What is the condition that $e^x=ax+a-1$ has non-negative solution?












1












$begingroup$


Consider the equation $e^x=ax+a-1$ for $aleq 2$. What is the condition that there exists a $xgeq 0$ as the solution of this equation? What are the solutions?






calculus systems-of-equations roots







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$endgroup$












  • $begingroup$
    Can you please tell what methods have you tried to solve this question?
    $endgroup$
    – Jaideep Khare
    Dec 28 '18 at 9:53






  • 1




    $begingroup$
    Minimize the function over $x geq 0$. You can see that there is non-negative zero iff $aln , a geq 1$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 28 '18 at 9:58


















1












$begingroup$


Consider the equation $e^x=ax+a-1$ for $aleq 2$. What is the condition that there exists a $xgeq 0$ as the solution of this equation? What are the solutions?






calculus systems-of-equations roots







share|cite|improve this question









$endgroup$












  • $begingroup$
    Can you please tell what methods have you tried to solve this question?
    $endgroup$
    – Jaideep Khare
    Dec 28 '18 at 9:53






  • 1




    $begingroup$
    Minimize the function over $x geq 0$. You can see that there is non-negative zero iff $aln , a geq 1$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 28 '18 at 9:58
















1












1








1





$begingroup$


Consider the equation $e^x=ax+a-1$ for $aleq 2$. What is the condition that there exists a $xgeq 0$ as the solution of this equation? What are the solutions?






calculus systems-of-equations roots







share|cite|improve this question









$endgroup$




Consider the equation $e^x=ax+a-1$ for $aleq 2$. What is the condition that there exists a $xgeq 0$ as the solution of this equation? What are the solutions?






calculus systems-of-equations roots




calculus systems-of-equations roots






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 28 '18 at 9:51









Math_YMath_Y

605




605












  • $begingroup$
    Can you please tell what methods have you tried to solve this question?
    $endgroup$
    – Jaideep Khare
    Dec 28 '18 at 9:53






  • 1




    $begingroup$
    Minimize the function over $x geq 0$. You can see that there is non-negative zero iff $aln , a geq 1$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 28 '18 at 9:58




















  • $begingroup$
    Can you please tell what methods have you tried to solve this question?
    $endgroup$
    – Jaideep Khare
    Dec 28 '18 at 9:53






  • 1




    $begingroup$
    Minimize the function over $x geq 0$. You can see that there is non-negative zero iff $aln , a geq 1$.
    $endgroup$
    – Kavi Rama Murthy
    Dec 28 '18 at 9:58


















$begingroup$
Can you please tell what methods have you tried to solve this question?
$endgroup$
– Jaideep Khare
Dec 28 '18 at 9:53




$begingroup$
Can you please tell what methods have you tried to solve this question?
$endgroup$
– Jaideep Khare
Dec 28 '18 at 9:53




1




1




$begingroup$
Minimize the function over $x geq 0$. You can see that there is non-negative zero iff $aln , a geq 1$.
$endgroup$
– Kavi Rama Murthy
Dec 28 '18 at 9:58






$begingroup$
Minimize the function over $x geq 0$. You can see that there is non-negative zero iff $aln , a geq 1$.
$endgroup$
– Kavi Rama Murthy
Dec 28 '18 at 9:58












3 Answers
3






active

oldest

votes


















-6












$begingroup$

Hint: Thw solution can be written using the LambertW function:
$$x=-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{
frac {a-1}{a}}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It is not related to amount of $a$?
    $endgroup$
    – Math_Y
    Dec 28 '18 at 10:48










  • $begingroup$
    Sorry it was only an exemple the solution has to be $$-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{ frac {a-1}{a}} $$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 28 '18 at 10:51






  • 2




    $begingroup$
    IMO, this brings you farther from the solution than the original setting of the question.
    $endgroup$
    – Yves Daoust
    Dec 28 '18 at 12:45



















3












$begingroup$

Hint :



Let $f(x)$ be a function $f : [0,+infty) to mathbb R$ such that :
$$f(x) = e^x-ax-a+1$$



The function $f$ is differentiable over $[0,+infty)$ with :
$$f'(x) = e^x -a implies f'(x)=0 Leftrightarrow x = ln a$$



Now, investigate cases to see what happens for $a leq 2$. Be careful for $a<0$ though, as the solution to the derivative won't exist and a different case holds.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    you must also consider the case if $$a<0$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 28 '18 at 10:08










