Why does this iterative way of solving of equation work? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Solving a Recurrence Relation/Equation, is there more than 1 way to solve this?Iterative method for matrix differential equationHow does this simplification work?Solution of $x^2(y')^2-2(xy-4)y'+y^2=0$For which values does this series converge?Why does changing variables work?Why does this method for solving recurrence relations work in some cases and not in others?What's wrong in this method of solving a difference equation?Solving a non-linear recurrence equationWhy variation of constant work to solve first order ODE?

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Why does this iterative way of solving of equation work?



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Solving a Recurrence Relation/Equation, is there more than 1 way to solve this?Iterative method for matrix differential equationHow does this simplification work?Solution of $x^2(y')^2-2(xy-4)y'+y^2=0$For which values does this series converge?Why does changing variables work?Why does this method for solving recurrence relations work in some cases and not in others?What's wrong in this method of solving a difference equation?Solving a non-linear recurrence equationWhy variation of constant work to solve first order ODE?










1












$begingroup$


I was solving some semiconductor physics problem and in order to get the temperature I got this nasty equation:



$$ T = dfrac7020dfrac32ln(T)+12.$$



I was thougt that I can solve this kind of equation simply by guessing solution for $T$ and then substituting that answer back into equation and then again substituting answer back into equation and so on until I am satisfied by precision of result. Somehow this method works.



Concretly for my example, my first guess was $T=1$ and I got this sequance of numbers $(585.0, 325.6419704169386, 339.4797907885183, 338.4580701961562, 338.53186591337385,338.52652733834424, ...)$ and they really seem to solve equation better and better.



Questions.



1) What is intuitive way to see why this method works?



2) How can I show rigoursly that this method actualy converges to solution of equation?



3) Obvious generalization for which the method will works seems to be:
$$ x = dfracabln(x)+c. $$ For which $a,b,c$ will this method work? Is this equation special case of some natural generalization of this equation? What are some similar equations which I can solve via this described method?



4) When will sequance of numbers in iteratiton process be finite to exacly solve equatiton? Does that case exist? Is solution to equation:
$$ x = dfracabln(x)+c $$
always (for every $a,b,c$) irational? Is it transcendental? If not, for which $a,b,c$ will that be the case?



Thank you for any help.










share|cite|improve this question









$endgroup$







  • 2




    $begingroup$
    Looks like the solution happens to be an attractive fixed point.
    $endgroup$
    – Brevan Ellefsen
    2 hours ago







  • 4




    $begingroup$
    It's called fixed point iteration; it is one of the classical topics of numerical analysis. When the method works there is some geometric intuition in terms of a "web plot"; the geometric intuition shown there essentially captures the main idea of the proof, which is called the contraction mapping principle or the Banach fixed point theorem. You can use the statement of the fixed point theorem to answer your question 3, although resolving the dependence on the initial guess is usually intractable in practice. Your question 4 is also usually intractable to answer in practice.
    $endgroup$
    – Ian
    2 hours ago







  • 1




    $begingroup$
    There is a proof for existence and uniqueness of fixed points for contraction mappings on certain domains using iteration. I'll find you some links to previous Questions here.
    $endgroup$
    – hardmath
    2 hours ago















1












$begingroup$


I was solving some semiconductor physics problem and in order to get the temperature I got this nasty equation:



$$ T = dfrac7020dfrac32ln(T)+12.$$



I was thougt that I can solve this kind of equation simply by guessing solution for $T$ and then substituting that answer back into equation and then again substituting answer back into equation and so on until I am satisfied by precision of result. Somehow this method works.



Concretly for my example, my first guess was $T=1$ and I got this sequance of numbers $(585.0, 325.6419704169386, 339.4797907885183, 338.4580701961562, 338.53186591337385,338.52652733834424, ...)$ and they really seem to solve equation better and better.



Questions.



1) What is intuitive way to see why this method works?



2) How can I show rigoursly that this method actualy converges to solution of equation?



3) Obvious generalization for which the method will works seems to be:
$$ x = dfracabln(x)+c. $$ For which $a,b,c$ will this method work? Is this equation special case of some natural generalization of this equation? What are some similar equations which I can solve via this described method?



4) When will sequance of numbers in iteratiton process be finite to exacly solve equatiton? Does that case exist? Is solution to equation:
$$ x = dfracabln(x)+c $$
always (for every $a,b,c$) irational? Is it transcendental? If not, for which $a,b,c$ will that be the case?



