Solving $e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$ for $x$












3














I need help with solving this difficult fluid dynamic expression. I have tried using rules of logs, symbolab algebra calculator and Wolfram Alpha calculator, and I have got no solution.




How would you solve the following expression for $x$?
$$e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$$




When solving this numerically, the solution is: $x=2.2418sqrt{vt}$




I want to know how you could solve the first expression to get the solution. So could someone please provide a step-by-step solution please?






algebra-precalculus exponential-function







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Alan Glenn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 1




    If you move everything to one side and ask Wolfram Alpha, you get a rather nasty-looking result involving the Lambert W function: wolframalpha.com/input/…
    – Eevee Trainer
    yesterday






  • 1




    The expression you get is also much nastier than the numeric solution, so I think solving for $x$ might not be the way to go, because of how nontrivial it is.
    – Eevee Trainer
    yesterday










  • Would you have to use use the Newton-Raphson formula?
    – Alan Glenn
    yesterday






  • 1




    x = 0 is a solution.
    – William Elliot
    yesterday










  • What are the parameter values for $v$,$t$? Looks like it could be a perturbation problem if $v t $ is large.
    – DavidG
    yesterday
















3














I need help with solving this difficult fluid dynamic expression. I have tried using rules of logs, symbolab algebra calculator and Wolfram Alpha calculator, and I have got no solution.




How would you solve the following expression for $x$?
$$e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$$




When solving this numerically, the solution is: $x=2.2418sqrt{vt}$




I want to know how you could solve the first expression to get the solution. So could someone please provide a step-by-step solution please?






algebra-precalculus exponential-function







share|cite|improve this question









New contributor




Alan Glenn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 1




    If you move everything to one side and ask Wolfram Alpha, you get a rather nasty-looking result involving the Lambert W function: wolframalpha.com/input/…
    – Eevee Trainer
    yesterday






  • 1




    The expression you get is also much nastier than the numeric solution, so I think solving for $x$ might not be the way to go, because of how nontrivial it is.
    – Eevee Trainer
    yesterday










  • Would you have to use use the Newton-Raphson formula?
    – Alan Glenn
    yesterday






  • 1




    x = 0 is a solution.
    – William Elliot
    yesterday










  • What are the parameter values for $v$,$t$? Looks like it could be a perturbation problem if $v t $ is large.
    – DavidG
    yesterday














3












3








3


2





I need help with solving this difficult fluid dynamic expression. I have tried using rules of logs, symbolab algebra calculator and Wolfram Alpha calculator, and I have got no solution.




How would you solve the following expression for $x$?
$$e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$$




When solving this numerically, the solution is: $x=2.2418sqrt{vt}$




I want to know how you could solve the first expression to get the solution. So could someone please provide a step-by-step solution please?






algebra-precalculus exponential-function







share|cite|improve this question









New contributor




Alan Glenn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I need help with solving this difficult fluid dynamic expression. I have tried using rules of logs, symbolab algebra calculator and Wolfram Alpha calculator, and I have got no solution.




How would you solve the following expression for $x$?
$$e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$$




When solving this numerically, the solution is: $x=2.2418sqrt{vt}$




I want to know how you could solve the first expression to get the solution. So could someone please provide a step-by-step solution please?






algebra-precalculus exponential-function




algebra-precalculus exponential-function






share|cite|improve this question









New contributor




Alan Glenn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Alan Glenn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 22 hours ago









dmtri

1,4341521




1,4341521






New contributor




Alan Glenn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked yesterday









Alan Glenn

253




253




New contributor




Alan Glenn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Alan Glenn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Alan Glenn is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    If you move everything to one side and ask Wolfram Alpha, you get a rather nasty-looking result involving the Lambert W function: wolframalpha.com/input/…
    – Eevee Trainer
    yesterday






  • 1




    The expression you get is also much nastier than the numeric solution, so I think solving for $x$ might not be the way to go, because of how nontrivial it is.
    – Eevee Trainer
    yesterday










  • Would you have to use use the Newton-Raphson formula?
    – Alan Glenn
    yesterday






  • 1




    x = 0 is a solution.
    – William Elliot
    yesterday










  • What are the parameter values for $v$,$t$? Looks like it could be a perturbation problem if $v t $ is large.
    – DavidG
    yesterday














