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Identities from The Treatise by Pascal

                                      
Again, let us recall that Pascal wrote the array down in the form A.W.F. Edwards calls the figurate numbers,  an infinite rectangular array of numbers written as follows:

  table
      l=0 l=1 l=2 l=3 l=4 l=5 …
k=0|  1   1   1   1   1   1
k=1|  1   2   3   4   5   6   
k=2|  1   3   6  10  15  21
k=3|  1   4  10  20  35  56
k=4|  1   5  15  35  70 126
k=5|  1   6  21  56 126 252  
 …
 
(Note: we have strayed slightly from Edwards’ notation, starting both the l and k subscripts at 0; Edwards’ starts the k at 0 and the l at 1.)

In this pattern, the top row for all values of l, the left-hand column  for all values of k, and elements are defined as the sum  , the sum of all the elements in the previous row that lie to the left or directly above the new element.  As noted earlier in discussing Pascal, this method of construction corresponds to the identity we call the Christmas Stocking Theorem or CST for short.  Given this numbering of the figurate numbers, we have these translations of figurate numbers into binomial coefficients and vice versa.

        

It was in this form that Pascal presented the numbers in his Treatise published in 1665; in the drawing from that work shown below, Pascal marks the upward diagonals to show they are related; in present day literature, these are exactly the rows of Pascal’s Triangle; he also marks the central downward diagonal, which are the binomial coefficients .  In this illustration, we see Pascal put symbols above the entries, some in Greek letters, some in the Roman alphabet; this is because he did not use subscripts or notation to identify the entries in the matrix the way we would now.  He wrote out an identity in French, then showed an example using the symbols he had written above the entries.  For example, if he wanted to show a Christmas Stocking Theorem example, he might have written A+B+C = F, which corresponds to 1+3+6 = 10, the first three entries of the third row adding up to the third entry in row 4.

               
                From the cover of the paperback edition of Edwards' Pascal's Arithmetical Triangle
                This image reproduced by the kind permission of Johns Hopkins University Press.

While Edwards in his very useful book “Pascal’s Arithmetical Triangle” gives us the identities of Pascal in terms of figurate numbers, we will translate these into equations using the binomial coefficients; while we will still refer to the figurate matrix at the top of this page, we have used the  symbols for the last time in this discussion. 

From his first drawing, Pascal clearly saw the binomial coefficients lying on the upward diagonals of the figurate matrix.  His identities are presented here in order; the original 19 are reduced to 12 because some are the symmetric notational versions of the same concept; a last identity from the second part of The Treatise is also presented.  

(If you would like to see Pascal's Treatise in its original form, Prof. Edwards has asked the Cambridge Library to have the pages of this rare text displayed on their website and they have graciously given their consent. Looking at even a few pages of this can give the reader an appreciation of the value of modern notation.)

Pascal’s 1^st Identity:
    This, of course, is the key additive identity when we think of the number in our modern standard form.

Pascal’s 2^nd  Identity:  (Corollaries 2 and 3)
    These are both versions of the result of the CST, one down a column of Pascal’s Triangle, the other the version of a column under the symmetry of Pascal’s Triangle; they are the formative methods of the figurate matrix, so it isn’t surprising that Pascal puts these near the top, and it isn’t hard to see that we have symmetry in this matrix, whether we sum across a row to get an element below or down a column to get an element to the right.  Note: in the first summand, we could start the count at j=k instead of j=0 without changing the total, since all the terms where j<k are 0.  In Concrete Mathematics, the two versions of the CST are called upper summation and parallel summation, respectively.

Pascal’s 3^rd Identity:  (Corollary 4)
    A very odd looking identity in the form of Pascal’s Triangle, it’s much easier to see in the figurate matrix. In the rectangle that is above and to the left of any entry in the figurate matrix, and add up all those numbers, then add 1, and you get the original entry back.  This is really just multiple applications of the CST in one direction, then adding 1 to let us use the CST in the other direction to get the value of our original entry.  This could be called Pascal’s Rectangular Identity or upper-parallel summation, if we coin a phrase borrowing from Concrete Mathematics.

Pascal’s 4^th Identity:    (Corollaries 5 and 6)
    When we look at the array as a triangle, this is very obvious and can be proven in many ways; even written as the figurate rectangular array, it’s not hard to see that row_j=column_j, which will give us the symmetry along the upwards diagonals.

Pascal’s 5^th Identity:  and  (Corollaries 7 and 8)
    By using the additive rule to make row n from the entries in row n-1, and noticing that every entry gets used twice, adding it to make both the entry below and to the left as well as below and to the right, it’s easy to see that the sum of any row is twice the sum of the row above, and since the (sum of row₀) = 1 = 2⁰, the second part of the identity follows immediately.

    Corollary 9 is the identity  , which while clearly useful, doesn’t prove anything directly about the triangle, but shows that Corollary 8 follows directly from Corollary 7.

Pascal’s 6^th Identity: (Corollaries 10 and 11)
    Here we are looking at a partial sum across a row, then using the additive rule backwards to get the entries in the row above that make up the parts that sum to each of the summands of the left side of the equation.

Pascal’s 7^th Identity:  (Corollary 12)
    This is the basic multiplicative identity of Pascal’s Triangle that lets us build a row across starting with 1 then multiplying by the fractions .

Pascal’s 8^th Identity:  and  (Corollaries 13 and 14)
    These multiplicative identities allow us to build a column of the Triangle using just the previous entries of that column or building down a mirror image of a regular column.  This identity is also known as absorption, and Knuth, Graham and Patashnik think enough of Absorption that it ranks in their top ten binomial coefficient identities, which is Table 174 in their book Concrete Mathematics; the other identites found in Pascal's work that make the top ten are the 1^st, the 4^th, the 7^th and the 2^nd twice, both for the parallel and upper summation forms. (The four that round out the top ten that are not found in Pascal's work are the binomial theorem, trinomial revision, upper negation and Vandermonde's convolution.)

Pascal’s 9^th Identity:  (Corollary 15)
    Both sides of this equation are equal to , as we see in the first part of the 2^nd Identity (CST) and the first part of the 8^th Identity.

Pascal’s 10^th Identity: (Corollary 16)
    Both sides are again equal to , using first part of the 2^nd Identity (CST) and the second part of the 8^th Identity.

Pascal’s 11^th Identity: (Corollaries 17 and 18)

    Let two Christmas Stockings, one hung left and one hung right, have the same “heel”, the last number in the sum; they will add up to consecutive entries in the row below, and we will use the 7^th Identity to multiply by a fraction to set them equal; the only difference between these two is the fraction, where we multiply by the reciprocal of the fraction to go from one identity to the other.

Pascal’s 12^th Identity: (Corollary 19)
    Using the additive property backwards from  to get to the (2n-2)^nd row, we get

. Since the term  is surrounded by two terms equal to each other, we then use the 7^th Identity to write one of these terms as a multiple of the middle term, and with a little algebra, we are done.

    In the second part of the Treatise, Pascal concerns himself with the Problem of the Game of Points, and the identities he has come up with in Part 1 are used to solve this probability problem, with very few new insights into the Triangle added.  There is one nice identity is in this section, which I will call Pascal’s Last Identity.

Pascal’s Last Identity:  (from Part 2 of the Treatise)

    If we write , then also factor in 2n copies of 2 into the denominator, the fraction becomes  ; we can cancel out the even terms from the top with one of the sets of even terms in parentheses from the bottom, and we get the fractional form Pascal saw. 
Note: dividing binomial coefficients  by 2ⁿ is an important idea in both statistics and probability.

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