Ah, the 'real' rwg stood up. My Respects, Sir! The following generalisation pops up in my mind: ... can an algorithm be described that similarly simplifies any ratio of two (order n)polynomes, given that their coefficients of x^k in numerator and denominator are (* of closed form *) :: replace (* .. *) with "a non-recursive expression of n and k" as needed. Or will it always remain a job to be done by hand & eye? Application (for the mathematicaly underprivileged): There are loads of sequences whose GF can be approached by such f_n(x)/g_n(x) ratios. They occur as (I - x A)^-1 where I is the appropriate Identity matrix, and A is a substitution matrix symbolising for example (in the case at hand) the substitution game k->(1,2,...,k,k+1) that generates the Catalans.

-----Original Message----- From: R. William Gosper [mailto:rwg@tc.spnet.com] Sent: vrijdag 7 mei 2004 7:27 To: math-fun@mailman.xmission.com Subject: Re: [math-fun] FW: funny limit to well known Generating Function

Wouter asked

...

...

f[x_] := Sum[(-1)^(n-j) Binomial[j,n-j] x^(n-j),{j,1,n}] / Sum[(-1)^(n+2-j) Binomial[j,n+2-j] x^(n+2-j),{j,1,n+2}] produces a well known Generating Function for the limit n->Infinity.

...

...

Q: What is needed to force Limit[f[x], n->Infinity] into submission? it equals (1-2x - Sqrt[1 - 4*x]) / (2 x^2)

...

Mathematica's package Calculus`Limit` doesn't feel like tackling it. Should it?

and I blathered

...

Both Mma and Macsyma need a little help here. Just send the sum limits to infinity first. It might also help to start from j=0, which you may have intended. If not, Mma may need you to "manually" subtract the j=0 term. Then you can even start from j=-inf. The resulting sums are simple algebraics(n), and n->inf is a snap.

Bull. This isn't even a limit problem. But it exposes several defects in my personally enhanced Macsyma. All that should have been necessary was (declare(n,integer), hypersimp(sum(...))). Instead, I have to sum from floor(n/2) instead of 1: (c225) (declare(n,integer),sum((-1)^(n-j)*binomial(j,n-j)*x^(n-j),j,f loor(n/2),n) /sum((-1)^(n+2-j)*binomial(j,n+2-j)*x^(n+2-j),j,floor(n/2),n+2))

n ==== \ n + j n - j > binomial(j, n - j) (- 1) x / ==== n j = floor(-) 2 (d225) --------------------------------------------------------- n + 2 ==== \ n + j n - j + 2 > binomial(j, n - j + 2) (- 1) x / ==== n j = floor(-) 2

then manually reverse the summation direction, (c226) rangereverse(%)

n floor(-) - 1 2 ==== \ n + j n - j > binomial(j, n - j) (- 1) x / ==== j = n + 1 (d226) --------------------------------------------------------- n floor(-) - 1 2 ==== \ n + j n - j + 2 > binomial(j, n - j + 2) (- 1) x / ==== j = n + 3

and then manually change the summation variable: (c227) linchange(%,n+1-j)

n n - floor(-) 2 ==== j \ binomial(n - j, j) x > (- ---------------------) / j ==== (- 1) j = 0 (d227) ---------------------------------------------- n n - floor(-) 2 ==== j + 2 \ binomial(n - j, j + 2) x > (- -----------------------------) / j ==== (- 1) j = - 2

Finally, (c228) hypersimp(%)

4 (d228) -------------------- 2 (sqrt(1 - 4 x) + 1)

I assume this is the square of what you'd have gotten with f(n)/f(n+1) instead of f(n)/f(n+2). --rwg

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