Mixing
Solving $e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$ for $x$
$begingroup$
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 Jan 3 at 7:28
dmtri
1,4522521
asked Jan 3 at 1:59
Alan GlennAlan Glenn
253
$endgroup$
|
show 1 more comment
$begingroup$
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 Jan 3 at 7:28
dmtri
1,4522521
asked Jan 3 at 1:59
Alan GlennAlan Glenn
253
$endgroup$
1
$begingroup$
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/…
$endgroup$
– Eevee Trainer
Jan 3 at 2:05
1
$begingroup$
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.
$endgroup$
– Eevee Trainer
Jan 3 at 2:06
$begingroup$
Would you have to use use the Newton-Raphson formula?
$endgroup$
– Alan Glenn
Jan 3 at 2:30
1
$begingroup$
x = 0 is a solution.
$endgroup$
– William Elliot
Jan 3 at 2:54
$begingroup$
What are the parameter values for $v$,$t$? Looks like it could be a perturbation problem if $v t $ is large.
$endgroup$
– DavidG
Jan 3 at 3:54
|
show 1 more comment
$begingroup$
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 Jan 3 at 7:28
dmtri
1,4522521
asked Jan 3 at 1:59
Alan GlennAlan Glenn
253
$endgroup$
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 Jan 3 at 7:28
dmtri
1,4522521
asked Jan 3 at 1:59
Alan GlennAlan Glenn
253
edited Jan 3 at 7:28
dmtri
1,4522521
asked Jan 3 at 1:59
Alan GlennAlan Glenn
253
edited Jan 3 at 7:28
dmtri
1,4522521
edited Jan 3 at 7:28
dmtri
1,4522521
edited Jan 3 at 7:28
dmtri
1,4522521
1,4522521
asked Jan 3 at 1:59
Alan GlennAlan Glenn
253
asked Jan 3 at 1:59
Alan GlennAlan Glenn
253
asked Jan 3 at 1:59
Alan GlennAlan Glenn
253
253
1
$begingroup$
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/…
$endgroup$
– Eevee Trainer
Jan 3 at 2:05
1
$begingroup$
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.
$endgroup$
– Eevee Trainer
Jan 3 at 2:06
$begingroup$
Would you have to use use the Newton-Raphson formula?
$endgroup$
– Alan Glenn
Jan 3 at 2:30
1
$begingroup$
x = 0 is a solution.
$endgroup$
– William Elliot
Jan 3 at 2:54
$begingroup$
What are the parameter values for $v$,$t$? Looks like it could be a perturbation problem if $v t $ is large.
$endgroup$
– DavidG
Jan 3 at 3:54
|
show 1 more comment
1
$begingroup$
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/…
$endgroup$
– Eevee Trainer
Jan 3 at 2:05
1
$begingroup$
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.
$endgroup$
– Eevee Trainer
Jan 3 at 2:06
$begingroup$
Would you have to use use the Newton-Raphson formula?
$endgroup$
– Alan Glenn
Jan 3 at 2:30
1
$begingroup$
x = 0 is a solution.
$endgroup$
– William Elliot
Jan 3 at 2:54
$begingroup$
What are the parameter values for $v$,$t$? Looks like it could be a perturbation problem if $v t $ is large.
$endgroup$
– DavidG
Jan 3 at 3:54
1
1
$begingroup$
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/…
$endgroup$
– Eevee Trainer
Jan 3 at 2:05
$begingroup$
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/…
$endgroup$
– Eevee Trainer
Jan 3 at 2:05
1
1
$begingroup$
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.
$endgroup$
– Eevee Trainer
Jan 3 at 2:06
$begingroup$
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.
$endgroup$
– Eevee Trainer
Jan 3 at 2:06
$begingroup$
Would you have to use use the Newton-Raphson formula?
$endgroup$
– Alan Glenn
Jan 3 at 2:30
$begingroup$
Would you have to use use the Newton-Raphson formula?
$endgroup$
– Alan Glenn
Jan 3 at 2:30
1
1
$begingroup$
x = 0 is a solution.
$endgroup$
– William Elliot
Jan 3 at 2:54
$begingroup$
x = 0 is a solution.
$endgroup$
– William Elliot
Jan 3 at 2:54
$begingroup$
What are the parameter values for $v$,$t$? Looks like it could be a perturbation problem if $v t $ is large.