Thank you for any help.










share|cite|improve this question









$endgroup$







  • 2




    $begingroup$
    Looks like the solution happens to be an attractive fixed point.
    $endgroup$
    – Brevan Ellefsen
    2 hours ago







  • 4




    $begingroup$
    It's called fixed point iteration; it is one of the classical topics of numerical analysis. When the method works there is some geometric intuition in terms of a "web plot"; the geometric intuition shown there essentially captures the main idea of the proof, which is called the contraction mapping principle or the Banach fixed point theorem. You can use the statement of the fixed point theorem to answer your question 3, although resolving the dependence on the initial guess is usually intractable in practice. Your question 4 is also usually intractable to answer in practice.
    $endgroup$
    – Ian
    2 hours ago







  • 1




    $begingroup$
    There is a proof for existence and uniqueness of fixed points for contraction mappings on certain domains using iteration. I'll find you some links to previous Questions here.
    $endgroup$
    – hardmath
    2 hours ago













1












1








1





$begingroup$


I was solving some semiconductor physics problem and in order to get the temperature I got this nasty equation:



$$ T = dfrac7020dfrac32ln(T)+12.$$



I was thougt that I can solve this kind of equation simply by guessing solution for $T$ and then substituting that answer back into equation and then again substituting answer back into equation and so on until I am satisfied by precision of result. Somehow this method works.



Concretly for my example, my first guess was $T=1$ and I got this sequance of numbers $(585.0, 325.6419704169386, 339.4797907885183, 338.4580701961562, 338.53186591337385,338.52652733834424, ...)$ and they really seem to solve equation better and better.



Questions.



1) What is intuitive way to see why this method works?



2) How can I show rigoursly that this method actualy converges to solution of equation?



3) Obvious generalization for which the method will works seems to be:
$$ x = dfracabln(x)+c. $$ For which $a,b,c$ will this method work? Is this equation special case of some natural generalization of this equation? What are some similar equations which I can solve via this described method?



4) When will sequance of numbers in iteratiton process be finite to exacly solve equatiton? Does that case exist? Is solution to equation:
$$ x = dfracabln(x)+c $$
always (for every $a,b,c$) irational? Is it transcendental? If not, for which $a,b,c$ will that be the case?



Thank you for any help.










share|cite|improve this question









$endgroup$




I was solving some semiconductor physics problem and in order to get the temperature I got this nasty equation:



$$ T = dfrac7020dfrac32ln(T)+12.$$



I was thougt that I can solve this kind of equation simply by guessing solution for $T$ and then substituting that answer back into equation and then again substituting answer back into equation and so on until I am satisfied by precision of result. Somehow this method works.



Concretly for my example, my first guess was $T=1$ and I got this sequance of numbers $(585.0, 325.6419704169386, 339.4797907885183, 338.4580701961562, 338.53186591337385,338.52652733834424, ...)$ and they really seem to solve equation better and better.



Questions.



1) What is intuitive way to see why this method works?



2) How can I show rigoursly that this method actualy converges to solution of equation?



3) Obvious generalization for which the method will works seems to be:
$$ x = dfracabln(x)+c. $$ For which $a,b,c$ will this method work? Is this equation special case of some natural generalization of this equation? What are some similar equations which I can solve via this described method?



4) When will sequance of numbers in iteratiton process be finite to exacly solve equatiton? Does that case exist? Is solution to equation:
$$ x = dfracabln(x)+c $$
always (for every $a,b,c$) irational? Is it transcendental? If not, for which $a,b,c$ will that be the case?



Thank you for any help.







real-analysis sequences-and-series recurrence-relations irrational-numbers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 3 hours ago









ThomThom

381111




381111







  • 2




    $begingroup$
    Looks like the solution happens to be an attractive fixed point.
    $endgroup$
    – Brevan Ellefsen
    2 hours ago







  • 4




    $begingroup$
    It's called fixed point iteration; it is one of the classical topics of numerical analysis. When the method works there is some geometric intuition in terms of a "web plot"; the geometric intuition shown there essentially captures the main idea of the proof, which is called the contraction mapping principle or the Banach fixed point theorem. You can use the statement of the fixed point theorem to answer your question 3, although resolving the dependence on the initial guess is usually intractable in practice. Your question 4 is also usually intractable to answer in practice.
    $endgroup$
    – Ian
    2 hours ago







