How to determine the coefficients of hypergeometric functions?

Question

I just wonder how do you determine the parameters of hypergeometric function so that they coincide with their limiting functions?

$$_2F_1(\color{red}{1,1;1+\dfrac{1}{x};1})=1+\dfrac{x}{1+x}+\dfrac{1\cdot2x^2}{(1+x)(1+2x)}+\dfrac{1\cdot2\cdot 3x^3}{(1+x)(1+2x)(1+3x)}$$

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

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

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

How do you obtain these identities by determining the arguments of hypergeometric functions?

Answer 1

Example 1
$f(x) = x\;_2F_1(1,1;2;-x)$ satisfies differential equation $$ (1+x) f''(x) + f'(x) = 0,\quad f(0)=0, \quad f'(0)=1 $$ But $\log(1+x)$ also satisfies this. Therefore $x\;_2F_1(1,1;2;-x) = \log(1+x)$.

Example 2
$f(x) = x\;_2F_1\left(\frac{1}{2},\frac{1}{2};\frac{3}{2};x^2\right)$ satisfies differential equation $$ (x^2-1) f''(x) + x f'(x) = 0,\quad f(0)=0,\quad f'(0)=1 $$ But $\arcsin x$ also satisfies this. Therefore $x\;_2F_1\left(\frac{1}{2},\frac{1}{2};\frac{3}{2};x^2\right) = \arcsin x$.

Examle 3
$f(x) = {}_2F_1(1,1;1;x)$ satisfies differential equation $$ (x-1) f'(x) +f(x) = 0, \quad f(0)=1 $$ But $\frac{1}{1-x}$ also satisfies this.

Answer 2

Since the hypergeometric function ${}_2F_1(a,b;c;z)$ is defined as power series in $z$ as \begin{align*} {}_2F_1(a,b;c;z)=\sum_{n=0}^{\infty}\frac{a^{\overline{n}}b^{\overline{n}}}{c^{\overline{n}}}\frac{z^n}{n!}\qquad\quad |z|<1 \end{align*} with $q^{\overline{n}}=q(q+1)\cdots(q+n-1)$ we can use this definition given the power series representations of the stated functions are known. We calculate ordered by increasing complexity

Case: ${}_2F_1(1,1;1;z)$

We obtain \begin{align*} \color{blue}{{}_2F_1(1,1;1;z)}&=\sum_{n=0}^\infty\frac{1^{\overline{n}}1^{\overline{n}}}{1^{\overline{n}}}\frac{z^n}{n!}\\ &=\sum_{n=0}^\infty \frac{n!n!}{n!}\frac{z^n}{n!}\\ &=\sum_{n=0}^\infty z^n\\ &\,\,\color{blue}{=\frac{1}{1-z}\qquad\quad |z|<1} \end{align*}

Case: $z\,{}_2F_1(1,1;2;-z)$

We obtain \begin{align*} \color{blue}{z\,{}_2F_1(1,1;2;-z)}&=z\sum_{n=0}^\infty\frac{1^{\overline{n}}1^{\overline{n}}}{2^{\overline{n}}}\frac{(-z)^n}{n!}\\ &=z\sum_{n=0}^\infty (-1)^n \frac{n!n!}{(n+1)!}\frac{z^n}{n!}\\ &=\sum_{n=0}^\infty (-1)^n\frac{ z^{n+1}}{n+1}\\ &=\sum_{n=1}^\infty (-1)^{n-1}\frac{ z^{n}}{n}\tag{1}\\ &\,\,\color{blue}{=\ln(1+z)\qquad\quad |z|<1} \end{align*}

Comment:

• In (1) we shift the index to start with $n=1$.

Case: $z\,{}_2F_1\left(\frac{1}{2},\frac{1}{2};\frac{3}{2};z^2\right)$

We obtain \begin{align*} \color{blue}{z\,{}_2F_1\left(\frac{1}{2},\frac{1}{2};\frac{3}{2};z^2\right)}&=z\sum_{n=0}^\infty\frac{\left(\frac{1}{2}\right)^{\overline{n}}\left(\frac{1}{2}\right)^{\overline{n}}}{\left(\frac{3}{2}\right)^{\overline{n}}}\frac{z^{2n}}{n!}\\ &=\sum_{n=0}^\infty\frac{\frac{1}{2}\cdot\frac{3}{2}\cdots\left(\frac{2n-1}{2}\right)\cdot\frac{1}{2}\cdot\frac{3}{2}\cdots\left(\frac{2n-1}{2}\right)} {\frac{3}{2}\cdot\frac{5}{2}\cdots\left(\frac{2n+1}{2}\right)}\frac{z^{2n+1}}{n!}\\ &=\sum_{n=0}^\infty\frac{\frac{1}{2}} {\frac{2n+1}{2}}\,\frac{(2n-1)!!}{2^n}\,\frac{z^{2n+1}}{n!}\tag{2}\\ &=\sum_{n=0}^\infty\frac{1} {2n+1}\,\frac{(2n)!}{2^n(2n)!!}\,\frac{z^{2n+1}}{n!}\tag{3}\\ &=\sum_{n=0}^\infty \frac{1}{4^n}\binom{2n}{n}\,\frac{z^{2n+1}}{2n+1}\tag{4}\\ &\,\,\color{blue}{=\arcsin(z)\qquad\quad |z|<1} \end{align*}

Comment:

• In (2) we cancel terms and from the $n$ right-most terms of the numerator we factor out $\frac{1}{2^n}$ and use double factorials $(2n-1)!!=(2n-1)(2n-3)\cdots 3\cdot 1$.

• In (3) we use double factorials $(2n)!!=(2n)(2n-2)\cdots4\cdot2$ and the identity $(2n)!=(2n)!!(2n-1)!!$.

• In (4) we use $(2n)!!=2^nn!$ and binomial coefficient notation.

Case: ${}_2F_1\left(1,1;1+\frac{1}{x};1\right)$:

We start with \begin{align*} {}_2F_1\left(1,1;1+\frac{1}{x};z\right)&=\sum_{n=0}^\infty\frac{1^{\overline{n}}1^{\overline{n}}}{\left(1+\frac{1}{x}\right)^{\overline{n}}}\frac{z^n}{n!}\\ &=\sum_{n=0}^\infty \frac{n!n!}{\left(1+\frac{1}{x}\right)\left(2+\frac{1}{x}\right)\cdots\left(n+\frac{1}{x}\right)}\frac{z^n}{n!}\\ &=\sum_{n=0}^\infty \frac{n!x^n}{\left(1+x\right)\left(1+2x\right)\cdots\left(1+nx\right)}z^n\qquad\quad |z|<1\tag{5} \end{align*}

Assuming the limit $z\to 1$ in (5) exists and the expression is defined for specific values of $x$ we have \begin{align*} \color{blue}{{}_2F_1\left(1,1;1+\frac{1}{x};1\right)} &=\lim_{z\to 1}{}_2F_1\left(1,1;1+\frac{1}{x};z\right)\\ &\,\,\color{blue}{=\sum_{n=0}^\infty \frac{n!x^n}{\left(1+x\right)\left(1+2x\right)\cdots\left(1+nx\right)}} \end{align*}