Expressing $1-exp{(lambda_1p+lambda_3q)}$ as $x+y$, where $x$ is in terms of $lambda_1$ and $y$ is in terms...












2












$begingroup$


I have this simple equation
$$c = 1 - expleft(lambda_1 R^2 left(frac{2pi}{3}-frac{sqrt{3}}{2}right) + lambda_3 R^2 left(frac{2pi}{3}-frac{sqrt{3}}{2}right)right)$$



I will like to express this in the form $c = x + y$, where $x$ is in terms of $lambda_1$, and $y$ is in terms of $lambda_3$. How can I go about this?



I have tried basic expansion. However, it seems that there might be a mathematical rule to get out of this.






exponential-function







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Cannot be done.
    $endgroup$
    – Gerry Myerson
    Jan 11 at 14:55
















2












$begingroup$


I have this simple equation
$$c = 1 - expleft(lambda_1 R^2 left(frac{2pi}{3}-frac{sqrt{3}}{2}right) + lambda_3 R^2 left(frac{2pi}{3}-frac{sqrt{3}}{2}right)right)$$



I will like to express this in the form $c = x + y$, where $x$ is in terms of $lambda_1$, and $y$ is in terms of $lambda_3$. How can I go about this?



I have tried basic expansion. However, it seems that there might be a mathematical rule to get out of this.






exponential-function







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Cannot be done.
    $endgroup$
    – Gerry Myerson
    Jan 11 at 14:55














2












2








2


1



$begingroup$


I have this simple equation
$$c = 1 - expleft(lambda_1 R^2 left(frac{2pi}{3}-frac{sqrt{3}}{2}right) + lambda_3 R^2 left(frac{2pi}{3}-frac{sqrt{3}}{2}right)right)$$



I will like to express this in the form $c = x + y$, where $x$ is in terms of $lambda_1$, and $y$ is in terms of $lambda_3$. How can I go about this?



I have tried basic expansion. However, it seems that there might be a mathematical rule to get out of this.






exponential-function







share|cite|improve this question











$endgroup$




I have this simple equation
$$c = 1 - expleft(lambda_1 R^2 left(frac{2pi}{3}-frac{sqrt{3}}{2}right) + lambda_3 R^2 left(frac{2pi}{3}-frac{sqrt{3}}{2}right)right)$$



I will like to express this in the form $c = x + y$, where $x$ is in terms of $lambda_1$, and $y$ is in terms of $lambda_3$. How can I go about this?



I have tried basic expansion. However, it seems that there might be a mathematical rule to get out of this.






exponential-function




exponential-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 12 at 2:48







Abdulhameed

















asked Jan 11 at 14:26









AbdulhameedAbdulhameed

153115




153115








  • 4




    $begingroup$
    Cannot be done.
    $endgroup$
    – Gerry Myerson
    Jan 11 at 14:55














  • 4




    $begingroup$
    Cannot be done.
    $endgroup$
    – Gerry Myerson
    Jan 11 at 14:55








4




4




$begingroup$
Cannot be done.
$endgroup$
– Gerry Myerson
Jan 11 at 14:55




$begingroup$
Cannot be done.
$endgroup$
– Gerry Myerson
Jan 11 at 14:55










1 Answer
1






active

oldest

votes


















2












$begingroup$

Write $$c(lambda_1,lambda_3)=1 - e^{(lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) + lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}))}$$



Note first that $c(0,0)=1-e^0=0$.



Suppose we had the desired expression $c(lambda_1,lambda_3)=x(lambda_1)+y(lambda_3)$. Then:



$$x(lambda_1)+y(0)=c(lambda_1,0)=1 - e^{lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) }$$



$$x(0)+y(lambda_3)=c(0,lambda_3)=1 - e^{ lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2})}$$



Adding the above, we get $$x(0)+y(0)+x(lambda_1)+y(lambda_3)=1 - e^{lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) }+1 - e^{ lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2})}$$
But also we have $$x(0)+y(0)+x(lambda_1)+y(lambda_3)=0+c(lambda_1,lambda_3)=1 - e^{(lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) + lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}))}$$



Hence $$1+e^{(lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) + lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}))}=e^{lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) }+e^{ lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2})}$$



Note that this can be written as $1+xy=x+y$, which can be rewritten as $0=xy-x-y+1=(x-1)(y-1)$ and therefore has solutions $x=1$ and $y=1$. Hence, IF we can decompose $c=x+y$ as desired, then either $1=e^{lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) }$ or $1=e^{ lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2})}$. In the first case, $lambda_1=0$ (or $R=0$); in the second case, $lambda_3=0$ (or $R=0$). These can be all thought of as trivial.



If $Rlambda_1lambda_3neq 0$, then it is impossible to have a decomposition as desired.






share|cite|improve this answer









$endgroup$














    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    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%2f3069885%2fexpressing-1-exp-lambda-1p-lambda-3q-as-xy-where-x-is-in-terms-of%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









    2












    $begingroup$

    Write $$c(lambda_1,lambda_3)=1 - e^{(lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) + lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}))}$$



    Note first that $c(0,0)=1-e^0=0$.



