# Approximating the exponential function

I have found experimentally something that seems graphically like an approximation of the exponential function. However, it is totally experimental and I have no idea whether it really converges towards the $\exp$.

Let : $$f\left(x,h,c\right)=\left(1+\frac{x}{c^h}\right)^{c^h}\text{(A quite understandable approximation for the exponential function)}$$ $$q\left(x,h,c\right)=\sum \limits_{p=0}^h\frac{c^{\frac{p^2+p}{2}}}{\left(\prod\limits _{i=1}^p\left(c^i-1\right)\right)\prod\limits _{i=1}^{h-p}\left(1-c^i\right)}f\left(x,p,c\right)$$ My approximation is : $$\exp(x)=\underset{h,c\rightarrow+\infty}{\lim}q(x,h,c)$$

Desmos shows that it s indeed really close near $1$, and that for low $h,c$ it starts diverging afterwards (but this might be because of computational errors on huge numbers (?) ).

Is that a known approximation for the exponential ? If not, is it an approximation of the exponential at all ?

Additional question that sparked from the comments : How can the comportment for low $h,c$ be analyzed ?

If you wonder where that formula comes from, I can't give a full explanation (it's really experimental work) but you might be interested by one of my previous questions An expression for $U_{h,0}$ given $U_{n,k}=\frac{c^n}{c^n-1}(U_{n-1,k+1})-\frac{1}{c^n-1}(U_{n-1,k})$.

• I assume you realise that $\left(1+\frac{x}{c^h}\right)^{c^h} \to \exp(x)$ as $c^h \to +\infty$. – Henry Aug 9 '14 at 13:00
• You also seem to have some confusion between $g$ and $q$ and between $m$ and $h$ – Henry Aug 9 '14 at 13:02
• @Henry Of course I do, but I am not using $f$ but $g$ – Hippalectryon Aug 9 '14 at 13:06
• @Henry I edited the definiton of $g$, is it better ? – Hippalectryon Aug 9 '14 at 13:07
• @Hippalectryon I think what Henry is getting at is that $h$ and $m$ together and $g$ and $q$ together are each pretty much the same thing. You could probably dispense with $g$ and $m$. – Jam Aug 9 '14 at 13:14

As shown in equation $(6)$ from this answer, $$\prod_{k=1}^h\frac{c^kx-1}{c^k-1}=\sum_{p=0}^h\frac{\displaystyle c^{p(p+1)/2}}{\displaystyle\prod_{k=1}^p(c^k-1)\prod_{k=1}^{h-p}(1-c^k)}x^p\tag{1}$$ Plugging $x=1$ into $(1)$, we get that $$\sum_{p=0}^h\frac{\displaystyle c^{p(p+1)/2}}{\displaystyle\prod_{k=1}^p(c^k-1)\prod_{k=1}^{h-p}(1-c^k)}=1\tag{2}$$ For $x\ge0$, $0\le e^x-\left(1+\frac xn\right)^n\le\dfrac{x^2e^x}{2n}$. Therefore, \begin{align} |e^x-f(x,p,c)| &=\left|\,e^x-\left(1+\frac{x}{c^p}\right)^{c^p}\,\right|\\ &\le\frac{x^2e^x}{2}c^{-p}\tag{3} \end{align} Since the coefficients of $x^p$ in $(1)$ alternate sign, we get \begin{align} &\left|\,e^x-\sum_{p=0}^h\frac{\displaystyle c^{p(p+1)/2}}{\displaystyle\prod_{k=1}^p(c^k-1)\prod_{k=1}^{h-p}(1-c^k)}f(x,p,c)\,\right|\\[6pt] &=\left|\,\sum_{p=0}^h\frac{\displaystyle c^{p(p+1)/2}}{\displaystyle\prod_{k=1}^p(c^k-1)\prod_{k=1}^{h-p}(1-c^k)}(e^x-f(x,p,c))\,\right|\tag{4}\\[6pt] &\le\frac{x^2e^x}{2}\left|\,\sum_{p=0}^h\frac{\displaystyle c^{p(p+1)/2}}{\displaystyle\prod_{k=1}^p(c^k-1)\prod_{k=1}^{h-p}(1-c^k)}\left(-\frac1c\right)^p\,\right|\tag{5}\\[18pt] &=\frac{x^2e^x}{2}\prod_{k=1}^h\frac{c^{k-1}+1}{c^k-1}\tag{6} \end{align} Explanation:
$(4)$: apply $(2)$
$(5)$: use $(3)$
$(6)$: apply $(1)$
For both a fixed $c\gt1$ as $h\to\infty$ and a fixed $h\ge1$ as $c\to\infty$, $(6)$ vanishes.
• Can you try to see what happens when $c$ is fixed ($>1$) and $h\rightarrow+\infty$ ? – Hippalectryon Aug 10 '14 at 11:04
• Under $(5)$: '$|a0|max(1,|ex−1−x|)≤ϵ$'; since $\lim\limits_{x\rightarrow+\infty}|ex−1−x|=+\infty$, how can we have $|a_0|\max(1,|e^x-1-x|)\le\epsilon$ for $c_0,\epsilon$ fixed ? – Hippalectryon Aug 10 '14 at 11:11
• What I give answers the question. A slight modification of the argument above works for a fixed $c\gt1$ as $h\to\infty$. As $h\to\infty$, we have $|a_0|$ to $|a_k|$ are small and $|e^x-f(x,p,c)|$ is small for $p\gt k$. Split the sum at $k$ instead of $1$. The argument is similar, but a lot longer and a bit messier. – robjohn Aug 10 '14 at 11:17
• Why are you looking at $x\to\infty$? I don't think that the convergence here is uniform in $x$. It is uniform on compact sets, but not on all of $\mathbb{R}$. Note that the convergence of $\left(1+\frac xn\right)^n$ to $e^x$ is only uniform on compact sets, not all of $\mathbb{R}$. – robjohn Aug 10 '14 at 11:20