, 2009 Joint work with F. Chyzak and M. Kauers 1 / 28 I Introduction 2 / 28 Examples of Identities: Definite Sums, q-sums, Integrals n k=0 + 0 1 2πi n=0 ( ) H n
More informationMath 1B, lecture 15: Taylor Series
Math B, lecture 5: Taylor Series Nathan Pflueger October 0 Introduction Taylor s theorem shows, in many cases, that the error associated with a Taylor approximation will eventually approach 0 as the degree
More information2t t dt.. So the distance is (t2 +6) 3/2
Math 8, Solutions to Review for the Final Exam Question : The distance is 5 t t + dt To work that out, integrate by parts with u t +, so that t dt du The integral is t t + dt u du u 3/ (t +) 3/ So the
More informationAPPM/MATH 4/5520 Solutions to Problem Set Two. = 2 y = y 2. e 1 2 x2 1 = 1. (g 1
APPM/MATH 4/552 Solutions to Problem Set Two. Let Y X /X 2 and let Y 2 X 2. (We can select Y 2 to be anything but when dealing with a fraction for Y, it is usually convenient to set Y 2 to be the denominator.)
More informationMTH103 Section 065 Exam 2. x 2 + 6x + 7 = 2. x 2 + 6x + 5 = 0 (x + 1)(x + 5) = 0
Absolute Value 1. (10 points) Find all solutions to the following equation: x 2 + 6x + 7 = 2 Solution: You first split this into two equations: x 2 + 6x + 7 = 2 and x 2 + 6x + 7 = 2, and solve each separately.
More informationNever leave a NEGATIVE EXPONENT or a ZERO EXPONENT in an answer in simplest form!!!!!
1 ICM Unit 0 Algebra Rules Lesson 1 Rules of Exponents RULE EXAMPLE EXPLANANTION a m a n = a m+n A) x x 6 = B) x 4 y 8 x 3 yz = When multiplying with like bases, keep the base and add the exponents. a
More informationPhysics 6303 Lecture 22 November 7, There are numerous methods of calculating these residues, and I list them below. lim
Physics 6303 Lecture 22 November 7, 208 LAST TIME:, 2 2 2, There are numerous methods of calculating these residues, I list them below.. We may calculate the Laurent series pick out the coefficient. 2.
More informationApproximations - the method of least squares (1)
Approximations - the method of least squares () In many applications, we have to consider the following problem: Suppose that for some y, the equation Ax = y has no solutions It could be that this is an
More informationACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER /2018
ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER 1 2017/2018 DR. ANTHONY BROWN 1. Arithmetic and Algebra 1.1. Arithmetic of Numbers. While we have calculators and computers
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 informationPractice Calculus Test without Trig
Practice Calculus Test without Trig The problems here are similar to those on the practice test Slight changes have been made 1 What is the domain of the function f (x) = 3x 1? Express the answer in interval
More informationarxiv: v3 [math.nt] 9 May 2011
ON HARMONIC SUMS AND ALTERNATING EULER SUMS arxiv:02.592v3 [math.nt] 9 May 20 ZHONG-HUA LI Department of Mathematics, Tongji University, No. 239 Siping Road, Shanghai 200092, China Graduate School of Mathematical
More informationCh. 11 Solving Quadratic & Higher Degree Inequalities
Ch. 11 Solving Quadratic & Higher Degree Inequalities We solve quadratic and higher degree inequalities very much like we solve quadratic and higher degree equations. One method we often use to solve quadratic
More information1 5 π 2. 5 π 3. 5 π π x. 5 π 4. Figure 1: We need calculus to find the area of the shaded region.
