I have never used the Lambert W() function because it is not on my calculator, but I have iteratively solved for all three real roots of this and similar equations, either by using fixed-point or Newton-raphson iteration. Everyone should become proficient in using these iterative methods to solve a wide range of nonlinear systems of equations.
Numerical methods are very helpful indeed (I should know, numeric computing was my Master’s specialisation). But we only use them if we can’t solve the problem directly, “statically” (i.e. on paper). Otherwise, you’re 1/ inefficient and 2/ burdened with the unavoidable numeric error. Still, kudos for knowing them… :-)
Your method gives two solutions 2 and -0.766 there should be three solutions using Lambert W function, where is the 4? I know if you solve it by not using lambert W function then you will get again two solutions as in your past video but this time it would be 2 and 4. The question now is: can you solve it for 2, 4 and - -0.766 using only One method?
Your first solution is wrong because. Your solution is only valid when x=2 so you could say that a=m when a^m=m^a . But actually when exponents are same by comparison we can say bases are same but in question the base and exponents values are completely altered so you cannot use that logic a^m=m^a .
(x > 0!) - ln x / x = - ln x / e^(ln x) = - ln x e^(- ln x) => W_n(- ln x / x) = - ln x, as long als ln x / x < 1 / e (for W_0 branch). Now, x=2 is positive and ln 2 / 2 is 0.69314... / 2 = 0.34657... and 1 / e = 0.36788..., so ln 2 / 2 < 1 / e. Thus follows W(- ln 2 / 2) = - ln 2 = -0.69314... and x_1 = 1 / e^W_0(-ln 2 / 2) = e^(ln 2) = 2. So much for the W_0 branch! ln |x| / x = ln 2 / 2 => W_k(- ln |x| e^(- ln |x|)) = W_l(- ln 2 e^(- ln 2)) = W_m(- ln 4 e^(- ln 4)), l = 0. Probably m=-1, have to check. The problem starts with x=4. There are reasons, why W_0 is not sufficient here! One is that ln 2 / 2 = 2 ln 2 / 4 = ln 2² / 4 = ln 4 / 4. There is no difference in ln x / x for x=2 and x=4. Thus you need another branch to explicitly select the solution. For x < 0, it's even more complicated: ln x is not available, nor is x = e^(ln x) true. You must circumvent this with a special case. As you said, something with ln |x|. To my own shame, I saw x=2 and x=4, but I oversaw the negative solution (that is only possible because of the even exponent!)... shit happens!
e^x is the inverse function of ln(x) ln(x) is the inverse function of e^x so e^ln(x) = x and ln(e^x) = x The same principle applies to Lambert's W function, which is the inverse function of xe^x so W(xe^x) = x
I have never used the Lambert W() function because it is not on my calculator, but I have iteratively solved for all three real roots of this and similar equations, either by using fixed-point or Newton-raphson iteration. Everyone should become proficient in using these iterative methods to solve a wide range of nonlinear systems of equations.
Numerical methods are very helpful indeed (I should know, numeric computing was my Master’s specialisation). But we only use them if we can’t solve the problem directly, “statically” (i.e. on paper). Otherwise, you’re 1/ inefficient and 2/ burdened with the unavoidable numeric error. Still, kudos for knowing them… :-)
x²=2^x --> 2ln(x)=xln(2)
[ln(x)]/x= [ln(2)]/2
But ln(2)/2=ln[(2²)]/2²
=l[n(4)]/4
Hence [ln(x)]/x=[ln(4)]/4
Therefore x={2 4}
Your method gives two solutions 2 and -0.766 there should be three solutions using Lambert W function, where is the 4? I know if you solve it by not using lambert W function then you will get again two solutions as in your past video but this time it would be 2 and 4.
The question now is: can you solve it for 2, 4 and -
-0.766 using only One method?
