Sums ITT9132 Concrete Mathematics Lecture 5 25 February 2019
Chapter Two Finite and Innite Calculus Derivative and Dierence Operators Integrals and Sums Summation by Parts Innite Sums Contents
1 Finite and Innite Calculus Derivative and Dierence Operators Integrals and Sums Summation by Parts
2 Innite Sums Next section
1 Finite and Innite Calculus Derivative and Dierence Operators Integrals and Sums Summation by Parts
2 Innite Sums Next subsection
1 Finite and Innite Calculus Derivative and Dierence Operators Integrals and Sums Summation by Parts
2 Innite Sums Derivative and Dierence Operators
Innite calculus: derivative Finite calculus: dierence
Euler's notation ∆f (x) = f (x + 1) − f (x) f (x + h) − f (x) Df (x) = lim In general, if h ∈ (or h ∈ ), then h→0 h R C Forward dierence Lagrange's notation ∆h [f ](x) = f (x + h) − f (x) 0 f (x) = Df (x) Backward dierence [f ](x) = f (x) − f (x − h) Leibniz's notation If y = f (x), then Oh dy df df (x) dx = dx (x) = dx = Df (x) Central dierence δ [f ](x) = Newton's notation h f (x + 1 h) − f (x − 1 h) y˙ = f 0(x) 2 2
∆ [f ](x) Df (x) = lim h h→0 h Derivative of Power function
Example: f (x) = x3 In this case,
∆h [f ](x) = f (x + h) − f (x) = (x + h)3 − x3 = x3 + 3x2h + 3xh2 + h3 − x3 = h · (3x2 + 3xh + h2)
Hence, 2 2 h · (3x + 3xh + h ) 2 2 2 Df (x) = lim = lim 3x + 3xh + h = 3x h→0 h h→0
In general, for m ≥ 1 integer:
1 D(xm) = mxm− (Forward) Dierence of Power Function
Example: f (x) = x3 In this case, 2 ∆f (x) = ∆1 [f ](x) = 3x + 3x + 1
In general, for m ≥ 1 integer:
m m ∆(xm) = ∑ xm−k k=1 k
because of Newton's binomial theorem. Falling and Rising Factorials
Denition
The falling factorial (power) is dened for m > 0 by:
xm = x(x − 1)(x − 2)···(x − m + 1)
The rising factorial (power) is dened for m > 0 by:
xm = x(x + 1)(x + 2)···(x + m − 1)
Properties
xm = (−1)m(−x)m xm+n = xm(x − m)n n! = nn = 1n xm+1 xm = if x 6= m n nk x − m = 1 1 k k! x−m = = (x + 1)m (x + 1)(x + 2)···(x + m) Dierence of falling factorial with positive exponent
∆(xm) = (x + 1)m − xm = (x + 1) · (x ···(x − m + 2)) − (x ···(x − m + 2)) · (x − m + 1) = (x + 1 − (x − m + 1)) · (x ···(x − m + 2)) = m · xm−1
Hence: m m−1 ∆(x ) = mx ∀m > 1 Dierence of falling factorial with negative exponent: Example
Let's check this formula for negative power:
2 ∆x−2 = (x + 1)− − x−2 1 1 = − (x + 2)(x + 3) (x + 1)(x + 2) (x + 1) − (x + 3) = (x + 1)(x + 2)(x + 3) −2 = (x + 1)(x + 2)(x + 3) = −2 · x−3 Dierence of falling factorial with negative exponent
∆x−m = (x + 1)−m − x−m 1 1 = − (x + 2)···(x + m)(x + m + 1) (x + 1)(x + 2)···(x + m) (x + 1) − (x + m + 1) = (x + 1)(x + 2)···(x + m)(x + m + 1) −m = (x + 1)(x + 2)···(x + m)(x + m + 1) 1 = −mx−(m+ ) = −mx−m−1 Next subsection
