How to use a summation in a tikz plot












4















I need to plot this function using tikz, but I have no idea how to go about using the summation in tikz. If someone could help me I would really appreciate it.
Function
The code I have is this:



documentclass{article}

usepackage{pgfplots}
usepackage{tikz}

usepackage[paperwidth=20cm, paperheight=10cm, margin=0.0cm]{geometry}

begin{document}
hspace{-0.4cm}
begin{tikzpicture}[
declare function={p(k,n) =((n!*(k*(k-1)))/((n^(k))*((n-(k-
1))!)));}
]
begin{axis}[axis lines=middle, grid=both, width=20cm, height=10cm, grid
style={line width=.2pt, draw=gray!10},major grid style={line
width=.3pt,draw=gray!50},
xlabel=$k$, ylabel=$y$,
ylabel style={anchor=south},
xlabel style={anchor=west},
ymax=1, xmax=21.5,
domain=0:21.5, samples at={0,...,21},
]
addplot+ [blue, thick, mark size={1.6pt}, mark options={draw=blue,
fill=white!75!cyan}] {p(x,20)};
end{axis}
end{tikzpicture}

end{document}


This plots the function of x, but I don't know how to sum the function from k=2 to x+1.










share|improve this question


















  • 1





    possibly helpful? tex.stackexchange.com/questions/291404/…

    – cmhughes
    Nov 23 '17 at 15:01











  • Thanks, I saw that earlier but couldn't figure it out. I've only been using latex for a short time and Im struggling to work out what to do.

    – John 545
    Nov 23 '17 at 15:08











  • Is this not possible in tikz?

    – John 545
    Nov 23 '17 at 20:03
















4















I need to plot this function using tikz, but I have no idea how to go about using the summation in tikz. If someone could help me I would really appreciate it.
Function
The code I have is this:



documentclass{article}

usepackage{pgfplots}
usepackage{tikz}

usepackage[paperwidth=20cm, paperheight=10cm, margin=0.0cm]{geometry}

begin{document}
hspace{-0.4cm}
begin{tikzpicture}[
declare function={p(k,n) =((n!*(k*(k-1)))/((n^(k))*((n-(k-
1))!)));}
]
begin{axis}[axis lines=middle, grid=both, width=20cm, height=10cm, grid
style={line width=.2pt, draw=gray!10},major grid style={line
width=.3pt,draw=gray!50},
xlabel=$k$, ylabel=$y$,
ylabel style={anchor=south},
xlabel style={anchor=west},
ymax=1, xmax=21.5,
domain=0:21.5, samples at={0,...,21},
]
addplot+ [blue, thick, mark size={1.6pt}, mark options={draw=blue,
fill=white!75!cyan}] {p(x,20)};
end{axis}
end{tikzpicture}

end{document}


This plots the function of x, but I don't know how to sum the function from k=2 to x+1.










share|improve this question


















  • 1





    possibly helpful? tex.stackexchange.com/questions/291404/…

    – cmhughes
    Nov 23 '17 at 15:01











  • Thanks, I saw that earlier but couldn't figure it out. I've only been using latex for a short time and Im struggling to work out what to do.

    – John 545
    Nov 23 '17 at 15:08











  • Is this not possible in tikz?

    – John 545
    Nov 23 '17 at 20:03














4












4








4


2






I need to plot this function using tikz, but I have no idea how to go about using the summation in tikz. If someone could help me I would really appreciate it.
Function
The code I have is this:



documentclass{article}

usepackage{pgfplots}
usepackage{tikz}

usepackage[paperwidth=20cm, paperheight=10cm, margin=0.0cm]{geometry}

begin{document}
hspace{-0.4cm}
begin{tikzpicture}[
declare function={p(k,n) =((n!*(k*(k-1)))/((n^(k))*((n-(k-
1))!)));}
]
begin{axis}[axis lines=middle, grid=both, width=20cm, height=10cm, grid
style={line width=.2pt, draw=gray!10},major grid style={line
width=.3pt,draw=gray!50},
xlabel=$k$, ylabel=$y$,
ylabel style={anchor=south},
xlabel style={anchor=west},
ymax=1, xmax=21.5,
domain=0:21.5, samples at={0,...,21},
]
addplot+ [blue, thick, mark size={1.6pt}, mark options={draw=blue,
fill=white!75!cyan}] {p(x,20)};
end{axis}
end{tikzpicture}

end{document}


This plots the function of x, but I don't know how to sum the function from k=2 to x+1.