  • 1




    If you move everything to one side and ask Wolfram Alpha, you get a rather nasty-looking result involving the Lambert W function: wolframalpha.com/input/…
    – Eevee Trainer
    yesterday






  • 1




    The expression you get is also much nastier than the numeric solution, so I think solving for $x$ might not be the way to go, because of how nontrivial it is.
    – Eevee Trainer
    yesterday










  • Would you have to use use the Newton-Raphson formula?
    – Alan Glenn
    yesterday






  • 1




    x = 0 is a solution.
    – William Elliot
    yesterday










  • What are the parameter values for $v$,$t$? Looks like it could be a perturbation problem if $v t $ is large.
    – DavidG
    yesterday








1




1




If you move everything to one side and ask Wolfram Alpha, you get a rather nasty-looking result involving the Lambert W function: wolframalpha.com/input/…
– Eevee Trainer
yesterday




If you move everything to one side and ask Wolfram Alpha, you get a rather nasty-looking result involving the Lambert W function: wolframalpha.com/input/…
– Eevee Trainer
yesterday




1




1




The expression you get is also much nastier than the numeric solution, so I think solving for $x$ might not be the way to go, because of how nontrivial it is.
– Eevee Trainer
yesterday




The expression you get is also much nastier than the numeric solution, so I think solving for $x$ might not be the way to go, because of how nontrivial it is.
– Eevee Trainer
yesterday












Would you have to use use the Newton-Raphson formula?
– Alan Glenn
yesterday




Would you have to use use the Newton-Raphson formula?
– Alan Glenn
yesterday




1




1




x = 0 is a solution.
– William Elliot
yesterday




x = 0 is a solution.
– William Elliot
yesterday












What are the parameter values for $v$,$t$? Looks like it could be a perturbation problem if $v t $ is large.
– DavidG
yesterday




What are the parameter values for $v$,$t$? Looks like it could be a perturbation problem if $v t $ is large.
– DavidG
yesterday










2 Answers
2






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oldest

votes


















5














$$e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$$
Let $y=frac{x^2}{4vt}$
$$e^y=1+2y$$
$$e^{-y}=frac{1}{1+2y}$$
$$(1+2y)e^{-y}=1$$
$$(frac12+y)e^{-y}=frac12$$
$$(-frac12-y)e^{-y}=-frac12$$
$$ (-frac12-y)e^{-y}e^{-frac12}=-frac12 e^{-frac12}$$
$$ (-frac12-y)e^{-frac12-y}=-frac12 e^{-frac12}=-frac{1}{2sqrt{e}}$$
$X=(-frac12-y)$
$$Xe^X=-frac{1}{2sqrt{e}}$$
From the definition of the Lambert W function : http://mathworld.wolfram.com/LambertW-Function.html
$$X=Wleft(-frac{1}{2sqrt{e}}right)$$
$$y=-frac12-X=-frac12-Wleft(-frac{1}{2sqrt{e}}right)$$
$$x=sqrt{4vty}=sqrt{-2-4Wleft(-frac{1}{2sqrt{e}}right)}sqrt{vt}$$
The Lambert W(z) function is a multi valuated function in $-frac{1}{e} <z<0$ , real $z$.



This is presently the case where $z=-frac{1}{2sqrt{e}}$ since $-frac{1}{e} <-frac{1}{2sqrt{e}}<0$



First root :



$W_0left(-frac{1}{2sqrt{e}}right)=-frac12 quad;quad {-2-4W_0left(-frac{1}{2sqrt{e}}right)}=-2-4(-1/2)=0 quad;quad x=0$



Second root :



$W_{-1}left(-frac{1}{2sqrt{e}}right)simeq -1.756431... $



https://www.wolframalpha.com/input/?i=lambertw(-1,-1%2F(2+sqrt(e)))



One can use series expansion to compute it approximately. See page 13 in https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function . The convergence is very slow. See the numerical calculus in the below addition. In practice, it is certainly faster to solve directly $e^y=1+2y$ with Newton-Raphson or similar iterative method.



Finally, an approximate value is :



$$xsimeqsqrt{-2-4(-1.756431)}sqrt{vt}simeq 2.2418128 sqrt{vt}$$



With more digits : https://www.wolframalpha.com/input/?i=sqrt(-2-4+lambertw(-1,-1%2F(2+sqrt(e))))



IN ADDITION :



Example of recursive numerical calculus of $W_{-1}(x)$ in the range $-frac{1}{e}<x<0$



enter image description here






share|cite|improve this answer































    1














    As mentioned above, clearly $x=0$ is a solution.



