Solving a 'Harvard' University entrance exam

[@bagenstb]

I should've tried this technique on tests where I couldn't figure things out. "The answer is B(5), where B is a function I'm defining right now that will solve this problem."

[@empathogen75]

Once you have it in that form, you can calculate the answer numerically using newton approximation to any level of precision you want. It’s time consuming but you can do it.

[@JohnSmith-nx7zj]

⁠@@empathogen75you can calculate the answer numerically to arbitrary precision without any knowledge of lambert W functions etc.

[@DergaZuul]

Well it is exactly how this work and that B function might be numerically approximated easier than W. But of course standardized methods are preferred.

[@luisfilipe2023]

I’ll never not be amazed by mathematicians ability to just make stuff up and call it the day

[@Aker811]

Exactly my thoughts, its fascinating and frustrating at the same time that i have no idea how it works.

[@ir2001]

LambertW function is not a hack. It's a well-defined and researched function that can be numerically approximated. I understand why it may feel otherwise, particularly when you're seeing it for the first time. You may consider the situation as similar to how sqrt(-1) may have once felt to you before recognizing the vast world of complex numbers.

[@luisfilipe2023]

@@ir2001 yeah but it’s made up they just said this is now the inverse of that because I say so kind of like imaginary numbers they were just defined as the solution to negative square roots

[@ir2001]

@@luisfilipe2023 True, but I beg to disagree with the characterization. Keeping LambertW(x) aside for a moment so as to keep my explanation understandable by means of an analogy, how about ln(x)? You may call it merely an inverse of the exponential function, but on further analysis you would realise that it can be expressed as an integral, which can in turn be computed via numerical approximation methods. Therefore, you get an additional weapon for your Math arsenal. Essentially, resourceful abstractions help simplify our expressions without loss of precision.

[@twwc960]

All of mathematics is "just made up". The so-called elementary functions, such as exp, ln, sin, cos, tan, etc. were all made up at one time to solve problems, either purely mathematical or practical. Assigning a name to a particular function which is made up to solve some class of problems makes it easy to then study that function in detail. Such study can involve finding larger classes of problems which it solves, finding efficient numerical methods to find approximations, plotting graphs, studying its domain, range, etc., working out derivatives and integrals, finding a power series, etc., etc. Just look at the Wikipedia page for the Lambert W function to see how much it has been studied, for example.

[@chuckw4680]

So it still can't be solved by hand and needs a computer/calculator and I still don't know what a Lambert function is. I'll call it a day.

[@dlevi67]

Can you solve by hand sin(2.71828)? W is simply defined as the inverse function of z(e^z). Nothing more, nothing less. Just like (one) definition of sin(x) is to consider a unit-radius circle centered at the origin and looking at the relationship between an angle and the vertical coordinate of the point on the circle at that angle.

[@MadaraUchihaSecondRikudo]

If we were to replace W with ln or with sqrt in the solution, do you think you'd have been able to get a number without a calculator then?

[@Programmable_Rook]

⁠@@MadaraUchihaSecondRikudoIt’s actually surprisingly easy to calculate square roots (At least of whole numbers). If you convert the number to base 2, there’s a pretty simple pattern that can find the square root by hand. (There are technically patterns that work for higher bases to find square roots, but they’re fiendishly complicated. The base 2 pattern could be done by the average fifth grader)

[@MadaraUchihaSecondRikudo]

@@Programmable_Rook Yeah, but this isn't the sqrt of a whole number, just like this isn't the W of a whole number. My point stands, it's a less well-known but no less well-defined function, whose values you generally need a calculator to find.

[@michaelz6555]

Learning about the “Goal Seek” feature in Excel alone was worth the cost of admission. Thanks!

[@angrytedtalks]

I'd never seen that function either.

[@michaelwisniewski6047]

Cool. I learned it in 1999. But good to see that people are still discovering the program’s features. Let me give you something bigger. Goal Seek can accommodate only one variable, but you can project backward for more variables by using the Solver add-in. With Solver you can get a solution that works for multiple variables and you can even set constraints for them.

[@meateaw]

@@michaelwisniewski6047 at which point I've gone and gotten my LP solving library ;) (which is probably what excel is doing anyway)

[@verkuilb]

If we’re going to create “magical new” functions (as Presh refers to them in the video), and then use those functions as part of the answer-why not just define a “magical new” function Z, which “undoes” 2^x + x ? Then the answer is simply x = Z(5).

[@PhilBoswell]

Maybe this might help: en.wikipedia.org/wiki/Lambert_W_function#Applications

[@christianbohning7391]

It's about usefulness, and Lambert-W seems to be useful, whereas your Z probably isn't. And you could ask the same question about the natural log function.

[@Yiryujin]

You need to prove that the function will always work under specified conditions. they are not arbitrarily undoing something, its actually undoing it so it becomes a function. Every math function you had to memorize to plug numbers into in order to pass math class has been proven to work. I'm sure people had to memorize Pythagorean theory C^2 = A^2 + B^2 to path geometry class. this function will always work as long as triangle has right angle. if you can prove that function Z(#) = #^x + x than you can have your own magical new function Z and be known as mathematician who found function Z.

[@Gem-In_Eye]

W(x) is just inverse for f(x) = x•e^x We just didn't know the algebraic form of W(x) so we use it in only symbol, we essentially know what it does, it exists in reality not made up. Just a plain formless inverse of x•e^x. Math's main tool is abstraction.

[@leif1075]

​@@christianbohning7391YES BUT THIS IS THE KEY POINT LAMBERT IS CONTRIVED AND THEREFORE A CHEAT..Since no one wpuld ever think of it organically..obky maybe if youve seen ut before..andcevenbthen maybe not..not even Ramanujan or anyone..Do you not agree with me? I don't see how anyone could disagree

[@martinhertsius9282]

What's the point of all this when there is no explanation of what the W function does??

[@Gem-In_Eye]

W(x) is just a inverse of f(x) = x•e^x. As we don't know how to write it in the algebraic form so we just use symbols.

