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!
Wow randomly scrolling through RUclips and seeing this guy used to be my math teacher a year ago glad to see you are successful on RUclips you are a great professor.
As an Engineering student and having to deal with these problem solving, I am glad that I am already done with calculus with the help of this guy's math marathon. Thanks man.
@@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 ?
This video gives us a very important lesson that many ML practioners overlook- Context is important in Math. When it comes to fields like Machine Learning, people sometimes blindly apply techniques without evaulating context. The part about 1 not being a valid solution encapsulates that perfectly.
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.. ?
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 was just playing around with number guesses yesterday in a sleep deprived state, and happened to find that pi^(1/3) is a very close approximation to the value you get for x. Not sure what this means, but it's cool!
If you plug y = (x^2 -1)/ln(x) and y = 3 into a graphing calculator and then use the intersect function you get 1.464. This video is still super cool because he got an exact answer using some math I didn't know existed, so keep up the great work!
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!
@@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.
The different branches of Lambert W correspond to the complex branches of the logarithm. An easier way to identify the solution, and the fact that it's unique, is graphically: we are looking for where the function f(x) = x^2 - 1 - 3 ln x takes the value 0 in the range x>0. This function tends to infinity as x-> 0+ and as x-> infinity, and by elementary calculus has a unique local minimum at x=(3/2)^(1/2), where f is negative. Hence (by the intermediate value theorem) it must have a zero on either side of the minimum. On the left, there is the easy solution x=1, which can be ruled out because (as noted in the video) it doesn't solve the original problem, which requires ln x to be non-zero. So the required solution must be bigger than square root of 1.5, and also less than 2, since f(2)>0. To get a better approximation, set x=1+y and expand as a Taylor series about y=0: f = - y ÷ 5/2 y^2 - y^3 + O(y^4). (This is a convergent series for |y|
Fascinating video. Frankly, I did a naive geometrical interpretation and got 1.5. I pretty much went back to the definition of what an integral was as the "area under a curve" and realized if I just made the "curve" a straight line and "integrated" it's just the area of a box with a base of 2. Using X = (1.5)^(1/t) obviously just cancels out the t's and makes it a constant being integrated - thereby giving me 1.5t from 0 to 2 which gives 3. Apparently, that's a fairly close approximation, but I don't know if I just go lucky.
I'm starting to understand things from your videos, which (for me) is incredible because before i didn't understand at all. Keep up the good work sir 👍
The answer checks out intuitively, as integral of 1^t from 0 to 2 would give 2, and integral of 2^t from 0 to 2 would give (4 - 1) / ln 2 =~ 4.32. So the answer feels like it should be just under 1.5.
At x2-1= 3lnx , you can also use graph method(plotting the equation graph and checking for any intersection of the two equation) also to find the solution.
I was so close to solving it! I knew I had to use the product log function and I knew there would be some tricky branch stuff and I got it so close to the required form but I couldn't work out to raise both sides to the -2/3 will certainly keep it in mind for next time
I just used the Newton method. Although, the xo approximation of 1.1 converges at 1, you get the other root of y=x^2-3lnx-1 by starting with xo=1.5. It converges to 1.464251632 with just a couple of iterations
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 :)
I always forget about the fish, and even if I'd remembered, I'd have given up on it if the principal branch hadn't worked. But I rapidly came up with a reasonable answer using Newton's method, and I was satisfied. It is always fun to see the fish, though.
by multiplying with ln(x) we actually introduced the wrong answer (1) into the equation. Thats why I gotten into the habbit of always excluding Zeros, I could potentially multiply with while doing stuff to equations, at the Time I'm introducing them.
The desired function f=e*(3/(e^2-1))^1/t The solution is obtained without using the Lambert W-function. To find the primitive of f^t, you need to answer the question, which function, when differentiating, will give a power function: (f^t)'=f^t*(ln(f)+t/f*f') It remains to solve the differential equation ln(f)+t/f*f'=1
@@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.
A simple function could also be x=(3/2)^(1/t), since the exponent simplifies. But I guess the video implicitly asks for x to be a constant, non a function of t, which makes the Lambert function necessary.