    Suppose we had the desired expression $c(lambda_1,lambda_3)=x(lambda_1)+y(lambda_3)$. Then:



    $$x(lambda_1)+y(0)=c(lambda_1,0)=1 - e^{lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) }$$



    $$x(0)+y(lambda_3)=c(0,lambda_3)=1 - e^{ lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2})}$$



    Adding the above, we get $$x(0)+y(0)+x(lambda_1)+y(lambda_3)=1 - e^{lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) }+1 - e^{ lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2})}$$
    But also we have $$x(0)+y(0)+x(lambda_1)+y(lambda_3)=0+c(lambda_1,lambda_3)=1 - e^{(lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) + lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}))}$$



    Hence $$1+e^{(lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) + lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}))}=e^{lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) }+e^{ lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2})}$$



    Note that this can be written as $1+xy=x+y$, which can be rewritten as $0=xy-x-y+1=(x-1)(y-1)$ and therefore has solutions $x=1$ and $y=1$. Hence, IF we can decompose $c=x+y$ as desired, then either $1=e^{lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) }$ or $1=e^{ lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2})}$. In the first case, $lambda_1=0$ (or $R=0$); in the second case, $lambda_3=0$ (or $R=0$). These can be all thought of as trivial.



    If $Rlambda_1lambda_3neq 0$, then it is impossible to have a decomposition as desired.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Write $$c(lambda_1,lambda_3)=1 - e^{(lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) + lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}))}$$



      Note first that $c(0,0)=1-e^0=0$.



      Suppose we had the desired expression $c(lambda_1,lambda_3)=x(lambda_1)+y(lambda_3)$. Then:



      $$x(lambda_1)+y(0)=c(lambda_1,0)=1 - e^{lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) }$$



      $$x(0)+y(lambda_3)=c(0,lambda_3)=1 - e^{ lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2})}$$



      Adding the above, we get $$x(0)+y(0)+x(lambda_1)+y(lambda_3)=1 - e^{lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) }+1 - e^{ lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2})}$$
      But also we have $$x(0)+y(0)+x(lambda_1)+y(lambda_3)=0+c(lambda_1,lambda_3)=1 - e^{(lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) + lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}))}$$



      Hence $$1+e^{(lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) + lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}))}=e^{lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) }+e^{ lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2})}$$



      Note that this can be written as $1+xy=x+y$, which can be rewritten as $0=xy-x-y+1=(x-1)(y-1)$ and therefore has solutions $x=1$ and $y=1$. Hence, IF we can decompose $c=x+y$ as desired, then either $1=e^{lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) }$ or $1=e^{ lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2})}$. In the first case, $lambda_1=0$ (or $R=0$); in the second case, $lambda_3=0$ (or $R=0$). These can be all thought of as trivial.



      If $Rlambda_1lambda_3neq 0$, then it is impossible to have a decomposition as desired.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Write $$c(lambda_1,lambda_3)=1 - e^{(lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) + lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}))}$$



        Note first that $c(0,0)=1-e^0=0$.



        Suppose we had the desired expression $c(lambda_1,lambda_3)=x(lambda_1)+y(lambda_3)$. Then:



        $$x(lambda_1)+y(0)=c(lambda_1,0)=1 - e^{lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) }$$



        $$x(0)+y(lambda_3)=c(0,lambda_3)=1 - e^{ lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2})}$$



        Adding the above, we get $$x(0)+y(0)+x(lambda_1)+y(lambda_3)=1 - e^{lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) }+1 - e^{ lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2})}$$
        But also we have $$x(0)+y(0)+x(lambda_1)+y(lambda_3)=0+c(lambda_1,lambda_3)=1 - e^{(lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) + lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}))}$$



        Hence $$1+e^{(lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) + lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}))}=e^{lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) }+e^{ lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2})}$$



        Note that this can be written as $1+xy=x+y$, which can be rewritten as $0=xy-x-y+1=(x-1)(y-1)$ and therefore has solutions $x=1$ and $y=1$. Hence, IF we can decompose $c=x+y$ as desired, then either $1=e^{lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) }$ or $1=e^{ lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2})}$. In the first case, $lambda_1=0$ (or $R=0$); in the second case, $lambda_3=0$ (or $R=0$). These can be all thought of as trivial.



        If $Rlambda_1lambda_3neq 0$, then it is impossible to have a decomposition as desired.






        share|cite|improve this answer









        $endgroup$



        Write $$c(lambda_1,lambda_3)=1 - e^{(lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) + lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}))}$$



        Note first that $c(0,0)=1-e^0=0$.



        Suppose we had the desired expression $c(lambda_1,lambda_3)=x(lambda_1)+y(lambda_3)$. Then:



        $$x(lambda_1)+y(0)=c(lambda_1,0)=1 - e^{lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) }$$



        $$x(0)+y(lambda_3)=c(0,lambda_3)=1 - e^{ lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2})}$$



        Adding the above, we get $$x(0)+y(0)+x(lambda_1)+y(lambda_3)=1 - e^{lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) }+1 - e^{ lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2})}$$
        But also we have $$x(0)+y(0)+x(lambda_1)+y(lambda_3)=0+c(lambda_1,lambda_3)=1 - e^{(lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) + lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}))}$$



        Hence $$1+e^{(lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) + lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}))}=e^{lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) }+e^{ lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2})}$$



        Note that this can be written as $1+xy=x+y$, which can be rewritten as $0=xy-x-y+1=(x-1)(y-1)$ and therefore has solutions $x=1$ and $y=1$. Hence, IF we can decompose $c=x+y$ as desired, then either $1=e^{lambda_1 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2}) }$ or $1=e^{ lambda_3 R^2 (frac{2pi}{3}-frac{sqrt{3}}{2})}$. In the first case, $lambda_1=0$ (or $R=0$); in the second case, $lambda_3=0$ (or $R=0$). These can be all thought of as trivial.



        If $Rlambda_1lambda_3neq 0$, then it is impossible to have a decomposition as desired.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 12 at 2:22









        vadim123vadim123

        76.6k897191




        76.6k897191






























            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%2f3069885%2fexpressing-1-exp-lambda-1p-lambda-3q-as-xy-where-x-is-in-terms-of%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

            Bressuire

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
            Read more

            Cabo Verde

            Image
              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
            Read more

            Gyllenstierna

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
            Read more