  • 1




    $begingroup$
    There is a proof for existence and uniqueness of fixed points for contraction mappings on certain domains using iteration. I'll find you some links to previous Questions here.
    $endgroup$
    – hardmath
    2 hours ago












  • 2




    $begingroup$
    Looks like the solution happens to be an attractive fixed point.
    $endgroup$
    – Brevan Ellefsen
    2 hours ago







  • 4




    $begingroup$
    It's called fixed point iteration; it is one of the classical topics of numerical analysis. When the method works there is some geometric intuition in terms of a "web plot"; the geometric intuition shown there essentially captures the main idea of the proof, which is called the contraction mapping principle or the Banach fixed point theorem. You can use the statement of the fixed point theorem to answer your question 3, although resolving the dependence on the initial guess is usually intractable in practice. Your question 4 is also usually intractable to answer in practice.
    $endgroup$
    – Ian
    2 hours ago







  • 1




    $begingroup$
    There is a proof for existence and uniqueness of fixed points for contraction mappings on certain domains using iteration. I'll find you some links to previous Questions here.
    $endgroup$
    – hardmath
    2 hours ago







2




2




$begingroup$
Looks like the solution happens to be an attractive fixed point.
$endgroup$
– Brevan Ellefsen
2 hours ago





$begingroup$
Looks like the solution happens to be an attractive fixed point.
$endgroup$
– Brevan Ellefsen
2 hours ago





4




4




$begingroup$
It's called fixed point iteration; it is one of the classical topics of numerical analysis. When the method works there is some geometric intuition in terms of a "web plot"; the geometric intuition shown there essentially captures the main idea of the proof, which is called the contraction mapping principle or the Banach fixed point theorem. You can use the statement of the fixed point theorem to answer your question 3, although resolving the dependence on the initial guess is usually intractable in practice. Your question 4 is also usually intractable to answer in practice.
$endgroup$
– Ian
2 hours ago





$begingroup$
It's called fixed point iteration; it is one of the classical topics of numerical analysis. When the method works there is some geometric intuition in terms of a "web plot"; the geometric intuition shown there essentially captures the main idea of the proof, which is called the contraction mapping principle or the Banach fixed point theorem. You can use the statement of the fixed point theorem to answer your question 3, although resolving the dependence on the initial guess is usually intractable in practice. Your question 4 is also usually intractable to answer in practice.
$endgroup$
– Ian
2 hours ago





1




1




$begingroup$
There is a proof for existence and uniqueness of fixed points for contraction mappings on certain domains using iteration. I'll find you some links to previous Questions here.
$endgroup$
– hardmath
2 hours ago




$begingroup$
There is a proof for existence and uniqueness of fixed points for contraction mappings on certain domains using iteration. I'll find you some links to previous Questions here.
$endgroup$
– hardmath
2 hours ago










3 Answers
3






active

oldest

votes


















2












$begingroup$

This method works because you are looking at a discrete dynamical of the form



$$x_n+1 = f(x_n)$$



where $f$ is a contraction. The rigorous proof is the Banach fixed point theorem.






share|cite|improve this answer









$endgroup$




















    1












    $begingroup$

    The equation $$T=fracab log (T)+c$$ has explicit solution(s) in terms of Lambert function.



    The result is given by
    $$T=fracab, Wleft(fraca be^fraccbright)$$



    In the linked page, you will see the different steps.



    Applied to your case, this will immeditely give
    $$T=frac4680Wleft(4680 e^8right)=338.526887451390053458527935852$$ If you do not access this function, for large values of the argument, use the expansion given in the linked page
    $$W(x)=L_1-L_2+fracL_2L_1+frac(L_2-2) L_22 L_1^2+frac(2 L_2^2-9L_2+6) L_26 L_1^3+ ...$$ with $L_1=log (x)$ and $L_2=log (L_1)$






    share|cite|improve this answer









    $endgroup$












    • $begingroup$
      That's equation (2.4.4) from Asymptotic Methods in Analysis by N. G. de Bruijn. A most excellent book.
      $endgroup$
      – marty cohen
      26 mins ago










    • $begingroup$
      @martycohen. Thanks for the information ! Fortunately, we can find it in many places. A few years ago, one of my PhD students extended it to very high orders. Cheers :-)
      $endgroup$
      – Claude Leibovici
      9 mins ago


















    0












    $begingroup$

    In general, a fixed point $p$ of a function $f(x)$ is an attractor for the iteration $x_n+1 = f(x_n)$ if $|f'(p)| < 1$. Then, if your initial guess is close enough to the fixed point, the iterations will eventually converge to it.
    If $|f'(p)| > 1$, the fixed point is a repeller, and the only way to converge to the fixed point is to start exactly there (or happen to land there after a finite
    number of iterations).