Some of my math teachers were obsessed with giving extremely long and tedious problems, so that I had to work on each problem for like 30 minutes, and some math teachers were much more forgiving and gave problems that could be solved in like 5 minutes if I knew what I was doing (like concise u-substitutions in Calculus 2, for example).
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.
Whenever the math has exponentials as key part, integers become just as messy as e is by the integers standards. It's like oil and water. It makes sense that the answer is this crazy.
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.
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.
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.
Just curious, which particular math class is Lambert W function taught? This is the first time I've ever heard of this function, and I've been in college for 3 years working with math classes.
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
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.
What my approach is... integrate then.. on right hand side lnx and other will populate on left hand side . Make graphs of both and find interesting points.
Cool example on "multiple" inversion of non-monotonic functions. I see there are eterogeneous comments about the option between the use of " standard" formal w function and the "direct" evaluation by means of numerical methods. Anyhow, if you study real calculus through a theoretical approach, you will see that the inverse of a continuos function is still continuos and therefore closed intervals are mapped to closed intervals (intermediate value theorem); as a matter of fact, the inverse function g(y) of a continuos function f(x) has to be always evaluated by solving equations of the type f(x)=y , i.e. F(x,y)=f(x)-y=0, with respect to the x symbol; the Dini's theorem provides all the theretical background (for instance, it describes also the derivability properties). As a matter of fact, it is due to a merely conventional tradition the fact we call the inverse function ln(y) of exp(x) "elementary", and the inverse function w(y) of x exp(x) "non elementary"... in both cases the theory states their continuity and derivability properties that can be exploited for efficient and reliable computation. For instance, for any "smooth" function (like x exp(x) ) we can simply express its inverse as a Taylor power series expansion by means of the Lagrange theorem. Obviously we can alternatively employ iterative methods (e g. Newton, dicotomic algorithms, Caccioppoli-Banach contractions...) whose correctness is founded on the continuity and completeness properties of the real topology.
@@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)...
Funny I actually was able to solve this as I encountered the Lambert W function as a young lass while I was personally interested in the equation nᵏ = kⁿ, n ≠ k. Was pretty surprising to me that there's such deep maths behind a seemingly innocent equation.
\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))
ln(1 + a) = a - a^2/2 + ... . So keeping the first two terms, (1 + a)^2 - 1 = 3 (a - a^2/2) . The solutions to this are a = 0, 2/5. So this approximate method yields x = 1.4.
I didn't get as far as the ^-2/3 step. As such, my application of W didn't help. I tried taking a quadratic in terms of x, and that rearranged back to the same equation so I did a quadratic in terms of 1. That made it worse.
Thank you. I did need in an exam the Lambert function, concerning the final exam of hydraulic centrals in engineering. No one reached me this function before. Hard to believe.
I saw the thumbnail and tried it on paper for myself. I haven't studied the Lambert W function before, so I didn't make it that far. Instead, I actually did use Newton's method but my pocket calculator was imprecise and when plugging the result (something like 1.3ish) in to the integral's answer, I got 2.something rather than 3. That's when I went "wait, what if the question mark is a function and not some constant?". But then it's easy to put something like x=at^(1/t) and get some value for a that solves the original question. Would be interesting to try to find the families of all functions for which this can work I suppose. Great question, though, I had fun!
@@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.
I like your videos, the math is somewhere around the top of what I can understand and some solutions I wouldn't find at all but this way I learn a lot and am usually watching them whole with 100% focus. Some solutions and ideas are pretty brilliant and clever.
When the index of the lambert w function is changed from 0 to -1, would this affect the left side in any way, or no matter what index you have on the left you will always get -2/3x²?
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 .)
This was my introduction to the Lambert W function. Thanks! 👍 😊 I took a summer course in Special Functions years ago but the whole course was on Laplace transforms! 😠
If you get a problem like this on a standardized test, you can approximate the answer very closely in 20 seconds with the regular power rule: Derivative of the function: t(x)^(t-1) (integral from 0 to 2) = 2x^1 = 3. X = 2/3 or 1.5, which is very close to the correct answer of 1.46. If that's your closest option by a wide ballpark on a multiple choice test, you'll know what to choose. Some tests obviously will make the problem much more difficult to approximate in your head like that, but cut corners where you can if you want to finish in time.
