so you want a VERY HARD math question?! 

Have you ever made a lesson-like video where you prove the rules of derivation and integration? I think it would be interesting

Lol 😆 that 1 guy want hard problem

Here’s an impossible question of trigonometry: Find the value of x: (1/sin2x) + (square root 3/cos3x) = -1/square root of 3 Looks fine but none of our school maths teachers could solve so do give it a go!

Where is sandwich guy

I wonder if there's a way to solve polynomial-trig or exponential-trig function

Bruh he's been successful on youtube for years

@@thebigbradwolf me when nothing ever happens🙄

You went to pierce huh

@@joshuagonzalez3880 still go yeah

I assumed that he was great in class, but it's cool to hear it confirmed!

😂😂😂

😂😂😂😂

😂😂😂

😂😂😂

😂😂😂😂

🤣🤣🤣

Bruh

B r u h

Same 😂

*B R U H*

Oh this particular video is almost entirely algebra

ill pray for you

Integration isn't too bad, good luck nevertheless

@Del Squared - دل تربيع Masya Allah, brother👍

@Del Squared - دل تربيع yes i am also learning it... 😀😀 Thanks brother ❤️❤️

Don't think this is your typical engineering problem dude... 😂

@@Vase0I0 yeah, they'll be dealing with more excel sheets than actual problem solving

@@w花b hey man excel sheets are great for problem solving

​@@w花b How many excel sheets do you think it took to design and develop the platform and device you are using to watch your favourite math videos and comment your "educated" opinion ?

@@w花b bu-but... excel sheets are beautiful 🥲 **hugs my excel sheets tightly**

😂

well the problem is perhaps actually even simpler as simple can be. at 1.54 bprp says: ..and maybe we can multiply with ln(x) on both sides .. (!), without excluding explicitly this factor being zero. right here he himself introduces an extra root, namely the case ln(x) = 0. well may we think that leaving a denominator the denominator if including a function of the unknown, that's algebra for beginners.. ?

Same With logarithmic Function. And actually same with the solution for x+2=4

i mean, there is a point between "Inventing a new function will help us with other fields" and "hiding an approximation behind a new function that is created for the sake of solving an equation" right?

I hate the Lambert function. It's just BS trying to find a 'solution' for something not possible otherwise.

@@chimaeria6887 so you're saying logarithm and inverse trigonometric function is BS too, what a fool.

@@chimaeria6887 What, you mean like logarithms?

This equals (1/sqrt(pi))^(-2/3). IIRC this has to do with the Gamma function

There is probably an approximation formula lurking around

@@blackbomber72 oh maybe sterlings formula!

I like pie.

@@opinionshurt2905 same

And you, sir, were using the "good enough" approach, finding the practical solution fast. Math nerds may hate it but engineering of any kind (except maybe nuclear and space tech) is about the "good enough" values. Enjoyers of PI = 22/7, all aboard!

@@DungeonNumber5 pi=3

So we just need to carry around graphing calculators? As a math teacher, I know that isn't true.

@@donmoore7785You can do it with a normal calculator as well. Just use like Newton’s method or something. Most scientific calculator’s have an iteration function so that once you set up the equation u can just spam the equal sign until the decimal places dont change.

​@@donmoore7785 As a math teacher, you should know that desmos is a thing that exists. So yes, not only should we, we already do.

Thanks. You should also check out the channel “just calculus” for calc 1 tutorials. That guy is okay. 😆

Checkout Proffesor Leonard. I am Electrical Engineering major and I have learnt calculus 1 from him. I am currently learning calc 2

Thanks! I am teaching calc 1 this semester and the last time I taught it was 2 years ago. So my mind is full of calc 1 these months.

🤓👈🤣🤣🤣

@@aca4262 What's wrong? You're in the comments of a math video. No idea if you were overwhelmed by the topics or something else.

@@Nino-eo8ey what's wrong with what's wrong of my comment?

Interesting, I never thought of looking at it like that.

yeahh of course

Nice intuitive approximation for 3/ln2 XD

that is exactly the reason why i say 2.7182818284590452 when introducing the thing to my friend

​@@TNTErick You're missing a decimal 7 there, my friend.

@@yurenchuThank you, my friend. I did miss a 7 there

thats why the fishy has a evil face

I disliked having to do u-sub after u-sub. Like 3 times. Though all of the technical math did teach me to stay weary of online calculators as there were rare occasions where they would get the wrong answer I think.

@@Speed001 Doing Calc I and II, i never encountered a problem where the online calculators gave me an incorrect answer; I only encountered problems which they couldn't solve. Those were interesting, but frustrating, because those problems couldn't be solved by bluntly applying the antiderivative techniques as explained in a textbook, but required some slight deviation that the calculator didn't account for. This was really rare, though. I only found one or two problems it actually couldn't solve.

One thing that I like to do is to try to use MatLab for as many math problems as possible; this is a great way to practise that program, while at the same time studying the math courses themselves.

You should always have a separate case for them, too. Because they could still be solutions, but would need to be treated differently to solve.

exactly. Newton's method is better in that it also gives an approximate result, but in a much simpler way. Anyway, the W function is very interesting :)

You’re so good wow

No idea what the fuck you just said

W(x) is also calculated numerically, for example with the Newton method.

