# Majoring sum with factorials

EDIT: sorry for all the crappy formulas, I tried to add the more context I can, I hope it is better.

I am currently trying to analytically bound the following term: $$\left\vert\left\vert e^{\lambda (H_0 + H_1)} - S_2(\lambda)\right\vert\right\vert$$ with $$S_2(\lambda) = e^{\frac{\lambda}{2}H_0}e^{\lambda H_1}e^{\frac{\lambda}{2}H_0}$$ for $$H_0$$ and $$H_1$$ two non-commuting matrices with the following properties:

• $$H_0^2 = H_1^2 = I$$
• $$\Lambda = \max \left\{ \vert\vert H_0\vert\vert, \vert\vert H_1\vert\vert \right\} = \vert\vert H_0\vert\vert = \vert\vert H_1\vert\vert = 1$$.

My final goal is to bound the following error $$\left\vert\left\vert e^{\lambda (H_0 + H_1)} - S_2\left(\frac{\lambda}{r}\right)^r\right\vert\right\vert \leqslant g\left(\lambda, \Lambda, r\right) \leqslant \epsilon$$ by a function $$g$$ and to solve $$g\left( \lambda, \Lambda, r \right) \leqslant \epsilon$$ for $$r$$ to obtain an expression of $$r$$ as: $$r = h\left( \lambda, \Lambda, \epsilon \right).$$

I currently have 2 bounds that are far from optimal from [1] (Section F.1, page 24-26): $$r^{\text{analytic}} = \left\lceil \max \left\{ 4 \Lambda \vert\lambda\vert, \sqrt{\frac{\left(4 \Lambda \vert\lambda\vert\right)^3}{3\epsilon}} \right\} \right\rceil$$ and $$r^{\text{minimized}} = \min \left\{ r \in \mathbb{N}^* : \frac{\left(4 \Lambda \vert\lambda\vert\right)^3}{3r^2} \exp\left(\frac{4 \Lambda \vert\lambda\vert}{r}\right) \leqslant \epsilon \right\}.$$

During the mathematical development of the error term I stumbled upon a mathematical expression that I failed to major finely enough to obtain a usable error bound: $$\left\vert\left\vert e^{\lambda (H_0 + H_1)} - S_2(\lambda)\right\vert\right\vert \leqslant \sum_{n=0}^{+\infty} \vert\lambda\vert^n \Lambda^n \left[ \frac{2^n-1}{n!} - \frac{n(n+1)}{2n!} + f(n)\right]$$ with $$f(n) = \sum_{j=1}^{n} \sum_{i=0}^{n-j} \left\vert \frac{1}{n!} - \frac{1}{2^{n-j} i! j! (n-j-i)!} \right\vert.$$

Using the triangle inequality, I ended up with

$$f(n) \leqslant \frac{n(n+1)}{2 n!} + \frac{2^n - 1}{n!}$$ by using the identity $$\sum_{i=0}^n \frac{n!}{i! (n-i)!} = 2^n$$ which leads to $$\left\vert\left\vert e^{\lambda (H_0 + H_1)} - S_2(\lambda)\right\vert\right\vert \leqslant 2 \left[ e^{2\vert\lambda\vert\Lambda} - e^{\vert\lambda\vert \Lambda} \right].$$ and finally to \begin{align} \left\vert\left\vert e^{\lambda (H_0 + H_1)} - S_2\left(\frac{\lambda}{r}\right)^r\right\vert\right\vert &\leqslant r \left\vert\left\vert e^{\lambda (H_0 + H_1)} - S_2\left(\frac{\lambda}{r}\right)\right\vert\right\vert\\ &\leqslant g(\lambda, \Lambda, r) = 2r e^{\vert\lambda\vert\Lambda / r} \left( e^{\vert\lambda\vert\Lambda / r} - 1 \right). \\ \end{align}

The main problem with $$g(\lambda, \Lambda, r)$$ is that its limit when $$r \to +\infty$$ is not $$0$$ (it is $$2\Lambda\vert\lambda\vert$$), so I will not be able to solve $$g\left( \lambda, \Lambda, r \right) \leqslant \epsilon$$ for $$r$$ as this equation might have no solution.

