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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            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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            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
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