I did the exact same thing

No, u can’t apply different branches of W to both sides of an equation. Like above clearly W(0) [-2/3 e^-2/3] isn’t equal to W(-1) [-2/3 e^-2/3] so u should treat W(0) as a different function to W(n) for any other branch n

I think that the LHS with the other branch will resolve to the same thing because it is a symbolic expression, but I am not entirely sure. Really, this should be repeated for every branch. Then, each yielded x should be plugged into the integral and verified. But any nonreal x can be immediately discarded due to the restriction of the domain assumed at the beginning (and the fact that we are not using complex analysis techniques in the actual integration). So, I suspect that there are only a finite handful of real-valued solutions x to the W equation, all of which I further suspect were shown here. We eliminated the bad ones, so only the good ones remain, if so.

Yeah, -1/e ≤ x ⇐ (W(x) ∈ ℝ, & x ∈ ℝ). Moreover, x < 0 ⇒ y ∈ {W₋₁(x), W₀(x)}, where: y e^y = x.

The W function can be understood as a multifunction, just like arcsin. There are infinitely many values of z that solve the equation sin z = 1/2 (for instance), but only one is between -π/2 and π/2, so we pick that one as the principle branch and say arcsin 1/2 = π/6. But if all we know is that x is a real number and that sin x = 1/2, it is invalid to conclude that x = π/6, because it could be π/6 plus any multiple of 2π. Suppose you were given the problem "solve sin x = 1/2, 1 < x < 3." You take the arcsine of both sides, and find that the only solution it gives is out of range. Where did my solution go? The correct solution is x = 5π/6, but your arcsine function didn't give it to you, because it picked the wrong branch. Picking a branch literally just means choosing one solution and discarding the rest. So the same thing is going on here. Well actually even before that, we introduce two spurious solutions. We know that x² - 1 = 3 log x is well-defined and true if and only if the original integral equation holds. But after that point, he exponentiates both sides of the equation, neglecting the cases where log x is undefined. That's how we get the invalid solutions. At this point, we should note that the original equation clearly can't hold for any x ≤ 1 and from that point on assume x > 1. Doing so makes the next few steps valid. We know that -2/3 x² e^(-2/3 x²) = -2/3 e^(-2/3) if and only if the original integral equation holds (because x > 1). In the next step, he applies the W function to both sides of the equation. That step is valid, but only in one direction. It is true that if a = b, then W(a) = W(b) (using the same branch on both sides), but the reverse is not necessarily true, because the W function is not injective. Note that if you plug in x = 1 or x = -1 to the equation at the start of this paragraph, it does hold, so these spurious solutions came in earlier like I said. The problem here is not that new wrong solutions are introduced but that old right solutions are discarded because we are only looking at one branch.

@@curtiswfranks You can just look at the graph and see that there are exactly two real solutions when -1/e < α < 0, which is what he was trying to explain at the end. So you only need to look at the 0 and -1 branches.

The Lambert W "function" is not a real function

It's just make belief

@Adrian Martinez Dorsett I was thinking the same thing... I am not sure what his point is.

@@anshumanagrawal346In your opinion shall the function f(x)= x*exp(x) map a real interval of x to non-real sets🤔? Perhaps in this context, we are not interested in the complex polidromic inverse function of z*exp(z)...

@@peterdecupis8296 I don't understand what you're saying

Yea but isn't that cheating since if you don't know about this function there's no way to really derive it?

@@leif1075 This is just one of the options that came to my mind right away. I found a whole class of functions suitable for this integral. For example: f=[(3*n/2^n)*t^(n-1)]^1/t; n>0 f=[3/tan(2) * cos(t)^-2]^1/t ... etc.

0:45 wise guy.

first time i came across it was actually in physics for quantum (schrödingers) but it was just additional information that wouldnt be tested so i never really cared about it

it’s not really in the curriculum I don’t think, maybe it comes up in some graduate level numerical diff eq courses but I haven’t seen it

even throughout a math degree program this function isn't used much. ive only seen it widely used in this channel.

Not even all mathematicians encounter it. At least, I don't recall encountering it during my formal master or phd education, but I have to say I was specializing in a very different field (related to combinatorial optimization). I only came across it when I was just fooling around with some equations (nothing to do with my work whatsoever) in my free time in a computer algebra system and it gave me a solution in terms of the Lambert function, which was when I decided to check it out.

We were told about it in 11th grade maths, just as the inverse of xe^x. Dunno why it's so rarely taught

Hehe, a famous one

Yikes

@@thecuriouskid4481 can you elaborate on this a bit more ?

@@iyannazarian866 because you can't really integrate a t^k/(t+1) dt, when k is arbitrary. But if you notice that 15/7 is quite close to pi-1, you can find an easy solution to this integral. Besides, even wolfram alpha proofs that the brute force result using hypergeometric functions is oddly accurate to our quirky substitution.

😂 good work!

You have to factor out (x-1), since x²-1 = (x+1)(x-1)

@@GSenna37 Brother ,That's all I know !

