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In the book A=B on p126 7.2 (I) using WZ method to prove Gauss's 2F1 identity it is stated. $$F(n,k)=\frac{(n+k)!(b+k)!(c-n-1)!(c-b-1)!}{(c+k)!(n-1)!(c-n-b-1)!(k+1)!(b-1)!}$$ and $\sum_kF(n,k)=1$ for all and any integer n>-1 or maybe n>0 ,not sure about this detail. Anyway for n=1 this implies $$\sum_{k=-\infty}^{-b-1} \frac{(b+k)!(c-2)!(c-b-1)}{(c+k)!(b-1)!}=1$$ and can assume $b$ is nonpositive integer and $c$ is not an integer. I want to know if it is an error in the text or not. If not error think would try reflection formula and perhaps half angle or double angle or sin or cos of sum in terms of ... or something like that but I can't seem to work out the details. Can anyone prove it is true or false even taking n equal any integer convenient. I think it must be a text error for I do direct calculations for various values of c equal half an odd integer and various nonpostive integers b and start summing k beginning at -11 and going down and does not seem to be converging to 1. One can read the book by google 'book A=B'.

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2 Answers 2

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Gauss's identity says the value $\sideset{_2}{_1}F(a,b;c;1)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}$ for all $\Re(c)>\Re(a+b)$, which you can prove from the integral representation $$ \sideset{_2}{_1}F(a,b;c;z)=\frac1{B(b,c-b)}\int_0^1 x^{b-1}(1-x)^{c-b-1}(1-zx)^{-a}\,\mathrm{d}x. $$ So moving all factors to one side, letting $a=n$ and plug in the definition of $\sideset{_2}{_1}F$, this is $$ \sum_{k=0}^\infty\frac{(n+k-1)!(b+k-1)!(c-1)!}{(n-1)!(b-1)!(c+k-1)! k!}\frac{(c-n-1)!(c-b-1)!}{(c-1)!(c-n-b-1)!}=1 $$ Cancelling the common $(c-1)!$ and shifting the $k$ to match the form of $F(n,k)$ given in text, we arrive at $$ \sum_{k=-1}^\infty\frac{(n+k)!(b+k)!(c-n-1)!(c-b-1)!}{(n-1)!(b-1)!(c+k)!(k+1)!(c-n-b-1)!}=1. $$