  • $begingroup$
    @Dr.SonnhardGraubner Yes, in that case the equation $f'(x) = 0$ has no solutions and is also $f'(x) >0$.
    $endgroup$
    – Rebellos
    Dec 28 '18 at 10:10










  • $begingroup$
    Thank you very much. There exists a closed form relation for the solutions?
    $endgroup$
    – Math_Y
    Dec 28 '18 at 10:29










  • $begingroup$
    @Math_Y: There is a closed form: $age e^{W(1)}$ where $W$ is the Lambert W function. This comes from the equation $aln a=1$, or equivalently $a^a=e$.
    $endgroup$
    – Rory Daulton
    Dec 28 '18 at 15:34





















0












$begingroup$

Hint:



The equation can be written



$$frac{e^x+1}{x+1}=a$$ and a plot clearly shows where the positive solutions for $ale2$ lie.



enter image description here



Now you have to show that the LHS decreases from $2$ at $x=0$ to a single minimum, which delimits the useful range.



The location of the minimum cannot be expressed with a closed-form expression, but you can relate it to the function value, and draw a simple condition.






share|cite|improve this answer











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    -6












    $begingroup$

    Hint: Thw solution can be written using the LambertW function:
    $$x=-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{
    frac {a-1}{a}}
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      It is not related to amount of $a$?
      $endgroup$
      – Math_Y
      Dec 28 '18 at 10:48










    • $begingroup$
      Sorry it was only an exemple the solution has to be $$-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{ frac {a-1}{a}} $$
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 28 '18 at 10:51






    • 2




      $begingroup$
      IMO, this brings you farther from the solution than the original setting of the question.
      $endgroup$
      – Yves Daoust
      Dec 28 '18 at 12:45
















    -6












    $begingroup$

    Hint: Thw solution can be written using the LambertW function:
    $$x=-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{
    frac {a-1}{a}}
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      It is not related to amount of $a$?
      $endgroup$
      – Math_Y
      Dec 28 '18 at 10:48










    • $begingroup$
      Sorry it was only an exemple the solution has to be $$-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{ frac {a-1}{a}} $$
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 28 '18 at 10:51






    • 2




      $begingroup$
      IMO, this brings you farther from the solution than the original setting of the question.
      $endgroup$
      – Yves Daoust
      Dec 28 '18 at 12:45














    -6












    -6








    -6





    $begingroup$

    Hint: Thw solution can be written using the LambertW function:
    $$x=-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{
    frac {a-1}{a}}
    $$






    share|cite|improve this answer











    $endgroup$



    Hint: Thw solution can be written using the LambertW function:
    $$x=-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{
    frac {a-1}{a}}
    $$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 28 '18 at 10:52

























    answered Dec 28 '18 at 10:36









    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

    79k42867




    79k42867












    • $begingroup$
      It is not related to amount of $a$?
      $endgroup$
      – Math_Y
      Dec 28 '18 at 10:48










    • $begingroup$
      Sorry it was only an exemple the solution has to be $$-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{ frac {a-1}{a}} $$
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 28 '18 at 10:51






    • 2




      $begingroup$
      IMO, this brings you farther from the solution than the original setting of the question.
      $endgroup$
      – Yves Daoust
      Dec 28 '18 at 12:45


















    • $begingroup$
      It is not related to amount of $a$?
      $endgroup$
      – Math_Y
      Dec 28 '18 at 10:48










    • $begingroup$
      Sorry it was only an exemple the solution has to be $$-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{ frac {a-1}{a}} $$
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 28 '18 at 10:51






    • 2




      $begingroup$
      IMO, this brings you farther from the solution than the original setting of the question.
      $endgroup$
      – Yves Daoust
      Dec 28 '18 at 12:45
















    $begingroup$
    It is not related to amount of $a$?
    $endgroup$
    – Math_Y
    Dec 28 '18 at 10:48




    $begingroup$
    It is not related to amount of $a$?
    $endgroup$
    – Math_Y
    Dec 28 '18 at 10:48












    $begingroup$
    Sorry it was only an exemple the solution has to be $$-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{ frac {a-1}{a}} $$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 28 '18 at 10:51




    $begingroup$
    Sorry it was only an exemple the solution has to be $$-{rm W} left(-{frac {1}{a}{{rm e}^{-{frac {a-1}{a}}}}}right)-{ frac {a-1}{a}} $$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 28 '18 at 10:51




    2




    2




    $begingroup$
    IMO, this brings you farther from the solution than the original setting of the question.
    $endgroup$
    – Yves Daoust
    Dec 28 '18 at 12:45




