This series converges for all x∈(−∞,∞), thus the function is analytic on the real line and defined by its Taylor series.
However, unlike the exponential function, this one is very elusive - I can't find any other representation for it - no integral, no differential equation, nothing.
Its derivatives exist (an infinite number of them), but it seems they can't be connected to the function itself in any other way.
I know that s(1)−1 is called 'Sophomore's Dream', because it has integral representation:
∞∑k=11kk=∫101ttdt
I don't know if the same method can be used to find a general integral representation for s(x), but I have some hope.
The most interesting (in my opinion) property of this function - s(x) and its derivatives all have exactly one zero (for x0<0) and one minimum (xm<0 and s(xm)<0). They are connected in a sense, since obviously if s(x) has a minimum, then s′(x) has a zero.
Here is a plot of s(x) for x<0:
I added 1 to the sum in the definition of s(x) for consistency - it can be seen from the form of its derivatives:
s(x)=1+x+x222+x333+⋯=1+∞∑k=1xkkk
s′(x)=1+x2+x29+x364+⋯=∞∑k=1k xk−1kk
s″(x)=12+2x9+3x264+4x3625+⋯=∞∑k=1k(k−1) xk−2kk
s‴(x)=29+3x32+12x2625+5x31944+⋯=∞∑k=1k(k−1)(k−2) xk−3kk
It is apparent, that lim, however, the zero on the negative line actually moves to the left with each differentiation (see the position of the minimum). So I'm not sure, how the 'infinite derivative' of s(x) would look.
I'd like to know if someone studied this function, and get a reference for it. But the main question is - what other definitions are possible for s(x), except for the series?
Thanks to this great answer, I'm able to write the integral definition:
s(x)=1+\int_0^1 x ~u^{-u~x} du
So I consider my question answered.
By the way, if we define the function:
p(x)=\frac{s(x)-1}{x}=\int_0^1 u^{-u~x} du=\sum_{k=1}^{\infty} \frac{x^{k-1}}{k^k}
We get a very monotone (exponential-like) behavior, without any zeroes or minima. The function and several of its derivatives are plotted below.
No comments:
Post a Comment