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I tried solving following partial difference equation in Mathematica by typing

RSolve[{C[n, k] == C[n-1, k] + C[n-1, k-1], C[n, 0] == 1, C[n, n] == 1}, C[n, k], {n, k}]

but returns the following error message

RSolve::conarg: The arguments should be ordered consistently. >>

but it does not work, please someone help me in this regard. Thanks and regards in advance.

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  • $\begingroup$ And when it failed it did not got you thinking ... ? $\endgroup$
    – Sektor
    Commented Mar 17, 2014 at 17:55
  • $\begingroup$ @Sektor Do you imply that there is a trivial error? I fail to see it, could you please be more specific? $\endgroup$
    – István Zachar
    Commented Mar 17, 2014 at 19:10
  • $\begingroup$ @IstvánZachar No, imply that he ran into some kind of trouble and the first thing to do is: Help me ! $\endgroup$
    – Sektor
    Commented Mar 17, 2014 at 19:17
  • $\begingroup$ @IstvánZachar C[n, n] == 1 isn't an init.cond nor a rec.rel. $\endgroup$
    – Dr. belisarius
    Commented Mar 17, 2014 at 20:23
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    $\begingroup$ @IstvánZachar To me, most Mma errors (including this one) sound more like an invitation to exegesis. $\endgroup$
    – Dr. belisarius
    Commented Mar 17, 2014 at 21:16

1 Answer 1

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It seems that Mathematica's RSolve[] simply can't solve this rather simple type of partial recurrence. However, the problem itself can easily be solved completely with Mathematica as follows. This is also a hint to try alternative formulations in mathematica if a specific one does not succeed. See Why does Mathematica not evaluate the difference recurrence using `RSolve` or `RecurrenceTable`? for a similar problem and its solution.

As a first step to solve the given recurrence we rewrite the recurrence relations as function definitions (henceforth writing a[n,k] instead of C[n,k] because C is a reserved symbol in Mathematica)

Clear[a]
a[n_, k_] := a[n, k] = a[n - 1, k] + a[n - 1, k - 1]
a[n_, n_] := a[n, n] = 1
a[n_, 0] := a[n, 0] = 1

The first few a[n,k] are then

nn = 10;
t = Table[a[n, k], {n, 0, nn}, {k, 0, n}]

(*
Out[217]= {
{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}, 
{1, 7, 21, 35, 35, 21, 7, 1}, 
{1, 8, 28, 56, 70, 56, 28, 8, 1}, 
{1, 9, 36, 84, 126, 126, 84, 36, 9, 1}, 
{1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1}}
*)

Now we proceed to find the general formula for the elements. We could guess it immediately from the numbers obtained to be a binomial coefficient. But let us proceed differently.

The generating function of the a[n,k] is easily found to be

Clear[g]

g[x_, y_] := 1/(1 - (x +  x y))

In fact the coefficients of the series expansion of g, given by

Clear[b]

b[n_, k_] := 1/(n! k!) D[g[x, y], {x, n}, {y, k}] /. {x -> 0, y -> 0}

are identical to a[n,k] as can be seen from the first few values.

Expansion of g in a geometrical series and using the binomial theorem leads to the explicit formula for the solution of the recurrence equation:

c[n_, k_] := Binomial[n, k]

which we probably have already recognized earlier from the numbers.

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