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 TH-cam and seeing this guy used to be my math teacher a year ago glad to see you are successful on TH-cam you are a great professor.
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.. ?
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 ?
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|
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.
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 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.
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.
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.
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.
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'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 👍
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.
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.
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.
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.
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)...
X=2*((ln2) ^1/t)) So (2*((ln2) ^1/t))) ^t=2^t*ln 2 Integral of 2^t*ln2 limits from 2 to 0 is 3.Sorry if you have not understood what I have written. 😅I hope , sir you will read my comment. 😂😂😂😂.
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 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!
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.
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! 😠
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.
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
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.
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.
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
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.
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.
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
Ok i haven't looked at the video yet but this is my approach: 1. Rewrite x^t as e^ln(x)*t 2. Solve the integral: x^2/ln(x)-1/ln(x) = 3 3. Simplify: x^2+3*ln(x)-1=0 4. This is where i'm stuck. Idk how to solve for x here.
I mean the calculus is trivial, and you solve the transcendental equation using a special function which whose properties require some complex analysis to understand (so well-above the paygrade of most calculus students). Altogether, a terrible question. It is easy to make math questions hard, the trick is to make them both hard and meaningfully instructive. The Bernoulli integral is my favorite example of a problem that actually stretches almost every important muscle in Calc 1-2 and challenges students fundamental understanding of limits.
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.
VERY HARD math equation?. Straightforwardly: int from 0 to 2 (ln 2 · 2^t) dt = (2^t) from 0 to 2 = 2² - 2^0 = 4 - 1 = 3. Despite I'm not so good ar math!.
@@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 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
Correct me if I am wrong (I am not very good at this) After integration with limits we get x^2-1=3lnx Then differentiate with respect to x both sides We get 2x=3/x 2x^2=3 x^2=3/2 x=root of(3/2)=1.2247 Edit:Most probably I am wrong.I didn’t consider that x is a constant
\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))
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
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
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
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.
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.
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 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 ?
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
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.
I would try to do this: equation f(x)=x^2-3lnx-1=0; if I take the first derivative I get f'(x)=2x-3/x which has the root of sqrt(3/2) and thus is negative between 0 and sqrt(3/2) and positive after. So original function (f(x)) is decreasing between 0 and sqrt(3/2) and increasing after. f(0) = +infinity; f(sqrt(3/2)) < 0 and f(+infinity) = +infinity. Thus I can have at most 2 solutions: one between 0 and sqrt(3/2) and one greater than sqrt(3/2). It is very simple to see that x=1 is the solution that is between 0 and sqrt(3/2), BUT it does not satisfy the original question as you shown in the video and that the solution greater than sqrt(3/2) works fine. I would just stop there as I see no value in putting it under the form of a function of W inverse as it brings no further clarity to anyone.
I might add that this type of problem is something we would study in the last year of high school here and would be considered slightly above average but by no means hard.
I dont know a lot of calculus, but I got the answer, if u say e^ln(x ) = x. U can sustitute in 3ln = x ^ 2 - 1, and also say ln(x) = z. You get 3z = e ^(2z) -1. I dont rlly know how to get Z. But both sides are very easy curves on the plane. So just graph both and get the intersección. Calculator gives z*= 0.3813. Reverse log z and get x. X=1.4642
I didn't like to take W_0 on the left side and W_1 on the right side. Ok, there are a lot of branches, you choose the best one. If you ended doing real numerical computations, why didn't you begin doing them before? A good point to start the numerical computation is when you get: e^(x^2) e^(-1) = x^3
![BOWKYLION - วิงวอน (ex-change) [Official MV]](http://i.ytimg.com/vi/cLdnZWDaO-w/mqdefault.jpg)
Try this one next: th-cam.com/video/hztqL8104d0/w-d-xo.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 TH-cam and seeing this guy used to be my math teacher a year ago glad to see you are successful on TH-cam 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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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 absolutely love math when I'm not the one doing it.
🤣🤣🤣
Bruh
B r u h
Same 😂
*B R U H*
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 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 ❤️❤️
Nothing more ominous then 36 boxes of expo markers sitting in the background.
😂
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|
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
What a fun little problem. It was nice to see the non-principle branch get utilized for a change!
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.
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?
Great question! Loved your introduction of Lambert W function here..
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
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.
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.
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.
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
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?
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
Thank you for making these videos. I watch daily just to enjoy the beauty of integrals. Well explained and engaging!
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.
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 👍
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.
8:10 "negative one is just outrageous" xD
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.
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
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.
I have never done this math but I’m interested
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
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.
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.
I don't know what is more impressive, the equation itself or the way you switch your marker.
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
X=2*((ln2) ^1/t))
So (2*((ln2) ^1/t))) ^t=2^t*ln 2
Integral of 2^t*ln2 limits from 2 to 0 is 3.Sorry if you have not understood what I have written. 😅I hope , sir you will read my comment.
😂😂😂😂.
0:33 are you implying that 80 million people know how to solve this?
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.
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?"
😆
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!
I had a 99%-sure hunch Mr. Lambert was hiding somewhere behind one of those curtains in this problem!
Thanks, this was fun!