    You have three parameters $a,b,c$, but there are really just two because you can multiply numerator and denominator by the same constant. So let's suppose $b=1$. As Claude remarked, the fixed point is
    $$ p = fracaW(a e^c)$$
    and this is the only real fixed point if $a,c>0$ (this is easy to see because $f(x)$ is decreasing where it is positive). The curve $f'(p) = -1$ in the $a,c$ plane looks like this:



    enter image description here



    Above the curve, the fixed point is an attractor. In particular that is always true for $a > e$. However, $a=c=1$ is right on the curve, and it's not clear whether the fixed point would be an attractor in that case (it turns out that it isn't, by taking higher derivatives into account). If $(a,c)$ is below the curve, the fixed point is a repeller.






    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









      2












      $begingroup$

      This method works because you are looking at a discrete dynamical of the form



      $$x_n+1 = f(x_n)$$



      where $f$ is a contraction. The rigorous proof is the Banach fixed point theorem.






      share|cite|improve this answer









      $endgroup$

















        2












        $begingroup$

        This method works because you are looking at a discrete dynamical of the form



        $$x_n+1 = f(x_n)$$



        where $f$ is a contraction. The rigorous proof is the Banach fixed point theorem.






        share|cite|improve this answer









        $endgroup$















          2












          2








          2





          $begingroup$

          This method works because you are looking at a discrete dynamical of the form



          $$x_n+1 = f(x_n)$$



          where $f$ is a contraction. The rigorous proof is the Banach fixed point theorem.






          share|cite|improve this answer









          $endgroup$



          This method works because you are looking at a discrete dynamical of the form



          $$x_n+1 = f(x_n)$$



          where $f$ is a contraction. The rigorous proof is the Banach fixed point theorem.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          Tony S.F.Tony S.F.

          3,51421030




          3,51421030





















              1












              $begingroup$

              The equation $$T=fracab log (T)+c$$ has explicit solution(s) in terms of Lambert function.



              The result is given by
              $$T=fracab, Wleft(fraca be^fraccbright)$$



              In the linked page, you will see the different steps.



              Applied to your case, this will immeditely give
              $$T=frac4680Wleft(4680 e^8right)=338.526887451390053458527935852$$ If you do not access this function, for large values of the argument, use the expansion given in the linked page
              $$W(x)=L_1-L_2+fracL_2L_1+frac(L_2-2) L_22 L_1^2+frac(2 L_2^2-9L_2+6) L_26 L_1^3+ ...$$ with $L_1=log (x)$ and $L_2=log (L_1)$






              share|cite|improve this answer









              $endgroup$












              • $begingroup$
                That's equation (2.4.4) from Asymptotic Methods in Analysis by N. G. de Bruijn. A most excellent book.
                $endgroup$
                – marty cohen
                26 mins ago










              • $begingroup$
                @martycohen. Thanks for the information ! Fortunately, we can find it in many places. A few years ago, one of my PhD students extended it to very high orders. Cheers :-)
                $endgroup$
                – Claude Leibovici
                9 mins ago















              1












              $begingroup$

              The equation $$T=fracab log (T)+c$$ has explicit solution(s) in terms of Lambert function.



              The result is given by
              $$T=fracab, Wleft(fraca be^fraccbright)$$



              In the linked page, you will see the different steps.



              Applied to your case, this will immeditely give
              $$T=frac4680Wleft(4680 e^8right)=338.526887451390053458527935852$$ If you do not access this function, for large values of the argument, use the expansion given in the linked page
              $$W(x)=L_1-L_2+fracL_2L_1+frac(L_2-2) L_22 L_1^2+frac(2 L_2^2-9L_2+6) L_26 L_1^3+ ...$$ with $L_1=log (x)$ and $L_2=log (L_1)$






              share|cite|improve this answer









              $endgroup$












              • $begingroup$
                That's equation (2.4.4) from Asymptotic Methods in Analysis by N. G. de Bruijn. A most excellent book.
                $endgroup$
                – marty cohen
                26 mins ago










              • $begingroup$
                @martycohen. Thanks for the information ! Fortunately, we can find it in many places. A few years ago, one of my PhD students extended it to very high orders. Cheers :-)
                $endgroup$
                – Claude Leibovici
                9 mins ago













              1












              1








              1





              $begingroup$

              The equation $$T=fracab log (T)+c$$ has explicit solution(s) in terms of Lambert function.



