On nonnegative-order first-kind Bessel functions with large argument

IridescentRain
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Hello.

I'm not terribly proficient with Bessel functions, but I know that those of the first kind are given by
<br /> \begin{eqnarray}<br /> J_n(x) &amp; = &amp; \left(\frac{x}{2}\right)^n\,\sum_{\ell=0}^\infty\frac{(-1)^\ell}{\ell!\,\Gamma(n+\ell+1)}\,\left(\frac{x}{2}\right)^{2\ell},<br /> \end{eqnarray}<br />where \Gamma is the gamma function.

I found in a book that for large x the following is true:
<br /> \begin{eqnarray}<br /> J_n(x) &amp; = &amp; \sqrt{\frac{2}{\pi x}}\,\cos\left(x-\frac{n\pi}{2}-\frac{\pi}{4}\right)\,\left[1+\mathcal{O}(x^{-1})\right]\\<br /> &amp; \approx &amp; \sqrt{\frac{2}{\pi x}}\,\cos\left[x-(2n+1)\,\frac{\pi}{4}\right].<br /> \end{eqnarray}<br />The second line of that last equation is obvious (they just removed every term except the first one from the first line because 1/x^\ell is negligible for large x and \ell\ge1), but where does the cosine come from? How would I begin to prove this?

I'm working with n\in\mathbb{N}\cup\{0\} only, so perhaps this expression is only true for Bessel functions of natural (or zero) order; I wouldn't know. All I know is that for such values of n the gamma function becomes \Gamma(n+\ell+1)=(n+\ell)!.

Thanks in advance for any help!
 
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\cos(x) = \sum^{\infty}_{k=0} (-1)^{k} \frac{x^{2k}}{(2k)!}

Try using that.
 
Thanks, Millennial. I will.
 
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