How to solve $S = x + xn + xn^2 + cdots + xn^{y-1}$ for $n$












0












$begingroup$


I need to come up with a formula to calculate the coefficient from this formula




$$S = x + xn + xn^2 + cdots + xn^{y-1} tag{1}$$




Variables:





  • $S$ - total prize pool


  • $x$ - amount the last place receives


  • $y$ - number of players


  • $n$ - coefficient



How do I solve for $n$?




Thank you






geometric-progressions







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$endgroup$












  • $begingroup$
    Read this: en.wikipedia.org/wiki/Geometric_series#Formula
    $endgroup$
    – Matti P.
    Jan 9 at 13:45










  • $begingroup$
    The $n$ in your equation corresponds to $r$ in Wikipedia's notation. But I'm afraid there isn't an exact formula to calculate $r$. You'll have to resort to numerical methods for solving it. You can, for example, plot $S$ as a function of $n$ and then see what the approximate answer is.
    $endgroup$
    – Matti P.
    Jan 9 at 13:46










  • $begingroup$
    I set the values for S, x, y. I just need to understand how to solve for n
    $endgroup$
    – Yuri Vaskovski
    Jan 9 at 13:56












  • $begingroup$
    Could you precise if $n$ is supposed to be small (such as $n=1+epsilon$ ? If this is the case, we can derive quite nice approximations.
    $endgroup$
    – Claude Leibovici
    Jan 13 at 9:32
















0












$begingroup$


I need to come up with a formula to calculate the coefficient from this formula




$$S = x + xn + xn^2 + cdots + xn^{y-1} tag{1}$$




Variables:





  • $S$ - total prize pool


  • $x$ - amount the last place receives


  • $y$ - number of players


  • $n$ - coefficient



How do I solve for $n$?




Thank you






geometric-progressions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Read this: en.wikipedia.org/wiki/Geometric_series#Formula
    $endgroup$
    – Matti P.
    Jan 9 at 13:45










  • $begingroup$
    The $n$ in your equation corresponds to $r$ in Wikipedia's notation. But I'm afraid there isn't an exact formula to calculate $r$. You'll have to resort to numerical methods for solving it. You can, for example, plot $S$ as a function of $n$ and then see what the approximate answer is.
    $endgroup$
    – Matti P.
    Jan 9 at 13:46










  • $begingroup$
    I set the values for S, x, y. I just need to understand how to solve for n
    $endgroup$
    – Yuri Vaskovski
    Jan 9 at 13:56












  • $begingroup$
    Could you precise if $n$ is supposed to be small (such as $n=1+epsilon$ ? If this is the case, we can derive quite nice approximations.
    $endgroup$
    – Claude Leibovici
    Jan 13 at 9:32














0












0








0





$begingroup$


I need to come up with a formula to calculate the coefficient from this formula




$$S = x + xn + xn^2 + cdots + xn^{y-1} tag{1}$$




Variables:





  • $S$ - total prize pool


  • $x$ - amount the last place receives


  • $y$ - number of players


  • $n$ - coefficient



How do I solve for $n$?




Thank you






geometric-progressions







share|cite|improve this question











$endgroup$




I need to come up with a formula to calculate the coefficient from this formula




$$S = x + xn + xn^2 + cdots + xn^{y-1} tag{1}$$




Variables:





  • $S$ - total prize pool


  • $x$ - amount the last place receives


  • $y$ - number of players


  • $n$ - coefficient



How do I solve for $n$?




Thank you






geometric-progressions




geometric-progressions






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 at 14:19









Blue

47.8k870152




47.8k870152










asked Jan 9 at 13:38









Yuri VaskovskiYuri Vaskovski

1




1












  • $begingroup$
    Read this: en.wikipedia.org/wiki/Geometric_series#Formula
    $endgroup$
    – Matti P.
    Jan 9 at 13:45










  • $begingroup$
    The $n$ in your equation corresponds to $r$ in Wikipedia's notation. But I'm afraid there isn't an exact formula to calculate $r$. You'll have to resort to numerical methods for solving it. You can, for example, plot $S$ as a function of $n$ and then see what the approximate answer is.
    $endgroup$
    – Matti P.
    Jan 9 at 13:46










  • $begingroup$
    I set the values for S, x, y. I just need to understand how to solve for n
    $endgroup$
    – Yuri Vaskovski
    Jan 9 at 13:56