    $begingroup$
    IMO, this brings you farther from the solution than the original setting of the question.
    $endgroup$
    – Yves Daoust
    Dec 28 '18 at 12:45











    3












    $begingroup$

    Hint :



    Let $f(x)$ be a function $f : [0,+infty) to mathbb R$ such that :
    $$f(x) = e^x-ax-a+1$$



    The function $f$ is differentiable over $[0,+infty)$ with :
    $$f'(x) = e^x -a implies f'(x)=0 Leftrightarrow x = ln a$$



    Now, investigate cases to see what happens for $a leq 2$. Be careful for $a<0$ though, as the solution to the derivative won't exist and a different case holds.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      you must also consider the case if $$a<0$$
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 28 '18 at 10:08










    • $begingroup$
      @Dr.SonnhardGraubner Yes, in that case the equation $f'(x) = 0$ has no solutions and is also $f'(x) >0$.
      $endgroup$
      – Rebellos
      Dec 28 '18 at 10:10










    • $begingroup$
      Thank you very much. There exists a closed form relation for the solutions?
      $endgroup$
      – Math_Y
      Dec 28 '18 at 10:29










    • $begingroup$
      @Math_Y: There is a closed form: $age e^{W(1)}$ where $W$ is the Lambert W function. This comes from the equation $aln a=1$, or equivalently $a^a=e$.
      $endgroup$
      – Rory Daulton
      Dec 28 '18 at 15:34


















    3












    $begingroup$

    Hint :



    Let $f(x)$ be a function $f : [0,+infty) to mathbb R$ such that :
    $$f(x) = e^x-ax-a+1$$



    The function $f$ is differentiable over $[0,+infty)$ with :
    $$f'(x) = e^x -a implies f'(x)=0 Leftrightarrow x = ln a$$



    Now, investigate cases to see what happens for $a leq 2$. Be careful for $a<0$ though, as the solution to the derivative won't exist and a different case holds.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      you must also consider the case if $$a<0$$
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 28 '18 at 10:08










    • $begingroup$
      @Dr.SonnhardGraubner Yes, in that case the equation $f'(x) = 0$ has no solutions and is also $f'(x) >0$.
      $endgroup$
      – Rebellos
      Dec 28 '18 at 10:10










    • $begingroup$
      Thank you very much. There exists a closed form relation for the solutions?
      $endgroup$
      – Math_Y
      Dec 28 '18 at 10:29










    • $begingroup$
      @Math_Y: There is a closed form: $age e^{W(1)}$ where $W$ is the Lambert W function. This comes from the equation $aln a=1$, or equivalently $a^a=e$.
      $endgroup$
      – Rory Daulton
      Dec 28 '18 at 15:34
















    3












    3








    3





    $begingroup$

    Hint :



    Let $f(x)$ be a function $f : [0,+infty) to mathbb R$ such that :
    $$f(x) = e^x-ax-a+1$$



    The function $f$ is differentiable over $[0,+infty)$ with :
    $$f'(x) = e^x -a implies f'(x)=0 Leftrightarrow x = ln a$$



    Now, investigate cases to see what happens for $a leq 2$. Be careful for $a<0$ though, as the solution to the derivative won't exist and a different case holds.






    share|cite|improve this answer











    $endgroup$



    Hint :



    Let $f(x)$ be a function $f : [0,+infty) to mathbb R$ such that :
    $$f(x) = e^x-ax-a+1$$



    The function $f$ is differentiable over $[0,+infty)$ with :
    $$f'(x) = e^x -a implies f'(x)=0 Leftrightarrow x = ln a$$



    Now, investigate cases to see what happens for $a leq 2$. Be careful for $a<0$ though, as the solution to the derivative won't exist and a different case holds.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 28 '18 at 10:21

























    answered Dec 28 '18 at 10:03









    RebellosRebellos

    15.7k31250




    15.7k31250












    • $begingroup$
      you must also consider the case if $$a<0$$
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 28 '18 at 10:08










    • $begingroup$
      @Dr.SonnhardGraubner Yes, in that case the equation $f'(x) = 0$ has no solutions and is also $f'(x) >0$.
      $endgroup$
      – Rebellos
      Dec 28 '18 at 10:10










    • $begingroup$
      Thank you very much. There exists a closed form relation for the solutions?
      $endgroup$
      – Math_Y
      Dec 28 '18 at 10:29