[@meateaw]

W(x)=Xe^X is it's definiton. Do you know precisely what log does? do you know what sine does? do you know what cosh does? At the end of the day, those functions are defined by what they do, and what they do is well known. W doesn't evaluate to a nice rational number, because it is based off the number "e", which is a mathematical constant. (like Pi) W(x) = x*e^X

[@deadpark121]

All you have to do to solve the equation is set the calculator to Wumbo

[@verkuilb]

Let me get this straight-you follow up a video about whether 3x5 is the same as 5x3…with this??? 🤯

[@bjorneriksson2404]

Hahahaha 😂😂 Well, you can't deny that he's got some range to his videos... 😊

[@Yiryujin]

ahahhhaaahahah. love it. wish this type of videos were around when I went to high school. then I may have actually grew to like and enjoy math.

[@wesss9353]

Common Core...

[@Ninja20704]

It is not a follow up video, it is just two seperate/unrelated videos he is uploading

[@Yiryujin]

@@Ninja20704 Lol. Verkuilb meant to follow up a video, not follow-up a video. Lol. Follow up is verb meaning sequential action. The act of following of a video by releasing another video. Follow-up is noun or adjective used when describing what you are referring to. A follow-up is a prompt and relevant response to a situation often in context of addressing a problem or providing additional information. So if you make up a follow-up appointment with a doctor, it means to check up on the same thing again to see how you're doing. But if you make a follow up appointment with a doctor, it just means your next visit.

[@JonSebastianF]

*_U 2 to the Power of U_* ...sounds like a power ballad by Prince💜

[@otakurocklee]

Nothing Compares to U

[@JonSebastianF]

@@otakurocklee ...apart from 5 - _x_ 😆

[@exoplanet11]

You are so right. That should be a song. Shades of "2 divided by zero" by the Pet Shop Boys.

[@RGP_Maths]

Or "One and One is One" by Medicine Head: the greatest Boolean logic single of all time!

[@davidrosenfeld1373]

First thing I noticed about the answer is that it is very nearly sqrt(3), which is probably just a coincidence.

[@chanuldandeniya9120]

Not very nearly actually only up to 1 decimal place. √3 = 1.732050807568877...

[@thecatofnineswords]

I saw the same approximation, but with (e-1)=1.71828 Probably also a coincidence, but now with logarithms.

[@chrisarmstrong8198]

The Lambert W function was never mentioned in my High School or University maths subjects (in the 1970's !). Thanks for the info.

[@dlevi67]

Hallelujah! Finally someone who has a sane reaction to learning something new. Thank _you!_

[@krabkrabkrab]

In my head, I tried x=5/3 and realized it's a bit low. SO I went for 1.7. Then Newton's method: x_new= x- (x log(2)-log(5-x))/(log(2)+1/(5-x)) immediately gives 1.7156 (on a calculator that doesn't have a Lambert function).

[@MusicalEutopia]

😮😮😮

[@Yiryujin]

using logarithmic naturally reduces exponents. but no way I'm doing that in my head without scientific calculator or log chart. In the past, majority of these exams were calculator free. So whenever these type of video mentions Harvard entrance exam or something, assume you can't use calculator. but in modern times they allow use of calculators with limited functionality. Even ACT (American College Testing) and other Professional College assessment exams such as MCAT (medical college assessment test) provided their own none scientific calculators in the past. This magical function lets you solve this without calculator. If you use windows, open up your calculator and set it to standard. that's basically what you were allowed to use IF they allowed calculators.

[@angrytedtalks]

I remember something from school about Newton-Rapherson approximation of integrals from about 1980. I just did trial and error on a calculator and got 1.7156207 ish in no time. How do you suppose a calculator does logarithms?

[@danmerget]

I solved the problem in a slightly different way, and got x = log2( W(32 * ln(2)) / ln(2) ). When I plugged it into a calculator, I got the same result as Presh: 1.71562. I was a bit freaked out as to how two different-looking answers could give the same result without any obvious conversion between them, but then I noticed that both answers contain W(32 * ln(2)) / ln(2). If we call that quantity Y, then Presh's answer was x = 5 - Y, and mine was x = log2(Y). The only way these two answers could be the same is if Y = 5 - x = 2^x, which would imply that 2^x + x = 5, and oohhhh I get it now.

[@wernerviehhauser94]

ok, but could we just NOT do a Lambert W Function for a week or so? The videos on that topic are getting out of hand...

[@hangslang]

just.... watch a different video? lol

[@ShubhamKumar-re4zv]

How did you create that link which leads to search results?

[@SchildkroeteHundFisch]

​@@ShubhamKumar-re4zvI think KZbin does that automatically sometimes.

[@sadiqabbaszade4789]

I mean, he could have at least explained how the wolframalpha calculates LamW

[@ShubhamKumar-re4zv]

@@SchildkroeteHundFisch Yes I also think so as the search link is not clickable now

[@adamrussell658]

I always forget about the Lambert function because W(x) doesnt mean anything to me. Plus, minus, square root, etc all have common sense meanings but it seems to me that W is an implied logic function as opposed to a mechanical function. If you say the solution is W(32ln2) its not clear what that is in real numbers or even a ballpark guess.

[@ThreePointOneFou]

The Lambert W function is just a terrible function to work with. It's a mess to calculate, it has two separate branches on part of its domain (because x*e^x isn't one-to-one over its range), and it has sum and difference formulas that are a pain to remember. I can't believe a problem requiring its use appears on a college entrance exam.

[@dlevi67]

It's not "clear" because you are not familiar with the function. How much is sin(2.71828)? Someone not familiar with trigonometric functions would have no clue; that does not make it poorly defined. I don't understand what you mean by "mechanical function" - W is neither more nor less mechanical than (say) sin.

[@yurenchu]

Draw a graph of the relation/function y = f(x) = x*(e^x) . Since 32*ln(2) is real and positive, W( 32*ln(2) ) is the x-coordinate of the _only point_ on the graph for which the y-coordinate equals 32*ln(2) . In general, W(y) * e^W(y) = y .

[@yurenchu]

​@@ThreePointOneFou A simple approach to this problem would be to rewrite the equation as 2^x = 5 - x , then sketch the graphs of f(x) = 2^x and g(x) = (5 - x) into one diagram, and estimate the coordinates of the intersection point of f(x) and g(x) . No Lambert W Function needed. (This approach would also demonstrate clearly that there exists only one real solution.)

