# Hypergeometric equation

#### ssh

##### New member
Show that,

$\tan^{-1}x = xF\left(\frac{1}{2},\, 1,\, \frac{3}{2},\, -x^2\right)$

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#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Show that tan -1(X) = x F (1\2, 1, 3/2, -x^2)
First note that the expansion of $\arctan(x) = \sum^{\infty}_{0}\frac{(-1)^n(x)^{2n+1}}{2n+1}$ $$F (\frac{1}{2}, 1, \frac{3}{2}, -x^2) = 1+\frac{\frac{1}{2}\cdot 1}{\frac{3}{2}}(-x^2)+\frac{\frac{1}{2}\cdot \frac{3}{2}\cdot 1 \cdot 2}{\frac{3}{2}\cdot \frac{5}{2}\cdot 2!}(x^4)+\frac{\frac{1}{2}\cdot \frac{3}{2}\cdot \frac{5}{2}\cdot 1 \cdot 2\cdot 3}{\frac{3}{2}\cdot \frac{5}{2}\cdot\frac{7}{2}\cdot 3!}(-x^6)+....$$ $$xF (\frac{1}{2}, 1, \frac{3}{2}, -x^2)= x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+... =\sum^{\infty}_{n=0}\frac{(-1)^nx^{2n+1}}{2n+1}$$ which is exactly the expansion of the arctan(x)