How to bound the error for the Taylor expansion of the inverse of a mean of exponentials?

If $$|x| \leq R / 10$$ for some $$R\in \mathbb{N}$$, then it is easily shown that $$\left|e^{-x} - \sum_{k=0}^R \frac{(-1)^k x^k}{k!}\right| \leq e^{-R}.$$

I would like to have a similar result (i.e. with an error bound of the same form) for the multivariate function $$\frac{1}{\frac{1}{M} \sum_{i=1}^M e^{x_i}}$$, where the necessary control on $$M\in \mathbb{N}$$ and on each $$x_i$$'s can be imposed.

Any ideas ?

After playing with formal Taylor series, here's a partial answer. We can write $$\sum_{i=1}^M e^{x_i} = \sum_{\alpha\in \mathbb{N}_0^M} c_{\alpha} x^{\alpha}, \quad \text{with } c_{\alpha} = \begin{cases} \frac{1}{k!}, &\mbox{if } \alpha = k e_k, \\ 0, &\mbox{otherwise}, \end{cases}$$ where $$e_k$$ is the $$k$$-th standard basis vector, and $$x^{\alpha} = \prod_{i=1}^M x_i^{\alpha_i}$$. Now the goal is to find coefficients $$d_{\beta}$$ such that $$\left(\sum_{\alpha\in \mathbb{N}_0^M} c_{\alpha} x^{\alpha}\right) \cdot \left(\sum_{\beta\in \mathbb{N}_0^M} d_{\beta} x^{\beta}\right) = \sum_{\gamma\in \mathbb{N}_0^M} x^{\gamma} \left(\sum_{\alpha + \beta = \gamma} c_{\alpha} d_{\beta}\right) = 1.$$

• If $M = 2$, what is the form of error bound? Commented May 3, 2021 at 2:51
• something like $e^{-R}$ Commented May 3, 2021 at 4:18
• I mean in $|\frac{2}{\mathrm{e}^{x_1} + \mathrm{e}^{x_2}} - A| < \mathrm{e}^{-R}$, what is the desired form of $A$? Commented May 3, 2021 at 5:02
• $A$ would be a truncated version of $M \sum_{\beta\in \mathbb{N}_0^M} d_{\beta}x^{\beta}$. Commented May 3, 2021 at 5:09
• Is it something like the Taylor series for 2-d function $\frac{2}{\mathrm{e}^{x_1} + \mathrm{e}^{x_2}}$ around $(0, 0)$? Commented May 3, 2021 at 8:16

Note the natural generalization of this inequality may not be $$\frac{1}{\frac{1}{M} \sum_{i=1}^M e^{x_i}}$$, but $$\frac{1}{M} \sum_{i=1}^M e^{-x_i}$$.

Considering the second case with $$|x_i| \leq R/10$$ gives $$\left|\frac{1}{M} \sum_{i=1}^M e^{-x_i} - \frac{1}{M} \sum_{i=1}^M \sum_{k=0}^R \frac{(-1)^k x_i^k}{k!} \right| \leq e^{-R}$$ by the triangle inequality.

Let $$x_i, i=1,...,n$$ be positive numbers with maximum term $$x_m$$. We claim the difference of their arithmetic mean and harmonic mean is maximized when one term goes to $$0$$ and the other terms are equal. Consider

$$g(x_j) = \frac{x_1 + \cdots + x_n}{n} - \frac{n}{x_1^{-1} + \cdots + x_n^{-1}} \geq 0$$ with equality iff $$x_1 = \cdots = x_n$$ by the HM-AM-GM inequality. If $$x_j$$ is the smallest term

$$g'(x_j) = \frac{1}{n} - \frac{n}{(\frac{x_j}{x_1} + \cdots + \frac{x_j}{x_n})^2} < 0$$ since $$x_j/x_k \leq 1$$. In particular $$g(x_j)$$ increases further until $$x_j$$ decreases to $$0$$. This shows the max has at least one $$x_i$$ zero. It follows the absolute max is $$\frac{n-1}{n} x_m$$ For $$|x_i| \leq R/10$$ \begin{align} \left|\frac{1}{\frac{1}{M}\sum_{i=1}^M e^{x_i}} - \frac{1}{M} \sum_{i=1}^M \sum_{k=0}^R \frac{(-1)^k x_i^k}{k!}\right| \leq & \left|\frac{1}{\frac{1}{M}\sum_{i=1}^M e^{x_i}} - \frac{1}{M} \sum_{i=1}^M e^{-x_i}\right| \ \\ & \ \ \ + \left|\frac{1}{M} \sum_{i=1}^M e^{-x_i} - \frac{1}{M} \sum_{i=1}^M \sum_{k=0}^R \frac{(-1)^k x_i^k}{k!} \right| \\ \leq & \ e^{-R} + \left(1-\frac{1}{M}\right) \max \{e^{-x_i}\}_{i=1}^M \end{align}