[@empathogen75]

I actually think the lambert w function is a legitimate way to solve it, but if you just want a numerical answer, newton’s method would have been a lot faster.

[@lucabastianello9830]

Ok, but the W remain and it solved like a deus ex machina...

[@bjorneriksson2404]

The W is quite a bit like normal logarithms, you usually "solve" them as well by means of the deux ex machina that we call a calculator (except no ordinary calculator has the W function). Side note: I'm 50 with an MSc in applied physics, and I heard of the W function only a few years ago. Definitely never learned about it in school...

[@Yiryujin]

in math, you often answer with functions. its same as answering with x = sin (x) or fun(x) = x^2 as long as its actual function that works in that specific general instance, its acceptable answer. since it saves time on writing out the entire page of equations. would you rather write X = 5 - w(32ln2)/ln2 or x = 5 - {ln(x/lnx) - {ln(x/lnx)/[1+ln(x/lnx)]} ln(1-lnlnx/lnx)}(32ln2)/ln2 i

[@lucabastianello9830]

@@Yiryujin the second One. I don't Need elegance if not explained. Moreover in the video Is talked like and operator like sin and cos (without demonstration ok) but you associate It like a substitution (nothing special if you think It would have been the third One in the example)

[@lucabastianello9830]

@@bjorneriksson2404 I never had problema using adanced physical or mathematicians feaurre, still my First time hearing about W -function

[@robertp9297]

I believe my first comment "disappeared"... @Presh- Thanks very much. I'll look into Lambert W (I did attempt a guess at x=1.7; but it was a guess, and not a solution. Take good care, Presh. Thanks again !

[@christianbohning7391]

It is worth to mention that the Lambert-W function isn't exactly one function. To invert x * e^x in the real domain one needs two different branches of the Lambert-W functions, otherwise there would be two function values for x between -1/e and 0. Meaning that for x between -1/e and 0 only one of the two function branches might give you the desired solution, and in that case it's pretty tricky to know which one. Also, x < -1/e doesn't yield any real solution.

[@Gem-In_Eye]

As I was fighting Comment Wars, I also researched that, most of it went above my head as only this semester I'm going to study Complex Analysis so. But it was interesting. I enjoyed it.

[@yurenchu]

The equation we ended up with here, is u*(e^u) = 32*ln(2) where u = (5-x)*ln(2) . Since the righthandside , 32*ln(2) , is real and positive, this equation has only one real solution for u ; or in other words, only _one_ branch (of the infinitely many branches) of the Lambert W Function leads to a real solution, namely u = W₀( 32*ln(2) ) . In general, consider the equation u*(e^u) = y If y is real and positive, then only u = W₀(y) is real (and it's also positive); all other branches u = Wₖ(y) would be complex-valued. If y is real and between -1/e and 0, then both u = W₀(y) and u = W₋₁(y) are real (all other branches would be complex-valued), with W₀(y) being between -1 and 0 , and W₋₁(y) being less than -1 . If y is real and less than -1/e, then there are no real solutions; all branches u = Wₖ(y) would be complex-valued. In other words: there are two real branches for W(y) _only when_ y is real ánd between -1/e and 0 . (Please note: you seem to mix up x and y . If we think of x as the real variable of the real function f(x) = x*(e^x), as your comment seems to be suggesting, then it's y = f(x) that is between -1/e and 0 , for which there exist two real branches of inverses x = W(y) (namely one branch x < -1 , and one branch x between -1 and 0). And for real y > 0 , there is only one real branch x = W(y) , and it's also positive.)

[@Smallpriest]

For everyone complaining, consider ln(5) (natural log) If the answer was ln(5), would you say that it's an exact solution? If so, why would W(5) (lambert W) not also be an exact solution?

[@docsigma]

I am writing this comment before I watch the video, and will edit it after I watch it. My initial impression from just the thumbnail is... no way would a college entrance exam question involve the Lambert W function, right? Nobody would expect high school kids to know about the Lambert W function, right? EDIT: ...huh.

[@dlevi67]

Possibly not - but if this were an interview question (rather than a written one), the interviewer could ask something like "imagine that you have a function that is the inverse of x(e^x) - could you solve it then?"

[@DemoniqueLewis]

Never heard of Lambert W… should be added to the calculus class where logarithms and natural logs are covered.

[@dlevi67]

It definitely should. The "problem" with it is that it requires complex analysis to understand it properly, but that was never an issue with roots, so I don't see why not!

[@TitanOfClash]

I'd seen the function a lot before, but this really crystallised the solving algorithm for me. Thanks!

[@FerdiLouw]

Thanks. Very educational. The next question is: How does a calculator calculate W(x)? Similar to how is SQRT(x), SIN(x), LN(x), etc. calculated?

[@dlevi67]

Usually these are all approximated using series expansions for the functions. Which ones are used depends on the implementation; historically (40 years ago, when I had to write routines for those things as part of my education) it was a trade off between speed of convergence and amount of memory required to achieve the desired precision. Nowadays, I suspect people go for speed a lot more...

[@psolien]

All of Higher Math's videos are about this base use of the Lambert function😂 What a joke,lol. I doubt it has anything to do with any entrance exam ever!

[@AcaciaAvenue]

Lambert function is, imo, just a way to write x=something where you have an expression you can't analitically explicitate. It may be the way they wanted at that entrance exam. I would've just proceeded by writing it as 2^x = 5-x then plotting y=2x and y=5-x and figure out an approximate value by trials choosing the starting value of x by that graphic.

[@Blox117]

i doubt any of his videos are real entrance exams questions

[@johnpollard9828]

I love your videos. You do an excellent job of explaining everything!

[@1412-kaito]

You know, Initially I placed x=1, x=2 Looked at the values, and thought, Hmmm.. for eq.=5, x=2 is much closer than x=1, so answer should be approx. x=1.75. Had we taken the weighted diff. b/w the 2 points, x=1 & x=2, it would be the graphical approximation, but much simpler.

