There are dozens of these videos where they always have a solution of the Lampert W function. They are trying to make the W function be one of the primary functions. But you could do the same thing with the zeta, gamma or Bessel functions.
I'm also kind of new to this, but I checked Wikipedia and found this: "Although it was widely believed that the Lambert W function cannot be expressed in terms of elementary (Liouvillian) functions, the first published proof did not appear until 2008." Long story short, you can plot it and identify useful patterns with it, but the Lambert W function is its own thing that cannot be expressed in different terms. You can (and should) use it whenever you want to solve for a with a*e^(a).
I went about this one a different way and got a different way of representing the answer: xlnx=ln25 lnx=ln25/x x=e^(ln25/x) ln25=(ln25/x) e^(ln25/x) W(ln25)=ln25/x x=ln25/W(ln25) On (approximate) calculation this gives the same value as e^W(ln25) so it appears to be a valid solution.
And again, I chose a bit different approach: x^x = 25 / both sides to the power of 1/x gives x^1 = 25^(1/x) / times 1/x gives 1 = 1/x . 25^(1/x) / times ln25 and note that 25 = e^ln25 ln25 = ln25/x . (e^ln25)^(1/x) , i.e. ln25 = ln25/x . e^(ln25/x) / now apply Lambert’s W: W(ln25) = ln25/x , so x = ln25/W(ln25) (which is equal to your result.)
I gave this question to AI Claude 3.5 Sonnet and this was its answer: "The equation x^x = 25 can be solved analytically using the Lambert W function. Here's the step-by-step solution: Start with x^x = 25 Take the natural log of both sides: ln(x^x) = ln(25) Using the properties of logarithms: x * ln(x) = ln(25) Multiply both sides by 1/x: ln(x) = ln(25) / x Multiply both sides by -1: -ln(x) = -ln(25) / x Take e to the power of both sides: e^(-ln(x)) = e^(-ln(25) / x) Simplify the left side: 1/x = e^(-ln(25) / x) Take the reciprocal of both sides: x = e^(ln(25) / x) This is in the form of y = e^(1/y), which has the solution y = e^W(1), where W is the Lambert W function. In our case, we have x = e^(ln(25) / x), so the solution is: x = e^W(ln(25)) "
On Prime Newtons channel you can find several good clips, e.g. the one with the title "Lambert W Function". And you can also find information in the Wikipedia with the same title.
Lots of time wasted in this clip to stretch an otherwise 1-2 minutes story into 10 minutes of repeating the same trivialities again and again, without ever giving a proper explanation of Lambert's W function. Interesting subject, but very disappointing clip.
I saw no analytical solution, just a partial analytical solution. And you end up with approximating the solution numerically, which defeats the whole point of video you made. You could have solved this numerically with the initial equation.
e^ln x is equal to x. just try e^ln 7 is equal to 7. why use e^ln x? because it's like an identity that e^ln x is equal to x. if your equality in left side is ln x, just use e as the base for both side than e^ln x = x to get easy x.
@@themightyquinn100for me it looks like lazy solution IMO. Never heard about lambert function before so it's a black box for me. But if it's calculated already, then sure, why not. You should apply it
@@asdbanz316 I'm not knowledgeable on this topic but it might be the only general solution. And the calculation probably requires a numerical algorithm.
There are dozens of these videos where they always have a solution of the Lampert W function. They are trying to make the W function be one of the primary functions. But you could do the same thing with the zeta, gamma or Bessel functions.
Respected Sir, Good evening. Pls get us the knowledge of "lambert w function "
I'm also kind of new to this, but I checked Wikipedia and found this: "Although it was widely believed that the Lambert W function cannot be expressed in terms of elementary (Liouvillian) functions, the first published proof did not appear until 2008." Long story short, you can plot it and identify useful patterns with it, but the Lambert W function is its own thing that cannot be expressed in different terms. You can (and should) use it whenever you want to solve for a with a*e^(a).
I do not think it has anything to do with Miranda!!
I went about this one a different way and got a different way of representing the answer:
xlnx=ln25
lnx=ln25/x
x=e^(ln25/x)
ln25=(ln25/x) e^(ln25/x)
W(ln25)=ln25/x
x=ln25/W(ln25)
On (approximate) calculation this gives the same value as e^W(ln25) so it appears to be a valid solution.
It not only appears to be the same. It actually is the same.
@@YAWTon Took me a hot minute to work it out exactly but yeah i see it now
And again, I chose a bit different approach:
x^x = 25 / both sides to the power of 1/x gives
x^1 = 25^(1/x) / times 1/x gives
1 = 1/x . 25^(1/x) / times ln25 and note that 25 = e^ln25
ln25 = ln25/x . (e^ln25)^(1/x) , i.e.
ln25 = ln25/x . e^(ln25/x) / now apply Lambert’s W:
W(ln25) = ln25/x , so
x = ln25/W(ln25)
(which is equal to your result.)