$endgroup$
– DavidG
Jan 3 at 3:54
$begingroup$
What are the parameter values for $v$,$t$? Looks like it could be a perturbation problem if $v t $ is large.
$endgroup$
– DavidG
Jan 3 at 3:54
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
$$e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$$
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 Jan 3 at 6:42
JJacquelinJJacquelin
43k21751
$endgroup$
add a comment |
$begingroup$
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 Jan 3 at 2:59
David G. StorkDavid G. Stork
10.4k21332
$endgroup$
$begingroup$
Sorry pardon my ignorance, which special function is $W_n(x)$?
$endgroup$
– DavidG
Jan 3 at 3:55
1
$begingroup$
The Lambert W function or ProductLog. en.wikipedia.org/wiki/Lambert_W_function . mathworld.wolfram.com/ProductLogFunction.html
$endgroup$
– David G. Stork
Jan 3 at 3:58
$begingroup$
Cheers @David G. Stork
$endgroup$
– DavidG
Jan 3 at 3:59
$begingroup$
How would you expand the Lamber W function to get $x=2.2418sqrt{vt}$ @DavidG.Stork
$endgroup$
– Alan Glenn
Jan 3 at 4:07
$begingroup$
@AlanGlenn: I would merely evaluate it numerically, just as we do with innumerable other functions, such as $sin$, $log$, etc.
$endgroup$
– David G. Stork
Jan 3 at 4:15
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
$$e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$$
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 Jan 3 at 6:42
JJacquelinJJacquelin
43k21751
$endgroup$
add a comment |
$begingroup$
$$e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$$
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 Jan 3 at 6:42
JJacquelinJJacquelin
43k21751
$endgroup$
add a comment |
$begingroup$
$$e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$$
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 Jan 3 at 6:42
JJacquelinJJacquelin
43k21751
$endgroup$
$$e^{frac{x^2}{4vt}} = 1+frac{x^2}{2vt}$$
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 Jan 3 at 6:42
JJacquelinJJacquelin
43k21751
edited Jan 3 at 7:45
answered Jan 3 at 6:42
JJacquelinJJacquelin
43k21751
answered Jan 3 at 6:42
JJacquelinJJacquelin
43k21751
answered Jan 3 at 6:42
JJacquelinJJacquelin
43k21751
43k21751
add a comment |
add a comment |
$begingroup$
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 Jan 3 at 2:59
David G. StorkDavid G. Stork
10.4k21332
$endgroup$
$begingroup$
Sorry pardon my ignorance, which special function is $W_n(x)$?
$endgroup$
– DavidG
Jan 3 at 3:55
1
$begingroup$
The Lambert W function or ProductLog. en.wikipedia.org/wiki/Lambert_W_function . mathworld.wolfram.com/ProductLogFunction.html
$endgroup$
– David G. Stork
Jan 3 at 3:58
$begingroup$
Cheers @David G. Stork
$endgroup$
– DavidG
Jan 3 at 3:59
$begingroup$
How would you expand the Lamber W function to get $x=2.2418sqrt{vt}$ @DavidG.Stork
$endgroup$
– Alan Glenn
Jan 3 at 4:07
$begingroup$
@AlanGlenn: I would merely evaluate it numerically, just as we do with innumerable other functions, such as $sin$, $log$, etc.
$endgroup$
– David G. Stork
Jan 3 at 4:15
add a comment |
$begingroup$
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 Jan 3 at 2:59
David G. StorkDavid G. Stork
10.4k21332
$endgroup$
$begingroup$
Sorry pardon my ignorance, which special function is $W_n(x)$?
$endgroup$
– DavidG
Jan 3 at 3:55
1
$begingroup$
The Lambert W function or ProductLog. en.wikipedia.org/wiki/Lambert_W_function . mathworld.wolfram.com/ProductLogFunction.html
$endgroup$
– David G. Stork
Jan 3 at 3:58
$begingroup$
Cheers @David G. Stork
$endgroup$
– DavidG
Jan 3 at 3:59
$begingroup$
How would you expand the Lamber W function to get $x=2.2418sqrt{vt}$ @DavidG.Stork
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– Alan Glenn
Jan 3 at 4:07
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@AlanGlenn: I would merely evaluate it numerically, just as we do with innumerable other functions, such as $sin$, $log$, etc.