  • $begingroup$
    Could you precise if $n$ is supposed to be small (such as $n=1+epsilon$ ? If this is the case, we can derive quite nice approximations.
    $endgroup$
    – Claude Leibovici
    Jan 13 at 9:32


















  • $begingroup$
    Read this: en.wikipedia.org/wiki/Geometric_series#Formula
    $endgroup$
    – Matti P.
    Jan 9 at 13:45










  • $begingroup$
    The $n$ in your equation corresponds to $r$ in Wikipedia's notation. But I'm afraid there isn't an exact formula to calculate $r$. You'll have to resort to numerical methods for solving it. You can, for example, plot $S$ as a function of $n$ and then see what the approximate answer is.
    $endgroup$
    – Matti P.
    Jan 9 at 13:46










  • $begingroup$
    I set the values for S, x, y. I just need to understand how to solve for n
    $endgroup$
    – Yuri Vaskovski
    Jan 9 at 13:56












  • $begingroup$
    Could you precise if $n$ is supposed to be small (such as $n=1+epsilon$ ? If this is the case, we can derive quite nice approximations.
    $endgroup$
    – Claude Leibovici
    Jan 13 at 9:32
















$begingroup$
Read this: en.wikipedia.org/wiki/Geometric_series#Formula
$endgroup$
– Matti P.
Jan 9 at 13:45




$begingroup$
Read this: en.wikipedia.org/wiki/Geometric_series#Formula
$endgroup$
– Matti P.
Jan 9 at 13:45












$begingroup$
The $n$ in your equation corresponds to $r$ in Wikipedia's notation. But I'm afraid there isn't an exact formula to calculate $r$. You'll have to resort to numerical methods for solving it. You can, for example, plot $S$ as a function of $n$ and then see what the approximate answer is.
$endgroup$
– Matti P.
Jan 9 at 13:46




$begingroup$
The $n$ in your equation corresponds to $r$ in Wikipedia's notation. But I'm afraid there isn't an exact formula to calculate $r$. You'll have to resort to numerical methods for solving it. You can, for example, plot $S$ as a function of $n$ and then see what the approximate answer is.
$endgroup$
– Matti P.
Jan 9 at 13:46












$begingroup$
I set the values for S, x, y. I just need to understand how to solve for n
$endgroup$
– Yuri Vaskovski
Jan 9 at 13:56






$begingroup$
I set the values for S, x, y. I just need to understand how to solve for n
$endgroup$
– Yuri Vaskovski
Jan 9 at 13:56














$begingroup$
Could you precise if $n$ is supposed to be small (such as $n=1+epsilon$ ? If this is the case, we can derive quite nice approximations.
$endgroup$
– Claude Leibovici
Jan 13 at 9:32




$begingroup$
Could you precise if $n$ is supposed to be small (such as $n=1+epsilon$ ? If this is the case, we can derive quite nice approximations.
$endgroup$
– Claude Leibovici
Jan 13 at 9:32










3 Answers
3






active

oldest

votes


















3












$begingroup$

The equation can be written



$$frac{n^y-1}{n-1}=frac Sx$$ or in the polynomial form



$$n^y-frac Sxn+frac Sx-1=0.$$



Such an equation doesn't have a closed-form solution, except for a few small values of the degree $y$.



A numerical solution by Newton's method works well.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    The practical way to do it, assuming $ S > xy$. You take the following function
    $$
    f(n) = left(1+frac Sx n-frac Sx right)^{1/y}
    $$



    and then apply it to itself, until result no longer changes, starting with $n_0=1+1/y$: $n_1=f(n_0)$, $n_2=f(n_1)$ and so on.



    For example, for $y=9$, $S/x=15$, we have the following
    $$n=1.111, 1.115, 1.118, 1.120, 1.121, 1.122, 1.122dots$$






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      As Yves Daoust wrote, you are looking for the zero of function
      $$f(n)=n^y-frac{ S}{x}(n-1)-1$$ Consider its derivatives
      $$f'(n)=y, n^{y-1}-frac{S}{x}$$
      $$f''(n)=y,(y-1) , n^{y-2}$$ For $y>1$, the second derivative is always positive.



