含参数差为1/2的₁F₂超几何函数的无穷级数解析解求解

Hey there, let's break down this infinite sum problem you're tackling. First, let's restate the sum clearly for reference:

$$
\sum_{k=1}^{\infty} \frac{(-1)^k y^{2k} k}{(2k)!} , {}_1F_2\left(k+1;2,k+\frac{1}{2};-\frac{x^2}{4}\right)
$$

I've looked into the structure of this ${}_1F_2$ hypergeometric function with your specific parameters, and here are some actionable approaches to simplify the sum:

1. Convert the ${}_1F_2$ function to elementary functions first #

Notice that for your parameter set ($a=k+1$, $b=2$, $c=k+\frac{1}{2}$), this ${}_1F_2$ can actually be rewritten using elementary functions. Let's verify with small $k$ values first:

• For $k=1$, ${}_1F_2\left(2;2,\frac{3}{2};-\frac{x^2}{4}\right) = {}_0F_1\left(\frac{3}{2};-\frac{x^2}{4}\right) = \frac{\sin x}{x}$ (since ${}_1F_2(a;a,c;z) = {}_0F_1(c;z)$, and this ${}_0F_1$ simplifies directly to the sine ratio).
• For $k=2$, expanding the ${}_1F_2$ series and splitting coefficients shows it can be written as:
  $$
  \frac{(x^2+1)\sin x - x + \frac{x^3}{6}}{2x3}
  $$

For general $k$, you can use induction or coefficient splitting to prove this ${}_1F_2$ is always a combination of $\frac{\sin x}{x}$, polynomial terms, and their derivatives. Once you have this elementary form, the original sum becomes a series of elementary functions that's much easier to handle.

2. Swap the order of summation #

The ${}_1F_2$ function itself is an infinite series, so substitute its expansion into the original sum and swap the order of summation (valid due to absolute convergence for non-negative $x$):
$$
{}1F_2\left(k+1;2,k+\frac{1}{2};-\frac{x^2}{4}\right) = \sum{n=0}^\infty \frac{(k+1)_n \left(-\frac{x^2}{4}\right)n}{(2)_n \left(k+\frac{1}{2}\right)_n n!}
$$

Substituting this in, the sum becomes:
$$
\sum_{n=0}^\infty \frac{\left(-\frac{x^2}{4}\right)n}{(2)n n!} \sum{k=1}^\infty (-1)^k y^{2k} k \frac{(k+1)_n}{(2k)! \left(k+\frac{1}{2}\right)_n}
$$

You can simplify the inner $k$-sum using gamma function identities (like the duplication formula for $\Gamma(2k)$) and rewrite it as a hypergeometric series. From there, you can use known ${}_2F_2$ or ${}_3F_2$ identities to find a closed form for the inner sum, then sum over $n$.

3. Use hypergeometric function recurrence relations #

There are recurrence relations specific to ${}_1F_2$ that can simplify your parameter set. For example, the derivative recurrence:
$$
\frac{d}{dz} {}_1F_2(a;b,c;z) = \frac{a}{bc} {}_1F_2(a+1;b+1,c+1;z)
$$

Or parameter shift identities that relate ${}_1F_2\left(k+1;2,k+\frac{1}{2};z\right)$ to ${}_1F_2\left(k;2,k-\frac{1}{2};z\right)$. These can help you build a recurrence for the terms in your original sum, making it easier to find a closed-form solution.

4. Simplify the original sum's coefficient first #

Notice that $\frac{k}{(2k)!} = \frac{1}{2(2k-1)!}$, so the original sum can be rewritten as:
$$
\frac{1}{2} \sum_{k=1}^{\infty} \frac{(-1)^k y^{2k}}{(2k-1)!} {}_1F_2\left(k+1;2,k+\frac{1}{2};-\frac{x^2}{4}\right)
$$

This form highlights the connection to odd factorials, which often pair nicely with sine or hyperbolic sine series. Once you have the elementary form of the ${}_1F_2$ function, you can match terms to known sine-related series identities.

备注：内容来源于stack exchange，提问作者JahonEE123