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  • $\begingroup$ What do you get if substitute n=1 in the summand only of your last expression ? Would you not agree that it is equal $$\frac{(b+k)!(c-2)!(c-b-1)}{(c+k)!(b-1)!}$$. If so then according the text we must sum that over all k for which it is not 0 which is from k=-inf to -b-1. So what is wrong with that ?Though you may be correct in your derivation that is not exactly what the text is stating. So then you would agree it is error in the text ? $\endgroup$
    – user158293
    Oct 4, 2021 at 9:41
  • $\begingroup$ It isn't really an error. $n$ should be taken as a nonpositive integer, and the role of $a,b$ are switched around from the version stated in chapter 3. $\endgroup$ Oct 4, 2021 at 10:22
  • $\begingroup$ Strange u would assume a,b are switched around . Anyway if assume n=0 then the only non 0 term in your last sum would be k=-1 alone so then u could take $\sum_{k=-1}^{k=-1}....=1$ Since have a direct cancellation. But for that special case only! But how on earth would you assume it equals 1 in any other case such as when n is in general any other negative integer. It is anything but obvious that it would equal 1 ? What is your logic for assuming that it equals 1 in the general case of n a general negative integer because it seems nothing that cancels out to give 1. $\endgroup$
    – user158293
    Oct 4, 2021 at 20:21
  • $\begingroup$ What is also most mysterious is after u stated "moving all factors to one side...." how you ended up with a one because moving all to one side then you end up with a 0 NOT 1 on the other side. How on earth u ended up with a 1 makes no sense at all - it just does NOT compute. Could u derive this non-sensical result ? Did u expand $(1-zx)^{-a}$ into an infinite power series in $zx$ or something and eventually set z to 1? Even if you did it would seem there is no closed form expression for that integral in x 0 to 1. $\endgroup$
    – user158293
    Oct 5, 2021 at 1:29
  • $\begingroup$ Though regarding the prior comment as to why the sum is 1 for all n<1 is i guess it must because it telescopes to 0 if form $G=RF$ and then just form the summand only on lhs as fct of k but replace it by F(n,k)-F(n-1,k) then on rhs have it equals $G(k+1)-G(k)$ so now sum k=-inf to inf on both sides then it works but only provided one can show G goes to 0 at both k=-inf and k=+inf. $\endgroup$
    – user158293
    Oct 5, 2021 at 1:42
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With the help of user10354138 who pointed out that the $n$ param. in $F(n,k)$ in the book A=B on p126 7.2 (I) Gauss's 2F1 actually plays the role of $b$ on p42 3.5 I so in this case the assumption is that $n$ is a nonpositive integer. But despite the brilliance of user10354138 in using the integral representation, gamma and beta fct's etc. to show the pertinent sum $\sum_k F(n,k)=1$ the essence of the WZ method is in fact to show that we don't need to go thru all that complication if a fct. $R(n,k)$ can be found such that $G(n,k)=R(n,k)F(n,k)$ and esp. if it can be put in the very simple form of a 2 term recurrence with cst. coeff.'s such as p122 (7.1.2) which is $$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k)$$ and as he states p123 there is no assurance that such an $R(n,k)$ can be found such that the simple 2 term recurrence is true. But he states surprisingly that 99% of the time such can be found so to try that first. The confusion here is that instead of n>-1 we have n<1, a nonpositive integer though that does not necessarily mean that it will not work in this case if for $n$ a nonpositive integer we have $G(n,k)\rightarrow 0$ for $k=\pm \infty.$ The main point is in this situation of compact support we can sum both sides $k=-\infty\; to \; +\infty$ so the rhs telescopes to 0. So then $\sum_k F(n,k)$ must be independent of $n$ for all $n, n+1$ both nonpositive such that $G(n,k)\rightarrow 0$ for $k=\pm \infty.$ There is some more detail in the text in that usually one does not start off with $\sum_k F(n,k)=1$ but usually with $\sum_k F'(n,k)=r(n,k)$ for some arbitrary $r(n,k)$ depending on the particular problem and then as long as $r(n,k)$ does not equal 0 or 1 then we just divide both sides by $r(n,k)$ and set $F(n,k)=\frac{F'(n,k)}{r(n,k)}$. So the point is that esp if we can easily verify $\sum_k F(n,k)=1$ for some one convenient value of $n$ nonpositive then it is true for all nonpositive $n$. For example in this case if $n\!=\!0$ the term $k\!=\!-1$ is the only value of $k$ for which $F(0,k)$ is not 0 and clearly it equals 1. We may also want to verify that with the given text $R(n,k)=\frac{(k+1)(k+c)}{n(n+1-c)}$ that $F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k)$ for general $k,n$ which I have done by putting both sides over a common denominator which turns out to be the same on both sides and also the numerator is same on both sides so it is identity. Also assume $b,c,b\!-\!c$ are non integer to avoid complications of more infinities. Then can verify as $k$ approaches $\pm \infty$ that $G$ approaches 0 because of the infinity $(n-1)!$ in the denominator. Note also as $k$ goes to $-\infty$ even though $(n+k)!$ as $k$ goes to $-|arbitrarily\;large\;integer|$ is of magnitude $\infty$ still $\frac{(n+k)!}{(n-1)!}$ tends to 0 as $k$ approaches $-\infty.$ So this shows $G(n,k)\rightarrow 0$ also as $k\rightarrow -\infty.$ Though in fact if $n<0$ and assume $c$ not integer and $b\!-\!c$ not integer or else $c>b$ then we only need sum over $k\;-2\;to\;-n\!-\!1.$ on rhs which is $G(n,k+1)-G(n,k)$. And if $b\!-\!c$ not integer or else $c>b$ can write $$\sum_{k=-1}^{-n-1}F(n,k)=1$$ for $n<1.$ Actually the requirement $b\!-\!c$ not integer or else $c>b$ can be relaxed if one knows how to cancel infinities though if using a botched full of bugs and errors such as maxima it won't compute.

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