00:02
We are given the sequence catalog numbers, and we are asked to, in part a, find the generating function, and in part b, to find the value of the catalog number, cn.
00:27
So first of all, in part a, you have that the catalon numbers are given by your recursive formula, cn is equal to sum from k equals zero to n minus one, of ck times cn minus k minus one, with initial condition.
00:41
C0 equals c1 equals 1.
00:49
If we take g of x to be the generating function of the catalan numbers, this is the sum from k equals 0 to infinity of c0 x to the k, or c of k x to the k, well, then we have that g of x squared is going to be the sum from k equals 0 to infinity and then the sum from m equals 0 to k of c k c m times c k minus m x to the k so the idea is we want to get this somehow back into the form of this record recursive formula so we have that x times g of x squared x squared minus g of x plus 1.
03:09
This is equal to sum from k equals 0 to infinity, sum from m equals 0 to k, of cm, ck minus m, x to the k plus 1, minus sum from k equal 0 to infinity of ck, x to the k, plus 1, we have that this is equal to the sum from k equals 0 to infinity, of the sum from m equals 0 to k, of cm times ck minus m, x to the k plus first, minus sum from k equals 1 to infinity of c k, x to the k, and then this is going to be plus 1 minus, c0, and this reduces 2.
05:12
You have that c0 is 1, so the ones cancel out.
05:16
This is going to be the sum from k equals 0 to infinity, of the sum from m equals 0 to k, of c sub m, c sub k minus m, x to the k plus 1 minus, and by the recursion formula, this is going to be the sum from, k equals 1 to infinity of the sum from m equals 0 to k of c m or 0 to k minus 1 i should say and this is going to be c m times k minus m plus 1 or minus 1 i mean times x to the k making a substitution we have that this is the same as the sum from k equals zero to infinity of the sum from m equals zero to k of c m c k minus m instead of k equals zero we can instead of k equal zero we can make substitution k minus one for k so this becomes k equals one to infinity the sum from m equals zero to k minus one of c m c k minus m which is k minus m minus 1, then x to k minus 1 plus 1 is simply x to the k minus the same thing.
07:52
This is going to give us 0.
07:54
And so it's shown that g of x does satisfy this equation here.
08:08
And then we're asked to conclude the form of g of x from this.
08:15
So it follows that g of x is equal to 1 plus or minus the square root of 1 ,000, minus 4 times 1 times 1 all over 2 times 1, or minus 4 times x times 1, all over 2x.
09:25
And we have that initial conditions tell us c0 equals 1 and c1 equals 1.
09:33
So we have that g of 0 is going to be c of 0, which is 1.
09:40
This has to be equal to, well we don't know what this is, because we have an x in the denominator.
09:55
But say we take if g is 1, we have 1 plus or minus square root of 1 minus 4 times 1 over 2.
10:38
That's one way to look at it.
10:47
Another way to see this is, again, if we try g of 0, this is supposed to be c of 0, which is 1.
10:54
But if we take 1 plus the square root of 1 minus 4 times 0 is just 1 over 2 times 0, we get 2 over 0, which is infinity.
11:09
So we don't have a first term in this case.
11:11
But if we take 1 minus root 1 over 2 times 0, then we have 0 over 0.
11:19
And there's a possibility that this could be equal to 1 in the limit.
11:25
And so we'll take a negative sign here.
11:30
So we have that g of x is equal to 1 minus square root of 1 minus 4x over 2x.
11:41
And so the way you would do this is maybe using a computer algebra system, verify that the limit as g approaches 0 when we have a plus sign does not exist.
11:53
The limit as g approaches 0 when a negative sign does exist and is equal to 1.
12:03
In part b, we're asked to use a previous exercise to conclude the form of g of x and then the form of the catalan numbers.
12:19
We have by the extended binomial theorem that 1 minus 4x to the negative 1 half is equal 2, 1 plus negative 4x to the negative 1 half.
12:42
And this is the sum from k equals 0 to infinity of negative 1 half, choose k, negative 4 x to the k and this is equal to by the previous exercise sum from k equals zero to infinity of 2k over k over negative 4k to k choose k divided by negative 4 to the k and this is of course going to be negative 4 to the k times x to the k this reduces to the sum from k equals zero to infinity of 2k choose k x to the k.
14:03
And now notice that if we take the derivative of g of x from the previous problem, this is where the calculus is involved.
14:17
G prime of x is going to be the bottom, which is 2x times the derivative of the top, which is negative 1 minus 4x to the negative 1ā2x to the negative 1ā2 times negative 4.
14:38
So this is simply 4 times 1 minus 4x to the negative 1.
14:41
Negative one half.
14:43
The bottom times the top minus the top, which is 1 minus 4x to the 1 half, or 1 minus square root of 1 minus 4x, times the derivative of the bottom, which is simply 2, all over the bottom squared, which is 4x squared.
15:17
This is equal to we had 8x times 1 minus 4x to the negative 1 half minus 2.
15:41
Plus 2 times 1 minus 4x to the positive 1 1 1ā2 for 4x squared...