    Also (Mathematica):



    $$x = pm sqrt{2} sqrt{-2 t v W_{-1}left(-frac{1}{2 sqrt{e}}right)-t v}$$






    share|cite|improve this answer





















    • Sorry pardon my ignorance, which special function is $W_n(x)$?
      – DavidG
      yesterday






    • 1




      The Lambert W function or ProductLog. en.wikipedia.org/wiki/Lambert_W_function . mathworld.wolfram.com/ProductLogFunction.html
      – David G. Stork
      yesterday










    • Cheers @David G. Stork
      – DavidG
      yesterday










    • How would you expand the Lamber W function to get $x=2.2418sqrt{vt}$ @DavidG.Stork
      – Alan Glenn
      yesterday












    • @AlanGlenn: I would merely evaluate it numerically, just as we do with innumerable other functions, such as $sin$, $log$, etc.
      – David G. Stork
      yesterday











    Your Answer





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    2 Answers
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    2 Answers
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    5














    $$e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$$
    Let $y=frac{x^2}{4vt}$
    $$e^y=1+2y$$
    $$e^{-y}=frac{1}{1+2y}$$
    $$(1+2y)e^{-y}=1$$
    $$(frac12+y)e^{-y}=frac12$$
    $$(-frac12-y)e^{-y}=-frac12$$
    $$ (-frac12-y)e^{-y}e^{-frac12}=-frac12 e^{-frac12}$$
    $$ (-frac12-y)e^{-frac12-y}=-frac12 e^{-frac12}=-frac{1}{2sqrt{e}}$$
    $X=(-frac12-y)$
    $$Xe^X=-frac{1}{2sqrt{e}}$$
    From the definition of the Lambert W function : http://mathworld.wolfram.com/LambertW-Function.html
    $$X=Wleft(-frac{1}{2sqrt{e}}right)$$
    $$y=-frac12-X=-frac12-Wleft(-frac{1}{2sqrt{e}}right)$$
    $$x=sqrt{4vty}=sqrt{-2-4Wleft(-frac{1}{2sqrt{e}}right)}sqrt{vt}$$
    The Lambert W(z) function is a multi valuated function in $-frac{1}{e} <z<0$ , real $z$.



    This is presently the case where $z=-frac{1}{2sqrt{e}}$ since $-frac{1}{e} <-frac{1}{2sqrt{e}}<0$



    First root :



    $W_0left(-frac{1}{2sqrt{e}}right)=-frac12 quad;quad {-2-4W_0left(-frac{1}{2sqrt{e}}right)}=-2-4(-1/2)=0 quad;quad x=0$



    Second root :



    $W_{-1}left(-frac{1}{2sqrt{e}}right)simeq -1.756431... $



    https://www.wolframalpha.com/input/?i=lambertw(-1,-1%2F(2+sqrt(e)))



    One can use series expansion to compute it approximately. See page 13 in https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function . The convergence is very slow. See the numerical calculus in the below addition. In practice, it is certainly faster to solve directly $e^y=1+2y$ with Newton-Raphson or similar iterative method.



    Finally, an approximate value is :



    $$xsimeqsqrt{-2-4(-1.756431)}sqrt{vt}simeq 2.2418128 sqrt{vt}$$



    With more digits : https://www.wolframalpha.com/input/?i=sqrt(-2-4+lambertw(-1,-1%2F(2+sqrt(e))))



    IN ADDITION :



    Example of recursive numerical calculus of $W_{-1}(x)$ in the range $-frac{1}{e}<x<0$