share|improve this question














I need to plot this function using tikz, but I have no idea how to go about using the summation in tikz. If someone could help me I would really appreciate it.
Function
The code I have is this:



documentclass{article}

usepackage{pgfplots}
usepackage{tikz}

usepackage[paperwidth=20cm, paperheight=10cm, margin=0.0cm]{geometry}

begin{document}
hspace{-0.4cm}
begin{tikzpicture}[
declare function={p(k,n) =((n!*(k*(k-1)))/((n^(k))*((n-(k-
1))!)));}
]
begin{axis}[axis lines=middle, grid=both, width=20cm, height=10cm, grid
style={line width=.2pt, draw=gray!10},major grid style={line
width=.3pt,draw=gray!50},
xlabel=$k$, ylabel=$y$,
ylabel style={anchor=south},
xlabel style={anchor=west},
ymax=1, xmax=21.5,
domain=0:21.5, samples at={0,...,21},
]
addplot+ [blue, thick, mark size={1.6pt}, mark options={draw=blue,
fill=white!75!cyan}] {p(x,20)};
end{axis}
end{tikzpicture}

end{document}


This plots the function of x, but I don't know how to sum the function from k=2 to x+1.







tikz-pgf plot






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 23 '17 at 14:01









John 545John 545

545




545








  • 1





    possibly helpful? tex.stackexchange.com/questions/291404/…

    – cmhughes
    Nov 23 '17 at 15:01











  • Thanks, I saw that earlier but couldn't figure it out. I've only been using latex for a short time and Im struggling to work out what to do.

    – John 545
    Nov 23 '17 at 15:08











  • Is this not possible in tikz?

    – John 545
    Nov 23 '17 at 20:03














  • 1





    possibly helpful? tex.stackexchange.com/questions/291404/…

    – cmhughes
    Nov 23 '17 at 15:01











  • Thanks, I saw that earlier but couldn't figure it out. I've only been using latex for a short time and Im struggling to work out what to do.

    – John 545
    Nov 23 '17 at 15:08











  • Is this not possible in tikz?

    – John 545
    Nov 23 '17 at 20:03








1




1





possibly helpful? tex.stackexchange.com/questions/291404/…

– cmhughes
Nov 23 '17 at 15:01





possibly helpful? tex.stackexchange.com/questions/291404/…

– cmhughes
Nov 23 '17 at 15:01













Thanks, I saw that earlier but couldn't figure it out. I've only been using latex for a short time and Im struggling to work out what to do.

– John 545
Nov 23 '17 at 15:08





Thanks, I saw that earlier but couldn't figure it out. I've only been using latex for a short time and Im struggling to work out what to do.

– John 545
Nov 23 '17 at 15:08













Is this not possible in tikz?

– John 545
Nov 23 '17 at 20:03





Is this not possible in tikz?

– John 545
Nov 23 '17 at 20:03










1 Answer
1






active

oldest

votes


















5














You can use Lua to compute the sum. Then you have to typeset using LuaLaTeX of course.



documentclass{article}

usepackage{pgfplots}
pgfplotsset{compat=newest}

usepackage{luacode}
begin{luacode*}
function factorial(n)
assert(n >= 0, "Factorial is only valid for positive integers")
if n == 0 then
return 1
end
return n*factorial(n-1)
end

function p(x)
assert(x == math.floor(x), "x must be an integer")
res = 0
for k = 2, x+1 do
res = res + factorial(x)/factorial(x-(k-1)) * k*(k-1)/(x^k)
end
tex.sprint(res)
end
end{luacode*}

begin{document}

begin{tikzpicture}[
declare function={p(n) = directlua{p(n)};}
]
begin{axis}[
use fpu=false, % very important!
xlabel=$x$, ylabel=$p(x)$,
samples at={0,...,21},
only marks,
]
addplot {p(x)};
end{axis}
end{tikzpicture}

end{document}


enter image description here






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    1 Answer
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    active

    oldest

    votes









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    oldest

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    active

    oldest

    votes









    5














    You can use Lua to compute the sum. Then you have to typeset using LuaLaTeX of course.



