Constructing a Hypergeometric Function

Question

I am asked to find values for $a,b$ and $c$ such that $$ \frac{1}{2} ((1+x)^{2\alpha}-(1-x)^{2\alpha}) = 2\alpha x\ _2F_1(a,b;c;x^2)$$

I have attempted the following: $$\frac{1}{2} ((1+x)^{2\alpha}-(1-x)^{2\alpha}) = \frac{1}{2}\sum^{\infty}_{k=0}\binom{2\alpha}{k}x^k-\sum^{\infty}_{k=0}\binom{2\alpha}{k}(-x)^k \\ = \frac{1}{2} \sum^{\infty}_{k=0}\binom{2\alpha}{k}(x^k-(-x)^k). $$

My lecturer then advised that I should try to rewrite the binomial as Pochammer Symbols which led me to: $$\frac{1}{2}\sum^{\infty}_{k=0}\frac{(2\alpha -k +1)_k}{k!}(x^k-(-x)^k)$$

We can also see here that for values $k= 2n$ we will recieve a term equal to 0, hence we can rewrite our equation as:

$$\frac{1}{2}\sum^{\infty}_{n=0}\frac{(2\alpha -(2n+1) +1)_{(2n+1)}}{(2n+1)!}(x^{(2n+1)}-(-x)^{(2n+1)})$$

We simpify to:

$$x\frac{1}{2}\sum^{\infty}_{n=0}\frac{(2\alpha-2n)_{(2n+1)}}{(2n+1)!}(x^{2n}-(-x)^{2n})$$

Now at this point I'm not really sure where to go... I feel like I'm really close and not seeing something or maybe really far away and not aware! Either way any help is greatly appreciated!

Answer 1

In the expansion \begin{align} f(x)&=\frac{1}{2}\sum_{k=0}^\infty \binom{2\alpha}{k}\left( x^k-(-x)^k \right)\\ &=x\sum_{n=0}^\infty \binom{2\alpha}{2n+1}x^{2n}\\ &=x\sum_{n=0}^\infty c_nX^{n} \end{align} with $X=x^2$, the ratio of two successive terms of the series is \begin{align} \frac{c_{n+1}}{c_n}\frac{X^{n+1}}{X^n}&=\frac{\binom{2\alpha}{2n+3}}{\binom{2\alpha}{2n+1}}X\\ &=\frac{(n-\alpha+1/2)(n-\alpha+1)}{(n+3/2)}\frac{X}{n+1} \end{align} then, with $a_1=\frac{1}{2}-\alpha$, $ a_2=1-\alpha$ and $ b_1=\frac{3}{2}$ and with $c_0=\binom{2\alpha}{1}=2\alpha$, it gives \begin{align} f(x)&=2\alpha x\sum_{n=0}^\infty \frac{(a_1)_n(a_2)_n}{(b_1)_n}\frac{X^n}{n!} \\ &=2\alpha x \ _2F_1\left(\frac{1}{2}-\alpha,1-\alpha;\frac{3}{2};x^2 \right) \end{align}

Answer 2

Using Maple I looked at the Maclaurin series of the difference of the two sides to order $x^{20}$, and solved with Groebner basis methods. It found $15$ solution families: $$ \matrix{\alpha = 0, \; a,b,c\ \text{arbitrary} \cr \alpha = -1,\; a=0, \; b,c\ \text{arbitrary} \cr \alpha = -1,\; b=0, \; a,c\ \text{arbitrary}\cr \alpha = -1,\; b=2,\; a=c,\; c \ \text{arbitrary}\cr \alpha = -1,\; b=c,\; a=2,\; c \ \text{arbitrary}\cr \alpha = 3/2,\; b=-1,\; a=-c/3,\; c \ \text{arbitrary}\cr \alpha = 3/2,\; b=-c/3,\; a=-1,\; c \ \text{arbitrary}\cr \alpha = 1/2,\; a=0, \; b,c\ \text{arbitrary}\cr \alpha = 1/2,\; b=0, \; a,c\ \text{arbitrary}\cr \alpha = -1/2,\; b = 1, \; a=c,\; c \ \text{arbitrary}\cr \alpha = -1/2,\; a = 1, \; b=c,\; c \ \text{arbitrary}\cr \alpha = 2,\; b=-1,\; a=-c,\; c \ \text{arbitrary}\cr \alpha = 2,\; a=-1,\; b=-c,\; c \ \text{arbitrary}\cr a = 1/2 - \alpha, \; b = 1-\alpha,\; c = 3/2,\; \alpha \ \text{arbitrary}\cr a = 1-\alpha,\; b = 1/2-\alpha,\; c = 3/2,\; \alpha \ \text{arbitrary}\cr } $$ Maple confirms that each of these works.