For the case : The integral of x^t wrt t. It implies that the integral of such form yielding the variable part-> x^t/Ln(x) is defined or valid for all non-zero (x≠0) positive integral (or generally positive reals) values of x except 1 or if Ln|x| is to replace Ln(x) so that the function is continuous for all negative integers( generally negative reals) as well, it must still be the case such that the domain of the function is restricted as 1
Aha,😂 I was confident until the first half , until you told thn answer was wrong. Needless to say I am not yet familiar with W Lambert functions. Yet, a good stretch. Thank you
I would also like to see a graphical proof, simply integrate, use a power series perhaps, is the following wrong: 2x^2-1 - 0x^0-1 = 3; 2x = 3; x = 3/2 or 1.5 ?
The integral is equal to (?^2 - 1)/ln(?), so the equation can be rewritten as (?^2 - 1)/ln(?) = 3. This is equivalent to ?^2 - 1 = 3·ln(?) = 3/2·2·ln(?) = 3/2·ln(?^2). Let x = ?^2. Hence x - 1 = 3/2·ln(x), which is equivalent to 2/3·x - 2/3 = ln(x), which is equivalent to e^(-2/3)·exp(2/3·x) = x, which is equivalent to exp(-2/3) = x·exp(-2/3·x), which is equivalent to -2/3·exp(-2/3) = -2/3·x·exp(-2/3·x), which is equivalent to -2/3·x = W[-1, -2/3·exp(-2/3)] or -2/3·x = W[0, -2/3·exp(-2/3)] = -2/3, which means x = -3/2·W[-1, -2/3·exp(-2/3)], or x = 1.
Hard question: solve for pi without "approximating" it and/or assuming it can't be done. Think about using an annulus composed of 4a units squared and reversing the square per quadrant.
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.
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
I feel good after seeing that you got x as + - 1 and I instantly thought of picking another branch of lambdaW function after using numerical methods to find the answer
We should allow X to be a function of t :D Let x(t) = 2*(ln(2))^(1/t) (x(t))^t = ln(2)*2^t => the integration is just 2^t evaluated from 0 to 2 = 4 - 1 = 3 :D
Such kinda equations involving exponential or log can also be solved by using Newton-Ralphson method. The same solution I have got using Newton-Ralphson method, by solving the equation x^2 - 1 -3lnx = 0 => x = 1.46.
I've not idea about maths. Literally. I only did 1 bach (spain) and i forgot everything. But i watched a lot of your videos. Dunno why. Dunno if you're doing it well or you're failing. I'm just enjoying this. Its like maths are incredible when you don't do exams about maths
First thoughts: (x^2-1)/lnx=3 x^2-1=3lnx (x-1)(x+1)=3lnx Finding the zeros.of LHS and RHS we get that x=1 Maybe there're more roots, these are the first things that came to my mind
Damn that’s crazy. Its one part about math ir really like, to explore questions and solutions like this and explore the meaning behind things like that with the professor. Not the useless, repetitive same questions for homework that just take time to solve but could by done by any idiot. Not the annoying thing of being forced to write so much and solve everything by hand. Its about exploring new interesting things instead of being bored to death with.. yeah solve this shit over and over until you can do it in your sleep. We‘re starting to learn some new things at uni thankfully but the bs is that we still have to do the same old shit in the so called practice groups we‘re actually graded on.
No !! You mistakely put the limits at the place of x. It will give eqn (X^2 - 1)/log x = 3 for which x have to find. It can't be calculated to the absolute value of x.
Hi (sorry for my bad English), I accidentally saw your video and got interested in the task. As I understand it, the value to be found does not necessarily have to be a constant. Therefore, I derived the function by iteration. This function: (3t/2)^(1/t). When searching, I started from the property of degrees. We need to get rid of from t. And then choose an expression that, when integrated, will give 3. I spent about 50 minutes. ((3t/2)^(1/t))^t => 3t/2 => (after integration) ((3t^2)/4) from 0 to 2 => 3*4/4 - 3*0/4 => 3
What I didn't understand is that you use principle brach in one side and (-1) brach of Lambert W function on other side on next step. Can anyone tell me that how is that possible?