    enter image description here






    share|cite|improve this answer




























      5














      $$e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$$
      Let $y=frac{x^2}{4vt}$
      $$e^y=1+2y$$
      $$e^{-y}=frac{1}{1+2y}$$
      $$(1+2y)e^{-y}=1$$
      $$(frac12+y)e^{-y}=frac12$$
      $$(-frac12-y)e^{-y}=-frac12$$
      $$ (-frac12-y)e^{-y}e^{-frac12}=-frac12 e^{-frac12}$$
      $$ (-frac12-y)e^{-frac12-y}=-frac12 e^{-frac12}=-frac{1}{2sqrt{e}}$$
      $X=(-frac12-y)$
      $$Xe^X=-frac{1}{2sqrt{e}}$$
      From the definition of the Lambert W function : http://mathworld.wolfram.com/LambertW-Function.html
      $$X=Wleft(-frac{1}{2sqrt{e}}right)$$
      $$y=-frac12-X=-frac12-Wleft(-frac{1}{2sqrt{e}}right)$$
      $$x=sqrt{4vty}=sqrt{-2-4Wleft(-frac{1}{2sqrt{e}}right)}sqrt{vt}$$
      The Lambert W(z) function is a multi valuated function in $-frac{1}{e} <z<0$ , real $z$.



      This is presently the case where $z=-frac{1}{2sqrt{e}}$ since $-frac{1}{e} <-frac{1}{2sqrt{e}}<0$



      First root :



      $W_0left(-frac{1}{2sqrt{e}}right)=-frac12 quad;quad {-2-4W_0left(-frac{1}{2sqrt{e}}right)}=-2-4(-1/2)=0 quad;quad x=0$



      Second root :



      $W_{-1}left(-frac{1}{2sqrt{e}}right)simeq -1.756431... $



      https://www.wolframalpha.com/input/?i=lambertw(-1,-1%2F(2+sqrt(e)))



      One can use series expansion to compute it approximately. See page 13 in https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function . The convergence is very slow. See the numerical calculus in the below addition. In practice, it is certainly faster to solve directly $e^y=1+2y$ with Newton-Raphson or similar iterative method.



      Finally, an approximate value is :



      $$xsimeqsqrt{-2-4(-1.756431)}sqrt{vt}simeq 2.2418128 sqrt{vt}$$



      With more digits : https://www.wolframalpha.com/input/?i=sqrt(-2-4+lambertw(-1,-1%2F(2+sqrt(e))))



      IN ADDITION :



      Example of recursive numerical calculus of $W_{-1}(x)$ in the range $-frac{1}{e}<x<0$



      enter image description here






      share|cite|improve this answer


























        5












        5








        5






        $$e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$$
        Let $y=frac{x^2}{4vt}$
        $$e^y=1+2y$$
        $$e^{-y}=frac{1}{1+2y}$$
        $$(1+2y)e^{-y}=1$$
        $$(frac12+y)e^{-y}=frac12$$
        $$(-frac12-y)e^{-y}=-frac12$$
        $$ (-frac12-y)e^{-y}e^{-frac12}=-frac12 e^{-frac12}$$
        $$ (-frac12-y)e^{-frac12-y}=-frac12 e^{-frac12}=-frac{1}{2sqrt{e}}$$
        $X=(-frac12-y)$
        $$Xe^X=-frac{1}{2sqrt{e}}$$
        From the definition of the Lambert W function : http://mathworld.wolfram.com/LambertW-Function.html
        $$X=Wleft(-frac{1}{2sqrt{e}}right)$$
        $$y=-frac12-X=-frac12-Wleft(-frac{1}{2sqrt{e}}right)$$
        $$x=sqrt{4vty}=sqrt{-2-4Wleft(-frac{1}{2sqrt{e}}right)}sqrt{vt}$$
        The Lambert W(z) function is a multi valuated function in $-frac{1}{e} <z<0$ , real $z$.



        This is presently the case where $z=-frac{1}{2sqrt{e}}$ since $-frac{1}{e} <-frac{1}{2sqrt{e}}<0$



        First root :



        $W_0left(-frac{1}{2sqrt{e}}right)=-frac12 quad;quad {-2-4W_0left(-frac{1}{2sqrt{e}}right)}=-2-4(-1/2)=0 quad;quad x=0$



        Second root :



        $W_{-1}left(-frac{1}{2sqrt{e}}right)simeq -1.756431... $



        https://www.wolframalpha.com/input/?i=lambertw(-1,-1%2F(2+sqrt(e)))



        One can use series expansion to compute it approximately. See page 13 in https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function . The convergence is very slow. See the numerical calculus in the below addition. In practice, it is certainly faster to solve directly $e^y=1+2y$ with Newton-Raphson or similar iterative method.



