Gosper’s Algorithm
By Zachary Vogel
Binomial Coefficients
• Binomial Coefficients
n
n!
 
 k  k ! n  k  !
• The Binomial Theorem
 n  nk k
(a  b)    a b
k 0  k 
n
k
Pascal’s Triangle
• Base identity for Pascal’s Triangle
 n   n  1   n  1
 


k
k

1
  
  k 
n n
with       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
Binomial Identities
• Parallel Summation identity
n
 r  k   r  n  1




k
n

1
k 0 
 

• Negation identity
n
 n 1 
k
m 1
(

1)

(

1)

 


k
m

1
k m  


Binomial Identities
• There are volumes of identities with binomial
coefficients.
• Here is one taken from a book:
• Nanjundiah’s identity
 m  x  y  n  x  y  x  k   x  y 



    

k
k 0 
 n  k  m  n   m  n 
n
Visual of Parallel Summation
r  k 
 r  n  1




k 
k 0 
 n 1 
n
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
Visual of Parallel Summation
r  k 
 r  n  1




k 
k 0 
 n 1 
n
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
Visual of Parallel Summation
r  k 
 r  n  1




k 
k 0 
 n 1 
n
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=2
Visual of Parallel Summation
r  k 
 r  n  1




k 
k 0 
 n 1 
n
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=3
Hypergeometrics
• How to find some order in all these
identities with binomial coefficients?
• Hypergeometric notation can be used to
standardize identities.
Notation
k !  k ( k  1)( k  2)...(1)
n  n(n  1)(n  2)...(n  k  1)
k
k
n
n!
(n  k  1)

 
k ! n  k  !
k!
k 
Hypergeometric Series
k
 a1 , a2 ...am

a1k a2k ...am
zk
F
| z   k k
k
b
,
b
...
b
b
b
...
b
k
n
 1 2

1 2
n k!
Special Cases
• Exponential series
1

F  | z 
1

1k z k


k
1 k!
k

k
zk
 ez
k!
• Geometric series
k k k
k
1,1


11 z
k !z
F
| z   k

k!
1 k!
k
k
 1

1
 z 
1 z
k
k
Hypergeometric Terms
• General Form of a hypergeometric term
k
 a1 , a2 ...am

a1k a2k ...am
zk
F
| z 
k
k
k
b
,
b
...
b
b
b
...
b
n
 1 2
k
1
2
n k!
1,1

1k1k z k
k !zk
k
F
| z 


z
k!
1k k !
 1
k
 n  k  1,1 
n
(n  k  1) k
F
| 1 
 
1
k!

k
k 
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.
Gosper’s Algorithm Overview
• Takes a hypergeometric term and sums it
indefinitely
• Example
 n
k
m 1  n  1 



k 
 1   1
 m  1

k m 



Gosper’s Algorithm Overview
• The algorithm determines if the sum is a multiple of
another hypergeometric term
t (k )  cT (k ) C
–OR –
• It determines that the sum cannot be put in this form.
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)
Term Ratio
• Write the term ratio for t(k)
t (k  1) p(k  1) q(k )

t (k )
p(k ) r (k  1)
• where initially p(k) = 1
• We will pull out some factors of q and r into p.
Gosper’s Algorithm Step 1
• We require that q(k) and r(k) must have no
factors
(k   ) \ q(k )
and (k   ) \ r (k )
• such that
    N Z

Gosper’s Algorithm Step 1
• For if     N Z
• then divide out the factors from q(k) and r(k) and
absorb those terms into p(k) as follows

q(k )
q(k ) 
(k   )
r (k  1)
r (k  1) 
(k    1)
p(k )  p(k )(k   1) N 1
• p(k+1)/p(k) telescopes nicely.
p(k  1)
(k   )(k    1)...(k    2)
k 


p(k )
(k    1)(k    2)...(k    1) k    1
Gosper’s Algorithm Step 2
r (k ) s (k )t (k )
T (k ) 
p(k )
• 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).
Gosper’s Algorithm Step 2
• By applying
• we get
t (k )  T (k  1)  T (k )
p(k )  q(k ) s(k  1)  r (k ) s (k )
• We look to solve for s(k).
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)
s(k) is a rational function of k
• By substitution
r (k ) s (k )
T (k ) 
(T (k  1)  T (k ))
p(k )
• and
T (k  1)  T (k )
p(k )

T (k )
r (k ) s (k )
• 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
s(k) is a polynomial
•
Knowing s(k) is a rational function of k, we can
write it as the quotient of polynomials
s(k ) 
f (k )
g (k )
•
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
•
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
f (k  1)
f (k )
p(k )  q(k )
 r (k )
g (k  1)
g (k )
• We have
p(k ) g (k ) g (k  1)  q(k ) f (k  1) g (k )  r (k ) f (k ) g (k  1)
s(k) is a polynomial
• Substitute a=k+1, and separately b = k, we get:
0  r (a  1) f (a) g (a  1)  q(b) f (b) g (b  1)
• Since f and g have no common root,
0  r (a  1) g (a  1)  q(b) g (b  1)
• 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.
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.
s (k )   d k d   d 1k d 1  ... 0
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:
p(k )  q(k ) s(k  1)  r (k ) s (k )
Bounding degree of s(k)
• By manipulating our previous equations, it can be seen
that
2 p(k )  Q(k )[ s(k  1)  s(k )]  R(k )[ s(k  1)  s(k )]
• With
R(k )  q(k )  r (k )   k d '  ...
Q(k )  q(k )  r (k )   ' k d '1  ...
We also Know
deg[ s (k )]  d
deg[ s (k  1)  s (k )]  d
deg[ s(k  1)  s(k )]  d  1
change to ≤
Bounding degree of s(k)
• Now if deg[Q(k )]  deg[ R(k )]
Then the degree of the RHS will be deg[Q( k )]  d
Therefore,
d  deg[ P(k )]  deg[Q(k )]
Otherwise, one of two options will occur
i) 2 ' d  0: d  deg[ p(k )]  deg[ R(k )]  1
ii)
2 ' d  0: d 
2 '

 deg[ p(k )]  deg[ R(k )]  1
remove RHS
Solving s(k)
• Knowing the degree of s(k), solve
p(k )  q(k ) s(k  1)  r (k ) s (k )
• Then simply plug the known s(k) into
r (k ) s(k )t (k )
T (k ) 
p(k )
Example of Gosper’s Algorithm
• To provide an example of Gosper’s algorithm at
work we will attempt to solve the negation
identity
n
 n 1 
k
m 1
(

1)

(

1)

 


k m  k 
 m  1
• To begin we set our t(k) to the summand
n
k
t (k )    (1)
k 
Negation Identity
• Now by setting up a term ratio we will
arrive at values for r(k), q(k) and p(k):
p(k )  1
r (k )  k q(k )  k  n
• This satisfies the conditions on r(k) and
q(k), so long as n is non-negative.
Negation Identity
• The next step is to determine s(k).
• We can bound the degree by calculating
R(k) and Q(k)
Q(k )  n
R ( k )  2k  n
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
1  (k  n) 0  k 0
1
 0    s(k )
n
Solution to negation identity
• Now we know our s(k) = -1/n, so we plug
in to get T(k):
 1  n
k
T (k )  k        (1)
 n  k 
 k  n
T (k )       (1) k 1
 n  k 
 n  1
k 1
T (k )  
 (1)
 k  1
Solution to negation identity
• So Gosper’s algorithm gives the identity
n
 n 1 
k
m 1
(

1)

(

1)

 


k
m

1
k m  