I gave this question to AI Claude 3.5 Sonnet and this was its answer:
"The equation x^x = 25 can be solved analytically using the Lambert W function. Here's the step-by-step solution:
Start with x^x = 25
Take the natural log of both sides: ln(x^x) = ln(25)
Using the properties of logarithms: x * ln(x) = ln(25)
Multiply both sides by 1/x: ln(x) = ln(25) / x
Multiply both sides by -1: -ln(x) = -ln(25) / x
Take e to the power of both sides: e^(-ln(x)) = e^(-ln(25) / x)
Simplify the left side: 1/x = e^(-ln(25) / x)
Take the reciprocal of both sides: x = e^(ln(25) / x)
This is in the form of y = e^(1/y), which has the solution y = e^W(1), where W is the Lambert W function.
In our case, we have x = e^(ln(25) / x), so the solution is: x = e^W(ln(25)) "
I have a L func, L(x)=x^x, so the solution of the above equation is x=inverse L(25).
Hello sir
Sir, I request can you please make a video for lamber W function with its concept
On Prime Newtons channel you can find several good clips, e.g. the one with the title "Lambert W Function". And you can also find information in the Wikipedia with the same title.
Lots of time wasted in this clip to stretch an otherwise 1-2 minutes story into 10 minutes of repeating the same trivialities again and again, without ever giving a proper explanation of Lambert's W function. Interesting subject, but very disappointing clip.
i really love the lambert w function, also w(xlnx)=lnx just to simplify
Thanks for this.
By ispection 2
x^x = a, x = e^(W(ln(a))) for x > 0.
Does not make sense e^W(ln 25), how do you work this out?
Hello
you can solve the following equation:
xˣ=2x
(x ➖ 5x+5).
before watching: xln(x)=ln(25)
ln(x)e^ln(x)=ln(25)
ln(x)=W(ln(25)
x=e^W(2ln(5))
You showoff the lambert function too much
I saw no analytical solution, just a partial analytical solution. And you end up with approximating the solution numerically, which defeats the whole point of video you made. You could have solved this numerically with the initial equation.
其實可以不一定要用朗伯函數😅
Please explaine,why in e^lnx, e and ln cancel out??🤔🤔🙏🙏❤❤❤
e^ln x is equal to x.
just try e^ln 7 is equal to 7.
why use e^ln x? because it's like an identity that e^ln x is equal to x. if your equality in left side is ln x, just use e as the base for both side than e^ln x = x to get easy x.
@@ciathing4654 Thanks👍❤️, But can we prove it other than, y=e^ln x ??Like directy?🤔
Definition of the inverse function. You're welcome.
@@elessarstrider5210exactly, that's the "proof" needed
this is not intuitive. this type of stuff turns most people away from math.
Not everything can be understood intuitively in mathematics. The basic stuff yes, but not always the advanced topics.
@@themightyquinn100for me it looks like lazy solution IMO. Never heard about lambert function before so it's a black box for me. But if it's calculated already, then sure, why not. You should apply it
@@asdbanz316 I'm not knowledgeable on this topic but it might be the only general solution. And the calculation probably requires a numerical algorithm.
😮 маньяки!
x^2 = 3^x
ln(x^2) = ln(3^x)
2*ln|x| = x*ln(3) ===> two cases
1st case: x > 0
2*ln(x) = x*ln(3)
ln(x)*x^(-1) = ln(3)/2
ln(x)*e^ln(x^(-1)) = ln(3)/2
ln(x)*e^(-ln(x)) = ln(3)/2
-ln(x)*e^(-ln(x)) = -ln(3)/2
W(-ln(x)*e^(-ln(x))) = W(-ln(3)/2)
-ln(x) = W(-ln(3)/2)
ln(x) = -W(-ln(3)/2)
x = e^(-W(-ln(3)/2)) ===> -ln(3)/2 < -1/e ===> no real solutions
2nd case: x < 0
2*ln(-x) = x*ln(3)
ln(-x)*x^(-1) = ln(3)/2
-ln(-x)*x^(-1) = -ln(3)/2
ln(-x)*(-x)^(-1) = -ln(3)/2
ln(-x)*e^ln((-x)^(-1)) = -ln(3)/2
ln(-x)*e^(-ln(-x)) = -ln(3)/2
-ln(-x)*e^(-ln(-x)) = ln(3)/2
W(-ln(-x)*e^(-ln(-x))) = W(ln(3)/2)
-ln(-x) = W(ln(3)/2)
ln(-x) = -W(ln(3)/2)
-x = e^(-W(ln(3)/2)
x = -e^(-W(ln(3)/2)) ===> ln(3)/2 > 0 ===> 1 real solution
x = -e^(-W₀(ln(3)/2)) = -0.686026724536251319713006846182238159503324237762343402417671916...