@@nextgaming4666 sorry , but this is wrong , correct range is,{0

@@nextgaming4666 my answer is also same ! But it is wrong

\sqrt{\frac{x-1}{x^2-1}}=\sqrt{\frac{x-1}{(x+1)(x-1)}}=\sqrt{\frac{1}{x+1}}(x eq 1) f(x) has no real value for x-1(+), f(x)->infinity. The function monotonically decreases to 0 as x->infinity, and excludes 1/sqrt(2), so: range: (0, infinity)\(1/sqrt(2))

There are lots of ways to approximate pi. Archimides came up with some of the first (dividing up a regular polygon with 2^n sides into triangles and take increasingly large n), and this was taken quite far by ancient Chinese mathematicians. Then Ramanujan (early 20th century) came up with a much better approximation method based on modular forms, and later this was developed further into what I believe is the best algorithm due to the Chudnovsky brothers, who calculated many millions of digits on a supercomputer they built in their New York apartment. There has maybe been an additional improvement in the method (Borwein? Bailey?) but I haven't followed it closely. In any case, this amount of precision is irrelevant for any practical purposes - it is more used as a benchmark problem to test the performance of new computers.

@@andrewhone3346 You didn't read what was asked. Solve for pi WITHOUT approximating it and/or assuming this can not be done. It is possible to use the Pythagorean theorem to do it & express the answer as a ratio of integers/non-. Archimedes' n-gon method produces the wrong answer & 3.14159... is not the circumference of a circle whose diameter is 1.

Thanks

imo this is not at the putnam level. It as a routine calculation if you know about the lambert W function, which forces there to be a solution. I think putnam tries to be much more elegant than this

1??

Just take ln and you get x^3 = x^2. You get 0=x^2 (x-1) so either x^2 or x-1 has to be 0 so you get x=0^x=1

I came up with it.

Also to (π+e)/4. Likely just coincidence.

No you can't. Because you are not saying the functions on both sides are the same. You are looking for specific values of x that satisfy the equation that contains two _different_ functions.

Good question! The index at the lefthandside is of no relevance, because really there is no index at the lefthandside. It can be considered similar to the case of solving the equation y * y = 49 This equation can be solved by applying the square-root function, sqrt(x) : y = sqrt₁(49) = +7 , or y = sqrt₂(49) = -7 where sqrt₁(x) is the principal branch (commonly notated as √x) and sqrt₂(x) is the secondary branch. (The reason sqrt(x) has two branches is because the function f(w) = w² has pairs (w1, w2) (with w1 ≠ w2) such that f(w1) = f(w2) .) However, we aren't really applying sqrt(x) to the lefthandside; we're directly solving y by applying (either branch of) sqrt(x) to the expression/value in the righthandside. In the video, the equation that we at some point seek to solve is y * e^y = (-2/3)e^(-2/3) where y = (-2/3)x² , and the righthandside is a fixed negative real value. If we figure out y, then we can proceed to calculate x . Obviously , one solution to this equation is y = -2/3 . However, the graph of f(w) = w*e^w shows that when f(w) < 0 , there exist _two_ different values of w that result in the same value of f(w) (with the exception of f(-1) = -1/e , which is the minimum of f(w) ). So w = -2/3 is not the only value of w that results in f(w) = (-2/3)e^(-2/3) ; there is another negative value . The two negative values of w that result in the same value of f(w) can be considered pairs; for each w1 between -1 and 0, there exists a w2 < -1, such that f(w2) = f(w1) . Now suppose that we know the value of f(w2) = f(w1) , let's call it K . The branches of the Lambert W Function have been defined such that the following holds: W₀(K) = w1 , in other words, W₀(K) renders the value w that's between -1 and 0 , W₋₁(K) = w2 , in other words, W₋₁(K) renders the value w that's less than -1 . (Provided that 0 > K ≥ -1/e . If K < -1/e , then W₀(K) and W₋₁(K) both render complex values; because f(w) has a minimum at f(-1) = -1/e .) So back to our equation: y * e^y = (-2/3)e^(-2/3) We can solve y by applying the Lambert W Function: y = W( {righthandside value} ). Since (-2/3) is between -1 and 0, we can see that y = W₀( {righthandside value} ) will give us (-2/3) . But we are interested in the other possible solution (because, as it later turns out, y = -2/3 does not yield a valid solution for x). In other words, we are interested in y = W₋₁( {righthandside value} ) which will give a different value than -2/3, namely some value < -1 . And this value turns out to be y = −1.4293552275... , and from there we can eventually solve x = 1.4642516318... . I hope this helps. (By the way, this explanation about the Lambert W Function applies to real numbers; values of w that are real and values of f(w) = K that are real. In that case, only the principal branch W₀ and the branch W₋₁ are relevant. But the Lambert W Function has more branches than just those two. When dealing with complex numbers (i.e. if f(w) = K is complex-valued, or if f(w) = K is real but we're looking for complex-valued w), then all of the infinitely many branches of the Lambert W Function render (different) solutions, not just W₀ and W₋₁ .)

But that solution is excluded at the start because you divide by ln x, so ln x cannot be zero

@@andrewhone3346 ohh I see

😂