My question is then, do you have an idea on how I could compute a better bound?

• That expression goes to $0$ very rapidly as $n$ increases, so perhaps you ought to tell us what values of $n$ you are interested in, and how small a bound you are hoping to achieve. – saulspatz Sep 3 '19 at 15:39
• I will add the whole context in the next few hours, sorry for the missing information. – Nelimee Sep 3 '19 at 16:19
• I added a few material based on quick and dirty regressions for your small numbers. – Claude Leibovici Sep 6 '19 at 4:06

This is too long for a comment.

Reading the post, I have been wondering about the meaning of the vertical bars in the definition of $$f(n)$$.

If it is $$f(n) = \sum_{j=1}^{n} \sum_{i=0}^{n-j} \left( \frac{1}{n!} - \frac{1}{2^{n-j} i! j! (n-j-i)!} \right)$$ we have $$\sum_{i=0}^{n-j} \left( \frac{1}{n!} - \frac{1}{2^{n-j} i! j! (n-j-i)!} \right)=\frac{n+1-j}{n!}-\frac{1}{j! (n-j)!}$$ which makes $$f(n)=\frac{n^2+n-2^{n+1}+2}{2 n!}$$ and then $$\left[ \frac{2^n-1}{n!} - \frac{n(n+1)}{2n!} + f(n)\right]=0$$

Edit

If we consider $$f(n) = \sum_{j=1}^{n} \sum_{i=0}^{n-j} \left\vert \frac{1}{n!} - \frac{1}{2^{n-j} i! j! (n-j-i)!} \right\vert$$ and compute the value of $$f(n)$$ for the range $$3 \leq n \leq 250$$, a quick and dirty regression for the model $$\log[f(n)]=a+b n^c +d n^{2c}$$ which is equivalent to the minimization of $$SSQ=\sum _{n=3}^{250} \left| \frac{\Delta f(n)}{f(n)}\right|^2$$ we have (with $$R^2=0.999998$$) $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & +6.666622 & 0.199625 & \{+6.273405,+7.059838\} \\ b & -0.671761 & 0.006089 & \{-0.683754,-0.659767\} \\ c & +1.329758 & 0.001918 & \{+1.325979,+1.333537\} \\ d & +0.000030 & \approx 0 & \{+0.000029,+0.000030\} \\ \end{array}$$ Similarly, using the same model for $$g(n)=\left[ \frac{2^n-1}{n!} - \frac{n(n+1)}{2n!} + f(n)\right]$$

$$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & +4.126499 & 0.156637 & \{+3.817960,+4.435038\} \\ b & -0.732346 & 0.005281 & \{-0.742749,-0.721943\} \\ c & +1.313991 & 0.001539 & \{+1.310961,+1.317022\} \\ d & +0.000032 & \approx 0 & \{+0.000031,+0.000032\} \\ \end{array}$$

• Thank you for your answer! The vertical bars are absolute values, that is why I tried the triangle inequality, without success. The absolute value comes from the fact that I normed the whole sum and used: $||A+B||\leqslant ||A||+||B||$, $||AB|| \leqslant ||A||||B||$ (for the spectral norm I am using) and $||\lambda A|| = |\lambda| ||A||$ (the coefficient in the sum is the $\lambda$ here). – Nelimee Sep 4 '19 at 7:47
• @Nelimee. I am sorry ! – Claude Leibovici Sep 4 '19 at 8:09
• Your comment gave me an idea! I tested it, and I will edit my question to include this idea. The result is not better than the 2 bounds from the paper linked in the question, but is way better than my previous bound! – Nelimee Sep 9 '19 at 11:11