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In this thread, How to derive this series

I have asked about how to derive:

$\dfrac{1}{1-x}=1+\dfrac{x}{1+x}+\dfrac{1\cdot2\cdot x^2}{(1+x)(1+2x)}+\dfrac{1\cdot2\cdot3\cdot x^3}{(1+x)(1+2x)(1+3x)}...$

Somos provided the excellent answer:

$$ S(x) = {}_2F_1(1,1;1+1/x;1) = \dfrac{1}{(1-x)} $$

To confirm it, I use the definition of the ordinary hypergeometric function to confirm it:

${}_2F_1(\alpha,\beta;\gamma;x)=1+\dfrac{\alpha\beta x}{\gamma\cdot1!}+\dfrac{\alpha(\alpha+1)\beta(\beta+1)x^2}{\gamma(\gamma+1)\cdot2!}+\dfrac{\alpha(\alpha+1)(\alpha+2)\beta(\beta+1)(\beta+2)x^3}{\gamma(\gamma+1)(\gamma+2)\cdot3!}...$

Everything is fine and correct until

I try to use Wolfram to confirm this identity but it doesn't give me the answer that I need.

https://www.wolframalpha.com/input/?i=2F1%281%2C1%3B1%2B1%2Fx%3B1%29

It doesn't prove that $${}_2F_1(1,1;1+1/x;1) = \dfrac{1}{1-x}$$

After tinkering with it for a while, I notice that

$${}_2F_1(1,1;1;x)= \dfrac{1}{1-x}$$

https://www.wolframalpha.com/input/?i=2F1%281%2C1%3B1%3Bx%29

My question is how to prove that ${}_2F_1(1,1;1;x)={}_2F_1(1,1;1+1/x;1)$

My second question how do you determine $\alpha, \beta, \gamma$ in ${}_2F_1(\alpha,\beta;\gamma;x)$ to determine the limiting function?

For example:

$$x{}_2F_1(\color{red}{1,1;2;-x})=\ln(x+1)$$

$$x{}_2F_1(\color{red}{\color{red}{\frac{1}{2},\frac{1}{2};\frac{3}{2};x^2})}=\arcsin(x)$$

(Is there a way to prove this without recoursing to gamma function, since I haven't learnt gamma function yet)

  • Removing the syntax sugar and noticing $_2F_1(1,1;1;x)={}_1F_0(1;x)$, I believe that this question is a not-so-exact duplicate. Anyway, it will be better to learn gamma function before playing with hypergeometric function. – Kemono Chen Jan 04 '20 at 19:42
  • Strictly, the two hypergeometric functions you have are not equal -- they have different domains. The domain of ${}_2F_1(1,1;1+\frac{1}{z};1)$ has interior ${z \in \Bbb{C} : |z-1/2|<1/2}$ (and may or may not include points on the boundary of that disk) and the other has domain $\Bbb{C} \smallsetminus {1}$. It is meaningful to ask whether they coincide on the intersection of their domains. – Eric Towers Jan 04 '20 at 20:02
  • @Eric Towers: how come they are all equal to $\dfrac{1}{1-x}$ – James Warthington Jan 04 '20 at 20:06
  • Try $x=2$ in ${}_2F_1(1,1;1+1/x;1)$: https://www.wolframalpha.com/input/?i=2F1%281%2C1%2C3%2F2%2C1%29 . You do not get $-1$; you get complex infinity because that function is not defined at $x = 2$. (This is basically because the series you wrote down does not converge for ${}_2F_1(1,1;3/2;1)$.) – Eric Towers Jan 04 '20 at 20:10
  • So this series is not equal to $\dfrac{1}{1-x}$? – James Warthington Jan 04 '20 at 20:12
  • A function is not just a recipe for converting inputs to outputs. A function is a pair: (domain, recipe). For two functions to be equal, they must have the same domain and the values produced by their recipes must agree at each point of the domain. If two functions do not have the same domain, they are not equal as functions. It is meaningful to ask whether they are equal when we shrink both domains to the intersection of their original domains. – Eric Towers Jan 04 '20 at 20:15

1 Answers1

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In using Hypergeometric series you have to be careful to distinguish between the sum of the series which may or may not be convergent depending on the parameters, and the Hypergeometric function which may be analytic except for poles or essential singularities or even branch points for logarithm. The important thing to know is that a Hypergeometric series is just a series and there are many different ways to proving convergence of series and not all of them apply in any particular case. For example, the ratio test may be applicable to prove convergence or divergence or may be inconclusive.

As a simple example, consider the geometric series $\, 1+x+x^2+\cdots \,$ which is convergent inside the unit circle. In this domain it agrees with $\,f(x) := 1/(1-x)$ which is a rational function with a simple pole at $\,1.$ The same function $\,f(x)\,$ has a geometric series expansion around any finite complex number. Thus, on some domains, $\,f(x)\,$ can be expressed in many different geometric series. These geometric series have different domains of convergence and are different as series yet they agree with $\,f(x)\,$ on their common domain.

Another example is the hypergeometric series $\, 1 + 1!x + 2!x^2 + 3!x^3 + \dots\,$ which is only convergent at $\,x=0.\,$ The great Euler was able to express this series in terms of the Exponential integral. From Wikipedia we have

However, there is a divergent series approximation that can be obtained by integrating $\,ze^zE_1(z)\,$ by parts $$ E_1(z)=\frac{\exp(-z)}z \sum_{n=0}^{N-1} \frac{n!}{(-z)^n} $$

and that was one of the ways that Euler used his hypergeometris series.

Somos
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  • So what does when Catalan mean(the series is taken from Catalan book) when he says that the function $\dfrac{1}{1-x}$ is also developable into $1+\dfrac{x}{x+1}+\dfrac{1\cdot 2x^2}{(1+x)(1+2x)}...$. I thought what he means is that $\dfrac{1}{1-x}$ is the final sum of the hypergeometric series? If this is not the case then how can I understand this? – James Warthington Jan 05 '20 at 00:06
  • @JamesWarthington Thanks for telling who the author was. What is meant, is, as usual, that if the series convergences, then it is equal to the value $1/(1-x)$. This is standard practice. – Somos Jan 05 '20 at 00:25
  • So is there a way to prove that the hypergeometric series is convergent towards $\dfrac{1}{1-x}$? – James Warthington Jan 05 '20 at 00:29
  • @JamesWarthington Please read the end of my answer to your original question. It converges for some values of $x$. Not all. – Somos Jan 05 '20 at 00:31
  • Is there a way that I can find the interval of convergence of the hypergeometric series? – James Warthington Jan 05 '20 at 00:39
  • @JamesWarthington Interval of convergence applies only to real power series. – Somos Jan 05 '20 at 01:19
  • I am confused, so how does one prove that the given hypergeometric series is going to converge to $\dfrac{1}{1-x}$ for some values of $x$? – James Warthington Jan 05 '20 at 01:48