        Finally, an approximate value is :



        $$xsimeqsqrt{-2-4(-1.756431)}sqrt{vt}simeq 2.2418128 sqrt{vt}$$



        With more digits : https://www.wolframalpha.com/input/?i=sqrt(-2-4+lambertw(-1,-1%2F(2+sqrt(e))))



        IN ADDITION :



        Example of recursive numerical calculus of $W_{-1}(x)$ in the range $-frac{1}{e}<x<0$



        enter image description here






        share|cite|improve this answer














        $$e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$$
        Let $y=frac{x^2}{4vt}$
        $$e^y=1+2y$$
        $$e^{-y}=frac{1}{1+2y}$$
        $$(1+2y)e^{-y}=1$$
        $$(frac12+y)e^{-y}=frac12$$
        $$(-frac12-y)e^{-y}=-frac12$$
        $$ (-frac12-y)e^{-y}e^{-frac12}=-frac12 e^{-frac12}$$
        $$ (-frac12-y)e^{-frac12-y}=-frac12 e^{-frac12}=-frac{1}{2sqrt{e}}$$
        $X=(-frac12-y)$
        $$Xe^X=-frac{1}{2sqrt{e}}$$
        From the definition of the Lambert W function : http://mathworld.wolfram.com/LambertW-Function.html
        $$X=Wleft(-frac{1}{2sqrt{e}}right)$$
        $$y=-frac12-X=-frac12-Wleft(-frac{1}{2sqrt{e}}right)$$
        $$x=sqrt{4vty}=sqrt{-2-4Wleft(-frac{1}{2sqrt{e}}right)}sqrt{vt}$$
        The Lambert W(z) function is a multi valuated function in $-frac{1}{e} <z<0$ , real $z$.



        This is presently the case where $z=-frac{1}{2sqrt{e}}$ since $-frac{1}{e} <-frac{1}{2sqrt{e}}<0$



        First root :



        $W_0left(-frac{1}{2sqrt{e}}right)=-frac12 quad;quad {-2-4W_0left(-frac{1}{2sqrt{e}}right)}=-2-4(-1/2)=0 quad;quad x=0$



        Second root :



        $W_{-1}left(-frac{1}{2sqrt{e}}right)simeq -1.756431... $



        https://www.wolframalpha.com/input/?i=lambertw(-1,-1%2F(2+sqrt(e)))



        One can use series expansion to compute it approximately. See page 13 in https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function . The convergence is very slow. See the numerical calculus in the below addition. In practice, it is certainly faster to solve directly $e^y=1+2y$ with Newton-Raphson or similar iterative method.



        Finally, an approximate value is :



        $$xsimeqsqrt{-2-4(-1.756431)}sqrt{vt}simeq 2.2418128 sqrt{vt}$$



        With more digits : https://www.wolframalpha.com/input/?i=sqrt(-2-4+lambertw(-1,-1%2F(2+sqrt(e))))



        IN ADDITION :



        Example of recursive numerical calculus of $W_{-1}(x)$ in the range $-frac{1}{e}<x<0$



        enter image description here







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 22 hours ago

























        answered 23 hours ago









        JJacquelin

        42.6k21750




        42.6k21750























            1














            As mentioned above, clearly $x=0$ is a solution.



            Also (Mathematica):



            $$x = pm sqrt{2} sqrt{-2 t v W_{-1}left(-frac{1}{2 sqrt{e}}right)-t v}$$






            share|cite|improve this answer





















            • Sorry pardon my ignorance, which special function is $W_n(x)$?
              – DavidG
              yesterday






            • 1




              The Lambert W function or ProductLog. en.wikipedia.org/wiki/Lambert_W_function . mathworld.wolfram.com/ProductLogFunction.html
              – David G. Stork
              yesterday










            • Cheers @David G. Stork
              – DavidG
              yesterday










            • How would you expand the Lamber W function to get $x=2.2418sqrt{vt}$ @DavidG.Stork
              – Alan Glenn
              yesterday












            • @AlanGlenn: I would merely evaluate it numerically, just as we do with innumerable other functions, such as $sin$, $log$, etc.
              – David G. Stork
              yesterday
















            1














            As mentioned above, clearly $x=0$ is a solution.