Ngl it took me like 2 minutes to figure out a way to do it. First notice that 2^2 - 2^0 = 3. This implies that the integral can be equal to 2^t between 0 and 2 Then take the derivative of 2^t (which is Ln(2)*2^t) Then arrange it so its in the form x^t Ln(2)*2^t = (2*Ln(2)^(1/t))^t (Even if the function isnt defined on 0 its integral (2^t) is so it isnt a problem) So a solution is x = 2*Ln(2)^(1/t)
Try this one: find solution x^4 + cos(x) + sin(x^3) = 5... And an obvious answer is MyPersonalF(5) where MyPersonalF(y) = root of x^4 + cos(x) + sin(x^3) = y.
i haven't seen the complete video i was just wondering can't we just log both sides then t would come out of the log then we would integrate it ,on the rhs we would get log 3 we would divide it by the integral of t ie 2 and then anti log it and we would get the answer i am not good at maths so idk if i am right or wrong and i haven't even seen the video to verify this btw my answer was 1.729 if we calculate in 3 digits and we in india take the base of the log as 10 ,it is a standard value so i have calculated accordingly
![Noob To Max With DRAGON REWORK In Blox Fruits [FULL MOVIE]](http://i.ytimg.com/vi/LnBMOinoOvA/mqdefault.jpg)
Try this one next: ruclips.net/video/hztqL8104d0/видео.html
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
Wow randomly scrolling through RUclips and seeing this guy used to be my math teacher a year ago glad to see you are successful on RUclips you are a great professor.
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!
Stop giving hard questions. You're making me feel like I'm bad at math.
😂😂😂
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I just started the integration chapter in calc 1 today, can't wait to understand most of the stuff on this channel!
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 ❤️❤️
As an Engineering student and having to deal with these problem solving, I am glad that I am already done with calculus with the help of this guy's math marathon. Thanks man.
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**
I absolutely love math when I'm not the one doing it.
🤣🤣🤣
Bruh
B r u h
Same 😂
*B R U H*
This video gives us a very important lesson that many ML practioners overlook- Context is important in Math. When it comes to fields like Machine Learning, people sometimes blindly apply techniques without evaulating context. The part about 1 not being a valid solution encapsulates that perfectly.
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.. ?
Nothing more ominous then 36 boxes of expo markers sitting in the background.
😂
I like functions like the Lambert W function. When there's an equation you can't solve, just invent a new function :D
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?
I was just playing around with number guesses yesterday in a sleep deprived state, and happened to find that pi^(1/3) is a very close approximation to the value you get for x. Not sure what this means, but it's cool!
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
Been learning cal one on my own watching this dudes vids is always an inspiration to keep on learning. Good luck with your maths chaps !!!
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
If you plug y = (x^2 -1)/ln(x) and y = 3 into a graphing calculator and then use the intersect function you get 1.464. This video is still super cool because he got an exact answer using some math I didn't know existed, so keep up the great work!
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.
Great question! Loved your introduction of Lambert W function here..
Woow this was sick!! Missed these crazy videos a bit on this channel between all the Calc 1 uploads recently!! :) great vid!!
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.
0:33 are you implying that 80 million people know how to solve this?
What a fun little problem. It was nice to see the non-principle branch get utilized for a change!
The different branches of Lambert W correspond to the complex branches of the logarithm. An easier way to identify the solution, and the fact that it's unique, is graphically: we are looking for where the function f(x) = x^2 - 1 - 3 ln x takes the value 0 in the range x>0. This function tends to infinity as x-> 0+ and as x-> infinity, and by elementary calculus has a unique local minimum at x=(3/2)^(1/2), where f is negative. Hence (by the intermediate value theorem) it must have a zero on either side of the minimum. On the left, there is the easy solution x=1, which can be ruled out because (as noted in the video) it doesn't solve the original problem, which requires ln x to be non-zero. So the required solution must be bigger than square root of 1.5, and also less than 2, since f(2)>0. To get a better approximation, set x=1+y and expand as a Taylor series about y=0: f = - y ÷ 5/2 y^2 - y^3 + O(y^4). (This is a convergent series for |y|
Thank you for making these videos. I watch daily just to enjoy the beauty of integrals. Well explained and engaging!