    • $begingroup$
      @Math_Y: There is a closed form: $age e^{W(1)}$ where $W$ is the Lambert W function. This comes from the equation $aln a=1$, or equivalently $a^a=e$.
      $endgroup$
      – Rory Daulton
      Dec 28 '18 at 15:34




















    • $begingroup$
      you must also consider the case if $$a<0$$
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 28 '18 at 10:08










    • $begingroup$
      @Dr.SonnhardGraubner Yes, in that case the equation $f'(x) = 0$ has no solutions and is also $f'(x) >0$.
      $endgroup$
      – Rebellos
      Dec 28 '18 at 10:10










    • $begingroup$
      Thank you very much. There exists a closed form relation for the solutions?
      $endgroup$
      – Math_Y
      Dec 28 '18 at 10:29










    • $begingroup$
      @Math_Y: There is a closed form: $age e^{W(1)}$ where $W$ is the Lambert W function. This comes from the equation $aln a=1$, or equivalently $a^a=e$.
      $endgroup$
      – Rory Daulton
      Dec 28 '18 at 15:34


















    $begingroup$
    you must also consider the case if $$a<0$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 28 '18 at 10:08




    $begingroup$
    you must also consider the case if $$a<0$$
    $endgroup$
    – Dr. Sonnhard Graubner
    Dec 28 '18 at 10:08












    $begingroup$
    @Dr.SonnhardGraubner Yes, in that case the equation $f'(x) = 0$ has no solutions and is also $f'(x) >0$.
    $endgroup$
    – Rebellos
    Dec 28 '18 at 10:10




    $begingroup$
    @Dr.SonnhardGraubner Yes, in that case the equation $f'(x) = 0$ has no solutions and is also $f'(x) >0$.
    $endgroup$
    – Rebellos
    Dec 28 '18 at 10:10












    $begingroup$
    Thank you very much. There exists a closed form relation for the solutions?
    $endgroup$
    – Math_Y
    Dec 28 '18 at 10:29




    $begingroup$
    Thank you very much. There exists a closed form relation for the solutions?
    $endgroup$
    – Math_Y
    Dec 28 '18 at 10:29












    $begingroup$
    @Math_Y: There is a closed form: $age e^{W(1)}$ where $W$ is the Lambert W function. This comes from the equation $aln a=1$, or equivalently $a^a=e$.
    $endgroup$
    – Rory Daulton
    Dec 28 '18 at 15:34






    $begingroup$
    @Math_Y: There is a closed form: $age e^{W(1)}$ where $W$ is the Lambert W function. This comes from the equation $aln a=1$, or equivalently $a^a=e$.
    $endgroup$
    – Rory Daulton
    Dec 28 '18 at 15:34













    0












    $begingroup$

    Hint:



    The equation can be written



    $$frac{e^x+1}{x+1}=a$$ and a plot clearly shows where the positive solutions for $ale2$ lie.



    enter image description here



    Now you have to show that the LHS decreases from $2$ at $x=0$ to a single minimum, which delimits the useful range.



    The location of the minimum cannot be expressed with a closed-form expression, but you can relate it to the function value, and draw a simple condition.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Hint:



      The equation can be written



      $$frac{e^x+1}{x+1}=a$$ and a plot clearly shows where the positive solutions for $ale2$ lie.



      enter image description here



      Now you have to show that the LHS decreases from $2$ at $x=0$ to a single minimum, which delimits the useful range.



      The location of the minimum cannot be expressed with a closed-form expression, but you can relate it to the function value, and draw a simple condition.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Hint:



        The equation can be written



        $$frac{e^x+1}{x+1}=a$$ and a plot clearly shows where the positive solutions for $ale2$ lie.



        enter image description here



        Now you have to show that the LHS decreases from $2$ at $x=0$ to a single minimum, which delimits the useful range.



        The location of the minimum cannot be expressed with a closed-form expression, but you can relate it to the function value, and draw a simple condition.






        share|cite|improve this answer











        $endgroup$



        Hint:



        The equation can be written



        $$frac{e^x+1}{x+1}=a$$ and a plot clearly shows where the positive solutions for $ale2$ lie.



        enter image description here



        Now you have to show that the LHS decreases from $2$ at $x=0$ to a single minimum, which delimits the useful range.



        The location of the minimum cannot be expressed with a closed-form expression, but you can relate it to the function value, and draw a simple condition.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 28 '18 at 12:54

























        answered Dec 28 '18 at 12:42









        Yves DaoustYves Daoust

        133k676231




        133k676231






























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            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
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            Can I use Tabulator js library in my java Spring + Thymeleaf project?

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            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
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