[@mohitrawat5225]

Once Presh went to an interview for the job at a Maths institute. The interviewer asked every candidate the same question - "Do you stand for humanity?" Many candidates replied yes and were rejected instantly. A candidate even replied no but was also rejected. When Presh was asked the same question, he said - "Your question has ambiguity. Does "U" mean the letter U or me as a person? Be clear cause the probability of my replies will change. Interviewer said - "Should we appoint you as the chairperson of this institute?"😂😂😂

[@vinaykumaryadav7013]

If it was the letter "U" he would have used "Does" instead of "Do". Hence the interviewer definitely meant "you". (Considering he had the basic language knowledge)

[@Aditya.7_7]

​@@vinaykumaryadav7013😂😂

[@verkuilb]

What about the interviewee who responded, “I’d rather sit for humanity. I already got my steps in for today.” ?

[@MrGeorge1896]

@@vinaykumaryadav7013 Actually the interviewer asked: Do "U"s stand for humanity? but in sloppy way of pronunciation. 🤠

[@Aditya.7_7]

@@verkuilb instantly rejected

[@TheJaguar1983]

Interesting, but this begs the obvious question: how do we calculate W(32 ln(2))?

[@dlevi67]

How do you calculate ln(32 + 2^π)? There are series expansion formulae for both...

[@TheJaguar1983]

@dlevi67 So, it's a function whose calculation is very complicated and, therefore, is beyond the scope of this video? Looking at other comments and their replies, it seems that I should see it just like sin or log: a well-defined function that isn't really practical to calculate by hand.

[@crimsoncanvas51]

High school maths to solve is assume f(x) = x^2+ x-5 and use Newton raphson method. xn1= xn0- f(xn0) /f'(xn0)

[@justinburley6000]

I got a degree in Engineering and can say with confidence that 99.9% of the stuff I learned has never shown up in my life again. The stupidity of reality becomes the real problems you gotta deal with.

[@ReginaldCarey]

The W function is cool. And it lets you carry around an exact form, it’s still approximate when reduced to numbers. It would be nice to include it in BLAS software

[@kytnCars]

This equation was actually easier than people think, Most of it is just ln.

[@TheChamp1971]

The solution to this reminds me of the Sydney Harris cartoon, "Then a Miracle Occurs..."

[@nushaerabrar7354]

you can also arrive at an approximate value using the Taylor Series at a=1.5. This simplifies the equation to a polynomial and we all can solve polynomials :3

[@nushaerabrar7354]

I used 1.5 as an estimate. Inserting x=1 is too small and x=2 is too large. So the actual answer might be around the middle. The higher order of derivatives you go, the more accurate answer you can get. But just the first derivative also approximates the answer quite well.

[@metalsplash310]

I estimated "about 1.7" in 10 seconds without a calculator by plugging in 2 and immediately roughly estimating that it was about 1.7.

[@metalsplash310]

I also immediately thought of using a graphing calculator, but I thought that would be cehating

[@metalsplash310]

I was able to get an exact number by just plugging in values of x. No one does this and its probably just cheating but it works really well and no one uses it.

[@dlevi67]

@@metalsplash310 Testing things "by inspection" is done by a lot of people in a lot of contexts. It's not 'cheating', but probably neither is it what the writers of the questions expected as an initial answer. BTW - you are _not_ getting an exact number. You are getting an approximation. The solution is an irrational (in fact, transcendental) number.

[@madcow3417]

I now remember why I quit math after differential equations.

[@mfhberg]

I have gone 42 years since looking at that function, our HS physics 2 teacher taught it in the last few weeks of class. Haven't seen it since.

[@mr9512]

@Blackpenredpen does a lot of videos (think a whole playlist's worth) re: Lambert W function and explains it rather well... Bonus - he also uses "fish" to explain it! 😂

[@dlevi67]

Alive without breath; As cold as death; Never thirsting, ever drinking; Clad in mail never clinking. Drowns on dry land, Thinks an island Is a mountain; Thinks a fountain Is a puff of air. So sleek, so fair! What a joy to meet! ***************** We only wish To catch a fish, So juicy-sweet!

[@somersetcace1]

The mathematical equivalency of unraveling a Rubik's cube.

[@NimrodClover]

The undeniable arbitrary declaration that my invented function W will undo the problem that standard mathematical equations can't just sounds a bit presumptive. As others have said, why not just say that my answer is x = V(5) where V is my function that "undoes" the problematic 2^x+x. Most rebuttals have said something to the extent that your V function must be rigorously tested / approved by the mathematical gatekeepers. But this ignores the fact that when Lambert first proposed this undoing but undefined function it lacked the rigorous mathematical review and that does not negate the fact that it produces a solution to the problem, even from the start. So, it is totally valid to state that my V function undoes the problematic 2^x+x since whomever set up this math question knew that conventional "undoing" actions are not applicable. This Lambert approach was never discussed in any of my math classes up and through collage and I had never heard of it till now. I don't know how applicable the solution would be in a test format, as it appears that the solution must still be run through some algorithms to get a numerical approximation. The closest I can recall would be in the first "solution" of the intersection lines of the two equations, which I did by hand using Analytical Geometry. Still the whole Lambert W approach seems very inapplicable in a test situation and I will not be using it. I really don't consider it a solution to the problem in the context it was given, unless the person grading the test accepts my V function... then we are fine, just like Lambert.

[@dlevi67]

1) The anecdotal fact that _you_ didn't know about the Lambert W function does not make it "poorly defined" or lacking "rigorous mathematical review". It was quite rigorously (for the times) defined by Lambert and Euler in the 18th century, and more study has been done on it more recently. 2) W it's only "unconventional" according to your limited mathematical knowledge. 3) W does not only produce a solution to this specific problem. It produces a general class of solutions to equations involving sums of exponentials and polynomials. Your _V_ function could do the same, but as it's a less general expression it would require transforming into the W function first anyway. 4) Prima facie, nobody is asking a numerical value as an answer to the question in the video. In any case, if they were, it's no more 'special' than attributing a numerical value to x in x^2 = 3.

[@Ankuhr_1]

My AP Calc teacher in high school had a hard enough time teaching us integration / derivation. Can't imagine teaching Lambert to high school kids to prep for a Harvard Entrance Exam lol

[@Potemkin2000]

It's just a way of rewriting an equation in terms of other predefined functions. Not an answer. You could also have done this with differential equations and limits. Or the NASA(x) function

[@mati74]

I don't know, but for some reason, I prefer the Excel approach. I actually made a quick approximation in my head before I started this video, and my conclusion was that it should be slightly less than 1.8. Not too bad. That was before I knew that there was no obvious way of solving this.

