Question

Show that: J_-1/2(x)=√(2/πx)cosx

Answer 1

Answer on Question #85174 – Math – Differential Equations

Question

Show that: $J_{\frac{1}{2}}(x) = \sqrt{\frac{2}{\pi x}} \cdot \cos(x)$

Solution

We know that

$$\begin{aligned} J_k(x) &= \sum_{n=0}^{\infty} \frac{(-1)^n}{n! \cdot \Gamma(n+k+1)} \cdot \left(\frac{x}{2}\right)^{2n+k} \Rightarrow J_{\frac{1}{2}}(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{n! \cdot \Gamma\left(n - \frac{1}{2} + 1\right)} \cdot \left(\frac{x}{2}\right)^{2n - \frac{1}{2}} = \\ &= \sqrt{\frac{2}{x}} \cdot \sum_{n=0}^{\infty} \frac{(-1)^n}{n! \cdot \Gamma\left(n + \frac{1}{2}\right)} \cdot \left(\frac{x}{2}\right)^{2n} = \left\{\Gamma\left(n + \frac{1}{2}\right) = \frac{1 \cdot 3 \cdot \ldots \cdot (2n-1)}{2^n} \sqrt{\pi} = \frac{(2n-1)!!}{2^n} \sqrt{\pi} \right\} = \\ &= \sqrt{\frac{2}{x}} \cdot \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n-1)!! \cdot \sqrt{\pi}} \cdot \frac{x^{2n}}{2^{2n}} = \sqrt{\frac{2}{\pi x}} \cdot \sum_{n=0}^{\infty} \frac{(-1)^n}{2^n \cdot n! \cdot (2n-1)!!} \cdot x^{2n} = \\ &= \{2^n \cdot n! \cdot (2n-1)!! = (2n)! \} = \sqrt{\frac{2}{\pi x}} \cdot \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} \cdot x^{2n} = \left\{\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!} \cdot x^{2n} = \cos(x) \right\} = \\ &= \sqrt{\frac{2}{\pi x}} \cdot \cos(x). \end{aligned}$$

Answer: $J_{\frac{1}{2}}(x) = \sqrt{\frac{2}{\pi x}} \cdot \cos(x)$.

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