Fascinating video. Frankly, I did a naive geometrical interpretation and got 1.5. I pretty much went back to the definition of what an integral was as the "area under a curve" and realized if I just made the "curve" a straight line and "integrated" it's just the area of a box with a base of 2. Using X = (1.5)^(1/t) obviously just cancels out the t's and makes it a constant being integrated - thereby giving me 1.5t from 0 to 2 which gives 3. Apparently, that's a fairly close approximation, but I don't know if I just go lucky.
I'm starting to understand things from your videos, which (for me) is incredible because before i didn't understand at all.
Keep up the good work sir 👍
I love it when answers to seemingly simple problems use special functions. Can you do more problems that can be solved using special functions?
🤓👈🤣🤣🤣
@@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?
7:38 It was at this exact point I went "Oh no, we are going into the multi-value complex logarithm stuff, aren't we?"
😆
The answer checks out intuitively, as integral of 1^t from 0 to 2 would give 2, and integral of 2^t from 0 to 2 would give (4 - 1) / ln 2 =~ 4.32. So the answer feels like it should be just under 1.5.
Interesting, I never thought of looking at it like that.
yeahh of course
Nice intuitive approximation for 3/ln2 XD
At x2-1= 3lnx , you can also use graph method(plotting the equation graph and checking for any intersection of the two equation) also to find the solution.
I was so close to solving it! I knew I had to use the product log function and I knew there would be some tricky branch stuff and I got it so close to the required form but I couldn't work out to raise both sides to the -2/3 will certainly keep it in mind for next time
I just used the Newton method. Although, the xo approximation of 1.1 converges at 1, you get the other root of y=x^2-3lnx-1 by starting with xo=1.5. It converges to 1.464251632 with just a couple of iterations
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
I literally screamed at 7:04 because I thought he was actually done, it's been at least 6 minutes since I knew what was going on, please send help
I always forget about the fish, and even if I'd remembered, I'd have given up on it if the principal branch hadn't worked. But I rapidly came up with a reasonable answer using Newton's method, and I was satisfied. It is always fun to see the fish, though.
by multiplying with ln(x) we actually introduced the wrong answer (1) into the equation.
Thats why I gotten into the habbit of always excluding Zeros, I could potentially multiply with while doing stuff to equations, at the Time I'm introducing them.
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.
I love how e is irrational, but the first 9 digits after the decimal misleads you to think e is rational and 1828 repeats forever.
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
The desired function f=e*(3/(e^2-1))^1/t
The solution is obtained without using the Lambert W-function.
To find the primitive of f^t, you need to answer the question, which function, when differentiating, will give a power function:
(f^t)'=f^t*(ln(f)+t/f*f')
It remains to solve the differential equation ln(f)+t/f*f'=1
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.
A simple function could also be x=(3/2)^(1/t), since the exponent simplifies. But I guess the video implicitly asks for x to be a constant, non a function of t, which makes the Lambert function necessary.
I had a 99%-sure hunch Mr. Lambert was hiding somewhere behind one of those curtains in this problem!
Thanks, this was fun!
Fred
8:10 "negative one is just outrageous" xD
Some of my math teachers were obsessed with giving extremely long and tedious problems, so that I had to work on each problem for like 30 minutes, and some math teachers were much more forgiving and gave problems that could be solved in like 5 minutes if I knew what I was doing (like concise u-substitutions in Calculus 2, for example).
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.
Whenever the math has exponentials as key part, integers become just as messy as e is by the integers standards. It's like oil and water. It makes sense that the answer is this crazy.
Just a non-maths student wondering: Is it legit to apply W(0) on LHS to get -2/3 x^2, while applying W(-1) on RHS?
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.
2:49
Can we differentiate both sides?
That way, we'll get (3/2)^1/2
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.