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– David G. Stork
Jan 3 at 4:15
add a comment |
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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 Jan 3 at 2:59
David G. StorkDavid G. Stork
10.4k21332
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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 Jan 3 at 2:59
David G. StorkDavid G. Stork
10.4k21332
answered Jan 3 at 2:59
David G. StorkDavid G. Stork
10.4k21332
answered Jan 3 at 2:59
David G. StorkDavid G. Stork
10.4k21332
answered Jan 3 at 2:59
David G. StorkDavid G. Stork
10.4k21332
10.4k21332
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Sorry pardon my ignorance, which special function is $W_n(x)$?
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– DavidG
Jan 3 at 3:55
1
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The Lambert W function or ProductLog. en.wikipedia.org/wiki/Lambert_W_function . mathworld.wolfram.com/ProductLogFunction.html
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– David G. Stork
Jan 3 at 3:58
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Cheers @David G. Stork
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– DavidG
Jan 3 at 3:59
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How would you expand the Lamber W function to get $x=2.2418sqrt{vt}$ @DavidG.Stork
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– Alan Glenn
Jan 3 at 4:07
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@AlanGlenn: I would merely evaluate it numerically, just as we do with innumerable other functions, such as $sin$, $log$, etc.
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– David G. Stork
Jan 3 at 4:15
add a comment |
$begingroup$
Sorry pardon my ignorance, which special function is $W_n(x)$?
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– DavidG
Jan 3 at 3:55
1
$begingroup$
The Lambert W function or ProductLog. en.wikipedia.org/wiki/Lambert_W_function . mathworld.wolfram.com/ProductLogFunction.html
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– David G. Stork
Jan 3 at 3:58
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Cheers @David G. Stork
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– DavidG
Jan 3 at 3:59
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How would you expand the Lamber W function to get $x=2.2418sqrt{vt}$ @DavidG.Stork
$endgroup$
– Alan Glenn
Jan 3 at 4:07
$begingroup$
@AlanGlenn: I would merely evaluate it numerically, just as we do with innumerable other functions, such as $sin$, $log$, etc.
$endgroup$
– David G. Stork
Jan 3 at 4:15
$begingroup$
Sorry pardon my ignorance, which special function is $W_n(x)$?
$endgroup$
– DavidG
Jan 3 at 3:55
$begingroup$
Sorry pardon my ignorance, which special function is $W_n(x)$?
$endgroup$
– DavidG
Jan 3 at 3:55
1
1
$begingroup$
The Lambert W function or ProductLog. en.wikipedia.org/wiki/Lambert_W_function . mathworld.wolfram.com/ProductLogFunction.html
$endgroup$
– David G. Stork
Jan 3 at 3:58
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The Lambert W function or ProductLog. en.wikipedia.org/wiki/Lambert_W_function . mathworld.wolfram.com/ProductLogFunction.html
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– David G. Stork
Jan 3 at 3:58
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Cheers @David G. Stork
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– DavidG
Jan 3 at 3:59
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Cheers @David G. Stork
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– DavidG
Jan 3 at 3:59
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How would you expand the Lamber W function to get $x=2.2418sqrt{vt}$ @DavidG.Stork
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– Alan Glenn
Jan 3 at 4:07
$begingroup$
How would you expand the Lamber W function to get $x=2.2418sqrt{vt}$ @DavidG.Stork
$endgroup$
– Alan Glenn
Jan 3 at 4:07
$begingroup$
@AlanGlenn: I would merely evaluate it numerically, just as we do with innumerable other functions, such as $sin$, $log$, etc.
$endgroup$
– David G. Stork
Jan 3 at 4:15
$begingroup$
@AlanGlenn: I would merely evaluate it numerically, just as we do with innumerable other functions, such as $sin$, $log$, etc.
$endgroup$
– David G. Stork
Jan 3 at 4:15
add a comment |
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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/…
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– Eevee Trainer
Jan 3 at 2:05
1
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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.
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– Eevee Trainer
Jan 3 at 2:06
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Would you have to use use the Newton-Raphson formula?
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– Alan Glenn
Jan 3 at 2:30
1
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x = 0 is a solution.
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– William Elliot
Jan 3 at 2:54
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What are the parameter values for $v$,$t$? Looks like it could be a perturbation problem if $v t $ is large.
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– DavidG
Jan 3 at 3:54