fix point solution or approximation available? logistic regression?












1












$begingroup$


please, is there a simple closed-form or approximation to the following fixed-point problem in $x$? $x$ is the value searched for. $m$, $g$ and $N$ are real parameters, all greater than 0,



begin{equation}
x=frac{1}{1+m g^{-N x}}
end{equation}



any insights on properties of $x$ are very welcome! In particular, comments about how $x$ evolves towards convergence, in case we use a fixed point algorithm to find $x$, are welcome as well. This problem looks a bit like logistic regression, but I'm not sure about how to further relate the two. thanks!






real-analysis roots logistic-regression







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    please, is there a simple closed-form or approximation to the following fixed-point problem in $x$? $x$ is the value searched for. $m$, $g$ and $N$ are real parameters, all greater than 0,



    begin{equation}
    x=frac{1}{1+m g^{-N x}}
    end{equation}



    any insights on properties of $x$ are very welcome! In particular, comments about how $x$ evolves towards convergence, in case we use a fixed point algorithm to find $x$, are welcome as well. This problem looks a bit like logistic regression, but I'm not sure about how to further relate the two. thanks!






    real-analysis roots logistic-regression







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      please, is there a simple closed-form or approximation to the following fixed-point problem in $x$? $x$ is the value searched for. $m$, $g$ and $N$ are real parameters, all greater than 0,



      begin{equation}
      x=frac{1}{1+m g^{-N x}}
      end{equation}



      any insights on properties of $x$ are very welcome! In particular, comments about how $x$ evolves towards convergence, in case we use a fixed point algorithm to find $x$, are welcome as well. This problem looks a bit like logistic regression, but I'm not sure about how to further relate the two. thanks!






      real-analysis roots logistic-regression







      share|cite|improve this question











      $endgroup$




      please, is there a simple closed-form or approximation to the following fixed-point problem in $x$? $x$ is the value searched for. $m$, $g$ and $N$ are real parameters, all greater than 0,



      begin{equation}
      x=frac{1}{1+m g^{-N x}}
      end{equation}



      any insights on properties of $x$ are very welcome! In particular, comments about how $x$ evolves towards convergence, in case we use a fixed point algorithm to find $x$, are welcome as well. This problem looks a bit like logistic regression, but I'm not sure about how to further relate the two. thanks!






      real-analysis roots logistic-regression




      real-analysis roots logistic-regression






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 20 '18 at 13:12







      Daniel S.

















      asked Dec 20 '18 at 11:23









      Daniel S.Daniel S.

      396




      396






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          I am afraid that there is no closed form of the solution.



          Without any information about $x$, let us consider the function
          $$f(x)=x left(1+m g^{-n x}right)-1$$ What we have is $f(0)=-1$ and $f(1)=m g^{-n} >0$.



          We also have to notice that
          $$f(0)=-1 qquad f'(0)=1+m qquad f''(0)=-2, m, n log (g)$$ So, if $g>1$, $f(0) times f''(0) >0$ and by Darboux theorem , starting with $x_0=0$, Newton method would converge without any overshoot of the solution. As a first iterate, Newton method will give $x_1=frac{1}{1+m}$ and just continue until convergence using
          $$x_{k+1}=frac{g^{n x_k}-m n x_k^2 log (g)}{g^{n x_k}-m n x_k log (g)+m}$$



          Edit



          Assuming that $x$ could be small, we could try to write
          $$y=x left(1+m g^{-n x}right)$$ Expand as a Taylor series built at $x=0$ to get
          $$y=(m+1) x-a m x^2+frac{a^2 m}{2} x^3 +Oleft(x^4right)$$ where $a=n log(g)$ and use series reversion to get
          $$x=frac{y}{m+1}+frac{a m }{(m+1)^3}y^2+frac{a^2m left(3 m-1right)}{2
          (m+1)^5}y^3+Oleft(y^4right)$$           and set $y=1$ to get
          $$xsimeqfrac{1}{m+1}+frac{a m }{(m+1)^3}+frac{a^2m left(3 m-1right)}{2
          (m+1)^5}$$