[@pentasquare]

WELL IF THERE IS NO EXACT ANSWER THEN x = 2 GIVES 6. CLOSE ENOUGH

[@WarmWeatherGuy]

A brute force method is to solve for one of the x's (leaving another x on the RHS) and then pick a value for x. Put that on the RHS and you get a new value for x on the LHS. Keep putting the new number into the RHS until it doesn't change. Often it diverges instead of converges so you have to solve for a different x. In this case x = 5 - 2^x diverges and x = ln(5 - x)/ln(2) converges to 1.715620733 after a bunch of iterations on my calculator. This method works for solving the Kepler equation for small eccentricities.

[@maxia2083]

You can also binary search the answer, as you know the value is between 1 and 2 and the function is strictly increasing.

[@flameb0w402]

So I spent 30 minutes trying to solve it on my own for nothing 😭

[@tvvt005]

9:09 at this step couldn’t we separate e^uln2 as e^u + e^ln2 . And e^ln2 is just 2 as we defined earlier right? So couldn’t we bring that over to the LHS and so it’s 16=ue^u?

[@Gideon_Judges6]

I learned about this in undergrad EE. I don't remember how it came up. Then in grad school Emag specifically, we also learned Hankel and Bessel functions of both the 1st and 2nd kinds.

[@jssamp4442]

That was a lot of hand waving at the end, before just entering it into Wolfram Alpha, which could have been done from the beginning without all of the intermediate work. What does that save us? You still didn't get a real answer, it's still a numerical approximation. How do you figure the result of the W function on (32 ln 2)?

[@dlevi67]

How do you figure out 'the result' of x = √3?

[@mydkarthikmecharena9010]

I put x=1.5, the result is 4.5, so I thought X is between 1.5 and 2. So I guessed 1.7 The answer was 1.7 lol

[@ansanch]

If X is 1.5, how is the result 4.5? it is more like 4.328...

[@LEDSlights]

I don't get how people can have a proble with the product log but not the natural log. They are created the same way. You just define a fucntion as the invers of an athoer one. The only difference is what function they are the inverse of and even then, it's almost the same. I honestly don't see why W(x) shouldn't be an elementary function. It's well defined with a specific domain and range, it's derivable and integratable and it has real world applications in physics. It's just as valid as the natural log.

[@dlevi67]

The definition of 'elementary function' can be seen as quite arbitrary, but there are good reasons for not considering W as 'elementary'. To start with, it's multi-branched even within its real co-domain. Other than that, I totally agree with you - it's purely a question of people reacting (with fairly strong emotions, sometimes) to the 'unknown'.

[@zam6323]

x = -3 is the answer, I didn’t even need to solve the math..😂

[@kenmore01]

Since you end up with an approximation after going through all that, why not be satisfied with your first approximation? Also, sitting doing an entrance exam (presumably without a calculator), you would be pretty hard pressed to come up with any answer. I came up with an approximation before even tuning in. It didn't go to more than one decimal place though.

[@dlevi67]

You are not ending with an approximation any more than you end up with an approximation solving x^2 = 3.

[@jdgower1]

I seriously doubt that equation is on a Harvard entrance exam. I know a ton of lawyers that can't even balance a checkbook - much less deal with natural logs and exponential functions.

[@dlevi67]

Harvard teaches mathematics and physics as well as law and business.

[@Nirakolov]

It's a little passed my bedtime right now, but why can't you just prolog that?

[@dlevi67]

Why not? You could python it as well. Or Fortran it. However, none of these will tell you that there is an analytical answer.

[@JannPoo]

Man, if in the end I need to resort using Wolfram Alpha I could have just done it from the beginning before the complicated transformations.

[@dlevi67]

How do you express the solution to x^2 = 3?

[@JannPoo]

@@dlevi67 If I'm using wolfram alpha I can just input that as it is and it will give me the solution. If I'm using a calculator I first need to change it into x = sqrt(3).

[@ir2001]

Perhaps a more neat form of the solution: log2(W(32*ln(2))/ln(2))

[@matthewedwards9423]

I've got it between 1.715 and 1.716 by brute force and I'm bored. Took me a minute or so. It's obviously between 1 and 2, I hoped at a glance it would be root 2 but I kinda expected that would be way too small, and anticipated a boring random answer above 1.5.

[@donm1547]

I'm sorry brother, but your answer was way BEYOND a 'Common Core' MATH solution , this one was extremely simple as 5 / 3

[@joeschmo622]

My lawyer is named Lambert W Function. He's wonderful.

[@thehoogard]

but what does Lambert W function do?

[@SOBIESKI_freedom]

It enables you to solve equations involving a mix of polynomials and exponentials.

[@DanielFSmith]

I defined the Daniel D function to be 2^x+x and took the inverse. The D function seems at least as useful as the W function (much moreso in binary), so Wolfram should know about it, right?

[@yurenchu]

Would the Daniel D function also be useful in finding/describing a solution to, for example, the equation x^x = 7 ?

[@michaelwisniewski6047]

@@yurenchuthat’s where the MW function comes in! The result is x=MW(7)

[@yurenchu]

@@michaelwisniewski6047 Wow, math becomes so much easier with the Wisniewski MW Function! They should definitely make this function available on every physical or online scientific calculator. :-) By the way, what's the value of MW(0) ?

[@toomanyhobbies2011]

I vaguely remember this kind of manipulation in undergrad mathematical physics. It's wonderful to use the result of a mathematician's hard work and inspiration, but it was little more to us than a sometimes useful technique. For our work, we did these problems with numerical techniques instead, given that computers were becoming more common. Most of use actually had our own PC-AT Clones at home! And Wolfram Research didn't even exist, but FORTRAN and C did.

[@dlevi67]

Well, to be honest, if a numerical solution is required, approximation is the only way here. Whether by dealing with the whole equation and using Newton-Raphson (or equivalent), or using series expansion to calculate approximate values for the analytical solution involving W and ln.