1 Finite and Innite Calculus Derivative and Dierence Operators Integrals and Sums Summation by Parts
2 Innite Sums Indenite Integrals and Sums
The Fundamental Theorem of Calculus
Z Df (x) = g(x) iff g(x)dx = f (x) + C
Denition
The indenite sum of the function g(x) is the class of functions f such that ∆f (x) = g(x):
∆f (x) = g(x) iff ∑g(x)δx = f (x) + C(x)
where C(x) is a function such that C(x + 1) = C(x) for any integer value of x. Denite Integrals and Sums
If g(x) = Df (x), then:
Z b b g(x)dx = f (x) = f (b) − f (a) a a
Similarly:
If g(x) = ∆f (x), then:
b b
∑g(x)δx = f (x) = f (b) − f (a) a a Denite sums
Some observations
a 0 ∑a g(x)δx = f (a) − f (a) = a+1 1 ∑a g(x)δx = f (a + ) − f (a) = g(a) b+1 b 1 ∑a g(x)δx − ∑a g(x)δx = f (b + ) − f (b) = g(b)
Hence,
b b−1 ∑g(x)δx = ∑ g(k) = ∑ g(k) a k=a a6k
If m 6= −1, then:
Z n xm+1 n nm+1 xmdx = = 0 m + 1 0 m + 1
Analogous nite case:
If m 6= −1, then:
n km+1 n nm+1 ∑kmδx = ∑ km = = 0 m + 1 0 m + 1 06k Case m = 1 n2 n(n − 1) ∑ k = 2 = 2 06k 3 2 2 n n ∑ k = 3 + 2 06k Taking n + 1 instead of n gives: (n + 1)n(2n + 1) n = 6 Sums of Powers (case m = −1) As a rst step, we observe that: ∆Hx = Hx+1 − Hx 1 1 1 1 1 = 1 + + ... + + − 1 + + ... + 2 x x + 1 2 x 1 = = x−1 x + 1 We conclude: b b −1 ∑x δx = Hx a a Sums of Discrete Exponential Functions We have: Dex = ex The nite analogue should have ∆f (x) = f (x). This means: f (x + 1) − f (x) = f (x), that is, f (x + 1) = 2f (x), only possible if f (x) = 2x For general base c > 0, the dierence of cx is: ∆(cx ) = cx+1 − cx = (c − 1)cx cx and the anti-dierence for c 6= 1 is . (c − 1) As an application, we compute the sum of the geometric progression: b cx b cb − ca cb−a − 1 ck = cx δx = = = ca · . ∑ ∑ c − 1 a c − 1 c − 1 a6k Differential equation Solution Difference equation Solution n m Dfn(x) = nfn−1(x) fn(x) = x ∆um(x) = mum−1(x) um(x) = x 0 0 0 0 0 0 fn( ) = [n = ] , n > um( ) = [m = ] , m > n m Dfn(x) = nfn−1(x) fn(x) = x ∆um(x) = mum−1(x) um(x) = x fn(1) = 1, n < 0 um(0) = 1/m!, m < 0 1 1 Df (x) = · [x > 0] f (x) = lnx ∆u(x) = · [x 1] u(x) = H x x > x f (1) = 1 u(1) = 1 Df (x) = f (x) f (x) = ex ∆u(x) = u(x) u(x) = 2x f (0) = 1 u(0) = 1 Df (x) = b · f (x) f (x) = ax ∆u(x) = b · u(x) u(x) = cx f (0) = 1 where b = lna u(0) = 1 where b = c − 1 l'Hôpital's rule and Stolz-Cesàro lemma l'Hôpital's rule: Hypotheses Stolz-Cesàro lemma: Hypotheses 1 f (x) and g(x) are both vanishing or both innite 1 u(n) and v(n) are dened at x0. for every value n ∈ N. 2 g 0(x) is always positive in 2 v(n) is positive, strictly some neighborhood of x0. increasing, and divergent. l'Hôpital's rule: Thesis Stolz-Cesàro lemma: Thesis 0 f (x) If ∆u(n) , If lim = L ∈ , limn→∞ = L ∈ R x→x0 g 0(x) R ∆v(n) f (x) u(x) then lim = L. then limn→∞ = L. x→x0 g(x) v(x) Next