          Edit



          Let $t=frac 1{m+1}$ and get
          $$x=t+a t^2+frac{a (3 a-2)}{2!} t^3+frac{a^2 (16 a-21)}{3!} t^4+frac{a^2(125 a^2-244 a+48 )}{4!} t^5+frac{a^3 left(1296 a^2-3355 a+1500right) } {5!} t^6+Oleft(t^7right)$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            thanks a lot, @ClaudeLeibovici Please, do you have a reference to Darboux-Fourier theorem? Also, does your solution possibly lead to an approximation of the root?
            $endgroup$
            – Daniel S.
            Dec 20 '18 at 15:20












          • $begingroup$
            @DanielS. Here is a link numdam.org/article/NAM_1869_2_8__17_0.pdf
            $endgroup$
            – Claude Leibovici
            Dec 20 '18 at 15:38












          • $begingroup$
            thanks a lot for the reference! so, $x_1$, $x_2$, and so on will be refined approximations of the root, right? maybe closed-form expressions for those quantities are good approximations for the root?
            $endgroup$
            – Daniel S.
            Dec 20 '18 at 16:35










          • $begingroup$
            thanks again, @ClaudeLeibovici
            $endgroup$
            – Daniel S.
            Dec 21 '18 at 11:25










          • $begingroup$
            @DanielS. Let me know how my last stuff works for approximation (run a few cases). Thanks & cheers. :-)
            $endgroup$
            – Claude Leibovici
            Dec 21 '18 at 11:27











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          1 Answer
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          $begingroup$

          I am afraid that there is no closed form of the solution.



          Without any information about $x$, let us consider the function
          $$f(x)=x left(1+m g^{-n x}right)-1$$ What we have is $f(0)=-1$ and $f(1)=m g^{-n} >0$.



          We also have to notice that
          $$f(0)=-1 qquad f'(0)=1+m qquad f''(0)=-2, m, n log (g)$$ So, if $g>1$, $f(0) times f''(0) >0$ and by Darboux theorem , starting with $x_0=0$, Newton method would converge without any overshoot of the solution. As a first iterate, Newton method will give $x_1=frac{1}{1+m}$ and just continue until convergence using
          $$x_{k+1}=frac{g^{n x_k}-m n x_k^2 log (g)}{g^{n x_k}-m n x_k log (g)+m}$$



          Edit



          Assuming that $x$ could be small, we could try to write
          $$y=x left(1+m g^{-n x}right)$$ Expand as a Taylor series built at $x=0$ to get
          $$y=(m+1) x-a m x^2+frac{a^2 m}{2} x^3 +Oleft(x^4right)$$ where $a=n log(g)$ and use series reversion to get
          $$x=frac{y}{m+1}+frac{a m }{(m+1)^3}y^2+frac{a^2m left(3 m-1right)}{2
          (m+1)^5}y^3+Oleft(y^4right)$$           and set $y=1$ to get
          $$xsimeqfrac{1}{m+1}+frac{a m }{(m+1)^3}+frac{a^2m left(3 m-1right)}{2
          (m+1)^5}$$



          Edit



          Let $t=frac 1{m+1}$ and get
          $$x=t+a t^2+frac{a (3 a-2)}{2!} t^3+frac{a^2 (16 a-21)}{3!} t^4+frac{a^2(125 a^2-244 a+48 )}{4!} t^5+frac{a^3 left(1296 a^2-3355 a+1500right) } {5!} t^6+Oleft(t^7right)$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            thanks a lot, @ClaudeLeibovici Please, do you have a reference to Darboux-Fourier theorem? Also, does your solution possibly lead to an approximation of the root?
            $endgroup$
            – Daniel S.
            Dec 20 '18 at 15:20












          • $begingroup$
            @DanielS. Here is a link numdam.org/article/NAM_1869_2_8__17_0.pdf
            $endgroup$
            – Claude Leibovici
            Dec 20 '18 at 15:38












          • $begingroup$
            thanks a lot for the reference! so, $x_1$, $x_2$, and so on will be refined approximations of the root, right? maybe closed-form expressions for those quantities are good approximations for the root?
            $endgroup$
            – Daniel S.
            Dec 20 '18 at 16:35










          • $begingroup$
            thanks again, @ClaudeLeibovici
            $endgroup$
            – Daniel S.
            Dec 21 '18 at 11:25










          • $begingroup$
            @DanielS. Let me know how my last stuff works for approximation (run a few cases). Thanks & cheers. :-)
            $endgroup$
            – Claude Leibovici
            Dec 21 '18 at 11:27
















          2












          $begingroup$

          I am afraid that there is no closed form of the solution.