Just curious, which particular math class is Lambert W function taught? This is the first time I've ever heard of this function, and I've been in college for 3 years working with math classes.
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
What my approach is... integrate then.. on right hand side lnx and other will populate on left hand side . Make graphs of both and find interesting points.
Cool example on "multiple" inversion of non-monotonic functions.
I see there are eterogeneous comments about the option between the use of " standard" formal w function and the "direct" evaluation by means of numerical methods. Anyhow, if you study real calculus through a theoretical approach, you will see that the inverse of a continuos function is still continuos and therefore closed intervals are mapped to closed intervals (intermediate value theorem); as a matter of fact, the inverse function g(y) of a continuos function f(x) has to be always evaluated by solving equations of the type f(x)=y , i.e. F(x,y)=f(x)-y=0, with respect to the x symbol; the Dini's theorem provides all the theretical background (for instance, it describes also the derivability properties). As a matter of fact, it is due to a merely conventional tradition the fact we call the inverse function ln(y) of exp(x) "elementary", and the inverse function w(y) of x exp(x) "non elementary"... in both cases the theory states their continuity and derivability properties that can be exploited for efficient and reliable computation. For instance, for any "smooth" function (like x exp(x) ) we can simply express its inverse as a Taylor power series expansion by means of the Lagrange theorem. Obviously we can alternatively employ iterative methods (e g. Newton, dicotomic algorithms, Caccioppoli-Banach contractions...) whose correctness is founded on the continuity and completeness properties of the real topology.
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
Funny I actually was able to solve this as I encountered the Lambert W function as a young lass while I was personally interested in the equation nᵏ = kⁿ, n ≠ k. Was pretty surprising to me that there's such deep maths behind a seemingly innocent equation.
thats why the fishy has a evil face
Its very easy. Just set the RHS to integral of (3/2)t. This resolves X to e^(ln(t)/t). Or simply Tth root of t. Simple. Ignoring 3/2.
Sir , please can you explain how to find range of the function f(x)=sqrt of (x-1)/(x²-1). Please I am very much frustrated 🥺
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))
ln(1 + a) = a - a^2/2 + ... . So keeping the first two terms, (1 + a)^2 - 1 = 3 (a - a^2/2) . The solutions to this are a = 0, 2/5. So this approximate method yields x = 1.4.
I didn't get as far as the ^-2/3 step. As such, my application of W didn't help. I tried taking a quadratic in terms of x, and that rearranged back to the same equation so I did a quadratic in terms of 1. That made it worse.
Thank you. I did need in an exam the Lambert function, concerning the final exam of hydraulic centrals in engineering. No one reached me this function before. Hard to believe.
this is amazing. To answer the title as well, I definitely could not solve it before watching :D
I saw the thumbnail and tried it on paper for myself. I haven't studied the Lambert W function before, so I didn't make it that far. Instead, I actually did use Newton's method but my pocket calculator was imprecise and when plugging the result (something like 1.3ish) in to the integral's answer, I got 2.something rather than 3. That's when I went "wait, what if the question mark is a function and not some constant?". But then it's easy to put something like x=at^(1/t) and get some value for a that solves the original question. Would be interesting to try to find the families of all functions for which this can work I suppose. Great question, though, I had fun!
Can you please try to find the integral of I={[x(π+49)]¹⁵/⁷}/π²(x^π+7)?
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.
I like your videos, the math is somewhere around the top of what I can understand and some solutions I wouldn't find at all but this way I learn a lot and am usually watching them whole with 100% focus. Some solutions and ideas are pretty brilliant and clever.
I really enjoyed watching that. Thank you.
Name a more iconic duo than blackpenredpen and lambert function.
How do calculators approximate values when using the Lambert W function? Is there some sort of infinite series or something of that sort?
Newton-Raphson method, probably.
Wikipedia has a page on the Lambert W function: en.wikipedia.org/wiki/Lambert_W_function
When the index of the lambert w function is changed from 0 to -1, would this affect the left side in any way, or no matter what index you have on the left you will always get -2/3x²?
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 .)
Comment: Stop giving 1+2
BPRP: Okay, so find √π!
