they don’t teach these kinds of expoential equations in algebra 

When writing Chinese characters, the order of each brush-stroke is very important.

Well that answers my question. What "dialect" of "Chinese" do you speak? They are in quotes because Chinese isn't just one language and things like Mandarin and Cantonese aren't dialects.

@@ozzymandius666 I speak and write japanese and I love how its simplified Kanji from the Chinese Dialect 😃 Bad part is Japanese read each Kanji with two or more readings called 音読み and 訓読み

i just know that you have other channel in mandarin. thanks a lot for making that channel!

Why don't you feature your two other channels?

College is not "scientific" enough...

Lol thanks!

@@blackpenredpen you are so cute 😍😍

Ok

Lambert function is so cool!

You thought right

Only pure, pure math. The things they make seniors study in Greece are outrageous...

@@erikkonstas, What's wrong with math education in Greece? Your country once pioneered mathematics.

Math by itself is fun but when you're forced to memorize stuff for exams, it loses its value

@@erik-ic3tp Man, you don't want to know, it's horrible. And, yes, this is a great shame some of us think could land the country in total despair very soon. Thales's, Pythagoras's, Archimedes's, Euclid's and the other great mathematicians' bones are shaking...

Yes

You can substitute x^3 as other variable and then solve

I'm 11

Yea it’s a new feature. I think it’s so cool!

Same , i think its cool too

The base change theorem doesn't work quite like that; 2 and x are different bases so if you were to change to base 2 you would get x^3 = 1/log_2(x). And if you changed to base x you would get x^3 = log_x(2). No base change gets you 2/x.

Year late reply but that doesn’t work because then it would be 1/3 x^3 in the exponent and would not cancel the cubed

@@Vivek-io3gj could you not swap around the exponents?

@@adamwright4634 you cant because it’s x^(x^3)^1/3 not x^((x^3)^1/3)

Mmm... yes and no. The Lambert W map is multivalued. So while one of the two branches -1 and 0 will evaluate to this, its objectively more useful to leave it in terms of the Lambert W expression because of the branches.

umm didn't he do this in the video ?

Thank you, I was looking for this kind of solution.

Dr. Peyam and Prof. Penn's channels : )

*an

Yes. Will release that this coming Friday

Important question. I took ln of both sides. Then i divided both sides by x^3 so i have ln x equals ln 2/x^3..Then i took thebderviative with respect to x of both sides to get rid of the ln x and be able to solve for x. Taking the derivative of both sides gets me 1/x equals ln 2(-3x^-4)..therefore solving for x you get x^3 equals -3(ln 2)..so taking the cibe root yiu get x equals the cube root of -3(ln2). Didnt a lot of people do it this way? it's totally valid i dont see why not? Especially if you dont k ow Lambert's function. It's still the same x variable even though i took the derivative so it's the correct value.

Leif ok after taking the derivative we have 1/x=-3ln(2)/x^4. After multiplying by x^4 we get x^3=-3ln(2) so: x=cbrt(-3ln(2)) The only things possible here are: 1. We did a mistake 2. There are more than one answer for this. 3. Both solutions equal to each other. When this is right, you've proven that e^(W(3ln(2))/3)=cbrt(-3ln(2))... But it's sadly not :( It's really weird. I can't spot the mistake here... bprp pls help

Leif No, it isn't valid. The fact that two functions are equal to each other at a point does not imply their derivatives are equal to each other at that point. In other words, f(x) = g(x) at x = c does not imply f'(x) = g'(x) at x = c. I have very simple counterexample to prove this. sin(x) = cos(x) if x = π/4. Taking the derivative yields cos(x) = -sin(x), and substituting x = π/4 implies 1/sqrt(2) = -1/sqrt(2), which is false. This just proves that taking the derivative is not a valid operation here.

Yea if u only have finite amount of x

It's the fish, and the smile, and lately, that little beard ... ... "isn't it?" You rarely get any of that with math teachers.

