Wow, brilliant demonstration. I completely forgot about that second solution, raising something to think about when Lambert's function is in negative territory.
1:51: you can see by observation that x=1 works. then, since exponentials grow faster than linear functions, unless ex is a tangent to e^x^2, there will also be a second solution. if we want to just find out umber of real solutions, graphing might be a good idea :)
A possible follow up video idea: I looked at this graph (f(x) = e^((x^2)-1) and f(x) = x) to see possible solutions (I checked 1 first so I knew 1 was a solution but plugging it into the problem) and saw that extra point. Then I began to watch your video and saw you warned not to multiply each side by X which is what I was planning on trying first. I got curious and put the following functions into desmos: f(x) = (e^x^2)/Ae f(x) = Aex And added A as a slider to see how it moves about. It seems that there is some value A where those two equations have one x,y value between them. It would be an interesting problem to add that caveat to the problem to determine what A is such that there is only one point where they cross each other. After, looking some more it seems like there are two values that A can be so maybe a constraint should be that A >= 0.
Great video. Worth pointing out that, in the end, you didn't actually need to use Lambert's W function to solve the problem. Simply putting both sides in the same format allows you to put like parts together, and might also lead you to think about whether there are any other solutions.
@@SweetSorrow777 It doesn't actually give the other solution(s), but if you sketch the graph taking into account slopes at x=1, it will suggest approximate value(s).
You can still continue solving it from this step: e^x^2 = xe ----eq1 Let e^x^2 = y Ln both sides , x^2= lny x= sqrt(ln y) By substitution in eq1, y= e sqrt(ln y) Square both sides and rearrange, ln y / y^2 = 1/ e^2 ----eq2 y^-2 ln y = e^-2 Multiply by -2 both sides and apply log properties , y^-2 ln y^-2 = -2e^-2 e^ln y^-2 * ln y^-2 = -0.27067 W( e^ln y^-2 * ln y^-2 )= W( -0.27067) ln y^-2 = W (-0.27067) y^-2 = e ^W (-0.27067) y^2 = 1.501366 y = 1.2253 x = sqrt( ln 1.2253) x = 0.45076 which is the first solution . From eq 2 : ln y / y^2 = 1/ e^2 We notice that : ln e / e^2 = 1/ e^2 So y = e e^x^2 = e Which implies that x^2 =1 x = 1 which is the second solution
If x < 0, eq. has no solutions. Square both sides: e^(2x² - 2) = x² t = - 2x² e^(- t - 2) = - ½t - 2 e^(-2) = t e^(t) t = W_-1 (- 2 e^(-2) ) = - 2 or t = W_0 (- 2 e^(-2) ) = - 0.4063757... Then x = 1 or x = 0.45076365...
Wow, brilliant demonstration. I completely forgot about that second solution, raising something to think about when Lambert's function is in negative territory.
I’m glad you got something from it! 😄
Excellent. Thank you syber
You’re welcome! 😍
Excellent!
When using the W function, I often just focus on the principal branch, but here's a case where you need both.
You're right, it's important to consider all the branches! 🔥
1:51: you can see by observation that x=1 works. then, since exponentials grow faster than linear functions, unless ex is a tangent to e^x^2, there will also be a second solution. if we want to just find out umber of real solutions, graphing might be a good idea :)
A possible follow up video idea:
I looked at this graph (f(x) = e^((x^2)-1) and f(x) = x) to see possible solutions (I checked 1 first so I knew 1 was a solution but plugging it into the problem) and saw that extra point. Then I began to watch your video and saw you warned not to multiply each side by X which is what I was planning on trying first. I got curious and put the following functions into desmos:
f(x) = (e^x^2)/Ae
f(x) = Aex
And added A as a slider to see how it moves about. It seems that there is some value A where those two equations have one x,y value between them. It would be an interesting problem to add that caveat to the problem to determine what A is such that there is only one point where they cross each other. After, looking some more it seems like there are two values that A can be so maybe a constraint should be that A >= 0.
Great video. Worth pointing out that, in the end, you didn't actually need to use Lambert's W function to solve the problem. Simply putting both sides in the same format allows you to put like parts together, and might also lead you to think about whether there are any other solutions.
Thanks, that's a great point! It's always good to look for alternative approaches and consider the broader implications of the solutions.
By inspection x=1. May or may not be the only solution.
Look at the derivatives!
@@pwmiles56 Will that give you the other solutions? I'll check that out.
@@SweetSorrow777 It doesn't actually give the other solution(s), but if you sketch the graph taking into account slopes at x=1, it will suggest approximate value(s).
Did not see that 2nd solution coming at the end. ☺
Nice!
Thanks!
how did you turned w(-2e> -2) to -2 ?
W(te^t) = t by definition
x=1 and x=0.4508... (not negative!)
You can still continue solving it from this step:
e^x^2 = xe ----eq1
Let e^x^2 = y
Ln both sides ,
x^2= lny
x= sqrt(ln y)
By substitution in eq1,
y= e sqrt(ln y)
Square both sides and rearrange,
ln y / y^2 = 1/ e^2 ----eq2
y^-2 ln y = e^-2
Multiply by -2 both sides and apply log properties ,
y^-2 ln y^-2 = -2e^-2
e^ln y^-2 * ln y^-2 = -0.27067
W( e^ln y^-2 * ln y^-2 )= W( -0.27067)
ln y^-2 = W (-0.27067)
y^-2 = e ^W (-0.27067)
y^2 = 1.501366
y = 1.2253
x = sqrt( ln 1.2253)
x = 0.45076 which is the first solution .
From eq 2 :
ln y / y^2 = 1/ e^2
We notice that :
ln e / e^2 = 1/ e^2
So y = e
e^x^2 = e
Which implies that
x^2 =1
x = 1 which is the second solution
Wow! Nice
x=√((e^(-W(-2/e^2))/e^2)=0,45076...
x = 1
If x < 0, eq. has no solutions.
Square both sides:
e^(2x² - 2) = x²
t = - 2x²
e^(- t - 2) = - ½t
- 2 e^(-2) = t e^(t)
t = W_-1 (- 2 e^(-2) ) = - 2 or
t = W_0 (- 2 e^(-2) ) =
- 0.4063757...
Then x = 1 or x = 0.45076365...