# A "relative" of the exponential function

It is well known that $$\mathrm{e}=\sum_{n=0}^\infty \frac{1}{n!}.$$ Recently, I was reminded that the volume of an $$n$$-simplex in $$n$$-dimensional space, with vertices $$v_0,v_1,\dots,v_n$$ is $$\left| \frac{1}{n!} \det \left(v_1-v_0,v_2-v_0,\dots,v_n-v_0 \right) \right|.$$ In particular, if for all $$i$$, $$v_i-v_0=e_i$$ is the $$i$$th standard basis vector, we find that the $$n$$-dimensional volume of the so-called "corner of the unit cube", $$\Delta_c^n$$ is $$1/n!$$.

Thus, one can say that Euler's number is the sum of the volumes of corners of unit cubes, taken over all dimensions, or as a formula $$\mathrm{e}=\sum_{n=0}^\infty \operatorname{vol_n}\left( \Delta_c^n \right).$$

Moreover, scaling each of the $$\Delta_c^ns$$ by the same nonnegative real number $$x$$, scales its volume by $$x^n$$, and one gets a geometrical formula for the exponential function:

$$\mathrm{e}^x =\sum_{n=0}^\infty \operatorname{vol_n} \left( x \cdot \Delta_c^n \right)=\sum_{n=0}^\infty \operatorname{vol_n} \left( \Delta_c^n \right) x^n.$$ In fact, the power series on the right can be continued analytically to a complex, entire function of $$x$$.

One can do the same for other families of shapes, for example it is known that the volume of an $$n$$-dimensional ball, of radius $$x$$, is $$x^n \pi^{n/2}/\Gamma(n/2+1)$$. Hence, the associated function takes the form $$\sum_{n=0}^\infty \frac{\pi^\frac{n}{2}}{\Gamma\left(\frac{n}{2}+1 \right)} x^n =e^{\pi x^2} \left(\text{erf}\left(\sqrt{\pi } x\right)+1\right).$$

I was wondering what happens if one replaces the corners of the cubes of the exponential function by the regular simplices, of side-length $$x$$. The associated function is then $$f(x)=\sum_{n=0}^\infty \frac{\sqrt{1+n}}{n! 2^\frac{n}{2}} x^n,$$ which is entire. I've tried expressing $$f(x)$$ in terms of other functions, to no avail. Moreover, I couldn't even get a closed form for the simple-looking sum $$f\left( \sqrt{2} \right)= \sum_{n=0}^\infty \frac{\sqrt{1+n}}{n!}.$$ I would appreciate any help understanding/simplifying $$f(x)$$ in general, and $$f(\sqrt{2})$$ in particular. Thank you!

• I'm going to throw it out there and say I doubt on any sort of closed form. Commented Jul 17, 2017 at 12:38
• One can write this as $\frac{1}{2\sqrt{\pi}}\int_0^{\infty} t^{-3/2} ( e^{-t+xe^{-t}}-e^x ) \, dt$, which doesn't look much like it has a closed form. Commented Jul 17, 2017 at 13:26
• If you let $x=t\sqrt 2$, it seems that $\log(f(t)) \sim a + bt+c \log(t)$ Commented Nov 30, 2023 at 9:32
• If its domain were extended, there would be a closed form via Bell polynomial Commented Dec 14, 2023 at 22:30

$$\sum_{n=0}^{\infty} \frac{\sqrt{1+n}}{n!}=\sum_{n=0}^{\infty} \frac{\sqrt{n(1+1/n)}}{n!}$$ $$\sum_{n=0}^{\infty} \frac{\sqrt{n(1+1/n)}}{n!}=\sum_{n=0}^{\infty} \frac{\sqrt{n}\sqrt{1+1/n}}{n!}$$ and then that square root can be expanded hypergeometrically.
• I'm intrigued. However, I don't know of the $3j/6j/9j/12j$ symbols. Could you please elaborate? Commented Nov 16, 2019 at 11:30