Finding a closed form for a Recurrence
$begingroup$
Let $ f : mathbb{N}^3 to mathbb{R} $, where $f$ is defined by the following recurrence,
$$ f(x,y,z) = frac{1}{x} left[ sum_{n=0}^{x-1} f(x,y-1,n) cdot H[|z-n|-2] right] $$
where $x>z$, $f(x,0,z)=1, (forall x,zinmathbb{N})$, and $H[cdot]$ is the Heaviside step function.
This expression arose as recursive programming solution to a dice rolling puzzle. It translates fairly cleanly to code (mathematica: H[n_]:=Piecewise[{{0,n<0},{1,n>=0}}];f[x_,y_,z_]:=f[x,y,z]=Piecewise[{{1,y==0},{(1/x)*Sum[f[x,y-1,n]*H[Abs[z-n]-2],{n,0,x-1}],True}}];
), but I am curious if a closed form or simpler recurrence exists.
Calculating one by hand shows a tree structure with depth $y$, where each node has $s-2$ or $s-3$ branches. This is due to the fact that $H[|z-n|-2]$ is $0$ only $2$ or $3$ times at each summation, but $1$ everywhere else.
Is there a closed form for this recurrence? If so, how do we find it? If not, why?
Note: Here's a plot of $g(x,y) = frac{1}{x} sum_{n=0}^{x-1}f(x,y-1,n)$,
recurrence-relations closed-form
$endgroup$
add a comment |
$begingroup$
Let $ f : mathbb{N}^3 to mathbb{R} $, where $f$ is defined by the following recurrence,
$$ f(x,y,z) = frac{1}{x} left[ sum_{n=0}^{x-1} f(x,y-1,n) cdot H[|z-n|-2] right] $$
where $x>z$, $f(x,0,z)=1, (forall x,zinmathbb{N})$, and $H[cdot]$ is the Heaviside step function.
This expression arose as recursive programming solution to a dice rolling puzzle. It translates fairly cleanly to code (mathematica: H[n_]:=Piecewise[{{0,n<0},{1,n>=0}}];f[x_,y_,z_]:=f[x,y,z]=Piecewise[{{1,y==0},{(1/x)*Sum[f[x,y-1,n]*H[Abs[z-n]-2],{n,0,x-1}],True}}];
), but I am curious if a closed form or simpler recurrence exists.
Calculating one by hand shows a tree structure with depth $y$, where each node has $s-2$ or $s-3$ branches. This is due to the fact that $H[|z-n|-2]$ is $0$ only $2$ or $3$ times at each summation, but $1$ everywhere else.
Is there a closed form for this recurrence? If so, how do we find it? If not, why?
Note: Here's a plot of $g(x,y) = frac{1}{x} sum_{n=0}^{x-1}f(x,y-1,n)$,
recurrence-relations closed-form
$endgroup$
add a comment |
$begingroup$
Let $ f : mathbb{N}^3 to mathbb{R} $, where $f$ is defined by the following recurrence,
$$ f(x,y,z) = frac{1}{x} left[ sum_{n=0}^{x-1} f(x,y-1,n) cdot H[|z-n|-2] right] $$
where $x>z$, $f(x,0,z)=1, (forall x,zinmathbb{N})$, and $H[cdot]$ is the Heaviside step function.
This expression arose as recursive programming solution to a dice rolling puzzle. It translates fairly cleanly to code (mathematica: H[n_]:=Piecewise[{{0,n<0},{1,n>=0}}];f[x_,y_,z_]:=f[x,y,z]=Piecewise[{{1,y==0},{(1/x)*Sum[f[x,y-1,n]*H[Abs[z-n]-2],{n,0,x-1}],True}}];
), but I am curious if a closed form or simpler recurrence exists.
Calculating one by hand shows a tree structure with depth $y$, where each node has $s-2$ or $s-3$ branches. This is due to the fact that $H[|z-n|-2]$ is $0$ only $2$ or $3$ times at each summation, but $1$ everywhere else.
Is there a closed form for this recurrence? If so, how do we find it? If not, why?
Note: Here's a plot of $g(x,y) = frac{1}{x} sum_{n=0}^{x-1}f(x,y-1,n)$,
recurrence-relations closed-form
$endgroup$
Let $ f : mathbb{N}^3 to mathbb{R} $, where $f$ is defined by the following recurrence,
$$ f(x,y,z) = frac{1}{x} left[ sum_{n=0}^{x-1} f(x,y-1,n) cdot H[|z-n|-2] right] $$
where $x>z$, $f(x,0,z)=1, (forall x,zinmathbb{N})$, and $H[cdot]$ is the Heaviside step function.
This expression arose as recursive programming solution to a dice rolling puzzle. It translates fairly cleanly to code (mathematica: H[n_]:=Piecewise[{{0,n<0},{1,n>=0}}];f[x_,y_,z_]:=f[x,y,z]=Piecewise[{{1,y==0},{(1/x)*Sum[f[x,y-1,n]*H[Abs[z-n]-2],{n,0,x-1}],True}}];
), but I am curious if a closed form or simpler recurrence exists.
Calculating one by hand shows a tree structure with depth $y$, where each node has $s-2$ or $s-3$ branches. This is due to the fact that $H[|z-n|-2]$ is $0$ only $2$ or $3$ times at each summation, but $1$ everywhere else.
Is there a closed form for this recurrence? If so, how do we find it? If not, why?
Note: Here's a plot of $g(x,y) = frac{1}{x} sum_{n=0}^{x-1}f(x,y-1,n)$,
recurrence-relations closed-form
recurrence-relations closed-form
asked Dec 8 '18 at 19:02
Dando18Dando18
4,67741235
4,67741235
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