          Without any information about $x$, let us consider the function
          $$f(x)=x left(1+m g^{-n x}right)-1$$ What we have is $f(0)=-1$ and $f(1)=m g^{-n} >0$.



          We also have to notice that
          $$f(0)=-1 qquad f'(0)=1+m qquad f''(0)=-2, m, n log (g)$$ So, if $g>1$, $f(0) times f''(0) >0$ and by Darboux theorem , starting with $x_0=0$, Newton method would converge without any overshoot of the solution. As a first iterate, Newton method will give $x_1=frac{1}{1+m}$ and just continue until convergence using
          $$x_{k+1}=frac{g^{n x_k}-m n x_k^2 log (g)}{g^{n x_k}-m n x_k log (g)+m}$$



          Edit



          Assuming that $x$ could be small, we could try to write
          $$y=x left(1+m g^{-n x}right)$$ Expand as a Taylor series built at $x=0$ to get
          $$y=(m+1) x-a m x^2+frac{a^2 m}{2} x^3 +Oleft(x^4right)$$ where $a=n log(g)$ and use series reversion to get
          $$x=frac{y}{m+1}+frac{a m }{(m+1)^3}y^2+frac{a^2m left(3 m-1right)}{2
          (m+1)^5}y^3+Oleft(y^4right)$$           and set $y=1$ to get
          $$xsimeqfrac{1}{m+1}+frac{a m }{(m+1)^3}+frac{a^2m left(3 m-1right)}{2
          (m+1)^5}$$



          Edit



          Let $t=frac 1{m+1}$ and get
          $$x=t+a t^2+frac{a (3 a-2)}{2!} t^3+frac{a^2 (16 a-21)}{3!} t^4+frac{a^2(125 a^2-244 a+48 )}{4!} t^5+frac{a^3 left(1296 a^2-3355 a+1500right) } {5!} t^6+Oleft(t^7right)$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            thanks a lot, @ClaudeLeibovici Please, do you have a reference to Darboux-Fourier theorem? Also, does your solution possibly lead to an approximation of the root?
            $endgroup$
            – Daniel S.
            Dec 20 '18 at 15:20












          • $begingroup$
            @DanielS. Here is a link numdam.org/article/NAM_1869_2_8__17_0.pdf
            $endgroup$
            – Claude Leibovici
            Dec 20 '18 at 15:38












          • $begingroup$
            thanks a lot for the reference! so, $x_1$, $x_2$, and so on will be refined approximations of the root, right? maybe closed-form expressions for those quantities are good approximations for the root?
            $endgroup$
            – Daniel S.
            Dec 20 '18 at 16:35










          • $begingroup$
            thanks again, @ClaudeLeibovici
            $endgroup$
            – Daniel S.
            Dec 21 '18 at 11:25










          • $begingroup$
            @DanielS. Let me know how my last stuff works for approximation (run a few cases). Thanks & cheers. :-)
            $endgroup$
            – Claude Leibovici
            Dec 21 '18 at 11:27














          2












          2








          2





          $begingroup$

          I am afraid that there is no closed form of the solution.



          Without any information about $x$, let us consider the function
          $$f(x)=x left(1+m g^{-n x}right)-1$$ What we have is $f(0)=-1$ and $f(1)=m g^{-n} >0$.