      The first derivative cancels at
      $$n_*=left(frac{S}{x y}right)^{frac{1}{y-1}}$$
      So, if $f(n_*)<0$ (remember that $n=1$ is a trivial solution to be discarded), to get a starting point for Newton method, expand $f(n)$ as a Taylor series to second order around $n=n_*$ to get
      $$f(n)=f(n_*)+frac 12 f''(n_*) (n-n*)^2+Oleft((n-n_*)^3right)$$ and, ignoring the high order tems
      $$n_0=n_*+sqrt{-2frac{f(n_*) }{f''(n_*) }}$$



      Now, iterate using
      $$n_{k+1}=n_k-frac{ f(n_k)} { f'(n_k)}$$



      For the values used by Vasily Mitch $(y=9,S=15x)$, $n_*=1.06594$ , $n_0=1.12738$ anf the iterates would be
      $$left(
      begin{array}{cc}
      k & n_k \
      0 & 1.127375420 \
      1 & 1.123698919 \
      2 & 1.123557057 \
      3 & 1.123556849
      end{array}
      right)$$



      Let us do the same with huge numbers : $y=123$, $S=123456789,x$. This will give $n_*=1.11994$ and $n_0=1.16505$. Newton iterates would be
      $$left(
      begin{array}{cc}
      k & n_k \
      0 & 1.165045878 \
      1 & 1.156842265 \
      2 & 1.150315554 \
      3 & 1.146560370 \
      4 & 1.145520226 \
      5 & 1.145454006 \
      6 & 1.145453756
      end{array}
      right)$$






      share|cite|improve this answer











      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        The equation can be written



        $$frac{n^y-1}{n-1}=frac Sx$$ or in the polynomial form



        $$n^y-frac Sxn+frac Sx-1=0.$$



        Such an equation doesn't have a closed-form solution, except for a few small values of the degree $y$.



        A numerical solution by Newton's method works well.






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          The equation can be written



          $$frac{n^y-1}{n-1}=frac Sx$$ or in the polynomial form



          $$n^y-frac Sxn+frac Sx-1=0.$$



          Such an equation doesn't have a closed-form solution, except for a few small values of the degree $y$.



          A numerical solution by Newton's method works well.






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            The equation can be written



            $$frac{n^y-1}{n-1}=frac Sx$$ or in the polynomial form



            $$n^y-frac Sxn+frac Sx-1=0.$$



            Such an equation doesn't have a closed-form solution, except for a few small values of the degree $y$.



            A numerical solution by Newton's method works well.






            share|cite|improve this answer









            $endgroup$



            The equation can be written



            $$frac{n^y-1}{n-1}=frac Sx$$ or in the polynomial form



            $$n^y-frac Sxn+frac Sx-1=0.$$



            Such an equation doesn't have a closed-form solution, except for a few small values of the degree $y$.



            A numerical solution by Newton's method works well.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 9 at 14:26









            Yves DaoustYves Daoust

            125k671222




            125k671222























                2












                $begingroup$

                The practical way to do it, assuming $ S > xy$. You take the following function
                $$
                f(n) = left(1+frac Sx n-frac Sx right)^{1/y}
                $$



                and then apply it to itself, until result no longer changes, starting with $n_0=1+1/y$: $n_1=f(n_0)$, $n_2=f(n_1)$ and so on.



                For example, for $y=9$, $S/x=15$, we have the following
                $$n=1.111, 1.115, 1.118, 1.120, 1.121, 1.122, 1.122dots$$






                share|cite|improve this answer











                $endgroup$


















                  2












                  $begingroup$

                  The practical way to do it, assuming $ S > xy$. You take the following function
                  $$
                  f(n) = left(1+frac Sx n-frac Sx right)^{1/y}
                  $$



                  and then apply it to itself, until result no longer changes, starting with $n_0=1+1/y$: $n_1=f(n_0)$, $n_2=f(n_1)$ and so on.



                  For example, for $y=9$, $S/x=15$, we have the following
                  $$n=1.111, 1.115, 1.118, 1.120, 1.121, 1.122, 1.122dots$$






                  share|cite|improve this answer











                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    The practical way to do it, assuming $ S > xy$. You take the following function
                    $$
                    f(n) = left(1+frac Sx n-frac Sx right)^{1/y}
                    $$



                    and then apply it to itself, until result no longer changes, starting with $n_0=1+1/y$: $n_1=f(n_0)$, $n_2=f(n_1)$ and so on.



                    For example, for $y=9$, $S/x=15$, we have the following
                    $$n=1.111, 1.115, 1.118, 1.120, 1.121, 1.122, 1.122dots$$






                    share|cite|improve this answer











                    $endgroup$



                    The practical way to do it, assuming $ S > xy$. You take the following function
                    $$
                    f(n) = left(1+frac Sx n-frac Sx right)^{1/y}
                    $$



                    and then apply it to itself, until result no longer changes, starting with $n_0=1+1/y$: $n_1=f(n_0)$, $n_2=f(n_1)$ and so on.