This was my introduction to the Lambert W function. Thanks! 👍 😊
I took a summer course in Special Functions years ago but the whole course was on Laplace transforms! 😠
Awesome as always!!
Thanks
I don't know what is more impressive, the equation itself or the way you switch your marker.
If you get a problem like this on a standardized test, you can approximate the answer very closely in 20 seconds with the regular power rule:
Derivative of the function:
t(x)^(t-1) (integral from 0 to 2) = 2x^1 = 3. X = 2/3 or 1.5, which is very close to the correct answer of 1.46. If that's your closest option by a wide ballpark on a multiple choice test, you'll know what to choose. Some tests obviously will make the problem much more difficult to approximate in your head like that, but cut corners where you can if you want to finish in time.
For the case : The integral of x^t wrt t. It implies that the integral of such form yielding the variable part-> x^t/Ln(x) is defined or valid for all non-zero (x≠0) positive integral (or generally positive reals) values of x except 1 or if Ln|x| is to replace Ln(x) so that the function is continuous for all negative integers( generally negative reals) as well, it must still be the case such that the domain of the function is restricted as 1
Aha,😂 I was confident until the first half , until you told thn answer was wrong. Needless to say I am not yet familiar with W Lambert functions. Yet, a good stretch. Thank you
Rewrite your equation as x = sqrt(3 * ln(x) + 1). Iterate on a spreadsheet and x settles around 1.464252 for x with an initial value > 1.
Lol I was watching your (infinity-infinity)^infinity video when this came out
I would also like to see a graphical proof, simply integrate, use a power series perhaps, is the following wrong: 2x^2-1 - 0x^0-1 = 3; 2x = 3; x = 3/2 or 1.5 ?
The integral is equal to (?^2 - 1)/ln(?), so the equation can be rewritten as (?^2 - 1)/ln(?) = 3. This is equivalent to ?^2 - 1 = 3·ln(?) = 3/2·2·ln(?) = 3/2·ln(?^2). Let x = ?^2. Hence x - 1 = 3/2·ln(x), which is equivalent to 2/3·x - 2/3 = ln(x), which is equivalent to e^(-2/3)·exp(2/3·x) = x, which is equivalent to exp(-2/3) = x·exp(-2/3·x), which is equivalent to -2/3·exp(-2/3) = -2/3·x·exp(-2/3·x), which is equivalent to -2/3·x = W[-1, -2/3·exp(-2/3)] or -2/3·x = W[0, -2/3·exp(-2/3)] = -2/3, which means x = -3/2·W[-1, -2/3·exp(-2/3)], or x = 1.
Watching him pause after he finished writing down the equation felt reassuring that people don't just immediately solve stuff like this.
This is very easy man , you are making me feel like I am super good at math 😂😂😂
😂 good work!
These is a easier solution.
(ln2^(1/t)*e^(ln2))^t
This simplifies
To
ln2*e^(ln2*t)
Which when you integrate from 0 to 2
You get
ln2/ln2 [4-1]
=3
Hard question: solve for pi without "approximating" it and/or assuming it can't be done.
Think about using an annulus composed of 4a units squared and reversing the square per quadrant.
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.
Hey! Thank you for video! Where did you buy the picture on the wall? I really like it :D
I designed it. You can check out my merch store in the description if you are interested
@@blackpenredpen Thank you! You're a master of design :3
I love it. You should submit a version of this in next year's Putnam.
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
I feel good after seeing that you got x as + - 1 and I instantly thought of picking another branch of lambdaW function after using numerical methods to find the answer
We should allow X to be a function of t :D
Let x(t) = 2*(ln(2))^(1/t)
(x(t))^t = ln(2)*2^t => the integration is just 2^t evaluated from 0 to 2 = 4 - 1 = 3 :D
Such kinda equations involving exponential or log can also be solved by using Newton-Ralphson method. The same solution I have got using Newton-Ralphson method, by solving the equation x^2 - 1 -3lnx = 0 => x = 1.46.
It was new to me (interesting).
Thank you so much *bP🖋️rP🖍️* ❤️
3 year old me searching up “hardest math question” for my dad to answer and then this dude comes up:
Did you create this question or did you find it somewhere. It's a good one.
