Seems simple? Lambert W Function.
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- Опубліковано 8 вер 2024
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To solve this equation I used the Lambert W Function. I really had fun solving it, thank you Meow for the question.
Equivalent (and imho nicer) solution is 2*sqrt(W(1/4)). You can get there from your solution with some W identities, but I found the algebra simpler when I started by exp-ing both sides of ln x = - x^2/8. Briefly:
ln x = -x^2/8
exp-ing:
x = exp(-x^2 / 8)
div by x:
1 = x^-1 exp(-x^2 / 8)
pow to -2 (*):
1 = x^2 exp(x^2 / 4)
div by 4 and W:
W(1/4) = x^2 / 4
Solve for x:
x = +- sqrt(W(1/4)).
An extra root was artificially introduced in the (*) step above so we can discard the - sign if we want to stay in R.
What I did was very similar, but before watching the video this is what I came up with.
What I did was:
x^2 + 8*ln(x) = 0
- 8*ln(x) - 8*ln(x)
x^2 = -8*ln(x)
divide x^2 on both sides
1 = -8*ln(x) / x^2 which is also equal to -8*ln(x)*x^-2
I then turned x^-2 in to e^ln(x) form and got
1 = - 8*ln(x)*e^(-2*ln(x))
I then divided everything by 4 to get the -8 to math the exponent of e and turned it into
1/4 = [-8*ln(x)*e^(-2*ln(x))]4 which simplifies the -8 to a -2
1/4 = -2*ln(x)*e^(-2*ln(x))
This is where I then used the Lambert W Function to solve the rest of the equation
W(1/4) = W( -2*ln(x)*e^(-2*ln(x))) This then cancels the form if we let ( -2*ln(x) be equal to y which gives us W(ye^y) just give us our y(or our ( -2*ln(x)) back straight to us.
After this we have
W(1/4) = -2*ln(x)
divide -2 on both sides to start getting the x by itself
W(1/4) = -2*ln(x)
/-2 /-2
which leaves us to the last two steps
(W(1/4))/-2 = ln(x)
exponentiate
e^((W(1/4))/-2) = x
and finally simplify
[e^(W(1/4))]^-1/2 = x
to get
sqrt(1/e^(W(1/4))) = x as the final solution.
which is roughly equal to 0.9030801227
Beautiful ! x^2+8 lnx = x^2+4 lnx^2.
X=x^2. X + 4 lnX = 0. More easy ! Same result of course. Have a nice day !
great explanation maam
Just an suggestion, when you did take the root you should explained why we do not take into account the negative part. I say this because some other problems will have both solutions, this is not the case but that need to be clear for us to understand...
Flow and way You explain is visually very clear. It keeps interesting. The written lines are very dim.
It seems so easy now. Thank you so much
Glad I could help 🙂
From lnx^(-2)=W(¼) --> -2lnx=W(¼)
lnx=-½W(¼)
Therefore x=e^[-½W(¼)]
Is this x value the same as the one you got?
I got same also
I got this also. Seems less messy
That's how I solved it before looking at the solution from the video.
yeah got the same, there isnt any misused logarithm rules therefore the answer should be exact same
You can also assume x=e^t in the beginning. After solving in terms of t, you can convert it back to x. I found this easier.
If x=e^t the equation becomes (e^t)²+8t=0
How do you solve for t?
とてもよく分かりました😊👏👏👏
Hmm interesting 😅
I like how I did it. Assuming we are looking for positive real solutions, exponentiate the original equation, you get
x^8e^(x^2)=1.
Take the principle fourth root to get
x^2e^((x^2)/4)=1.
Divide by 4.
((x^2)/4)e^((x^2)/4)=1/4
Using W,
(x^2)/4=W(1/4).
And multiplying by 4 and taking the principle square root we conclude
x=2sqrt(W(1/4))
where W is the principle branch.
we may have a graphical method : intersection of the parabola x^2 with -8 lnx since x is positive.
Can we evaluate W(y) if y = 1/x • e^x?
Mam u r great
Much easier: the solution in another form is 2sqrt(W(1/4)), just divide the original equation by 4, rise to the e power both sides, divide both sides by 4, apply W to get x^2/4 = W(1/4)
Shoot; that's what I was gonna say!
I think x= 2sqrt(W(1/4)) is also a solution. numerically it is the same as yours 0.90308
Yes, if you approached it differently you can get the same result, but different expression.
I just searched for this comment as i got the same . Its exactly the same.
sqrt( 1 / (e^ (W(1/4)) ) ) = 2 sqrt( W(1/4) )
|>>>> square both sides
1/( e^ (W(1/4)) ) = 4 (W(1/4))
|>>>> multiply by e^ ..blah.. And divide by 4
1/4 = W(1/4) • e^(W(1/4))
|>>>> W and e^W eliminate each other
1/4 = 1/4
In your previous video in which you got the question here from the Comments section, you had an identity [e^W(x) = x/W(x)] that you can use near your last steps.
I got x=e^[(-1/2)*w(1/4)] because, I took e^[-2lnx] instead of e^[lnx^(-2)] in the beginning.
yeah me too but isnt e^[-2lnx] equal to e^[lnx^(-2)]? so why does it gave me a different answer?
It's the exact same. You can bring out the 1/2 of the exponent and get [e^-(W(1/4))]^1/2 in which a negative exponent always flips it and gets you [1/e^(W(1/4))]^1/2 and the power of 1/2 is the same thing as a square root which gets the answer that appeared in the video as:
x = sqrt(1/e^(W(1/4))).
Dear teacher, your chalk is fade! difficult to see, it seems approximately all the chalk you use are fade!
I got 1/(eighth root(e)) (devided both sides by 8)
And I got eighth root(1/e) (used properties of logarithms)
Unlikely that you did the math right. Show your work so we can point out the error(s).
i have an another solution for it mam( namaste mam, you are briliant mam)
Feel free to write your solution.
Only a brilliant person sees brilliance in another ❤🙏
Please I need your Email because I have question in my paper... iam working in Some lifetime distributions and I need to find quantile function (inverse of cumulative function) but my distribution have complex cdf.... I reached to form but not direct function I know the w lambert function will solve my problem but need to attach to you my paper to know my question.... Thanx a lot and with all my respect
AND HOW TO SOLVE X^(X-1)=a ? OR (X-1)^X=b.THANK YOU.
Then x= ?