Comment: Unfortunately the difference between the HM and AM destroyed the accuracy of the inequality. It doesn't seem easy to fix this so we need a new approach.

Denote $$\phi(x_1) = \frac{1}{M} \sum_{i=1}^M \sum_{k=0}^R \frac{x_i^k}{k!}$$. Replace $$x_i$$ with $$-x_i$$ in the first inequality implies $$\phi(x_1)>e^{-R/10}-e^{-R} > 0$$. Dividing the first inequality by $$\phi(x_1) \sum_{i=1}^M e^{x_i}$$ gives

\begin{align} \left|\frac{1}{\phi(x_1)} - \frac{1}{\frac{1}{M}\sum_{i=1}^M e^{x_i}}\right| &\leq \frac{e^{-R}}{\phi(x_1) \frac{1}{M} \sum_{i=1}^M e^{x_i}} \\ &\leq \frac{e^{-R}}{(e^{-R/10}-e^{-R})e^{-R/10}} \\ & = e^{-4R/5}(1-e^{-9R/10})^{-1} \\ & \leq e^{-4R/5}(1+e^{-9R/10}) \end{align} If you can calculate $$\phi(x_1)$$, dividing at the end doesn't seem a significant extra computation. If you still want $$\frac{1}{\phi(x_1)}$$ to be in the form of a truncated multivariable Taylor series its probably most straightforward to redo the problem calculating the Taylor series.

$$f(\mathbf{x}) = e^{\mathbf{x} \cdot \nabla} f(y_1,...,y_M) \Bigg|_{y_1,...,y_M = 0}$$. For us $$\frac{M}{\sum_{i=1}^M e^{x_i}} = \sum_{k=0}^N \sum_{k_1 + \cdots + k_M = k} \frac{k!}{k_1!\cdots k_M!} \frac{\partial^k}{\partial y_1^{k_1} \cdots \partial y_M^{k_M}} \frac{M}{e^{y_1} + \cdots + e^{y_M}} \Bigg|_{y_1, ..., y_M = 0} x_1^{k_1} \cdots x_M^{k_M} + R_N$$

where $$R_N = \frac{\frac{d^{N+1}}{dt^{N+1}}f(t_0 \mathbf{x})}{(N+1)!}||\mathbb{x}||^{N+1}, \ \ t_0 \in [0,1].$$ The coefficients can be put in closed form, but might not be simplifiable.

One way to turn over $$1/\phi(x_1)$$ is to use Burmann's Theorem, but the remainder won't be fun and might involve one over the other (x_i). "Cayley's Formula" says $$\frac{1}{\phi(x_1)} = \frac{1}{\phi(0)} + \sum_{k=1}^\infty \frac{(-x_1)^k}{\phi(0)^{k+1}(k+1)!} (x \phi(x_1)^k)_{(k+1)x_1}$$ but this doesn't help either.

• How do you get $$\left|\frac{1}{\phi(x_1)} - \frac{1}{\frac{1}{M}\sum_{i=1}^M e^{x_i}}\right| \leq \frac{e^{-R}}{\phi(x_1) \frac{1}{M} \sum_{i=1}^M e^{x_i}} ?$$ Isn't there a $e^{-R/10}$ missing ? Commented May 8, 2021 at 3:45
• In the equality describing the second case, replace $-x_i$ with $x_i$ and divide by $\phi(x_1) \frac{1}{M} \sum e^{x_i}$. Commented May 8, 2021 at 3:51
• I understand that, but the algebra doesn't seem to work Commented May 8, 2021 at 3:51
• Ah I see, it's just that when you wrote "Dividing by", it wasn't clear from which equation you were starting. Commented May 8, 2021 at 3:58
• however $\frac{1}{M} \sum_{i=1}^M \sum_{k=1}^R \frac{x_i^k}{k!}$ is too big for this to work. Commented May 8, 2021 at 4:31