subsection 1 Finite and Innite Calculus Derivative and Dierence Operators Integrals and Sums Summation by Parts 2 Innite Sums Summation by Parts Innite analogue: integration by parts Z Z u(x)v 0(x)dx = u(x)v(x) − u0(x)v(x)dx Dierence of a product ∆(u(x)v(x)) = u(x + 1)v(x + 1) − u(x)v(x) = u(x + 1)v(x + 1) − u(x)v(x + 1) + u(x)v(x + 1) − u(x)v(x) = ∆u(x)v(x + 1)+ u(x)∆v(x) = u(x)∆v(x) + Ev(x)∆u(x) where E is the shift operator Ef (x) = f (x + 1). We then have the: Rule for summation by parts ∑u∆v δx = uv − ∑Ev∆u δx Why the shift? If we repeat our derivation with two continuous functions f and g of one real variable x, we nd for any increment h 6= 0: f (x + h)g(x + h) − f (x)g(x) = f (x + h)g(x + h) − f (x)g(x + h) + f (x)g(x + h) − f (x)g(x) = f (x)(g(x + h) − g(x)) + g(x + h)(f (x + h) − f (x)) The incremental ratio is thus: f (x + h)g(x + h) − f (x)g(x) g(x + h) − g(x) f (x + h) − f (x) = f (x) · + g(x + h) · h h h So there is a shift: but it is innitesimaland disappears by continuity of g. Summation by Parts (2) Example: n 2k S = ∑k=0 k Taking u(x) = x, v(x) = 2x and Ev(x) = 2x+1: 1 1 ∑x2x δx = x2x − ∑2x+ δx = x2x − 2x+ + C This yields: n n+1 ∑ k2k = ∑ x2x δx k=0 0 n+1 = (x2x − 2x+1) 0 = (n + 1)2n+1 − 2n+2 − (0 · 20 − 2) = (n − 1)2n+1 + 2 Summation by Parts (3) Example: n−1 S = ∑k=0 kHk Continuous analogue: Z x2 Z x2 1 x lnx dx = lnx − · dx 2 2 x x2 1 Z = 2 lnx − 2 x dx x2 1 x2 = 2 lnx − 2 · 2 x2 1 = 2 lnx − 2 Summation by Parts (3) Example: n−1 S = ∑k=0 kHk Taking and 2 2, we get −1 1 , u(x) = Hx vx = x / ∆u(x) = ∆Hx = x x+1 2 1 (x+1) ∆v(x) = x = x , and Ev(x) = 2 . Then: n−1 n n n kH = xHx δx = uv − Ev∆u δx ∑ k ∑ 0 ∑ k=0 0 0 2 2 n n 1 x (x + ) −1 = Hx − ∑ · x δx 2 0 0 2 x2 n 1 n = Hx − ∑x δx 2 0 2 0 x2 1 x2 n = Hx − · 2 2 2 0 n2 1 = 2 Hn − 2 Next section 1 Finite and Innite Calculus Derivative and Dierence Operators Integrals and Sums Summation by Parts 2 Innite Sums How to sum innite number sequences? Setting n seems meaningful . . . ∑k∈N ak = limn→∞ ∑k=0 ak Example 1 Let 1 1 1 1 1 1 1 1 S = + 2 + 4 + 8 + 16 + 32 + 64 + 128 + ··· . Then 1 1 1 1 1 1 2 2 1 2 S = + + 2 + 4 + 8 + 16 + 32 + 64 + ··· = + S, and S = 2 How to sum innite number sequences? Setting n seems meaningful . . . ∑k∈N ak = limn→∞ ∑k=0 ak Example 1 Let 1 1 1 1 1 1 1 1 S = + 2 + 4 + 8 + 16 + 32 + 64 + 128 + ··· . Then 1 1 1 1 1 1 2 2 1 2 S = + + 2 + 4 + 8 + 16 + 32 + 64 + ··· = + S, and S = 2 But can we manipulate such sums like we do with nite sums? How to sum innite number sequences? Example 2 Let T = 1 + 2 + 4 + 8 + 16 + 32 + 64 + ... Then 2T = 2 + 4 + 8 + 16 + 32 + 64 + 128... = T − 1 and T = −1 How to sum innite number sequences? Example 2 Let T = 1 + 2 + 4 + 8 + 16 + 32 + 64 + ... Then 2T = 2 + 4 + 8 + 16 + 32 + 64 + 128... = T − 1 and T = −1 Problem: The sum T is innite... and we cannot subtract an innite quantity from another innite quantity. How to sum innite number sequences? Example 3 Let ∑ (−1)k = 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + ... k>0 Dierent ways to sum (1 − 1) + (1 − 1) + (1 − 1) + (1 − 1) + ... = 0 + 0 + 0 + 0 + ... = 0 and 1−(1−1)−(1−1)−(1−1)−(1−1)−... = 1−0−0−0−0−0−... = 1 How to sum innite number sequences? Example 3 Let ∑ (−1)k = 1 − 1 + 1 − 1 + 1 − 1 + 1 − 1 + ... k>0 Dierent ways to sum (1 − 1) + (1 − 1) + (1 − 1) + (1 − 1) + ... = 0 + 0 + 0 + 0 + ... = 0 and 1−(1−1)−(1−1)−(1−1)−(1−1)−... = 1−0−0−0−0−0−... = 1 Problem: The sequence of the partial sums does not converge... and we cannot manipulate something that does not exist. Dening Innite Sums: Nonnegative Summands Denition 1 If 0 for every 0, then: ak > k > n ak = lim ak = sup ak ∑ n→∞ ∑ ∑ k>0 k=0 K⊆N,|K|<∞ k∈K Note that: The denition as a limit is (sort of) a Riemann integral. The denition as a least upper bound is a Lebesgue integral. The limit / least upper bound above can be nite or innite, but are always equal. Exercise: Prove this fact. Dening Innite Sums: Riemann Summation Denition 2 (Riemann sum of a series) A series ∑k 0 ak with real coecients converges to a real number S, called the sum of the series, if:> n lim ak = S . n→∞ ∑ k=0 In this case, we write: . ∑k>0 ak = S The values n are called the partial sums of the series. Sn = ∑k=0 ak The series converges absolutely if converges. ∑k>0 ak ∑k>0 |ak | If the series ∑k 0 ak = ∑k 0(bk + ick ) has complex coecients, we say that it converges to > if> converges to and converges to . S = T + iU ∑k>0 bk T ∑k>0 ck U A series that converges, but not absolutely (−1)k−1 Let a = [k > 0]. Then ∑ 0 a = ln2. k k k> k However, it is easy to prove by induction that 2n n for every 1. ∑k=0 |ak | = H2n > 2 n > Innite Sums: Associativity Associativity A series ∑k 0 ak has the associative property if for every two strictly increasing sequences > i0 = 0 < i1 < i2 < ... < ik < ik+1 < ... j0 = 0 < j1 < j2 < ... < jk < jk+1 < ... we have: ik+1−1 ! jk+1−1 ! ∑ ∑ ai = ∑ ∑ ai 0 0 k> i=ik k> j=jk We have seen that the series 1 k does not have the associative property. ∑k>0(− ) Theorem A series has the associative property if and only if it is convergent. Proof: Regrouping as in the denition means taking a subsequence of the sequence of partial sums, which can converge to any of the latter's limit point. Dening Innite Sums: Lebesgue Summation Every real number can be written as x = x+ − x−, where: x+ = x · [x > 0] = max(x,0) and x− = −x · [x < 0] = max(−x,0) + − + − Note that: x > 0, x > 0, and x + x = |x|. Denition 3 (Lebesgue sum of a series) Let {ak }k be an absolutely convergent sequence of