Solving $e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$ for $x$
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
edited 22 hours ago
dmtri
1,4341521
asked yesterday
Alan Glenn
253
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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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|
show 1 more comment
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
edited 22 hours ago
dmtri
1,4341521
asked yesterday
Alan Glenn
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.
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
|
show 1 more comment
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
edited 22 hours ago
dmtri
1,4341521
asked yesterday
Alan Glenn
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.
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
edited 22 hours ago
dmtri
1,4341521
asked yesterday
Alan Glenn
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.
edited 22 hours ago
dmtri
1,4341521
asked yesterday
Alan Glenn
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.
edited 22 hours ago
dmtri
1,4341521
edited 22 hours ago
dmtri
1,4341521
edited 22 hours ago
dmtri
1,4341521
1,4341521
asked yesterday
Alan Glenn
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.
asked yesterday
Alan Glenn
253
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
|
show 1 more comment
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
|
show 1 more comment
2 Answers
2
active
oldest
votes
$$e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$$
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 $-frac{1}{e}<x<0$
answered 23 hours ago
JJacquelin
42.6k21750
add a comment |
As mentioned above, clearly $x=0$ is a solution.
Also (Mathematica):
$$x = pm sqrt{2} sqrt{-2 t v W_{-1}left(-frac{1}{2 sqrt{e}}right)-t v}$$
answered yesterday
David G. Stork
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
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
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votes
$$e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$$
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 $-frac{1}{e}<x<0$
answered 23 hours ago
JJacquelin
42.6k21750
add a comment |
$$e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$$
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 $-frac{1}{e}<x<0$
answered 23 hours ago
JJacquelin
42.6k21750
add a comment |
$$e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$$
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 $-frac{1}{e}<x<0$
answered 23 hours ago
JJacquelin
42.6k21750
$$e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$$
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 $-frac{1}{e}<x<0$
answered 23 hours ago
JJacquelin
42.6k21750
edited 22 hours ago
answered 23 hours ago
JJacquelin
42.6k21750
answered 23 hours ago
JJacquelin
42.6k21750
answered 23 hours ago
JJacquelin
42.6k21750
42.6k21750
add a comment |
add a comment |
As mentioned above, clearly $x=0$ is a solution.
Also (Mathematica):
$$x = pm sqrt{2} sqrt{-2 t v W_{-1}left(-frac{1}{2 sqrt{e}}right)-t v}$$
answered yesterday
David G. Stork
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
add a comment |
As mentioned above, clearly $x=0$ is a solution.
Also (Mathematica):
$$x = pm sqrt{2} sqrt{-2 t v W_{-1}left(-frac{1}{2 sqrt{e}}right)-t v}$$
answered yesterday
David G. Stork
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
add a comment |
As mentioned above, clearly $x=0$ is a solution.
Also (Mathematica):
$$x = pm sqrt{2} sqrt{-2 t v W_{-1}left(-frac{1}{2 sqrt{e}}right)-t v}$$
answered yesterday
David G. Stork
9,81521232
As mentioned above, clearly $x=0$ is a solution.
Also (Mathematica):
$$x = pm sqrt{2} sqrt{-2 t v W_{-1}left(-frac{1}{2 sqrt{e}}right)-t v}$$
answered yesterday
David G. Stork
9,81521232
answered yesterday
David G. Stork
9,81521232
answered yesterday
David G. Stork
9,81521232
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
add a comment |
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
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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.
Alan Glenn is a new contributor. Be nice, and check out our Code of Conduct.
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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