. Area In order to quantify the size of a 2-dimensional object, we use area. Since we measure area in square units, we can think of the area of an object as the number of such squares it fills up. Using
More informationFeynman integrals and multiple polylogarithms. Stefan Weinzierl
Feynman integrals and multiple polylogarithms Stefan Weinzierl Universität Mainz I. Basic techniques II. Nested sums and iterated integrals III. Multiple Polylogarithms IV. Applications The need for precision
More information1 Roots of polynomials
CS348a: Computer Graphics Handout #18 Geometric Modeling Original Handout #13 Stanford University Tuesday, 9 November 1993 Original Lecture #5: 14th October 1993 Topics: Polynomials Scribe: Mark P Kust
More informationDIVISIBILITY OF BINOMIAL COEFFICIENTS AT p = 2
DIVISIBILITY OF BINOMIAL COEFFICIENTS AT p = 2 BAILEY SWINFORD Abstract. This paper will look at the binomial coefficients divisible by the prime number 2. The paper will seek to understand and explain
More informationPrecalculus Summer Assignment 2015
Precalculus Summer Assignment 2015 The following packet contains topics and definitions that you will be required to know in order to succeed in CP Pre-calculus this year. You are advised to be familiar
More informationProblem 1 Oh Snap... Look at the Denominator on that Rational
Problem Oh Snap... Look at the Denominator on that Rational Previously, you learned that dividing polynomials was just like dividing integers. Well, performing operations on rational epressions involving
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 informationSuper congruences involving binomial coefficients and new series for famous constants
Tal at the 5th Pacific Rim Conf. on Math. (Stanford Univ., 2010 Super congruences involving binomial coefficients and new series for famous constants Zhi-Wei Sun Nanjing University Nanjing 210093, P. R.
More informationREAL ANALYSIS II: PROBLEM SET 2
REAL ANALYSIS II: PROBLEM SET 2 21st Feb, 2016 Exercise 1. State and prove the Inverse Function Theorem. Theorem Inverse Function Theorem). Let f be a continuous one to one function defined on an interval,
More informationOn Recurrences for Ising Integrals
On Recurrences for Ising Integrals Flavia Stan Research Institute for Symbolic Computation (RISC-Linz) Johannes Kepler University Linz, Austria December 7, 007 Abstract We use WZ-summation methods to compute
More informationOBJECTIVE: To understand the relation between electric fields and electric potential, and how conducting objects can influence electric fields.
Name Section Question Sheet for Laboratory 4: EC-2: Electric Fields and Potentials OBJECTIVE: To understand the relation between electric fields and electric potential, and how conducting objects can influence
More informationTHE CAPORASO-HARRIS FORMULA AND PLANE RELATIVE GROMOV-WITTEN INVARIANTS IN TROPICAL GEOMETRY
THE CAPORASO-HARRIS FORMULA AND PLANE RELATIVE GROMOV-WITTEN INVARIANTS IN TROPICAL GEOMETRY ANDREAS GATHMANN AND HANNAH MARKWIG Abstract. Some years ago Caporaso and Harris have found a nice way to compute
More informationSYMBOL NAME DESCRIPTION EXAMPLES. called positive integers) negatives, and 0. represented as a b, where
EXERCISE A-1 Things to remember: 1. THE SET OF REAL NUMBERS SYMBOL NAME DESCRIPTION EXAMPLES N Natural numbers Counting numbers (also 1, 2, 3,... called positive integers) Z Integers Natural numbers, their
More informationIrrationality proofs, moduli spaces, and dinner parties
Irrationality proofs, moduli spaces, and dinner parties Francis Brown, IHÉS-CNRS las, Members Colloquium 17th October 2014 1 / 32 Part I History 2 / 32 Zeta values and Euler s theorem Recall the Riemann
More informationHONORS GEOMETRY Summer Skills Set
HONORS GEOMETRY Summer Skills Set Algebra Concepts Adding and Subtracting Rational Numbers To add or subtract fractions with the same denominator, add or subtract the numerators and write the sum or difference
More informationA Level Summer Work. Year 11 Year 12 Transition. Due: First lesson back after summer! Name:
A Level Summer Work Year 11 Year 12 Transition Due: First lesson back after summer! Name: This summer work is compulsory. Your maths teacher will ask to see your work (and method) in your first maths lesson,
More informationChapter 1A -- Real Numbers. iff. Math Symbols: Sets of Numbers
Fry Texas A&M University! Fall 2016! Math 150 Notes! Section 1A! Page 1 Chapter 1A -- Real Numbers Math Symbols: iff or Example: Let A = {2, 4, 6, 8, 10, 12, 14, 16,...} and let B = {3, 6, 9, 12, 15, 18,
More information