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Closed Forms for Summations. Lecture 19. Two categories (maybe 3) of Summations. Indefinite summation: S 1 · i · n f(i); f not dependent on n f dependent on n -- finite difference calculus has a long history, work done by Newton, Euler, Bernoulli, Boole.

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closed forms for summations

Closed Forms for Summations

Lecture 19

Richard Fateman CS 282 Lecture 19

two categories maybe 3 of summations
Two categories (maybe 3) of Summations
  • Indefinite summation:
    • S1 · i · n f(i); f not dependent on n

f dependent on n

-- finite difference calculus has a long history, work done by Newton, Euler, Bernoulli, Boole.

Definite summation (particular solutions): Zeilberger, Gosper, Ramanujan, etal

Richard Fateman CS 282 Lecture 19

indefinite summation parallels to integration
Indefinite summation parallels to integration
  • integration of polynomials
  • integration of rational functions
  • difference operator D parallels the derivative
  • S and s are similar
  • But not similar enough for some purposes!

Richard Fateman CS 282 Lecture 19

some simple examples macsyma in this case
Some simple examples (Macsyma, in this case)

Richard Fateman CS 282 Lecture 19

a more elaborate example
A more elaborate example

Richard Fateman CS 282 Lecture 19

start simply
Start simply.
  • if we need g(n) = åi=an f(i) we approach by

finding the indefinite summation

h(x) = åi=0x-1 f(i)

then

g(x) = h(n+1) – h(a).

Sidestepping any issues of singularities.

Note also that D h(x) defined by h(x+1)-h(x) is f(x)

Richard Fateman CS 282 Lecture 19

slide7
Also
  • D-1 f(x) = h(x) = åi=0x-1 f(i)
  • Note parallel: we can obtain an expression for the summation by anti-differencing; compare to integration by anti-differentiation.

Richard Fateman CS 282 Lecture 19

simple properties of d
Simple Properties of D
  • Unique up to addition of functions whose first difference is zero
    • Constants
    • functions with period 1, e.g. sin (p x)

Richard Fateman CS 282 Lecture 19

we also need the shift operation e
We also need the shift operation, E
  • E f(x) = f(x+1)
  • hence
    • D f(x) = Ef(x)-f(x)

Richard Fateman CS 282 Lecture 19

more properties of d
More properties of D

Richard Fateman CS 282 Lecture 19

occasionally useful property
Occasionally useful property
  • The chain rule
  • D f(g(x)) = D f(g(x))
  • where D f(x,y) = f(x+h,y)-f(x,y)

g(x)

x

1

D g(x)

x

h

Richard Fateman CS 282 Lecture 19

the simplest non trivial form to sum is a polynomial
The simplest non-trivial form to sum is a polynomial
  • A(x) = å ai xi
  • The analogy to differential calculus is to integrate, term by term:
    • easy since Dxn = nxn-1.
  • Differences of powers are not so concise:

D(xn) = (x+1)n-xn = å(binomial(n,i),i=0..n-1) and so neither is summation.

INSTEAD consider factorial functions.

[x]n = x(x-1)(x-2)....(x-n+1).

Richard Fateman CS 282 Lecture 19

what is the difference of a factorial function
What is the difference of a factorial function?
  • D [x]n = E [x]n – [x]n =n[x]n-1.
  • Proof:

E[x]n = (x+1)x(x-1)(x-2).... (x-n+2)

[x]n = x(x-1)(x-2)....(x-n+2)(x-n+1).

All the terms in red are the same, and one can factor them out. they are [x]n-1. The remaining factor is simply (x+1)-(x-n+1) = n.

Richard Fateman CS 282 Lecture 19

to sum a polynomial convert it to factorial form
To sum a polynomial, convert it to factorial form:
  • one way is to expand a bunch of factorial functions [x]1+x, [x]2= x2-x, etc, solve for powers of x, e.g. x2= [x]2-[x]1, and we are done.
  • Another is to use Newton’s divided difference interpolation formula in a table:
  • f(x)=sum([x]i/i! Di f(0)) where we use higher differences this way:

Richard Fateman CS 282 Lecture 19

divided difference table for f 3 x 3 2 x 1
Divided difference table for f= 3*x^3-2*x+1

Richard Fateman CS 282 Lecture 19

divided difference table for f 3 x 3 2 x 116
Divided difference table for f= 3*x^3-2*x+1

D-1 f =sum([x]i/i! Di f(0))=1*[x]1+ 1/2*[x]2 +18/3!*[x]3 +18/4!*[x]4 .