              The result is given by
              $$T=fracab, Wleft(fraca be^fraccbright)$$



              In the linked page, you will see the different steps.



              Applied to your case, this will immeditely give
              $$T=frac4680Wleft(4680 e^8right)=338.526887451390053458527935852$$ If you do not access this function, for large values of the argument, use the expansion given in the linked page
              $$W(x)=L_1-L_2+fracL_2L_1+frac(L_2-2) L_22 L_1^2+frac(2 L_2^2-9L_2+6) L_26 L_1^3+ ...$$ with $L_1=log (x)$ and $L_2=log (L_1)$






              share|cite|improve this answer









              $endgroup$



              The equation $$T=fracab log (T)+c$$ has explicit solution(s) in terms of Lambert function.



              The result is given by
              $$T=fracab, Wleft(fraca be^fraccbright)$$



              In the linked page, you will see the different steps.



              Applied to your case, this will immeditely give
              $$T=frac4680Wleft(4680 e^8right)=338.526887451390053458527935852$$ If you do not access this function, for large values of the argument, use the expansion given in the linked page
              $$W(x)=L_1-L_2+fracL_2L_1+frac(L_2-2) L_22 L_1^2+frac(2 L_2^2-9L_2+6) L_26 L_1^3+ ...$$ with $L_1=log (x)$ and $L_2=log (L_1)$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered 2 hours ago









              Claude LeiboviciClaude Leibovici

              125k1158135




              125k1158135











              • $begingroup$
                That's equation (2.4.4) from Asymptotic Methods in Analysis by N. G. de Bruijn. A most excellent book.
                $endgroup$
                – marty cohen
                26 mins ago










              • $begingroup$
                @martycohen. Thanks for the information ! Fortunately, we can find it in many places. A few years ago, one of my PhD students extended it to very high orders. Cheers :-)
                $endgroup$
                – Claude Leibovici
                9 mins ago
















              • $begingroup$
                That's equation (2.4.4) from Asymptotic Methods in Analysis by N. G. de Bruijn. A most excellent book.
                $endgroup$
                – marty cohen
                26 mins ago










              • $begingroup$
                @martycohen. Thanks for the information ! Fortunately, we can find it in many places. A few years ago, one of my PhD students extended it to very high orders. Cheers :-)
                $endgroup$
                – Claude Leibovici
                9 mins ago















              $begingroup$
              That's equation (2.4.4) from Asymptotic Methods in Analysis by N. G. de Bruijn. A most excellent book.
              $endgroup$
              – marty cohen
              26 mins ago




              $begingroup$
              That's equation (2.4.4) from Asymptotic Methods in Analysis by N. G. de Bruijn. A most excellent book.
              $endgroup$
              – marty cohen
              26 mins ago












              $begingroup$
              @martycohen. Thanks for the information ! Fortunately, we can find it in many places. A few years ago, one of my PhD students extended it to very high orders. Cheers :-)
              $endgroup$
              – Claude Leibovici
              9 mins ago




              $begingroup$
              @martycohen. Thanks for the information ! Fortunately, we can find it in many places. A few years ago, one of my PhD students extended it to very high orders. Cheers :-)
              $endgroup$
              – Claude Leibovici
              9 mins ago











              0












              $begingroup$

              In general, a fixed point $p$ of a function $f(x)$ is an attractor for the iteration $x_n+1 = f(x_n)$ if $|f'(p)| < 1$. Then, if your initial guess is close enough to the fixed point, the iterations will eventually converge to it.
              If $|f'(p)| > 1$, the fixed point is a repeller, and the only way to converge to the fixed point is to start exactly there (or happen to land there after a finite
              number of iterations).



