I want to know if this function: for any integer $n>1$ $$ f(t) = e^{1-1/(1-(2x)^{2n})}\cdot(\theta(t+1/2)-\theta(t-1/2))$$ with $\theta(t)$ the unitary step function, which behaves as having:
- $\max_t\{f(t)\}=1$
- $f(t) \geq 0, \forall t, \forall n >1$
- $f(t) = 0, \forall |t|> \frac{1}{2}$
- Develops a flat-top for $n \geq 2$, so adjacent points near $0$ have the same value and $\frac{df(t)}{dt}=0$ in this neighborhood, so its not analytic within.
- $\lim_{t \to \pm {\frac{1}{2}}^\pm} f(t) = 0$
- $\lim_{t \to \pm {\frac{1}{2}}^\pm} \frac{df(t)}{dt} = 0$
A) Is this $f(t)$ a "Bump Function"???
This is equivalent to prove that is smooth in the interval, i.e., every derivative of the function exist within, and also that $\lim_{t \to \pm {\frac{1}{2}}^\pm} \frac{d^nf(t)}{dt^n} = 0, \forall n\geq0,\, n \in \mathbb{Z}$.
Also, it happens that: $$f(t)= \frac{1}{2} \Rightarrow t^*= \pm \frac{1}{2}\cdot\left(\frac{\log(2)}{1+\log(2)}\right)^{\frac{1}{2n}} \therefore \lim_{n \to \infty} t^* = \pm \frac{1}{2}$$
So I want to know if it is true that the standard unitary rectangular function $\Pi(t)$ could be defined through a limit version of $f(t)$.
B) I true that $\lim_{n \to \infty} f(t) \cong \Pi(t)$???
PS1: I have seen a lot of "Bump functions"'s questions, but there is no related Tag. So, if someone can, please create it, I will be really grateful.
PS2: I verified the properties using the website Wolfram Alpha [1], so I am taking them as valid results without proving them by myself. If something is wrong please explain why.