Richard Fateman CS 282 Lecture 19

converting to conventional polynomial form is done by expanding x i and combining terms
Converting to conventional polynomial form is done by expanding [x]i and combining terms
  • [x]1 =x
  • ½ [x]2 = - ½ x+ ½x2
  • 3[x]3 = 6x –9 x2 +3x3
  • ¾[x]4 = -9/2x – 21/4 x2-9/2x3+ ¾ x4
  • total is 2x-55/4*x2-3/2x3+3/4x4

Richard Fateman CS 282 Lecture 19

sums of rational functions
Sums of rational functions
  • Define factorial operators on functions...
  • [f(x)]k = f(x) ¢ f(x-1) ¢ ... ¢ f(x-k+1) for k >0
  • extend the operator by noticing
  • [f(x)]k = [f(x)]r¢ [f(x-r)]k-r
  • Define [f(x)]0 to be 1 and use the previous line as an identity. Then for k=0 we get
  • [f(x)]-r = 1/[f(x+r)]r

Richard Fateman CS 282 Lecture 19

differences of factorials
Differences of factorials

R. Moenck, Macsyma Users’ Conf. 1977

Richard Fateman CS 282 Lecture 19

what does this mean for summation d 1
What does this mean for summation (D-1)?
  • If we can get rational expressions so they look like the RHS of that equation, we can find their summation, namely [f(x)]-l

Richard Fateman CS 282 Lecture 19

we need to use shift free decomposition to go further
We need to use Shift Free Decomposition to go further.

Given a product of functions, we can decompose it into a product of factorial functions.

Let S=a ¢ b ¢ c where a,b,c are mutually relatively prime and Ea=b. Then shift S:

ES = (Ea)¢(Eb)¢(Ec) = b ¢ Eb ¢ Ec

GCD(S,ES) = b

So we can divide out b and a from S and express

S=[b]2¢ c .

Richard Fateman CS 282 Lecture 19

if we apply this observation repeatedly we can get s to be shift free
If we apply this observation repeatedly, we can get S to be shift free
  • S=[s1]1¢ [s2]2¢ ... ¢ [sk]k where the individual sk are shift-free.
  • Analogous to partial fraction decomposition in the differential calculus Hermite integration process, we can form a shift-free partial fraction for some rational function we wish to sum. That is,
  • A(x)/S(x) = å (Ai/[si]i), i=1..k
  • and a “complete” decomposition
  • A(x)/S(x) = åå(Aij/[si]j), i=1..k,j=1..i

Richard Fateman CS 282 Lecture 19

shift 1 independence is not enough we need to show s x is k shift free
Shift-1 independence is not enough. We need to show S(x) is k-shift-free
  • Compute resultant of S(x) and S(x+k) with respect to k. If there is an integer k>1 shift, then fill in the terms for numerator and denominator. e.g. if S = x*(x+3), change it to

[x+3]4 and multiply numerator by (x+1)(x+2).

Richard Fateman CS 282 Lecture 19

summation by parts
Summation by parts
  • Similar to Hermite integration using

D-1(u¢D v) = u ¢ v - D (Ev ¢D u)

can be used to reduce denominators of the form [xi]j to [xi]1

  • EVENTUALLY... one gets a rational function plus an indefinite summation of terms with shift-free denominators of factorial degree 1.

Richard Fateman CS 282 Lecture 19

the transcendental part
The transcendental part
  • Define ym(x)= Dm(logG(x+1)), m>0 where n!=G(n+1) is the well-known gamma function
  • ym(x)=

Dm(D logG(x+1))=

Dm(log(G(x+2)/G(x+1)))=

Dm log(x+1) =

Dm-1(1/(x+1)) =

((-1)m-1¢ (m-1)!¢(x+1)-m.

Richard Fateman CS 282 Lecture 19

the sum of a negative power of x 1 finishes the task
The sum of a negative power of x+1 finishes the task
  • D-1(x+1)-m = (-1)m-1/(m-1)! ym(x)
  • The ym functions are known as polygamma functions and serve a role similar to logs in Hermite integration.
  • Rational summation is pretty much solved, though people still look for fast ways of doing some of the steps (shift-free decomposition).

Richard Fateman CS 282 Lecture 19

this is not the end of the story what about more elaborate summands
This is not the end of the story: what about more elaborate summands?
  • Gosper’s algorithm looks at å ai = D-1ai by seeking a “telescoping function” f(n).
  • Let an = D g(n) = g(n+1)-g(n)
  • then suppose g(n)=f(n)*an.
  • We have to solve the functional equation
  • C(n)=an+1/an = (f(n)+1)/f(n+1)
  • Only the ratio of 2 terms is used (easily computed). If C(n) is rational in n, then this is called hypergeometric summation.

Richard Fateman CS 282 Lecture 19

restrictions extensions
Restrictions/ Extensions
  • Note that the terms an can be far more general than rational; just an+1/an is rational
  • Gosper’s work is the basis for a decision procedure, widely used in computer algebra systems.
  • More recent work by Zeilberger (see refs) pushes this further.

Richard Fateman CS 282 Lecture 19