                    For example, for $y=9$, $S/x=15$, we have the following
                    $$n=1.111, 1.115, 1.118, 1.120, 1.121, 1.122, 1.122dots$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 9 at 14:37

























                    answered Jan 9 at 14:13









                    Vasily MitchVasily Mitch

                    1,62338




                    1,62338























                        0












                        $begingroup$

                        As Yves Daoust wrote, you are looking for the zero of function
                        $$f(n)=n^y-frac{ S}{x}(n-1)-1$$ Consider its derivatives
                        $$f'(n)=y, n^{y-1}-frac{S}{x}$$
                        $$f''(n)=y,(y-1) , n^{y-2}$$ For $y>1$, the second derivative is always positive.



                        The first derivative cancels at
                        $$n_*=left(frac{S}{x y}right)^{frac{1}{y-1}}$$
                        So, if $f(n_*)<0$ (remember that $n=1$ is a trivial solution to be discarded), to get a starting point for Newton method, expand $f(n)$ as a Taylor series to second order around $n=n_*$ to get
                        $$f(n)=f(n_*)+frac 12 f''(n_*) (n-n*)^2+Oleft((n-n_*)^3right)$$ and, ignoring the high order tems
                        $$n_0=n_*+sqrt{-2frac{f(n_*) }{f''(n_*) }}$$



                        Now, iterate using
                        $$n_{k+1}=n_k-frac{ f(n_k)} { f'(n_k)}$$



                        For the values used by Vasily Mitch $(y=9,S=15x)$, $n_*=1.06594$ , $n_0=1.12738$ anf the iterates would be
                        $$left(
                        begin{array}{cc}
                        k & n_k \
                        0 & 1.127375420 \
                        1 & 1.123698919 \
                        2 & 1.123557057 \
                        3 & 1.123556849
                        end{array}
                        right)$$



                        Let us do the same with huge numbers : $y=123$, $S=123456789,x$. This will give $n_*=1.11994$ and $n_0=1.16505$. Newton iterates would be
                        $$left(
                        begin{array}{cc}
                        k & n_k \
                        0 & 1.165045878 \
                        1 & 1.156842265 \
                        2 & 1.150315554 \
                        3 & 1.146560370 \
                        4 & 1.145520226 \
                        5 & 1.145454006 \
                        6 & 1.145453756
                        end{array}
                        right)$$






                        share|cite|improve this answer











                        $endgroup$


















                          0












                          $begingroup$

                          As Yves Daoust wrote, you are looking for the zero of function
                          $$f(n)=n^y-frac{ S}{x}(n-1)-1$$ Consider its derivatives
                          $$f'(n)=y, n^{y-1}-frac{S}{x}$$
                          $$f''(n)=y,(y-1) , n^{y-2}$$ For $y>1$, the second derivative is always positive.



                          The first derivative cancels at
                          $$n_*=left(frac{S}{x y}right)^{frac{1}{y-1}}$$
                          So, if $f(n_*)<0$ (remember that $n=1$ is a trivial solution to be discarded), to get a starting point for Newton method, expand $f(n)$ as a Taylor series to second order around $n=n_*$ to get
                          $$f(n)=f(n_*)+frac 12 f''(n_*) (n-n*)^2+Oleft((n-n_*)^3right)$$ and, ignoring the high order tems
                          $$n_0=n_*+sqrt{-2frac{f(n_*) }{f''(n_*) }}$$



                          Now, iterate using
                          $$n_{k+1}=n_k-frac{ f(n_k)} { f'(n_k)}$$



                          For the values used by Vasily Mitch $(y=9,S=15x)$, $n_*=1.06594$ , $n_0=1.12738$ anf the iterates would be
                          $$left(
                          begin{array}{cc}
                          k & n_k \
                          0 & 1.127375420 \
                          1 & 1.123698919 \
                          2 & 1.123557057 \
                          3 & 1.123556849
                          end{array}
                          right)$$



                          Let us do the same with huge numbers : $y=123$, $S=123456789,x$. This will give $n_*=1.11994$ and $n_0=1.16505$. Newton iterates would be
                          $$left(
                          begin{array}{cc}
                          k & n_k \
                          0 & 1.165045878 \
                          1 & 1.156842265 \
                          2 & 1.150315554 \
                          3 & 1.146560370 \
                          4 & 1.145520226 \
                          5 & 1.145454006 \
                          6 & 1.145453756
                          end{array}
                          right)$$






                          share|cite|improve this answer











                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            As Yves Daoust wrote, you are looking for the zero of function
                            $$f(n)=n^y-frac{ S}{x}(n-1)-1$$ Consider its derivatives
                            $$f'(n)=y, n^{y-1}-frac{S}{x}$$
                            $$f''(n)=y,(y-1) , n^{y-2}$$ For $y>1$, the second derivative is always positive.