Эволюция математика: от "да это устно решается!" до "неужели нет способа найти все корни без функции Ламберта?!".🤭
This is mental math. 😂😂
Mentally disturbing meth😂😂
Similar problem solved without Lambert function: ruclips.net/video/NGc2sDbke34/видео.html
X=2 because if a^m=m^a, then a=m
Another method
X^2=2^x
Take the log
logx^2=log 2^x
2.log x=x. log2
logx/x=log2/2
So,
X=2
logx/x=log2/2 how did you get that?
@ainamanicolleb , Taking the antilog of the equation....
4?
Better watch the video because you only found 1 solution of the 3.
Your first solution is wrong because. Your solution is only valid when x=2 so you could say that a=m when a^m=m^a . But actually when exponents are same by comparison we can say bases are same but in question the base and exponents values are completely altered so you cannot use that logic a^m=m^a .
Why divideb by 2x it's confusing
(x > 0!) - ln x / x = - ln x / e^(ln x) = - ln x e^(- ln x) => W_n(- ln x / x) = - ln x, as long als ln x / x < 1 / e (for W_0 branch).
Now, x=2 is positive and ln 2 / 2 is 0.69314... / 2 = 0.34657... and 1 / e = 0.36788..., so ln 2 / 2 < 1 / e.
Thus follows W(- ln 2 / 2) = - ln 2 = -0.69314... and x_1 = 1 / e^W_0(-ln 2 / 2) = e^(ln 2) = 2. So much for the W_0 branch!
ln |x| / x = ln 2 / 2 => W_k(- ln |x| e^(- ln |x|)) = W_l(- ln 2 e^(- ln 2)) = W_m(- ln 4 e^(- ln 4)), l = 0. Probably m=-1, have to check.
The problem starts with x=4. There are reasons, why W_0 is not sufficient here!
One is that ln 2 / 2 = 2 ln 2 / 4 = ln 2² / 4 = ln 4 / 4. There is no difference in ln x / x for x=2 and x=4. Thus you need another branch to explicitly select the solution.
For x < 0, it's even more complicated: ln x is not available, nor is x = e^(ln x) true. You must circumvent this with a special case. As you said, something with ln |x|.
To my own shame, I saw x=2 and x=4, but I oversaw the negative solution (that is only possible because of the even exponent!)... shit happens!
Why didn't you do the same thing on the right side - immediately? (I expect, you'll do it later in the video).
Okay, you didn't. My answer has moved to the comment root...
I just did the trial-and-error way, it's way simpler
its only work if you have simple answer as he show here for exemple but in real situation, it is way more complicated
X=2 X=4 X^2=2^X
Nice, A better way to shape this problem as of Lambert w function...
How x=e^ln x ?? I didn't get it
The e and the ln cancel out
e^x is the inverse function of ln(x)
ln(x) is the inverse function of e^x
so e^ln(x) = x
and ln(e^x) = x
The same principle applies to Lambert's W function, which is the inverse function of xe^x
so W(xe^x) = x
По какому алгоритму эта загадочная функция считается, так никто в итоге и не объяснил. %)
x=2
(x ➖ 2x+2). {2x+2x ➖}=4x^2 2^2x^2 1^1x^2 1x^2(x ➖ 2x+1).
x₁ = e^(-W(-ln(2)/2)) = e^(-W(-ln(2)*(2^(-1))) = e^(-W(-ln(2)*e^ln(2^(-1))) = e^(-W(-ln(2)*e^(-ln(2))) = e^(-(-ln(2))) = e^ln(2) = 2
x₂ = e^(-W(-ln(2)/2)) = e^(-W(-(2*ln(2))/(2*2))) = e^(-W(-ln(2^2)/4)) = e^(-W(-ln(4)*(4^(-1))) = e^(-W(-ln(4)*e^ln(4^(-1))) =
= e^(-W(-ln(4)*e^(-ln(4))) = e^(-(-ln(4))) = e^ln(4) = 4
x₃ = -0.766664695962123093111204422510314848006675346669832058460884376...
x=2