How can we solve for $w$ : $2we^{w^2}-2e^w=0$












0












$begingroup$


This was the first dervative obtained with respect to $w$ from equation: $e^x+e^y+e^{w^2}-2e^w-x-y$.



We had to find the maxima and minima for which the first derivative is equal to zero and got this above equation $2we^{w^2}-2e^w=0$ which i couldnot solve to get the value of $w$.






ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    The obvious solution is $w = 1$. With some CAS you see that there is no other solution.
    $endgroup$
    – Alpha001
    Dec 24 '18 at 16:10






  • 1




    $begingroup$
    @Alpha001 What is 'CAS'?
    $endgroup$
    – shwetha
    Dec 24 '18 at 16:18










  • $begingroup$
    @shwetha CAS = Computer Algebra System. For example WolframAlpha.
    $endgroup$
    – mrtaurho
    Dec 24 '18 at 16:20










  • $begingroup$
    @mrtauro Thank you!
    $endgroup$
    – shwetha
    Dec 24 '18 at 16:24
















0












$begingroup$


This was the first dervative obtained with respect to $w$ from equation: $e^x+e^y+e^{w^2}-2e^w-x-y$.



We had to find the maxima and minima for which the first derivative is equal to zero and got this above equation $2we^{w^2}-2e^w=0$ which i couldnot solve to get the value of $w$.






ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    The obvious solution is $w = 1$. With some CAS you see that there is no other solution.
    $endgroup$
    – Alpha001
    Dec 24 '18 at 16:10






  • 1




    $begingroup$
    @Alpha001 What is 'CAS'?
    $endgroup$
    – shwetha
    Dec 24 '18 at 16:18










  • $begingroup$
    @shwetha CAS = Computer Algebra System. For example WolframAlpha.
    $endgroup$
    – mrtaurho
    Dec 24 '18 at 16:20










  • $begingroup$
    @mrtauro Thank you!
    $endgroup$
    – shwetha
    Dec 24 '18 at 16:24














0












0








0


1



$begingroup$


This was the first dervative obtained with respect to $w$ from equation: $e^x+e^y+e^{w^2}-2e^w-x-y$.



We had to find the maxima and minima for which the first derivative is equal to zero and got this above equation $2we^{w^2}-2e^w=0$ which i couldnot solve to get the value of $w$.






ordinary-differential-equations







share|cite|improve this question











$endgroup$




This was the first dervative obtained with respect to $w$ from equation: $e^x+e^y+e^{w^2}-2e^w-x-y$.



We had to find the maxima and minima for which the first derivative is equal to zero and got this above equation $2we^{w^2}-2e^w=0$ which i couldnot solve to get the value of $w$.






ordinary-differential-equations




ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 16:06









mrtaurho

6,22571641




6,22571641










asked Dec 24 '18 at 16:05









Pritee SubediPritee Subedi

31




31








  • 3




    $begingroup$
    The obvious solution is $w = 1$. With some CAS you see that there is no other solution.
    $endgroup$
    – Alpha001
    Dec 24 '18 at 16:10






  • 1




    $begingroup$
    @Alpha001 What is 'CAS'?
    $endgroup$
    – shwetha
    Dec 24 '18 at 16:18










  • $begingroup$
    @shwetha CAS = Computer Algebra System. For example WolframAlpha.
    $endgroup$
    – mrtaurho
    Dec 24 '18 at 16:20










  • $begingroup$
    @mrtauro Thank you!
    $endgroup$
    – shwetha
    Dec 24 '18 at 16:24














  • 3




    $begingroup$
    The obvious solution is $w = 1$. With some CAS you see that there is no other solution.
    $endgroup$
    – Alpha001
    Dec 24 '18 at 16:10






  • 1




    $begingroup$
    @Alpha001 What is 'CAS'?
    $endgroup$
    – shwetha
    Dec 24 '18 at 16:18










  • $begingroup$
    @shwetha CAS = Computer Algebra System. For example WolframAlpha.
    $endgroup$
    – mrtaurho
    Dec 24 '18 at 16:20










  • $begingroup$
    @mrtauro Thank you!
    $endgroup$
    – shwetha
    Dec 24 '18 at 16:24








