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Published byWinfred Mathews Modified over 9 years ago
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Gosper’s Algorithm By Zachary Vogel
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Binomial Coefficients The Binomial Theorem
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Pascal’s Triangle Base identity for Pascal’s Triangle 1 1 2 1 1 3 3 1 1 4 6 4 1 15 10 10 5 1 1 6 15 20 15 6 1
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Binomial Identities Parallel Summation identity Negation identity
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Binomial Identities There are volumes of identities with binomial coefficients. Here is one taken from a book: Nanjundiah’s identity
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Visual of Parallel Summation 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
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Visual of Parallel Summation 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
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Visual of Parallel Summation 1 1 2 1 1 3 3 1 1 4 6 4 1 15 10 10 5 1 1 6 15 20 15 6 1 n=2
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Visual of Parallel Summation 1 1 2 1 1 3 3 1 1 4 6 4 1 15 10 10 5 1 1 6 15 20 15 6 1 n=3
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Hypergeometrics How to find some order in all these identities with binomial coefficients? Hypergeometric notation can be used to standardize identities.
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Notation
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Hypergeometric Series
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Special Cases Exponential series Geometric series
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Hypergeometric Terms General Form of a hypergeometric term
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Successive Terms 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.
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Gosper’s Algorithm Overview Takes a hypergeometric term and sums it indefinitely Example
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Gosper’s Algorithm Overview 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.
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Gosper’s Algorithm Step 1 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)
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Term Ratio Write the term ratio for t(k) where initially p(k) = 1 We will pull out some factors of q and r into p.
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Gosper’s Algorithm Step 1 We require that q(k) and r(k) must have no factors such that
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Gosper’s Algorithm Step 1 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.
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Gosper’s Algorithm Step 2 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).
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Gosper’s Algorithm Step 2 By applying we get We look to solve for s(k).
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Unknown function 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)
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s(k) is a rational function of 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
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s(k) is a polynomial 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
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s(k) is 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
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s(k) is a polynomial 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.
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s(k) is a polynomial 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.
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Bounding degree of s(k) 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:
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Bounding degree of s(k) By manipulating our previous equations, it can be seen that With We also Know change to ≤
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Bounding degree of s(k) Now if Then the degree of the RHS will be Therefore, Otherwise, one of two options will occur i) ii) remove RHS
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Solving s(k) Knowing the degree of s(k), solve Then simply plug the known s(k) into
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Example of Gosper’s Algorithm 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
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Negation Identity 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.
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Negation Identity The next step is to determine s(k). We can bound the degree by calculating R(k) and Q(k)
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Negation Identity Since deg(R(k)) > deg(Q(k)), we have two options: d=0 or d=n. We will try d = 0 first. So
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Solution to negation identity Now we know our s(k) = -1/n, so we plug in to get T(k):
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Solution to negation identity So Gosper’s algorithm gives the identity
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