Fred
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
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.
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! 😠
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.
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
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.
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.
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
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.
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.
Watching him pause after he finished writing down the equation felt reassuring that people don't just immediately solve stuff like this.
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
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
Ok i haven't looked at the video yet but this is my approach:
1. Rewrite x^t as e^ln(x)*t
2. Solve the integral: x^2/ln(x)-1/ln(x) = 3
3. Simplify: x^2+3*ln(x)-1=0
4. This is where i'm stuck. Idk how to solve for x here.
this is amazing. To answer the title as well, I definitely could not solve it before watching :D
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.
I really enjoyed watching that. Thank you.
I mean the calculus is trivial, and you solve the transcendental equation using a special function which whose properties require some complex analysis to understand (so well-above the paygrade of most calculus students). Altogether, a terrible question.
It is easy to make math questions hard, the trick is to make them both hard and meaningfully instructive. The Bernoulli integral is my favorite example of a problem that actually stretches almost every important muscle in Calc 1-2 and challenges students fundamental understanding of limits.
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
VERY HARD math equation?. Straightforwardly: int from 0 to 2 (ln 2 · 2^t) dt = (2^t) from 0 to 2 = 2² - 2^0 = 4 - 1 = 3. Despite I'm not so good ar math!.
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 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
Lol I was watching your (infinity-infinity)^infinity video when this came out
Correct me if I am wrong (I am not very good at this)
After integration with limits we get x^2-1=3lnx
Then differentiate with respect to x both sides
We get 2x=3/x
2x^2=3
x^2=3/2
x=root of(3/2)=1.2247
Edit:Most probably I am wrong.I didn’t consider that x is a constant
It is wrong: x^2 -1 -3lnx=0 doesn’t imply its derivative will be zero. Also if you take x as constant you can’t take its derivative with respect to x
@@user-kg2dn5jr6e ok…
I don’t know if x is constant if it is then I can’t
this only finds where the slopes are the same, not the actual values
@@dvoid4968 okay thank you
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))
omg this video literally taught me W function and its features
Comment: Stop giving 1+2
BPRP: Okay, so find √π!
Name a more iconic duo than blackpenredpen and lambert function.
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
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
Awesome as always!!
Thanks
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
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.
We didn't set 'x' is constant, so we can set x=f(t) such as x=(3/2)^(1/t).
0:45 wise guy.
This is very easy man , you are making me feel like I am super good at math 😂😂😂
😂 good work!
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.
It was new to me (interesting).
Thank you so much *bP🖋️rP🖍️* ❤️
Interesting to note that x is quite close to 2 • ( sqrt(3) - 1 ), if only for the numbers involved.
Also to (π+e)/4. Likely just coincidence.
The architect used the same equation to eliminate Neo but not 3
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)
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
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 ?
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
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.
Did you create this question or did you find it somewhere. It's a good one.
I came up with it.
I would try to do this: equation f(x)=x^2-3lnx-1=0; if I take the first derivative I get f'(x)=2x-3/x which has the root of sqrt(3/2) and thus is negative between 0 and sqrt(3/2) and positive after. So original function (f(x)) is decreasing between 0 and sqrt(3/2) and increasing after. f(0) = +infinity; f(sqrt(3/2)) < 0 and f(+infinity) = +infinity. Thus I can have at most 2 solutions: one between 0 and sqrt(3/2) and one greater than sqrt(3/2). It is very simple to see that x=1 is the solution that is between 0 and sqrt(3/2), BUT it does not satisfy the original question as you shown in the video and that the solution greater than sqrt(3/2) works fine. I would just stop there as I see no value in putting it under the form of a function of W inverse as it brings no further clarity to anyone.
I might add that this type of problem is something we would study in the last year of high school here and would be considered slightly above average but by no means hard.
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
Could you try solving f'(t)=2f(2t+1)-2f(2t-1), f(0)=1? (Non constant solutions)
1.46425….
But to be honest, solved using Desmos.
Now do :
Integral of ln(sin(x))ln(cos(x))/ tan(x) dx bette en 0 to pi/2
And intégral of xth root of tan(x), or tan(x)^(1/x) dx.
Damn maths is hard but there is always ways to make it easy.
Try 0->1 integral of x^(xe^x) dx
Bernoulli integral
FISH! FISH! FISH!
I tried to visualize it in my head and come up with a quick guess before watching and got sqrt(2) which is actually suprisingly close.
*I Love Fish🐠*
I dont know a lot of calculus, but I got the answer, if u say e^ln(x ) = x.
U can sustitute in 3ln = x ^ 2 - 1, and also say ln(x) = z.
You get 3z = e ^(2z) -1.
I dont rlly know how to get Z. But both sides are very easy curves on the plane. So just graph both and get the intersección.
Calculator gives z*= 0.3813.
Reverse log z and get x. X=1.4642
Don't touch my math
I didn't like to take W_0 on the left side and W_1 on the right side. Ok, there are a lot of branches, you choose the best one.
If you ended doing real numerical computations, why didn't you begin doing them before?
A good point to start the numerical computation is when you get:
e^(x^2) e^(-1) = x^3