3




3




$begingroup$
The obvious solution is $w = 1$. With some CAS you see that there is no other solution.
$endgroup$
– Alpha001
Dec 24 '18 at 16:10




$begingroup$
The obvious solution is $w = 1$. With some CAS you see that there is no other solution.
$endgroup$
– Alpha001
Dec 24 '18 at 16:10




1




1




$begingroup$
@Alpha001 What is 'CAS'?
$endgroup$
– shwetha
Dec 24 '18 at 16:18




$begingroup$
@Alpha001 What is 'CAS'?
$endgroup$
– shwetha
Dec 24 '18 at 16:18












$begingroup$
@shwetha CAS = Computer Algebra System. For example WolframAlpha.
$endgroup$
– mrtaurho
Dec 24 '18 at 16:20




$begingroup$
@shwetha CAS = Computer Algebra System. For example WolframAlpha.
$endgroup$
– mrtaurho
Dec 24 '18 at 16:20












$begingroup$
@mrtauro Thank you!
$endgroup$
– shwetha
Dec 24 '18 at 16:24




$begingroup$
@mrtauro Thank you!
$endgroup$
– shwetha
Dec 24 '18 at 16:24










1 Answer
1






active

oldest

votes


















4












$begingroup$

Maybe it is even possible to show that this equation only has got one real root without invoking a CAS. Therefore lets rewrite the equation in the following way



$$begin{align}
2we^{w^2}-2e^w&=0\
we^{w^2}&=e^w\
we^{w^2-w}&=1\
we^{w(1-w)}&=1
end{align}$$



From the last line we can directly conclude that $w=1$ satisfies the equation. Howsoever assuming only positive $w$ we may take the logarithm of both sides to further get



$$begin{align}
lnleft(we^{w(1-w)}right)&=ln(1)\
ln(w)+w(w-1)&=0\
-w^2+w&=log(w)\
-left(w-frac12right)^2+frac14&=log(w)
end{align}$$



The LHS describes as parabola with vertex at $(1/2,1/4)$ and roots at $w_1=0$ and $w_2=1$. Due the minus sign infront we can further conclude that the vertex is a maximum. Basically we are done now. The logarithm intersects with the $w$-axis at $w=1$ and tends to negative infinity as $w$ goes to $0$. But hence the parabola has got a root at $w=0$ they cannot intersect on the left hand side of $w=1$ whereas on the right hand side of $w=1$ the logarithm is strictly increasing and the parabola stricly decreasing.

So the only possible real root is $w=1$ due the above argumentation with regard to the graphs of both functions.






share|cite|improve this answer









$endgroup$














    Your Answer








    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3051396%2fhow-can-we-solve-for-w-2wew2-2ew-0%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Maybe it is even possible to show that this equation only has got one real root without invoking a CAS. Therefore lets rewrite the equation in the following way



    $$begin{align}
    2we^{w^2}-2e^w&=0\
    we^{w^2}&=e^w\
    we^{w^2-w}&=1\
    we^{w(1-w)}&=1
    end{align}$$



    From the last line we can directly conclude that $w=1$ satisfies the equation. Howsoever assuming only positive $w$ we may take the logarithm of both sides to further get



    $$begin{align}
    lnleft(we^{w(1-w)}right)&=ln(1)\
    ln(w)+w(w-1)&=0\
    -w^2+w&=log(w)\
    -left(w-frac12right)^2+frac14&=log(w)
    end{align}$$



    The LHS describes as parabola with vertex at $(1/2,1/4)$ and roots at $w_1=0$ and $w_2=1$. Due the minus sign infront we can further conclude that the vertex is a maximum. Basically we are done now. The logarithm intersects with the $w$-axis at $w=1$ and tends to negative infinity as $w$ goes to $0$. But hence the parabola has got a root at $w=0$ they cannot intersect on the left hand side of $w=1$ whereas on the right hand side of $w=1$ the logarithm is strictly increasing and the parabola stricly decreasing.

