Sums ITT9132 Concrete 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 ?

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 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 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 ( of a )

A series ∑k 0 ak with real coecients converges to a 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 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 . 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