[@deleted-something]

Idk why but I feel like phi will be have something to do with it

[@Lycos_dae]

No explanation of the Lambert W function?! 😭

[@dlevi67]

He's given you an explanation and a definition in the first half of the video - it's the inverse function of x(e^x) (in fact, z(e^z), but never mind). There is no simpler explanation than this. For more, the Wikipedia article has a lot of info.

[@pedrogarcia8706]

so how does wolfram alpha know what the decimal approximation of W(32ln2)? that's always been the part of lambert w videos that frustrate me.

[@dlevi67]

It uses some series expansion to approximate it. Just like it uses some series expansion to approximate ln(2) - quite which one (multiple options exist for both), I honestly don't know. You can find expressions for some of these series on Wikipedia.

[@KaivalyaChess]

Sir you are the best, can you make a collab vid with higher mathematics, he's a great guy

[@michael.a.covington]

How do you solve it? Numerically. And I don't think this is a real Harvard "entrance exam."

[@dlevi67]

Well... it could well be part of an entrance exam. And you don't need to solve it numerically.

[@deerh2o]

Somehow I got into Harvard without having ever heard of the Lambert-W function. Go figure. Thanks, Presh, for the introduction. I'll do some more research into it. 🤓

[@fredashay]

Well, right away (2**1) + 1 = 3, and (2**2) + 2 = 6, so x has to be a fraction between 1 and 2. Knowing that, I would just use a binary search to solve it: i.e., try x=1.5, then either x=1.25 or x=1.75, and so on and so on...

[@nevorotem6819]

You should use ^ instead of ** because that's a more commonly used symbol

[@verkuilb]

No-it doesn’t have to be a fraction. It could be irrational.

[@fredashay]

@@nevorotem6819 You're right, but as a retired software developer, I got used to using ** because that's what most programming languages use.

[@nevorotem6819]

@@fredashay I know I also code but still most people won't understand

[@dlevi67]

@@verkuilb It doesn't need to be a rational number - a bisection algorithm will approximate the value with arbitrary precision no matter what the true value. Of course, it will always approximate it with a rational number, but that is the nature of approximation!

[@emiliogrv]

The entrance tests requiring you to solve problems that THEY should TEACH YOU how to solve, to me, is just nonsense. If they continue on that track, people will get to Harward not to learn something (if they already know, they have no need to learn from them), but just to say they've got there. And the more people figure it out, the more these college will see their self-mad bubble explode.

[@dlevi67]

I think there is a lot of context missing from the question. For example, if this is a question asked during an interview, there could be a very good discussion (not requiring the use of W) about different ways of getting to the solution, demonstrating the uniqueness of it, and finally "given that there is a function W equal to the inverse of x(e^x)", how would you get another approach to solving the equation. As to Harvard (and other 'top schools') being a bubble - there I agree, but it's not a bubble that is going to burst any time soon or easily/painlessly.

[@KING-zi7qq]

If the computer can find W(5) then we should also be able to do it right can you make a detailed video on this function?

[@dlevi67]

The Wikipedia article is pretty detailed. I don't think there is anything dramatically interesting in the series expansion of W(z).

[@nekogod]

That's pretty cool, I guessed it'd be near the golden ratio ((1+sqrt(5))/2) because of the square and the 5

[@ItzD4nte]

we could have use log to solve this.

[@markwallen6570]

I straight away calculated that x is equal to negative three; can someone please explain to this senior, why that is incorrect?

[@RGP_Maths]

Because that would be a solution to 2^x + x = -2.875. You are supposing the first x is 3 and the second is -3.

[@ashishm]

I still don't get how function W is solved? Can students refer to some W table like log table in Harward exam?

[@dlevi67]

Probably not - but it wouldn't matter (the tables exist, if you require them). The solution involving W is analytically precise, and there is no indication that a numerical value is needed. W is _defined_ to be the inverse function of x(e^x). Just like _ln_ is (or can be) defined to be the inverse function of e^x. It's just that while there are other 'simple' definitions for ln (e.g. the integral of 1/x), and its series expansion is 'simple', there are no equivalently simple other definitions for W, and its series expansion(s) are quite gnarly.

[@seasong7655]

So there is no answer? This has to be a joke. Screw the harvard professor who came up with this.

[@dlevi67]

Is there an answer to x^2 = 3?

[@orionspur]

2^x - x = 5 is a far more interesting and instructive case.

[@yurenchu]

2^x - x = 5 That equation has two real solutions (because the graphs of f(x) = 2^x and g(x) = x+5 have two intersections; one for a positive value of x , and one for a negative value of x). It is easily seen that x = 3 is a solution; the other real solution lies somewhere between -5 and -4.5 . 2^x = x+5 1 = (x+5) / 2^x 1/32 = (x+5) / 2^(x+5) 1/32 = (x+5) * 2^[-(x+5)] -1/32 = [-(x+5)] * 2^[-(x+5)] -1/32 = [-(x+5)] * (e^ln(2))^[-(x+5)] -1/32 = [-(x+5)] * e^[-(x+5)*ln(2)] -ln(2)/32 = [-(x+5)*ln(2)] * e^[-(x+5)*ln(2)] ... substitute u = -(x+5)*ln(2) ... u*(e^u) = -ln(2)/32 Note that -ln(2)/32 is between -1/e and 0 , hence the above equation has two real solutions (that are also both negative): u = W₀(-ln(2)/32) > -1 OR u = W₋₁(-ln(2)/32) < -1 where Wₖ(y) is the k'th branch of the Lambert W Function. Note that -ln(2)/32 = = -ln(2) / (2^5) = -ln(2) * 2^(-5) = -ln(2) * (2^3) * 2^(-8) = -8*ln(2) * 2^(-8) = -8*ln(2) * e^(-8*ln(2)) and hence W₋₁(-ln(2)/32) = = W₋₁( -8*ln(2) * e^(-8*ln(2)) ) = -8*ln(2) (which is a real number and less than -1 .) which leads to u = W₋₁(-ln(2)/32) u = -8*ln(2) -(x+5)*ln(2) = -8*ln(2) (x+5) = 8 x = 3 The other solution leads to u = W₀(-ln(2)/32) > -1 -(x+5)*ln(2) = W₀(-ln(2)/32) (x+5) = -W₀(-ln(2)/32) / ln(2) x = -5 - W₀(-ln(2)/32)/ln(2) which is a solution between -5 and -5+1/ln(2) . By the way, using the original equation, this solution can be rewritten as: 2^x = 5 + x 2^x = 5 + (-5 - W₀(-ln(2)/32)/ln(2)) 2^x = -W₀(-ln(2)/32)/ln(2) x = log₂( -W₀(-ln(2)/32)/ln(2) )

[@dlevi67]

@@yurenchu Thank you. How do you get the super/subscripts in YT comments?

