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While playing around with the fractional calculus, I got stuck trying to show that two different ways of differintegrating the cosine give the same result. DLMF and the Wolfram Functions site don't seem to have this "identity" or something that can obviously be transformed into what I have, so I'm asking here.

The "identity" in question is

$(\alpha-1)\left({}_1 F_2 \left(1;\frac{1-\alpha}{2},\frac{2-\alpha}{2};-\frac{x^2}{4}\right)-{}_1 F_2 \left(-\frac{\alpha}{2};\frac12,\frac{2-\alpha}{2};-\frac{x^2}{4}\right)\cos(x)\right)\stackrel{?}{=}\alpha x \sin(x)\,{{}_1 F_2 \left(\frac{1-\alpha}{2};\frac32,\frac{3-\alpha}{2};-\frac{x^2}{4}\right)}$

Expanding the LHS minus the RHS in a Taylor series shows that the coefficients up to the 50th power are 0; trying out random complex values of $\alpha$ and $x$ seems to verify the identity. I would however like to see a way to confirm the identity analytically. How do I go about it?

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    $\begingroup$ The usual approach to such identities is to find a linear differential equation satisfied by both sides. Here you might get away without finding one explicitly; if you know one of a certain form exists and both sides are the same modulo a high power of $x$, possibly that would be enough to ensure that both sides are the same. $\endgroup$
    – Robin Chapman
    Commented Sep 5, 2010 at 12:55

2 Answers 2

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You can use the great HolonomicFunctions package by Christoph Koutschan to prove this identity in Mathematica. It automatically proves for you that both sides of your identity satisfy the sixth order differential equation \begin{eqnarray} 0=&&x^2 \left(2 a^2-11 a+18 x^2+14\right) D_x^6 -2 x \left(2 a^3-19 a^2+18 a x^2+58 a-54 x^2-56\right) D_x^5 \\\\ &&+\left(2 a^4-25 a^3+28 a^2 x^2+115 a^2-133 a x^2-230 a+90 x^4+154 x^2+168\right) D_x^4 \\\\ &&-4 x \left(4 a^3-37 a^2+36 a x^2+115 a-99 x^2-114\right) D_x^3 \\\\ &&+4 \left(2 a^4-23 a^3+20 a^2 x^2+96 a^2-71 a x^2-172 a+18 x^4+71 x^2+112\right) D_x^2 \\\\ &&+8 x \left(4 a^2-34 a+36 x^2+43\right) D_x +8 \left(2 a^2-17 a+18 x^2+35\right). \end{eqnarray} Together with your check that the first six Taylor coefficients (with respect to x) agree, this proves your identity.

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  • $\begingroup$ Awesome stuff, thanks a lot! This takes care of my particular problem, though I still can't help but feel there's a more general addition theorem for ${}_1 F_2$ lurking behind the scenes, of which what I have is a mere special case. I wonder how one might find this? $\endgroup$
    – J. M. isn't a mathematician
    Commented Sep 6, 2010 at 1:26
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As Robin pointed out already, it is sufficient to note that both sides satisfy a linear differential equation, since the hypergeometric functions, sine and cosine do so, and power series satisfying linear differential equations are closed under addition and multiplication.

You only have to find bounds for order and coefficient degree, and check appropriately many Taylor coefficients. gfun for maple and generatingFunctions for mathematica do it for you...

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  • $\begingroup$ I've already done the task of expanding the difference of the two sides of the "identity" in a Taylor series; how do I determine "appropriately many"? $\endgroup$
    – J. M. isn't a mathematician
    Commented Sep 5, 2010 at 14:24
  • $\begingroup$ Suppose f has order (maximal derivative) d1 and g has order d2. Then f+g has order at most d1+d2 and fg has order at most (d1+1)(d2+1)-1. Hm. I might have made a stupid mistake: I do not see a bound on the coefficient degrees right now. $\endgroup$
    – Martin Rubey
    Commented Sep 5, 2010 at 15:48

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