    So the only possible real root is $w=1$ due the above argumentation with regard to the graphs of both functions.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      Maybe it is even possible to show that this equation only has got one real root without invoking a CAS. Therefore lets rewrite the equation in the following way



      $$begin{align}
      2we^{w^2}-2e^w&=0\
      we^{w^2}&=e^w\
      we^{w^2-w}&=1\
      we^{w(1-w)}&=1
      end{align}$$



      From the last line we can directly conclude that $w=1$ satisfies the equation. Howsoever assuming only positive $w$ we may take the logarithm of both sides to further get



      $$begin{align}
      lnleft(we^{w(1-w)}right)&=ln(1)\
      ln(w)+w(w-1)&=0\
      -w^2+w&=log(w)\
      -left(w-frac12right)^2+frac14&=log(w)
      end{align}$$



      The LHS describes as parabola with vertex at $(1/2,1/4)$ and roots at $w_1=0$ and $w_2=1$. Due the minus sign infront we can further conclude that the vertex is a maximum. Basically we are done now. The logarithm intersects with the $w$-axis at $w=1$ and tends to negative infinity as $w$ goes to $0$. But hence the parabola has got a root at $w=0$ they cannot intersect on the left hand side of $w=1$ whereas on the right hand side of $w=1$ the logarithm is strictly increasing and the parabola stricly decreasing.

      So the only possible real root is $w=1$ due the above argumentation with regard to the graphs of both functions.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Maybe it is even possible to show that this equation only has got one real root without invoking a CAS. Therefore lets rewrite the equation in the following way



        $$begin{align}
        2we^{w^2}-2e^w&=0\
        we^{w^2}&=e^w\
        we^{w^2-w}&=1\
        we^{w(1-w)}&=1
        end{align}$$



        From the last line we can directly conclude that $w=1$ satisfies the equation. Howsoever assuming only positive $w$ we may take the logarithm of both sides to further get



        $$begin{align}
        lnleft(we^{w(1-w)}right)&=ln(1)\
        ln(w)+w(w-1)&=0\
        -w^2+w&=log(w)\
        -left(w-frac12right)^2+frac14&=log(w)
        end{align}$$



        The LHS describes as parabola with vertex at $(1/2,1/4)$ and roots at $w_1=0$ and $w_2=1$. Due the minus sign infront we can further conclude that the vertex is a maximum. Basically we are done now. The logarithm intersects with the $w$-axis at $w=1$ and tends to negative infinity as $w$ goes to $0$. But hence the parabola has got a root at $w=0$ they cannot intersect on the left hand side of $w=1$ whereas on the right hand side of $w=1$ the logarithm is strictly increasing and the parabola stricly decreasing.

        So the only possible real root is $w=1$ due the above argumentation with regard to the graphs of both functions.






        share|cite|improve this answer









        $endgroup$



        Maybe it is even possible to show that this equation only has got one real root without invoking a CAS. Therefore lets rewrite the equation in the following way



        $$begin{align}
        2we^{w^2}-2e^w&=0\
        we^{w^2}&=e^w\
        we^{w^2-w}&=1\
        we^{w(1-w)}&=1
        end{align}$$



        From the last line we can directly conclude that $w=1$ satisfies the equation. Howsoever assuming only positive $w$ we may take the logarithm of both sides to further get



        $$begin{align}
        lnleft(we^{w(1-w)}right)&=ln(1)\
        ln(w)+w(w-1)&=0\
        -w^2+w&=log(w)\
        -left(w-frac12right)^2+frac14&=log(w)
        end{align}$$



        The LHS describes as parabola with vertex at $(1/2,1/4)$ and roots at $w_1=0$ and $w_2=1$. Due the minus sign infront we can further conclude that the vertex is a maximum. Basically we are done now. The logarithm intersects with the $w$-axis at $w=1$ and tends to negative infinity as $w$ goes to $0$. But hence the parabola has got a root at $w=0$ they cannot intersect on the left hand side of $w=1$ whereas on the right hand side of $w=1$ the logarithm is strictly increasing and the parabola stricly decreasing.

        So the only possible real root is $w=1$ due the above argumentation with regard to the graphs of both functions.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 24 '18 at 16:37









        mrtaurhomrtaurho

        6,22571641




        6,22571641






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3051396%2fhow-can-we-solve-for-w-2wew2-2ew-0%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
            Read more

            Puebla de Zaragoza

            Image
            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc
            Read more

            Musa

            Image
            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
            Read more