I am struggling to express error function by confluent hypergeometric function
Which I found on wikipedia: $\displaystyle \operatorname{erf}(x)=\frac{2x}{\sqrt \pi} \Phi \left(\frac{1}{2};\frac{3}{2};-x^2\right)$
I tried to Taylor expand it then integrate it which results in $\displaystyle \frac{2}{\sqrt \pi} \sum_{n=0}^{\infty}\frac {(-1)^n x^{2n+1}}{(2n+1)n!}$
From my book, this problem provided a hint $2n+1= \dfrac{\Gamma(3/2)}{\Gamma(1/2)}$, which this hints seems doesn't make any sense to me as the numerical value should be $\frac{1}{2}$
Best Answer
Begin with $$ \begin{aligned} \operatorname{erf}(z)% &=\frac{2}{\sqrt\pi}\sum_{k=0}^\infty\frac{(-1)^k z^{2k+1}}{k! (2k+1)}\\ &=\frac{2z}{\sqrt\pi}\sum_{k=0}^\infty\frac{(-z^2)^k}{k! (2k+1)}. \end{aligned} $$ Now, $$ \begin{aligned} \frac{1}{2k+1}=\frac{1}{2}\frac{1}{k+1/2}% &=\frac{1}{2}\frac{\Gamma(k+1/2)}{\Gamma(k+3/2)}\\ &=\frac{1}{2}\frac{\Gamma(k+1/2)}{\Gamma(1/2)}\frac{\Gamma(1/2)}{\Gamma(3/2)}\frac{\Gamma(3/2)}{\Gamma(k+3/2)}\\ &=\frac{1}{2}\frac{(1/2)_k}{(3/2)_k}\frac{\Gamma(1/2)}{\Gamma(3/2)}\\ &=\frac{1}{2}\frac{(1/2)_k}{(3/2)_k}\frac{\Gamma(1/2)}{\frac{1}{2}\Gamma(1/2)}\\ &=\frac{(1/2)_k}{(3/2)_k}\\ \end{aligned} $$ Substituting into our Taylor series then gives $$ \begin{aligned} \operatorname{erf}(z)% &=\frac{2z}{\sqrt\pi}\sum_{k=0}^\infty\frac{(1/2)_k}{(3/2)_k}\frac{(-z^2)^k}{k!}\\ &=\frac{2z}{\sqrt\pi}{_1F_1}\left({1/2 \atop 3/2};-z^2\right), \end{aligned} $$ with the last equality being due to the definition of the generalized hypergeometric series.