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







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  • 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
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$begingroup$

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.






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    $begingroup$

    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

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