Lambert function is relaly silly stuff. You could come up with e.g. Sillyman function f(x) = x^2 * e^x, and then assume its some enumerated static lookup table. For me really nonsense this Lambert function
You shouldn't complain. If it weren't for videos like this I wouldn't have learned about the Lambert function. Also, please note that at the end he showed why the answer was complex. The grafs didn't intersect
I was offering constructive criticism. What's wrong with that? I'd say that people who still need such an expanded reduction of the original equation to the Lambert W function because they aren't comfortable with logarithms etc. yet, shouldn't start with the Lambert W function but should first get comfortable with logs.
You first spoon feed the viewer with simple steps to solve the equation, then you start about the Lambert W function, which is indeed higher maths, then you loosely comment that the equation doesn't have any real solutions, without showing how you come to that conclusion. Try to be more consistent and your videos will improve greatly.
@@janami-dharmam Like this: f(w) = we^w for real w has an absolute minimum -1/e for w = -1. -ln(5) < -1 < -1/e, so we^w can't get that value. Another way to see it is to view the graphs of the functions x-> x and x -> 5^x - they don't intersect.
@@florisv559 He referred to a previous video where he discussed the relationship of complex valued solutions and non-intersecting graphs. BUT WHICH VIDEO?
@@janami-dharmam W(x) is the inverse of f(x) = xe^x (x >= -1 for reasons explained below, but we will consider x < -1 anyway). Now f'(x) = (x+1)e^x, f'(x) = 0 when x = -1, and f(-1) = -1/e. f'(x) < 0 when x < -1 and f'(x) > 0 when x > -1, so [1] -1/e is the absolute minimum of f and [2] f takes all values between -1/e and 0 twice, which is why it doesn't have an inverse for all x in ℝ. Because of [1], f doesn't take any lower values than -1/e, like -ln5 (< -1), and therefore, W(-ln5) isn't a real number.
Lambert function is relaly silly stuff. You could come up with e.g. Sillyman function f(x) = x^2 * e^x, and then assume its some enumerated static lookup table. For me really nonsense this Lambert function
مشكلة أنتم تشرحون شيء بطريقة جيدة....لكن سر متعلم لازم تعطيه نقطة لكل موقف يصادفه يكون جوابه متشابه بنسبة 60٪ هو لا يحتاج يشاهدك هو يفكه لوحده....
{5x+5x ➖ }=10x^2 5^5x^2 2^3^2^3x^2 1^1^1^3x^2 3x^2 (x ➖ 3x+2).
what is this symbol? ➖ means what?
@janami-dharmam what symbol do you mean
@@RealQinnMalloryu4 this one: ➖
@@RealQinnMalloryu4 5x+5x ➖ the last one
You shouldn't complain. If it weren't for videos like this I wouldn't have learned about the Lambert function. Also, please note that at the end he showed why the answer was complex. The grafs didn't intersect
I was offering constructive criticism. What's wrong with that? I'd say that people who still need such an expanded reduction of the original equation to the Lambert W function because they aren't comfortable with logarithms etc. yet, shouldn't start with the Lambert W function but should first get comfortable with logs.
5^log x = x
⁵
Make videos on SAT maths questions (vote button)
You referred to a previous video where you discussed the relationship of complex valued solutions and non-intersecting graphs. BUT WHICH VIDEO?
You first spoon feed the viewer with simple steps to solve the equation, then you start about the Lambert W function, which is indeed higher maths, then you loosely comment that the equation doesn't have any real solutions, without showing how you come to that conclusion. Try to be more consistent and your videos will improve greatly.
Ask for your money back.
how do you know that W(-ln5) does not exist?
@@janami-dharmam Like this: f(w) = we^w for real w has an absolute minimum -1/e for w = -1. -ln(5) < -1 < -1/e, so we^w can't get that value. Another way to see it is to view the graphs of the functions x-> x and x -> 5^x - they don't intersect.
@@florisv559 He referred to a previous video where he discussed the relationship of complex valued solutions and non-intersecting graphs. BUT WHICH VIDEO?
@@janami-dharmam W(x) is the inverse of f(x) = xe^x (x >= -1 for reasons explained below, but we will consider x < -1 anyway).
Now f'(x) = (x+1)e^x, f'(x) = 0 when x = -1, and f(-1) = -1/e. f'(x) < 0 when x < -1 and f'(x) > 0 when x > -1, so
[1] -1/e is the absolute minimum of f and
[2] f takes all values between -1/e and 0 twice, which is why it doesn't have an inverse for all x in ℝ.
Because of [1], f doesn't take any lower values than -1/e, like -ln5 (< -1), and therefore, W(-ln5) isn't a real number.
Can you expkain the eular num rules. & wlambert function with avarious exercises..... ❤
5^0=0 X=0
Any number to the 0 power = 1
No. X⁰ = 1 ( X€ R)
And
5^x≠x (X€R)
So the solution is complex/ imaginary
You are solving questions for other people to see not for yourself. For God's sake, writing should be bigger so we can see easily .
Its written pretty well. Have you checked your eyes recently
x = -0.01075 - 0.98279 i
5^x = x
(e^ln 5)^x = x
e^(x ln 5) = x
x ln 5 = ln x
(ln x)/x = ln 5
(ln x)/(e^ln x) = ln 5
(ln x)e^(-ln x) = ln 5
(-ln x)e^(-ln x) = -ln 5
W((-ln x)e^(-ln x)) = W(-ln 5)
-ln x = W(-ln 5)
ln x = -W(-ln 5)
x = e^(-W(-ln 5))
how??
@@janami-dharmamWolfram Alpha or Mathematica, or one of many other computer algebra systems. Look at Wikipedia article on CAS.
Here's more solutions. There are countably many. Sheet refers to the Riemann surface, which is cut into sheets. Each sheet fills the complex plane.
All of the complex x values satisfy 5^x=x.
sheet=-5
x=1.7462696339565527+16.5264299576489i
x-5^x=10^-14 + 0.i
sheet=-4
x=1.5796262962043934+12.610451070467139i
x-5^x=10^-14 - 10^-15i
sheet=-3
x=1.3507162537895936+8.688085376179854i
x-5^x=10^-15 - 10^-15i
sheet=-2
x=0.9815502377624126+4.753430109599094i
x-5^x=10^-16+10^-16i
sheet=-1
x=-0.010750838472555747+0.9827872262285992i
x-5^x=-10^-17+0.i
sheet=0
x=-0.010750838472555783-0.9827872262285992i
x-5^x=10^-17+0.i
sheet=1
x=0.9815502377624126-4.753430109599094i
x-5^x=10^-16-10^-16i
sheet=2
x=1.3507162537895936-8.688085376179854i
x-5^x=10^-15+10^-15i
sheet=3
x=1.5796262962043934-12.610451070467139i
x-5^x=10^-14+10^-15i
sheet=4
x=1.7462696339565527-16.5264299576489i
x-5^x=10^-14+0.i
sheet=5
x=1.8774508360596127-20.43888917162955i
x-5^x=10^-14+10^-15i