To integrate W(x), you use a u-substitution. Let u = W(x). Hence x = ue^u, hence dx = (u + 1)e^u·du. Therefore, when you convert everything "to the u-world" and simplify the integrand, this becomes the antiderivative of u(u + 1)e^u with respect to u. Integrate this by parts by differentiating u(u + 1) to 2u + 1 and antidifferentiating e^u to e^u. This results in u(u + 1)e^u minus the antiderivative of (2u + 1)e^u. (2u + 1)e^u can itself be antidifferentiated by parts by differentiating 2u + 1 to 2 and antidifferentiating e^u to e^u, resulting in (2u + 1)e^u minus the antiderivative of 2e^u, which is just 2e^u. Therefore, the antiderivative of u(u + 1)e^u with respect to u is given by u(u + 1)e^u - (2u + 1)e^u + 2e^u. Notice that u = W(x) and x = ue^u. Therefore, u(u + 1)e^u - (2u + 1)e^u + 2e^u = [W(x) + 1]x - 2x - e^W(x) + 2e^W(x) = x[W(x) + 1] - 2x + e^W(x) = x[W(x) + 1] - 2x + x/W(x). With all the simplifications completed, don't forget to add your +C. Therefore, the antiderivative of W(x) is x[W(x) + 1] - 2x + x/W(x) + C.

You can use it in bith

Because it is! : ) I got it from his merch store!

Ideally, in a precalculus course.

Newton's method is quite fast 1.5 1.40 1.348 1.3371 1.33671

cos(x)^2 = 1 - sin(x)^2, hence 2·cos(x)^2 = 2 - 2·sin(x)^2. Therefore, 2·cos(x)^2 - sin(x)^2 - 2·sin(x) - 1 = 0 is equivalent to 2 - 2·sin(x)^2 - sin(x)^2 - 2·sin(x) - 1 = 1 - 2·sin(x) - 3·sin(x)^2 = 0, which is equivalent to 3·sin(x)^2 + 2·sin(x) - 1 = 0. Hopefully that helps you finish it.

That = cos(2x) + cos^2(x) -2sin(x) -1 = cos(2x) - sin^2(x) - 2sin(x) = 1-2sin^2(x)-sin^2(x) - 2sin(x) = 1-3sin^2(x)-2sin(x)=0 I’m pretty sure you know how to use the quadratic equation, so I’ll leave it here.

Plugged identity cos^2(x)=1-sin^2(x); you would solve for sin x = 1 or sin x = -1/3 The answer for it would be (in degree; it has periodic answer): x = 90 + 360*k or x = arcsin (-1/3) + 360*k (I cannot reverse sin 1/3 except using calculator, except your teacher allow it, I would leave it in this form)

I'm not bprb obviously but I suggest that u can turn the expression into à quadratic equation in terms of sin(x) then use the quadratic formula to seek for solutions. Meaning: 2cos²(x)-sin²(x)-2sin(x)-1=0 2(1-sin²(x))-sin²(x)-2sin(x)-1=0 -3sin²(x)-2sin(x)-1=0 We may notice that a-b+c=0 so sin(x)=-1 or sin(x)=-c/a=-1/3 So x=-pi/2 + 2kpi or x=arcsin(-1/3) + 2kpi where k is an integer.

Wow thanks guys.This community is the best!

yes it is working even you add more powers x^x^x^..x^c=c x=c^1/c

Yes.

I have that in the description.

He has several videos of it already, plus you can check the Wolfram Alpha page on the function and the Wikipedia page on it for an introduction to the topic. In essence, though, all it is, it is the inverse "function" of f(x) = xe^x.

Yes

Yes. You probably already see the pattern: exp(W[n·ln(n)]/n) = n^(1/n) for any positive integer n. This also implies that W(n·ln(n)) = ln(n) for any n. In fact, more generally, W[x·ln(x)] = ln(x) for any x satisfying the domain restrictions (and if you allow complex numbers, then for all nonzero complex x).

It works more or less the same.

But you need the solutions to be in real world.

@@Shreyas_Jaiswal they are in real world, they just happen to not fit in the set of real numbers.

@@MrRyanroberson1 real world means on earth. But complex numbers are on moon. 😂😂