Build a sequence $u_k $ N. 2
Build $ u_k in C^0(RR) nn L^1(RR)$ converging in $L^1(RR) $ to $e^(-x^2)chi(x)$, where $chi(x)$ is the characteristic function of $[0, oo)$.
Risposte
With a little bit of fantasy:
by the de Moivre - Laplace theorem we know that
$((n),(k))p^k(1-p)^(n-k) ~1/sqrt(2pinp(1-p)) e^(-(k-np)^2/(2np(1-p)))$.
Anyway, among the distributions, the Bernoulli one is discrete, so we need to make it continuous.
The only problem is the binomial coefficient $((n),(k))$, because $k$ has to belong to $NN$.
Anyway, we can generalize a little bit: it is known that, for complex $z$,
$((z),(k))=prod_(n=1)^k(z-k+n)/(n)=1/(k!)prod_(n=1)^k(z-k+n)$.
The problem persists: $k$ has to be natural. The problem can be eliminated if we generalize $k!$ with $Gamma(k+1)$.
So let's set $((z),(k))=1/(Gamma(k+1))prod_(n=1)^(|__k__|)(z-k+n)$ and see what comes out.
Going back to the theorem, to obtain $e^(-x^2)$, we need $2np(1-p)=1$, yielding $p=1/(2n)[n pm sqrt(n(n-2))]$.
Finally, with the substitution $k=x-np$ and multiplying both sides by $sqrt(2pinp(1-p))$, we obtain
$e^(-x^2)~(p^(x-np)(1-p)^(n-x+np) sqrt(2pinp(1-p)))/(Gamma(x-np+1)) prod_(j=1)^(|__x-np__|) (n-x+np+j)=phi_n(x)$.
To obtain the step effect given by $chi[0,+oo)$ and a continuous function, we may set
$u_n(x)={(0 mbox( if )x<0),(nx mbox( if )0<=x<=x_n),(phi_n(x) mbox( if ) x>x_n):}$
where $x_n$ is the abscissa of the intersection between the line $y=nx$ and the curve $y=phi_n(x)$.
by the de Moivre - Laplace theorem we know that
$((n),(k))p^k(1-p)^(n-k) ~1/sqrt(2pinp(1-p)) e^(-(k-np)^2/(2np(1-p)))$.
Anyway, among the distributions, the Bernoulli one is discrete, so we need to make it continuous.
The only problem is the binomial coefficient $((n),(k))$, because $k$ has to belong to $NN$.
Anyway, we can generalize a little bit: it is known that, for complex $z$,
$((z),(k))=prod_(n=1)^k(z-k+n)/(n)=1/(k!)prod_(n=1)^k(z-k+n)$.
The problem persists: $k$ has to be natural. The problem can be eliminated if we generalize $k!$ with $Gamma(k+1)$.
So let's set $((z),(k))=1/(Gamma(k+1))prod_(n=1)^(|__k__|)(z-k+n)$ and see what comes out.
Going back to the theorem, to obtain $e^(-x^2)$, we need $2np(1-p)=1$, yielding $p=1/(2n)[n pm sqrt(n(n-2))]$.
Finally, with the substitution $k=x-np$ and multiplying both sides by $sqrt(2pinp(1-p))$, we obtain
$e^(-x^2)~(p^(x-np)(1-p)^(n-x+np) sqrt(2pinp(1-p)))/(Gamma(x-np+1)) prod_(j=1)^(|__x-np__|) (n-x+np+j)=phi_n(x)$.
To obtain the step effect given by $chi[0,+oo)$ and a continuous function, we may set
$u_n(x)={(0 mbox( if )x<0),(nx mbox( if )0<=x<=x_n),(phi_n(x) mbox( if ) x>x_n):}$
where $x_n$ is the abscissa of the intersection between the line $y=nx$ and the curve $y=phi_n(x)$.
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