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# Gosper’s Algorithm - PowerPoint PPT Presentation

Gosper’s Algorithm. By Zachary Vogel. Binomial Coefficients. Binomial Coefficients The Binomial Theorem. Pascal’s Triangle. Base identity for Pascal’s Triangle. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 5 10 10 5 1 1 6 15 20 15 6 1. Binomial Identities.

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### Gosper’s Algorithm

By Zachary Vogel

• Binomial Coefficients

• The Binomial Theorem

• Base identity for Pascal’s Triangle

• 1

• 1 1

• 1 2 1

• 1 3 3 1

• 1 4 6 4 1

• 5 10 10 5 1

• 1 6 15 20 15 6 1

• Parallel Summation identity

• Negation identity

• There are volumes of identities with binomial coefficients.

• Here is one taken from a book:

• Nanjundiah’s identity

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

n=0

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

n=1

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

• 5 10 10 5 1

1 6 15 20 15 6 1

n=2

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

• 5 10 10 5 1

1 6 15 20 15 6 1

n=3

• How to find some order in all these identities with binomial coefficients?

• Hypergeometric notation can be used to standardize identities.

• Exponential series

• Geometric series

• General Form of a hypergeometric term

• If you take the ratio of successive terms of a hypergeometric series, the ratio is a rational polynomial of k.

• If the ratio of successive terms form a rational function of k, then the series is hypergeometric up to a constant multiple.

• Takes a hypergeometric term and sums it indefinitely

• Example

• The algorithm determines if the sum is a multiple of another hypergeometric term

–OR –

• It determines that the sum cannot be put in this form.

• We will assume that some such T(k) exists.

• If we get an impossible situation, then no such T(k) exists.

• The first step is to work out the term ratio of the summand t(k)

• Write the term ratio for t(k)

• where initially p(k) = 1

• We will pull out some factors of q and r into p.

• We require that q(k) and r(k) must have no factors

• such that

• For if

• then divide out the factors from q(k) and r(k) and absorb those terms into p(k) as follows

• p(k+1)/p(k) telescopes nicely.

• Cleverly set

• s(k) is some unknown function which will be the focus of the remainder of the algorithm

• If we can determine what s(k) is, we can determine the final summation T(k).

• By applying

• we get

• We look to solve for s(k).

• In order to determine T(k) we must solve for s(k). This requires a few steps

• Determine that s(k) is a rational function of k

• Determine that s(k) is a polynomial in k

• Determine a bound on the degree of s(k)

• By substitution

• and

• With the left hand side a rational function of k, and p(k) and r(k) are polynomials, s(k) must be a rational function of k

• Knowing s(k) is a rational function of k, we can write it as the quotient of polynomials

• such that f(k) and g(k) have no common factor

• we will also assume that g(k) has a root, then find a contradiction

• any polynomial without a root is just a constant, so s(k) will, itself, be a polynomial

• Suppose that g(a) = g(b) = 0, and b-a is a nonnegative integer. (In particular, we might have a = b).

• Since

• We have

• Substitute a=k+1, and separately b = k, we get:

• Since f and g have no common root,

• So either g(a-1)=0, or g(b+1)=0, or both r(a-1) = q(b) = 0.

• The last choice is impossible by construction.

• We now know that g(b+1) or g(a-1) is a root

• By repeating this argument with a-1 and b, or a and b+1, we get infinitely many roots for g(k).

• Therefore, g(k) has no root, thus is a constant. So s(k) is, itself a polynomial.

If we know a bound to the degree d of s(k), then we can solve it by a system of d+1 linear equations, as given by the equation:

Bounding degree of s(k)

By manipulating our previous equations, it can be seen that solve it by a system of d+1 linear equations, as given by the equation:

With

Bounding degree of s(k)

We also Know

change to ≤

Bounding degree of s(k) solve it by a system of d+1 linear equations, as given by the equation:

• Now if

Then the degree of the RHS will be

Therefore,

Otherwise, one of two options will occur

i)

ii)

remove RHS

Solving s(k) solve it by a system of d+1 linear equations, as given by the equation:

• Knowing the degree of s(k), solve

• Then simply plug the known s(k) into

Example of Gosper’s Algorithm solve it by a system of d+1 linear equations, as given by the equation:

• To provide an example of Gosper’s algorithm at work we will attempt to solve the negation identity

• To begin we set our t(k) to the summand

Negation Identity solve it by a system of d+1 linear equations, as given by the equation:

• Now by setting up a term ratio we will arrive at values for r(k), q(k) and p(k):

• This satisfies the conditions on r(k) and q(k), so long as n is non-negative.

Negation Identity solve it by a system of d+1 linear equations, as given by the equation:

• The next step is to determine s(k).

• We can bound the degree by calculating R(k) and Q(k)

Negation Identity solve it by a system of d+1 linear equations, as given by the equation:

• Since deg(R(k)) > deg(Q(k)), we have two options: d=0 or d=n. We will try d = 0 first.

• So

Solution to negation identity solve it by a system of d+1 linear equations, as given by the equation:

• Now we know our s(k) = -1/n, so we plug in to get T(k):

Solution to negation identity solve it by a system of d+1 linear equations, as given by the equation:

• So Gosper’s algorithm gives the identity