    documentclass{article}

    usepackage{pgfplots}
    pgfplotsset{compat=newest}

    usepackage{luacode}
    begin{luacode*}
    function factorial(n)
    assert(n >= 0, "Factorial is only valid for positive integers")
    if n == 0 then
    return 1
    end
    return n*factorial(n-1)
    end

    function p(x)
    assert(x == math.floor(x), "x must be an integer")
    res = 0
    for k = 2, x+1 do
    res = res + factorial(x)/factorial(x-(k-1)) * k*(k-1)/(x^k)
    end
    tex.sprint(res)
    end
    end{luacode*}

    begin{document}

    begin{tikzpicture}[
    declare function={p(n) = directlua{p(n)};}
    ]
    begin{axis}[
    use fpu=false, % very important!
    xlabel=$x$, ylabel=$p(x)$,
    samples at={0,...,21},
    only marks,
    ]
    addplot {p(x)};
    end{axis}
    end{tikzpicture}

    end{document}


    enter image description here






    share|improve this answer




























      5














      You can use Lua to compute the sum. Then you have to typeset using LuaLaTeX of course.



      documentclass{article}

      usepackage{pgfplots}
      pgfplotsset{compat=newest}

      usepackage{luacode}
      begin{luacode*}
      function factorial(n)
      assert(n >= 0, "Factorial is only valid for positive integers")
      if n == 0 then
      return 1
      end
      return n*factorial(n-1)
      end

      function p(x)
      assert(x == math.floor(x), "x must be an integer")
      res = 0
      for k = 2, x+1 do
      res = res + factorial(x)/factorial(x-(k-1)) * k*(k-1)/(x^k)
      end
      tex.sprint(res)
      end
      end{luacode*}

      begin{document}

      begin{tikzpicture}[
      declare function={p(n) = directlua{p(n)};}
      ]
      begin{axis}[
      use fpu=false, % very important!
      xlabel=$x$, ylabel=$p(x)$,
      samples at={0,...,21},
      only marks,
      ]
      addplot {p(x)};
      end{axis}
      end{tikzpicture}

      end{document}


      enter image description here






      share|improve this answer


























        5












        5








        5







        You can use Lua to compute the sum. Then you have to typeset using LuaLaTeX of course.



        documentclass{article}

        usepackage{pgfplots}
        pgfplotsset{compat=newest}

        usepackage{luacode}
        begin{luacode*}
        function factorial(n)
        assert(n >= 0, "Factorial is only valid for positive integers")
        if n == 0 then
        return 1
        end
        return n*factorial(n-1)
        end

        function p(x)
        assert(x == math.floor(x), "x must be an integer")
        res = 0
        for k = 2, x+1 do
        res = res + factorial(x)/factorial(x-(k-1)) * k*(k-1)/(x^k)
        end
        tex.sprint(res)
        end
        end{luacode*}

        begin{document}

        begin{tikzpicture}[
        declare function={p(n) = directlua{p(n)};}
        ]
        begin{axis}[
        use fpu=false, % very important!
        xlabel=$x$, ylabel=$p(x)$,
        samples at={0,...,21},
        only marks,
        ]
        addplot {p(x)};
        end{axis}
        end{tikzpicture}

        end{document}


        enter image description here






        share|improve this answer













        You can use Lua to compute the sum. Then you have to typeset using LuaLaTeX of course.



        documentclass{article}

        usepackage{pgfplots}
        pgfplotsset{compat=newest}

        usepackage{luacode}
        begin{luacode*}
        function factorial(n)
        assert(n >= 0, "Factorial is only valid for positive integers")
        if n == 0 then
        return 1
        end
        return n*factorial(n-1)
        end

        function p(x)
        assert(x == math.floor(x), "x must be an integer")
        res = 0
        for k = 2, x+1 do
        res = res + factorial(x)/factorial(x-(k-1)) * k*(k-1)/(x^k)
        end
        tex.sprint(res)
        end
        end{luacode*}

        begin{document}

        begin{tikzpicture}[
        declare function={p(n) = directlua{p(n)};}
        ]
        begin{axis}[
        use fpu=false, % very important!
        xlabel=$x$, ylabel=$p(x)$,
        samples at={0,...,21},
        only marks,
        ]
        addplot {p(x)};
        end{axis}
        end{tikzpicture}

        end{document}


        enter image description here







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 23 '17 at 23:40









        Henri MenkeHenri Menke

        73.9k8162274




        73.9k8162274






























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