                            The first derivative cancels at
                            $$n_*=left(frac{S}{x y}right)^{frac{1}{y-1}}$$
                            So, if $f(n_*)<0$ (remember that $n=1$ is a trivial solution to be discarded), to get a starting point for Newton method, expand $f(n)$ as a Taylor series to second order around $n=n_*$ to get
                            $$f(n)=f(n_*)+frac 12 f''(n_*) (n-n*)^2+Oleft((n-n_*)^3right)$$ and, ignoring the high order tems
                            $$n_0=n_*+sqrt{-2frac{f(n_*) }{f''(n_*) }}$$



                            Now, iterate using
                            $$n_{k+1}=n_k-frac{ f(n_k)} { f'(n_k)}$$



                            For the values used by Vasily Mitch $(y=9,S=15x)$, $n_*=1.06594$ , $n_0=1.12738$ anf the iterates would be
                            $$left(
                            begin{array}{cc}
                            k & n_k \
                            0 & 1.127375420 \
                            1 & 1.123698919 \
                            2 & 1.123557057 \
                            3 & 1.123556849
                            end{array}
                            right)$$



                            Let us do the same with huge numbers : $y=123$, $S=123456789,x$. This will give $n_*=1.11994$ and $n_0=1.16505$. Newton iterates would be
                            $$left(
                            begin{array}{cc}
                            k & n_k \
                            0 & 1.165045878 \
                            1 & 1.156842265 \
                            2 & 1.150315554 \
                            3 & 1.146560370 \
                            4 & 1.145520226 \
                            5 & 1.145454006 \
                            6 & 1.145453756
                            end{array}
                            right)$$






                            share|cite|improve this answer











                            $endgroup$



                            As Yves Daoust wrote, you are looking for the zero of function
                            $$f(n)=n^y-frac{ S}{x}(n-1)-1$$ Consider its derivatives
                            $$f'(n)=y, n^{y-1}-frac{S}{x}$$
                            $$f''(n)=y,(y-1) , n^{y-2}$$ For $y>1$, the second derivative is always positive.



                            The first derivative cancels at
                            $$n_*=left(frac{S}{x y}right)^{frac{1}{y-1}}$$
                            So, if $f(n_*)<0$ (remember that $n=1$ is a trivial solution to be discarded), to get a starting point for Newton method, expand $f(n)$ as a Taylor series to second order around $n=n_*$ to get
                            $$f(n)=f(n_*)+frac 12 f''(n_*) (n-n*)^2+Oleft((n-n_*)^3right)$$ and, ignoring the high order tems
                            $$n_0=n_*+sqrt{-2frac{f(n_*) }{f''(n_*) }}$$



                            Now, iterate using
                            $$n_{k+1}=n_k-frac{ f(n_k)} { f'(n_k)}$$



                            For the values used by Vasily Mitch $(y=9,S=15x)$, $n_*=1.06594$ , $n_0=1.12738$ anf the iterates would be
                            $$left(
                            begin{array}{cc}
                            k & n_k \
                            0 & 1.127375420 \
                            1 & 1.123698919 \
                            2 & 1.123557057 \
                            3 & 1.123556849
                            end{array}
                            right)$$



                            Let us do the same with huge numbers : $y=123$, $S=123456789,x$. This will give $n_*=1.11994$ and $n_0=1.16505$. Newton iterates would be
                            $$left(
                            begin{array}{cc}
                            k & n_k \
                            0 & 1.165045878 \
                            1 & 1.156842265 \
                            2 & 1.150315554 \
                            3 & 1.146560370 \
                            4 & 1.145520226 \
                            5 & 1.145454006 \
                            6 & 1.145453756
                            end{array}
                            right)$$







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                            edited Jan 13 at 15:14

























                            answered Jan 13 at 7:27









                            Claude LeiboviciClaude Leibovici

                            120k1157132




                            120k1157132






























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