          We also have to notice that
          $$f(0)=-1 qquad f'(0)=1+m qquad f''(0)=-2, m, n log (g)$$ So, if $g>1$, $f(0) times f''(0) >0$ and by Darboux theorem , starting with $x_0=0$, Newton method would converge without any overshoot of the solution. As a first iterate, Newton method will give $x_1=frac{1}{1+m}$ and just continue until convergence using
          $$x_{k+1}=frac{g^{n x_k}-m n x_k^2 log (g)}{g^{n x_k}-m n x_k log (g)+m}$$



          Edit



          Assuming that $x$ could be small, we could try to write
          $$y=x left(1+m g^{-n x}right)$$ Expand as a Taylor series built at $x=0$ to get
          $$y=(m+1) x-a m x^2+frac{a^2 m}{2} x^3 +Oleft(x^4right)$$ where $a=n log(g)$ and use series reversion to get
          $$x=frac{y}{m+1}+frac{a m }{(m+1)^3}y^2+frac{a^2m left(3 m-1right)}{2
          (m+1)^5}y^3+Oleft(y^4right)$$           and set $y=1$ to get
          $$xsimeqfrac{1}{m+1}+frac{a m }{(m+1)^3}+frac{a^2m left(3 m-1right)}{2
          (m+1)^5}$$



          Edit



          Let $t=frac 1{m+1}$ and get
          $$x=t+a t^2+frac{a (3 a-2)}{2!} t^3+frac{a^2 (16 a-21)}{3!} t^4+frac{a^2(125 a^2-244 a+48 )}{4!} t^5+frac{a^3 left(1296 a^2-3355 a+1500right) } {5!} t^6+Oleft(t^7right)$$






          share|cite|improve this answer











          $endgroup$



          I am afraid that there is no closed form of the solution.



          Without any information about $x$, let us consider the function
          $$f(x)=x left(1+m g^{-n x}right)-1$$ What we have is $f(0)=-1$ and $f(1)=m g^{-n} >0$.



          We also have to notice that
          $$f(0)=-1 qquad f'(0)=1+m qquad f''(0)=-2, m, n log (g)$$ So, if $g>1$, $f(0) times f''(0) >0$ and by Darboux theorem , starting with $x_0=0$, Newton method would converge without any overshoot of the solution. As a first iterate, Newton method will give $x_1=frac{1}{1+m}$ and just continue until convergence using
          $$x_{k+1}=frac{g^{n x_k}-m n x_k^2 log (g)}{g^{n x_k}-m n x_k log (g)+m}$$



          Edit



          Assuming that $x$ could be small, we could try to write
          $$y=x left(1+m g^{-n x}right)$$ Expand as a Taylor series built at $x=0$ to get
          $$y=(m+1) x-a m x^2+frac{a^2 m}{2} x^3 +Oleft(x^4right)$$ where $a=n log(g)$ and use series reversion to get
          $$x=frac{y}{m+1}+frac{a m }{(m+1)^3}y^2+frac{a^2m left(3 m-1right)}{2
          (m+1)^5}y^3+Oleft(y^4right)$$           and set $y=1$ to get
          $$xsimeqfrac{1}{m+1}+frac{a m }{(m+1)^3}+frac{a^2m left(3 m-1right)}{2
          (m+1)^5}$$



          Edit



          Let $t=frac 1{m+1}$ and get
          $$x=t+a t^2+frac{a (3 a-2)}{2!} t^3+frac{a^2 (16 a-21)}{3!} t^4+frac{a^2(125 a^2-244 a+48 )}{4!} t^5+frac{a^3 left(1296 a^2-3355 a+1500right) } {5!} t^6+Oleft(t^7right)$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 22 hours ago

























          answered Dec 20 '18 at 14:52









          Claude LeiboviciClaude Leibovici

          124k1158135




          124k1158135












          • $begingroup$
            thanks a lot, @ClaudeLeibovici Please, do you have a reference to Darboux-Fourier theorem? Also, does your solution possibly lead to an approximation of the root?
            $endgroup$
            – Daniel S.
            Dec 20 '18 at 15:20












          • $begingroup$
            @DanielS. Here is a link numdam.org/article/NAM_1869_2_8__17_0.pdf
            $endgroup$
            – Claude Leibovici
            Dec 20 '18 at 15:38