              You have three parameters $a,b,c$, but there are really just two because you can multiply numerator and denominator by the same constant. So let's suppose $b=1$. As Claude remarked, the fixed point is
              $$ p = fracaW(a e^c)$$
              and this is the only real fixed point if $a,c>0$ (this is easy to see because $f(x)$ is decreasing where it is positive). The curve $f'(p) = -1$ in the $a,c$ plane looks like this:



              enter image description here



              Above the curve, the fixed point is an attractor. In particular that is always true for $a > e$. However, $a=c=1$ is right on the curve, and it's not clear whether the fixed point would be an attractor in that case (it turns out that it isn't, by taking higher derivatives into account). If $(a,c)$ is below the curve, the fixed point is a repeller.






              share|cite|improve this answer









              $endgroup$

















                0












                $begingroup$

                In general, a fixed point $p$ of a function $f(x)$ is an attractor for the iteration $x_n+1 = f(x_n)$ if $|f'(p)| < 1$. Then, if your initial guess is close enough to the fixed point, the iterations will eventually converge to it.
                If $|f'(p)| > 1$, the fixed point is a repeller, and the only way to converge to the fixed point is to start exactly there (or happen to land there after a finite
                number of iterations).



                You have three parameters $a,b,c$, but there are really just two because you can multiply numerator and denominator by the same constant. So let's suppose $b=1$. As Claude remarked, the fixed point is
                $$ p = fracaW(a e^c)$$
                and this is the only real fixed point if $a,c>0$ (this is easy to see because $f(x)$ is decreasing where it is positive). The curve $f'(p) = -1$ in the $a,c$ plane looks like this:



                enter image description here



                Above the curve, the fixed point is an attractor. In particular that is always true for $a > e$. However, $a=c=1$ is right on the curve, and it's not clear whether the fixed point would be an attractor in that case (it turns out that it isn't, by taking higher derivatives into account). If $(a,c)$ is below the curve, the fixed point is a repeller.






                share|cite|improve this answer









                $endgroup$















                  0












                  0








                  0





                  $begingroup$

                  In general, a fixed point $p$ of a function $f(x)$ is an attractor for the iteration $x_n+1 = f(x_n)$ if $|f'(p)| < 1$. Then, if your initial guess is close enough to the fixed point, the iterations will eventually converge to it.
                  If $|f'(p)| > 1$, the fixed point is a repeller, and the only way to converge to the fixed point is to start exactly there (or happen to land there after a finite
                  number of iterations).



                  You have three parameters $a,b,c$, but there are really just two because you can multiply numerator and denominator by the same constant. So let's suppose $b=1$. As Claude remarked, the fixed point is
                  $$ p = fracaW(a e^c)$$
                  and this is the only real fixed point if $a,c>0$ (this is easy to see because $f(x)$ is decreasing where it is positive). The curve $f'(p) = -1$ in the $a,c$ plane looks like this:



                  enter image description here



                  Above the curve, the fixed point is an attractor. In particular that is always true for $a > e$. However, $a=c=1$ is right on the curve, and it's not clear whether the fixed point would be an attractor in that case (it turns out that it isn't, by taking higher derivatives into account). If $(a,c)$ is below the curve, the fixed point is a repeller.






                  share|cite|improve this answer









                  $endgroup$



                  In general, a fixed point $p$ of a function $f(x)$ is an attractor for the iteration $x_n+1 = f(x_n)$ if $|f'(p)| < 1$. Then, if your initial guess is close enough to the fixed point, the iterations will eventually converge to it.
                  If $|f'(p)| > 1$, the fixed point is a repeller, and the only way to converge to the fixed point is to start exactly there (or happen to land there after a finite
                  number of iterations).



                  You have three parameters $a,b,c$, but there are really just two because you can multiply numerator and denominator by the same constant. So let's suppose $b=1$. As Claude remarked, the fixed point is
                  $$ p = fracaW(a e^c)$$
                  and this is the only real fixed point if $a,c>0$ (this is easy to see because $f(x)$ is decreasing where it is positive). The curve $f'(p) = -1$ in the $a,c$ plane looks like this:



                  enter image description here



                  Above the curve, the fixed point is an attractor. In particular that is always true for $a > e$. However, $a=c=1$ is right on the curve, and it's not clear whether the fixed point would be an attractor in that case (it turns out that it isn't, by taking higher derivatives into account). If $(a,c)$ is below the curve, the fixed point is a repeller.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  Robert IsraelRobert Israel

                  331k23221477




                  331k23221477



























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