I came up with it.
I've not idea about maths. Literally. I only did 1 bach (spain) and i forgot everything. But i watched a lot of your videos. Dunno why. Dunno if you're doing it well or you're failing. I'm just enjoying this.
Its like maths are incredible when you don't do exams about maths
First thoughts:
(x^2-1)/lnx=3
x^2-1=3lnx
(x-1)(x+1)=3lnx
Finding the zeros.of LHS and RHS we get that x=1
Maybe there're more roots, these are the first things that came to my mind
But x = 1 doesn't work, because substituting it back into the original integral gives 2 = 3.
Fred
omg this video literally taught me W function and its features
Damn maths is hard but there is always ways to make it easy.
How are u calculating the lamda W function after the pricipal branch?
The architect used the same equation to eliminate Neo but not 3
Damn that’s crazy. Its one part about math ir really like, to explore questions and solutions like this and explore the meaning behind things like that with the professor. Not the useless, repetitive same questions for homework that just take time to solve but could by done by any idiot. Not the annoying thing of being forced to write so much and solve everything by hand. Its about exploring new interesting things instead of being bored to death with.. yeah solve this shit over and over until you can do it in your sleep.
We‘re starting to learn some new things at uni thankfully but the bs is that we still have to do the same old shit in the so called practice groups we‘re actually graded on.
1.46425….
But to be honest, solved using Desmos.
In 2:08 a^x/loga will be applied so it will be 3ln2??
No !! You mistakely put the limits at the place of x.
It will give eqn
(X^2 - 1)/log x = 3 for which x have to find.
It can't be calculated to the absolute value of x.
@@jayantsingh3078 yes you are right bro I thought it's dx
I am stuck at,
e(?)³ = e^(?)²
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
Hi (sorry for my bad English), I accidentally saw your video and got interested in the task. As I understand it, the value to be found does not necessarily have to be a constant. Therefore, I derived the function by iteration. This function: (3t/2)^(1/t).
When searching, I started from the property of degrees. We need to get rid of from t. And then choose an expression that, when integrated, will give 3. I spent about 50 minutes.
((3t/2)^(1/t))^t => 3t/2 => (after integration) ((3t^2)/4) from 0 to 2 => 3*4/4 - 3*0/4 => 3
Don't touch my math
What I didn't understand is that you use principle brach in one side and (-1) brach of Lambert W function on other side on next step. Can anyone tell me that how is that possible?
FISH! FISH! FISH!
Ngl it took me like 2 minutes to figure out a way to do it.
First notice that 2^2 - 2^0 = 3.
This implies that the integral can be equal to 2^t between 0 and 2
Then take the derivative of 2^t (which is Ln(2)*2^t)
Then arrange it so its in the form x^t
Ln(2)*2^t = (2*Ln(2)^(1/t))^t
(Even if the function isnt defined on 0 its integral (2^t) is so it isnt a problem)
So a solution is x = 2*Ln(2)^(1/t)
*I Love Fish🐠*
Try this one: find solution x^4 + cos(x) + sin(x^3) = 5... And an obvious answer is MyPersonalF(5) where MyPersonalF(y) = root of x^4 + cos(x) + sin(x^3) = y.
i haven't seen the complete video i was just wondering can't we just log both sides then t would come out of the log then we would integrate it ,on the rhs we would get log 3 we would divide it by the integral of t ie 2 and then anti log it and we would get the answer i am not good at maths so idk if i am right or wrong and i haven't even seen the video to verify this btw my answer was 1.729 if we calculate in 3 digits and we in india take the base of the log as 10 ,it is a standard value so i have calculated accordingly
Try 0->1 integral of x^(xe^x) dx
Bernoulli integral
5:08
I've managed to get only here and couldn't find any thoughts what to do next
Could you try solving f'(t)=2f(2t+1)-2f(2t-1), f(0)=1? (Non constant solutions)
Ohhh you do Black Pen Red Pen sooo goood. prettygood!!
thanks! please, explain more in detail the branches of the Lambert function, I got lost there