[@dlevi67]

💯!

[@yurenchu]

​@@dlevi67 I usually prepare my posts in a "notepad"-like app, before copy&paste-ing it into the youtube writing field. Furthermore, I have a "notepad" file where I have saved oft-used symbols, such as subscripts, superscripts and other math symbols. (I simply collected them from the internet, such as Wikipedia pages, also by copy&paste.)

[@dlevi67]

@@yurenchu A-ha! Thank you - I seem to be missing some of those from the standard character map (which I use frequently!)

[@corone8960]

how would you even use the lambert-W function on the entrance exam; would you leave it in terms of W? i dont see how it'd be done on a calculator

[@dlevi67]

Why would you need a calculator to work out the solution to x^2 = 3? The solution in terms of W is "qualitatively" exactly the same as saying that the solution of the equation above is ±√3

[@donleyp]

I guess we’re not meant to know the formula for LambertW(). 😂

[@dlevi67]

It's not expressible as a finite formula of simpler (elementary) functions or operators. There are series that converge to W, but they are infinite, and Wolframalpha is using one (or more) of these to calculate the numerical value truncated to some level of precision. You can find some of these series expansions on Wikipedia.

[@ryqueezy6327]

0:45 in. I’ll try to solve it right now. 2^x + x = 5 2^1.75 + 1.75 ≈ 5.113585661 Rounded to the nearest hundredth, the answer would be x = 1.75. I will now watch the rest of the video.

[@sidharthsidhu928]

When I saw the thumbnail I straight away used excel goal seek to do the same... usually when I watch these type of videos I do this. Goal seek is a nice tool saves lot of time ...only issue is if the parameters are not changing the value can be used ...but if it's a variable then solution is important as that is necessary for formula or something

[@pavlopanasiuk7297]

Either it's youtube crap who suddenly decided to drop all the LambertW videos on my head, or it's a math creators united crisis. "Harvard entry exam" "90% of the people couldn't solve etc" blahblah introduce this nonelementary function and act as if this was really surprising. Not much content behind that.

[@eddiekorkis]

I don’t understand most of this. But I got close to the answer. I correctly got 1.7. Getting the “exact” number is amazing.

[@sumanjangid1250]

Oh! so the whole idea is to convert the equation from its implicit form to its explicit form in x and then use a so called function or simply a computer program to get it solved for us. Interesting. Btw Good job Presh to bring up this problem!

[@markphc99]

Er ,I see what it does , so what is W , and how is one supposed to calculated the answer without a computer?

[@dlevi67]

W is defined as the inverse function of x(e^x). Just as ln is (also) defined as the inverse function of e^x. No difference. How is one supposed to calculate ln(37) without a computer?

[@markphc99]

@@dlevi67 log tables , plus my calculator has a ln button. Further,I know what ln is terms of infinite series and exponentiation.

[@dlevi67]

@@markphc99 Your calculator _is_ a computer. And there are tables for the W function as well (calculated - suprise - by using one of its series expansion, just like Wolframalpha does). So, now you know (nearly) as much about the W function as you do about logarithms.

[@bhanuchhabra7634]

My first guess, 1.72 .... Not far off tbh, within 5 iterations I can be in 1%

[@Loifleiss]

i used newton raphson method to solve this equation but i got confused and i ended up with the wrong approximation bc i miscalculated f(1.7) and 𝑓'(1.7) thinking my guess of x0 = 1.7 was reasonable, i got to the approximation around x≈1.3, watched this video if my ans was correct I find its between 1.7 and 1.8, converging approximately x=1.7. btw f means function.

[@yanminglui7077]

how can someone even use wolfram alpha during an entrance exam LOL

[@dlevi67]

Nobody would expect you to, and you wouldn't need to. Would you need to access Wolframalpha if the question were to find x in e^x-5 = 6873?

[@chrvberg]

If I were the examiner I would fail you using some non-canonical function that does the trick. -- Unless you define it and prove it's fitness of purpose. Actually, I expected you using Newton's method or another honest approach.

[@dlevi67]

It's not "non canonical". It's non elementary, and usually not introduced to maths students before university.

[@moonlight-td8ed]

it should be 1.6 or 1.7... but my best guess if root 3.. which is 1.732

[@adamrussell658]

I got as far as 1

[@Macion-sm2ui]

I actually knew this function thanks to Black Pen Red Pen, but I didn't know that it is acutally usefull in some real problems. If so, then indeed it should be tought at universities, but under one conditon - that they will find better symbol for that. For me placing big W in the middle of the equation seems pretty awfull. From the other side letters ln also aren't mathematical symbol, but we aren't disgusted by using them.

[@dlevi67]

You aren't "disgusted" simply because you are used to seeing ln, but you are not used to seeing W - there is nothing intrinsically disgusting in either symbol. FWIW, Lambert and Euler originally called this function "omega" (ω), but there are already a lot of mathematical and physical concepts using ω.

[@Macion-sm2ui]

@@dlevi67 Yeah, that's exactly what I mean in the second part of my comment. But still W doesn't mean anything. ln is for (n)atural (l)ogarythm, but W is simply for W

[@GDyoutube2022]

Lambert was the kind of mathematician who was too embarrassed to admit he could not solve unsolvable equations. Hence he made up one to solve them 😂

[@clayton97330]

To be fair, it has an infinite series approximation the same as sin and cos and nobody calls those fake

[@GDyoutube2022]

Well periodicity does help a lot ;)

[@animelover6643]

2^x + x =5 2^x + x =4 +1 2^x + x =2^2 +1 Call it a day .😂😂😂

[@prakashchandrajoshi2164]

Do Lamber W function have a table? Like log table?

