What do we wind up with if we try something similar but with the primes instead? Naturally there are plenty of possibilities. In what follows let $p_i$ be the i'th prime number.
How about just having products of primes in the denominator? $$\sum_i { {x^i} \over {\prod_{n=1}^i p_n}}$$
It falls to a minimum and then starts to rise again as you go negative. It doesn't rise as fast as $e^x$ for increasing $x$, unsuprisingly.
OK, suppose we use factorials instead of just the products of primes. $$\sum_i {{x^i} \over {p_i !}}$$
For $x$ increasing it also doesn't rise as fast as $e^x$, but for negative $x$ it climbs faster than before. Both this and the previous have a minimum: the one about about -2.7 and the previous at about -3.27.
One more, just for laughs. Pick out just the $e^x$ Taylor series expansion terms with prime powers of x.
$$ \sum_i {{x^{p_i} \over { {p_i} !}} $$It has 2 inflection points, and rises with $x$ increasing and falls with $x$ decreasing--sort of like a cubic would. Curious.(*)
And not obviously useful.
To wrap up, what started the exercise for me was, for $x \in [-1,1)$, $$ \sum_i { {x^i} \over {p_i} } $$.
Naturally this diverges almost everywhere, but it's cute.
If you've wondered why high school graphing calculators haven't changed in 20 years, this is why: it would do most of your algebra homework for you.
(*) It looks like Mathjax fails with this command--it boxed the raw LaTeX instead of processing it. I wonder why. And to make the text fall below the images, I had to add "style=clear:both;" inside the paragraph command.
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