real numbers. Then: + − ∑ak = ∑ak − ∑ak k k k The series ∑k ak : converges absolutely if + and − ; ∑k ak < +∞ ∑k ak < +∞ diverges positively if + and − ; ∑k ak = +∞ ∑k ak < +∞ diverges negatively if + and − . ∑k ak < +∞ ∑k ak = +∞ If both + and − then Bad Stu happens. ∑k ak = +∞ ∑k ak = +∞ Innite Sums: Bad Stu Riemann series theorem Let ∑k ak be a series with real coecients which converges, but not absolutely. For every real number L there exists a permutation p of N such that: n lim ap(k) = L n→∞ ∑ k=0 Example: The harmonic series 1 1 1 If we rearrange the terms of the series 1 as follows: − 2 + 3 − 4 + ... 1 1 1 1 1 1 1 1 1 − − + − − + ... = ... + − − + ... 2 4 3 6 8 2k − 1 2(2k − 1) 4k 1 1 1 = ... + − + ... 2 2k − 1 2k we obtain: 1 1 1 1 1 1 1 1 1 1 2 1 2 − 2 + 3 − 4 + ... = ln but − 2 − 4 + 3 − 6 − 8 + ... = 2 ln Innite Sums: Commutativity Commutativity A series has the commutative property if for every permutation of , ∑k>0 ak p N ∑ ap(k) = ∑ak k>0 k The Riemann series theorem says that any series which is convergent, but not absolutely convergent, does not have the commutative property. Theorem A convergent series has the commutative property if and only if it is absolutely convergent. Proof: (Sketch) Think of Lebesgue summation. Innite Sums: Commutativity Commutativity A series has the commutative property if for every permutation of , ∑k>0 ak p N ∑ ap(k) = ∑ak k>0 k The Riemann series theorem says that any series which is convergent, but not absolutely convergent, does not have the commutative property. Theorem A convergent series has the commutative property if and only if it is absolutely convergent. If we want to manipulate innite sums like nite ones, we must require absolute convergence. Multiple innite sums Denition: Double innite sums For every 0 let 0. j,k > aj,k > 1 If 0 for every and , then: aj,k > j k aj,k = sup aj,k = lim aj,k . ∑ ∑ n→∞ ∑ j,k K⊆N×N,|K|<∞ K 06j,k6n (Recall that 0 0 .) ∑06j,k6n aj,k = ∑j,k aj,k [ 6 j 6 n][ 6 k 6 n] 2 If ∑j,k |aj,k | < +∞, then: + − ∑aj,k = ∑aj,k − ∑aj,k . j,k j,k j,k Multiple innite sums Denition: Double innite sums For every 0 let 0. j,k > aj,k > 1 If 0 for every and , then: aj,k > j k aj,k = sup aj,k = lim aj,k . ∑ ∑ n→∞ ∑ j,k K⊆N×N,|K|<∞ K 06j,k6n (Recall that 0 0 .) ∑06j,k6n aj,k = ∑j,k aj,k [ 6 j 6 n][ 6 k 6 n] 2 If ∑j,k |aj,k | < +∞, then: + − ∑aj,k = ∑aj,k − ∑aj,k . j,k j,k j,k Can we use or instead? ∑j>0 ∑k>0 aj,k ∑k>0 ∑j>0 aj,k Multiple innite sums Denition: Double innite sums For every 0 let 0. j,k > aj,k > 1 If 0 for every and , then: aj,k > j k aj,k = sup aj,k = lim aj,k . ∑ ∑ n→∞ ∑ j,k K⊆N×N,|K|<∞ K 06j,k6n (Recall that 0 0 .) ∑06j,k6n aj,k = ∑j,k aj,k [ 6 j 6 n][ 6 k 6 n] 2 If ∑j,k |aj,k | < +∞, then: + − ∑aj,k = ∑aj,k − ∑aj,k . j,k