[@dlevi67]

Yep, they exist (or at least, I have them in my books from the late 1970s/early 1980s). Or you can copy or write your code based on series expansion, or you can use something like Wolframalpha.

[@filipsichrovsky]

Why does it use e in particular and not 2

[@dlevi67]

Because it keeps things simpler. It's a bit like "why do the trigonometric functions use radians and not degrees?"

[@engineboy_1449]

how can one be so humble :)

[@eduardocortes6340]

why in the definition of 2^u = e^(u ln2) at 8:54 it didnt just simplified to 2^u = 2e^u? so when subtituted in the main equation, then it could turn into a way simpler form of 16 = ue^u, thus without the need of v definition and get rid off the ln2?

[@yurenchu]

Because 2^u = e^(u*ln(2)) doesn't simplify to 2e^u . Just try random values for u , such as u = 0 or u = 1 . Clearly, 2^u does _not_ equal 2e^u .

[@eduardocortes6340]

​@@yurenchu ok, thats fair. but it doesnt algebraically explains why it cant be, since you could do e^(u ln2) = e^u • e^ln2 = e^u • 2. am i missing some important requirements to such simplification be valid? also, u already cant be 0 or 1 by definition (domain?) as a limitation from the equation, it would break if so at 32 = u2^u (this is also implied by x can't be 5 for the same reason)

[@eduardocortes6340]

thinking about it now, i guess that your contradiction isnt so valid since it may actually mean that, yes, u cant be 0 or 1, but as an incognito from the equation we are trying to find which value it may assume so the expression is valid, right?

[@yurenchu]

@@eduardocortes6340 You're mixing up e^[u * ln(2)] with e^[u + ln(2)] . e^[u + ln(2)] = e^[u] * e^[ln(2)] = e^u * 2 = 2(e^u) vs. e^[u * ln(2)] = (e^u)^[ln(2)] = (e^[ln(2)])^u = 2^u Also note that 2^u = e^[u*ln(2)] is an _identity_ , not an equation. An identity is _always_ true, hence an identity holds for any value of u (such as, for example, for u = 0 and u = 1 ; inserting random values of u is a way to check if a statement is an identity). An equation isn't always true, it's only true as a condition for the problem under consideration, which means a particular value of u must be derived for which the equation holds true. Other examples: (cos(x))² + (sin(x))² = 1 is an identity, not an equation. If you try to solve for x, you'll find that any value of x will do; because the statement is true for any value of variable x . x² = 14x - 40 is an equation. It's not generally true (for example, if x = 3 then LHS = 9 and RHS = 2 , which isn't a match, so the statement is not true for x=3), but the assignment here is to solve for the particular values of x for which the statement becomes true. Those particular values are x = 4 and x = 10 , hence x=4 and x=10 are solutions to this problem. In the video, Presh wants to solve the equation 32 = u*(2^u) One step towards a solution is by rewriting (part of) the RHS by using the identity 2^u = e^[u*ln(2)] ; because that way, the equation becomes 32 = u*e^[u*ln(2)] (which is a statement that is _equivalent_ to the previous equation 32 = u*(2^u) ) and so he's a step closer to an equation of the form c = v*(e^v) , which he can eventually solve by using the Lambert W Function. I hope that helps.

[@yurenchu]

@@eduardocortes6340 You're mixing up e^[u * ln(2)] with e^[u + ln(2)] . e^[u + ln(2)] = e^[u] * e^[ln(2)] = (e^u) * 2 = 2(e^u) vs. e^[u * ln(2)] = (e^u)^[ln(2)] = (e^[ln(2)])^u = 2^u Yes, we are trying to find the values of u for which the equation 32 = u * e^[u*ln(2)] [Equation 1] is valid. That means that we can't just substitute e^[u*ln(2)] in the righthandside with 2(e^u) , because the statement e^[u*ln(2)] = 2(e^u) is not generally true, it's only true for a particular value of u, and that particular value is probably _not_ the value(s) of u for which Equation 1 is true. Therefore, the substitution would be illegitimate. Likewise, you wouldn't solve the equation 13 = 3 + u + u by replacing u+u in the righthandside with (let's say) u^3 , right? No, you would replace u+u with 2u , because u+u = 2u is generally true (and hence useful here), while u+u = u^3 is not generally true (and hence illogical here).

[@Jack-h6m3p]

My only answers in my brain: 2² × 1 = 5

[@AcaciaAvenue]

One thing I truly don't like about the Lambert function is tthat, as far as I know (never studied it, I admit), there is no easy reliable way to figure out its numerical value. What use do I have of an operator that I need to use Wolphram alpha or similar to calculate anyway? I see barely any difference between using Lambert function to find the x or leave the implicit equation as it is.

[@yurenchu]

Your complaint about the Lambert W Function would also apply to (for example) the sine, cosine and tangent function; or to their inverses (arcsine, arccosine, arctangent), or to just the natural logarithm function. Would you say those functions/operators are useless too?

[@dlevi67]

Would you complain similarly about the equation x^2 = 73?

[@AcaciaAvenue]

@@yurenchu At least sine and cosine have certain set values, there are sheets, exercises usually end up having those values. You can do trials with the logarythms.

[@AcaciaAvenue]

@@dlevi67 at least there's a way to figure out the value by trials. You start with the highest perfect square lower than 73, then add decimals and progress that way. How do you even do trials with the Lambert function?

[@dlevi67]

@@AcaciaAvenue For x>0, W is a continuous, increasing, single-branched function, just like the square root, so you can interpolate exactly the same way (unlike - say - with sin(x)). For example, using the equation in the video, you can immediately "by inspection" see that the value of x is between 1 and 2, and with very little more effort, that it is somewhere between 1.5 and 2. It's just that you are familiar with the square root, so you can instantly recall some of the values - and maybe even know the Babylonian algorithm to calculate it by hand. As far as I know, there is no 'easy' algorithm to approximate values of W mentally or in writing, but that doesn't make it 'special' - it is perfectly well approximated using series expansion, just like the square root. Before scientific calculators were cheap and common (50+ years ago), we used tables to calculate trigonometric and logarithmic/exponential functions. Now, if you need W values, you either write/copy some code or use tables - or use Wolframalpha.

[@clayton97330]

I love Prime Newtons!