            Also (Mathematica):



            $$x = pm sqrt{2} sqrt{-2 t v W_{-1}left(-frac{1}{2 sqrt{e}}right)-t v}$$






            share|cite|improve this answer





















            • Sorry pardon my ignorance, which special function is $W_n(x)$?
              – DavidG
              yesterday






            • 1




              The Lambert W function or ProductLog. en.wikipedia.org/wiki/Lambert_W_function . mathworld.wolfram.com/ProductLogFunction.html
              – David G. Stork
              yesterday










            • Cheers @David G. Stork
              – DavidG
              yesterday










            • How would you expand the Lamber W function to get $x=2.2418sqrt{vt}$ @DavidG.Stork
              – Alan Glenn
              yesterday












            • @AlanGlenn: I would merely evaluate it numerically, just as we do with innumerable other functions, such as $sin$, $log$, etc.
              – David G. Stork
              yesterday














            1












            1








            1






            As mentioned above, clearly $x=0$ is a solution.



            Also (Mathematica):



            $$x = pm sqrt{2} sqrt{-2 t v W_{-1}left(-frac{1}{2 sqrt{e}}right)-t v}$$






            share|cite|improve this answer












            As mentioned above, clearly $x=0$ is a solution.



            Also (Mathematica):



            $$x = pm sqrt{2} sqrt{-2 t v W_{-1}left(-frac{1}{2 sqrt{e}}right)-t v}$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered yesterday









            David G. Stork

            9,81521232




            9,81521232












            • Sorry pardon my ignorance, which special function is $W_n(x)$?
              – DavidG
              yesterday






            • 1




              The Lambert W function or ProductLog. en.wikipedia.org/wiki/Lambert_W_function . mathworld.wolfram.com/ProductLogFunction.html
              – David G. Stork
              yesterday










            • Cheers @David G. Stork
              – DavidG
              yesterday










            • How would you expand the Lamber W function to get $x=2.2418sqrt{vt}$ @DavidG.Stork
              – Alan Glenn
              yesterday












            • @AlanGlenn: I would merely evaluate it numerically, just as we do with innumerable other functions, such as $sin$, $log$, etc.
              – David G. Stork
              yesterday


















            • Sorry pardon my ignorance, which special function is $W_n(x)$?
              – DavidG
              yesterday






            • 1




              The Lambert W function or ProductLog. en.wikipedia.org/wiki/Lambert_W_function . mathworld.wolfram.com/ProductLogFunction.html
              – David G. Stork
              yesterday










            • Cheers @David G. Stork
              – DavidG
              yesterday










            • How would you expand the Lamber W function to get $x=2.2418sqrt{vt}$ @DavidG.Stork
              – Alan Glenn
              yesterday












            • @AlanGlenn: I would merely evaluate it numerically, just as we do with innumerable other functions, such as $sin$, $log$, etc.
              – David G. Stork
              yesterday
















            Sorry pardon my ignorance, which special function is $W_n(x)$?
            – DavidG
            yesterday




            Sorry pardon my ignorance, which special function is $W_n(x)$?
            – DavidG
            yesterday




            1




            1




            The Lambert W function or ProductLog. en.wikipedia.org/wiki/Lambert_W_function . mathworld.wolfram.com/ProductLogFunction.html
            – David G. Stork
            yesterday




            The Lambert W function or ProductLog. en.wikipedia.org/wiki/Lambert_W_function . mathworld.wolfram.com/ProductLogFunction.html
            – David G. Stork
            yesterday












            Cheers @David G. Stork
            – DavidG
            yesterday




            Cheers @David G. Stork
            – DavidG
            yesterday












            How would you expand the Lamber W function to get $x=2.2418sqrt{vt}$ @DavidG.Stork
            – Alan Glenn
            yesterday






            How would you expand the Lamber W function to get $x=2.2418sqrt{vt}$ @DavidG.Stork
            – Alan Glenn
            yesterday














            @AlanGlenn: I would merely evaluate it numerically, just as we do with innumerable other functions, such as $sin$, $log$, etc.
            – David G. Stork
            yesterday




            @AlanGlenn: I would merely evaluate it numerically, just as we do with innumerable other functions, such as $sin$, $log$, etc.
            – David G. Stork
            yesterday










            Alan Glenn is a new contributor. Be nice, and check out our Code of Conduct.










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            Alan Glenn is a new contributor. Be nice, and check out our Code of Conduct.













            Alan Glenn is a new contributor. Be nice, and check out our Code of Conduct.












            Alan Glenn is a new contributor. Be nice, and check out our Code of Conduct.
















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