The discussion confirms that the hypergeometric function can be used to express the inverse tangent function as follows: \(\tan^{-1}x = xF\left(\frac{1}{2},\, 1,\, \frac{3}{2},\, -x^2\right)\). The series expansion of \(\arctan(x)\) is shown to match the series derived from the hypergeometric function, specifically \(F\left(\frac{1}{2}, 1, \frac{3}{2}, -x^2\right)\). This equivalence is established through the manipulation of series expansions, demonstrating the relationship between these two mathematical constructs.
PREREQUISITESMathematicians, students studying advanced calculus, and researchers exploring the connections between special functions and series expansions will benefit from this discussion.
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)ssh said:Show that tan -1(X) = x F (1\2, 1, 3/2, -x^2)