j,k j,k Can we use or instead? In general, no: ∑j>0 ∑k>0 aj,k ∑k>0 ∑j>0 aj,k One writing is the limit on j of a limit on k which is a function of j; The other writing is the limit on k of a limit on j which is a function of k. There are no guarantees that the double limits be equal! Multiple sums: An example of noncommutativity From Joel Feldman's notes1 Let aj,k = [j = k = 0] + [k = j + 1] − [k = j − 1]: 0 1 2 3 4 ... 0 1 1 0 0 0 ... 1 −1 0 1 0 0 ... 2 0 −1 0 1 0 ... 3 0 0 −1 0 1 ...... Then: for every 0, 2 0 ; j ≥ ∑k>0 aj,k = · [j = ] for every 0, 0; and k ≥ ∑j>0 aj,k = for every 0, 1. n ≥ ∑06j,k6n aj,k = Hence: aj,k = 2 ; aj,k = 0 ; and lim aj,k = 1. ∑ ∑ ∑ ∑ n→∞ ∑ j>0 k>0 k>0 j>0 06j,k6n 1 http://www.math.ubc.ca/£feldman/m321/twosum.pdf retrieved 21.02.2019. Multiple innite sums: Swapping indices Theorem For 0 let be real numbers. j,k > aj,k Tonelli If 0 for every and , then: aj,k > j k ∑ ∑ aj,k = ∑ ∑ aj,k = ∑aj,k , j>0 k>0 k>0 j>0 j,k regardless of the quantities above being nite or innite. Fubini If ∑j,k |aj,k | < +∞, then: ∑ ∑ aj,k = ∑ ∑ aj,k = ∑aj,k . j>0 k>0 k>0 j>0 j,k Fubini's theorem is proved in the textbook. Multiple innite sums: Swapping indices Theorem For 0 let be real numbers. j,k > aj,k Tonelli If 0 for every and , then: aj,k > j k ∑ ∑ aj,k = ∑ ∑ aj,k = ∑aj,k , j>0 k>0 k>0 j>0 j,k regardless of the quantities above being nite or innite. Fubini If ∑j,k |aj,k | < +∞, then: ∑ ∑ aj,k = ∑ ∑ aj,k = ∑aj,k . j>0 k>0 k>0 j>0 j,k Fubini's theorem is proved in the textbook. Again: If we want to manipulate innite sums like nite ones, we must require absolute convergence. Cesàro summation Given a series , consider the sequence n of the partial sums. ∑k ak Sn = ∑k=0 ak Put x−1 and . Then and 1. u(x) = ∑k=0 Sk v(x) = x ∆u(x) = Sx ∆v(x) = S Suppose a converges. Put L = a = lim n . ∑k k ∑k>0 k n→∞ 1 1 ∑n− S We then have by the Stolz-Cesàro lemma: lim k=0 k = L. n→∞ n Given a (not necessarily convergent) series ∑k ak , the quantity: n−1 ∑k=0 Sk C ak = lim ∑ n→∞ k n is called the Cesàro sum of the series ∑k ak . Cesàro summation Given a series , consider the sequence n of the partial sums. ∑k ak Sn = ∑k=0 ak Put x−1 and . Then and 1. u(x) = ∑k=0 Sk v(x) = x ∆u(x) = Sx ∆v(x) = S Suppose a converges. Put L = a = lim n . ∑k k ∑k>0 k n→∞ 1 1 ∑n− S We then have by the Stolz-Cesàro lemma: lim k=0 k = L. n→∞ n Given a (not necessarily convergent) series ∑k ak , the quantity: n−1 ∑k=0 Sk C ak = lim ∑ n→∞ k n is called the Cesàro sum of the series ∑k ak . k A series can have a Cesàro sum without being convergent. For example, if ak = (−1) , n + [n is odd] 1 then n−1 S = , so C (−1)k = . ∑k=0 k 2 ∑k 2 n 0 1 2 3 4 5 6 7 8 9 an 1 −1 1 −1 1 −1 1 −1 1 −1 Sn 1 0 1 0 1 0 1 0 1 0 n−1 0 1 1 2 2 3 3 4 4 5 ∑k=0 Sk