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We define an entire function on $\mathbb{C}^m$ by $$ f(z_1,\cdots,z_m)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}t^{2n}(z_1^2+\cdots+z_m^2)^n, $$ here $t$ is some (positive) real number. Of course, $f(x)=\text{cos} (t|x|)$ for $x=(x_1,\cdots,x_m)\in\mathbb{R}^m$ hence is bounded by the constant $1$. My question is:

For $x=(x_1,\cdots,x_m)\in\mathbb{R}^m$ and $z=(z_1,\cdots,z_m)\in \mathbb{C}^m$, if $|z_k-x_k|=1, k=1,\cdots, m$, then could we get a reasonable bound for $|f(z)|$ in terms of $t$ and $x=(x_1,\cdots,x_m)$?

I hope it can be bounded by a polynomial of $(t, x)$.

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  • $\begingroup$ That is certainly not possible. For example, if $n=1$ and $z=i$ (so $x=0$), then $f(z)=\cos it$. $\endgroup$ – Christian Remling Oct 20 '16 at 18:05
  • $\begingroup$ Oh, yes, good, so I only hope it to be bounded by a fuction of $(t, x)$ which is a polynomial of $x$. $\endgroup$ – Lao-tzu Oct 20 '16 at 18:09
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$f(z_1, \ldots, z_m) = \cos(t \sqrt{z_1^2 + \ldots + z_m^2})$ so if $s$ is a square root of $z_1^2 + \ldots + z_m^2$, $$\sinh^2(t \; \text{Im}(s)) \le |f(z_1, \ldots, z_m)|^2 \le \cosh^2(t\; \text{Im}(s))$$

If $z_k = x_k + e^{i \theta_k}$ for all $k$, then write $$ \sum_{k=1}^m z_k^2 = A+2iB = R e^{i\theta}$$ where $$\eqalign{ A &= \sum_k (x_k^2 + 2 x_k \cos(\theta_k) + \cos(2\theta_k))\cr B &= \sum_k (2 x_k \sin(\theta_k) + \sin(2 \theta_k))\cr R &= \sqrt{A^2 + B^2}\cr \sin \theta &= B/R\cr \cos \theta &= A/R\cr \text{Im}(s) &= \sqrt{R} \sin(\theta/2)\cr }$$ For example, we could take all $x_k = x > 1$, all $\theta_k = \pi/2$, so $$\eqalign{ \sum_{k=1}^m z_k^2 &= m (x+i)^2\cr \text{Im}(s) &= \sqrt{m}\cr \sinh^2&(t \sqrt{m}) \le |f(z_1, \ldots, z_m)|^2 \le \cosh^2(t \sqrt{m})}$$

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  • $\begingroup$ Shouldn't the first line have $\cos(t \sqrt{(z_1^2 + \ldots + z_m^2)})$ instead of $\cos(t (z_1^2 + \ldots + z_m^2))$? $\endgroup$ – Igor Khavkine Oct 20 '16 at 18:34
  • $\begingroup$ @Robert Israel Thanks a lot for the work! So what I hope is too much. $\endgroup$ – Lao-tzu Oct 20 '16 at 21:12

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