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Suppose we have the following function, where $s\in\mathbb{R}$ and $t_1,t_2,n\in\mathbb{N}\cup \{0\}$ are constants:

$$\mathbf{P}(r)=\left(t_1+\prod_{k=1}^{r}(t_2+k^{s})\right)^n$$

Question: What is the asymptotic expansion of $\mathbf{P}(r)$? (Is there a more accurate approximation using special functions?)

Attempt:

Here is what I tried with Mathematica:

First, I tried to compute the asymptotic expansion of $\prod_{k=1}^{r}(t_2+k^{s})$.

(*The product expression*)productExpr = Product[t + k^s, {k, 1, r}]; (*Take the logarithm to convert the product to a sum*) logProduct = Log[productExpr]; (*Find the asymptotic expansion of the sum*) asymptoticLogSum = Asymptotic[logProduct, r -> Infinity]; (*Exponentiate the result to get the asymptotic expansion of the \ product*) asymptoticProduct = Exp[asymptoticLogSum]; (*Display the result*) asymptoticProduct // FullSimplify

However, I get the same expression:

$$\prod_{k=1}^{r}(t_2+k^{s})$$

I'm not sure how to continue further. Any answer with code or math would be appreciated.

Edit: $s\in\mathbb{R}$

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  • $\begingroup$ @ClaudeLeibovici Did you figure it out? $\endgroup$ Commented Sep 22 at 4:09

1 Answer 1

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If $s$ is a positive integer and $t_2>0$, we can consider $$\Big(\mathbf{P}(r)\Big)^{\frac 1n}-t_1=\prod_{k=1}^{r}(t_2+k^{s})=\prod_{k=1}^{r}\,\prod_{m=1}^s (k-x_m)=\prod_{m=1}^s\,\prod_{k=1}^{r}(k-x_m)$$ where the $x_m$ are the complex roots of $(k^s+t_2)=0$.

Let $$A_m=\prod_{k=1}^{r}(k-x_m)=\frac{\Gamma (r+1-x_m)}{\Gamma (1-x_m)}$$ which is asymptotic to $$\small \frac{\sqrt{2 \pi }}{\Gamma (1-x_m)\, r^{x_m}}\,\frac {r^{r+\frac{1}{2}}}{e^r}\Bigg(1+\frac{6 x_m^2-6 x_m+1}{12 r}+\frac{36 x_m^4-24 x_m^3-24 x_m^2+12 x_m+1}{288 r^2}+O\left(\frac{1}{r^3}\right) \Bigg)$$

$$\Big(\mathbf{P}(r)\Big)^{\frac 1n}-t_1=\prod_{m=1}^s A_m$$

If we just use the front factor, we should have $$\prod_{m=1}^s A_m=\frac{\Bigg(\sqrt{2 \pi}\, \,\frac {r^{r+\frac{1}{2}}}{e^r}\Bigg)^m}{\prod_{m=1}^s r^{x_m}\, \Gamma (1-x_m) } $$ I suppose that this could simplify using the duplication formula of the gamma function.

Using $s=4$ $$\Big(\mathbf{P}(r)\Big)^{\frac 1n}-t_1=-\frac{4 \sin \left((-1)^{1/4} \pi \sqrt[4]{t_2}\right) \,\,\sin \left((-1)^{3/4} \pi \sqrt[4]{t_2}\right)}{\sqrt{t_2}}\,e^{-4 r}\, r^{4 r+2}$$

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  • $\begingroup$ It seems finding an asymptotic expansion $\mathbf{P}(r)$, when $s\in\mathbb{R}$ is too complicated. I'm better off leaving $\mathbf{P}(r)$ as it is. $\endgroup$ Commented Sep 22 at 20:36

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