2

While playing around with the fresnel integrals, I came across this tantalizing power series (it is actually a particular hypergeometric series) which looks really similar to cosine! I am calling this function $C_{\frac{1}{2}}$

$$ C_{\frac{1}{2}}(x) = \sum_{n = 0}^{\infty} \frac{(-1)^nx^{2n}}{(2n+\frac{1}{2})!} $$

My question is this: can you find any point $(x, C_{\frac{1}{2}}(x))$ that satisfies this function? The only thing that I can say for sure is that the function approaches $0$ as $x$ gets arbitrarily large. A particular solution however would be VERY VERY interesting.

Arctic Char
  • 16,007
guavas222
  • 554
  • By writing $(2n+\frac{1}{2})$ I mean $\Gamma(2n +\frac{3}{2})$. Also the value at $x=0$ is not 1 it is more like 1.128 – guavas222 Jul 22 '20 at 06:49

1 Answers1

2

Amazing function between $\cos(x)$ and $\frac{\sin (x)}{x}$ ! $$f(x)=\sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n+\frac{1}{2})!}x^{2n}=\frac{2 }{\sqrt{\pi }}\,\,\, _1F_2\left(1;\frac{3}{4},\frac{5}{4};-\frac{x^2}{4}\right)$$

For large values of $x$, it seems to be $$f(x)\sim\frac{\sin \left(x+\frac{\pi }{4}\right)}{\sqrt{x} }-\frac{1}{2 \sqrt{\pi } x^2}+\frac{15}{8 \sqrt{\pi } x^4}+\cdots$$ For $x=10$, the exact value is $-0.311997$ while the above truncated expansion gives $-0.311984$.

But going deeper in the simplification of the hypergeometric function $$\color{red}{f(x)=\sqrt {\frac 2 x}\left(C\left(\sqrt{\frac{2x}{\pi }} \right) \cos (x)+S\left(\sqrt{\frac{2x}{\pi }} \right) \sin (x) \right)}$$

Edit

You must be very careful if you just sum the terms for a given value of $x$. For example, the partials sums $$S_p=\sum_{n = 0}^{p} \frac{(-1)^n}{(2n+\frac{1}{2})!}10^{2n}$$ are given below to show the serious problems.

$$\left( \begin{array}{cc} p & S_p \\ 0 & +1.12838 \\ 1 & -28.9617 \\ 2 & +162.087 \\ 3 & -372.314 \\ 4 & +465.962 \\ 5 & -374.415 \\ 6 & +210.195 \\ 7 & -88.4566 \\ 8 & +28.3181 \\ 9 & -7.75129 \\ 10 & +1.27170 \\ 11 & -0.593517 \\ 12 & -0.269554 \\ 13 & -0.317495 \\ 14 & -0.311378 \\ 15 & -0.312058 \\ 16 & -0.311992 \\ 17 & -0.311997 \end{array} \right)$$ So, now, how many terms to be added for a given accuracy ?

Writing $$f(x)=\sum_{n = 0}^{p} \frac{(-1)^n}{(2n+\frac{1}{2})!}x^{2n}+\sum_{n = p+1}^{\infty} \frac{(-1)^n}{(2n+\frac{1}{2})!}x^{2n}$$ we need to find $p$ such that $$\frac{x^{2 (p+1)}}{\left(2p+\frac{5}{2}\right)!} \leq 10^{-k}$$ that we can rewrite as $$\left(2p+\frac{5}{2}\right)! \geq x^{2p+\frac{5}{2}} \frac {10^k}{\sqrt x} $$

Looking at this question of mine, you will notice a superb approximation proposed by @robjohn. Applied to this case, it will give

$$\color{blue}{p \sim \frac 12 \left(x\, e^{1+W(t)}-3 \right)}\qquad \text{where}\qquad \color{blue}{t=\frac{1}{2 e x}\log \left(\frac{10^{2 k}}{2 \pi x^2}\right)}$$ Using $k=6$ and $x=10$, this gives $p=16.6868$ so $p=17$ (just as in the above table).

Notice that the exact solution would be $p=16.6872$.

Claude Leibovici
  • 260,315