          • $begingroup$
            thanks a lot for the reference! so, $x_1$, $x_2$, and so on will be refined approximations of the root, right? maybe closed-form expressions for those quantities are good approximations for the root?
            $endgroup$
            – Daniel S.
            Dec 20 '18 at 16:35










          • $begingroup$
            thanks again, @ClaudeLeibovici
            $endgroup$
            – Daniel S.
            Dec 21 '18 at 11:25










          • $begingroup$
            @DanielS. Let me know how my last stuff works for approximation (run a few cases). Thanks & cheers. :-)
            $endgroup$
            – Claude Leibovici
            Dec 21 '18 at 11:27


















          • $begingroup$
            thanks a lot, @ClaudeLeibovici Please, do you have a reference to Darboux-Fourier theorem? Also, does your solution possibly lead to an approximation of the root?
            $endgroup$
            – Daniel S.
            Dec 20 '18 at 15:20












          • $begingroup$
            @DanielS. Here is a link numdam.org/article/NAM_1869_2_8__17_0.pdf
            $endgroup$
            – Claude Leibovici
            Dec 20 '18 at 15:38












          • $begingroup$
            thanks a lot for the reference! so, $x_1$, $x_2$, and so on will be refined approximations of the root, right? maybe closed-form expressions for those quantities are good approximations for the root?
            $endgroup$
            – Daniel S.
            Dec 20 '18 at 16:35










          • $begingroup$
            thanks again, @ClaudeLeibovici
            $endgroup$
            – Daniel S.
            Dec 21 '18 at 11:25










          • $begingroup$
            @DanielS. Let me know how my last stuff works for approximation (run a few cases). Thanks & cheers. :-)
            $endgroup$
            – Claude Leibovici
            Dec 21 '18 at 11:27
















          $begingroup$
          thanks a lot, @ClaudeLeibovici Please, do you have a reference to Darboux-Fourier theorem? Also, does your solution possibly lead to an approximation of the root?
          $endgroup$
          – Daniel S.
          Dec 20 '18 at 15:20






          $begingroup$
          thanks a lot, @ClaudeLeibovici Please, do you have a reference to Darboux-Fourier theorem? Also, does your solution possibly lead to an approximation of the root?
          $endgroup$
          – Daniel S.
          Dec 20 '18 at 15:20














          $begingroup$
          @DanielS. Here is a link numdam.org/article/NAM_1869_2_8__17_0.pdf
          $endgroup$
          – Claude Leibovici
          Dec 20 '18 at 15:38






          $begingroup$
          @DanielS. Here is a link numdam.org/article/NAM_1869_2_8__17_0.pdf
          $endgroup$
          – Claude Leibovici
          Dec 20 '18 at 15:38














          $begingroup$
          thanks a lot for the reference! so, $x_1$, $x_2$, and so on will be refined approximations of the root, right? maybe closed-form expressions for those quantities are good approximations for the root?
          $endgroup$
          – Daniel S.
          Dec 20 '18 at 16:35




          $begingroup$
          thanks a lot for the reference! so, $x_1$, $x_2$, and so on will be refined approximations of the root, right? maybe closed-form expressions for those quantities are good approximations for the root?
          $endgroup$
          – Daniel S.
          Dec 20 '18 at 16:35












          $begingroup$
          thanks again, @ClaudeLeibovici
          $endgroup$
          – Daniel S.
          Dec 21 '18 at 11:25




          $begingroup$
          thanks again, @ClaudeLeibovici
          $endgroup$
          – Daniel S.
          Dec 21 '18 at 11:25












          $begingroup$
          @DanielS. Let me know how my last stuff works for approximation (run a few cases). Thanks & cheers. :-)
          $endgroup$
          – Claude Leibovici
          Dec 21 '18 at 11:27




          $begingroup$
          @DanielS. Let me know how my last stuff works for approximation (run a few cases). Thanks & cheers. :-)
          $endgroup$
          – Claude Leibovici
          Dec 21 '18 at 11:27


















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          Verónica Boquete

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          Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
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          Ida-Boy-Ed-Garten

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          Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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