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.

[@TeKnOShEeP]

"The exact form of B(x) is an exercise left to the grader."

[@pk2712]

@@empathogen75 As I think you are saying it is much less work to just use newton's method on the original equation . I am not really impressed with this Lambert W jazz .

[@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)

[@eiyukabe]

I thought the same!

[@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.

[@Halfrida]

Ngl when I saw the question I start by guess it’s 2 and start using the calculator to make the number more specific by adding digits and actually got like 1.7158 sth lol within probably a minute

[@brendanward2991]

It's called the W function because in the end you need to use Wolfram-Alpha to solve the equation.

[@bjornfeuerbacher5514]

Not at all. You need Wolfram Alpha to get an _approximation_ to the solution. The equation itself already was solved in the video before Wolfram Alpha was used. Why do sooooo many people think that an equation is only solved after one gets a numerical value and don't care at all that this numerical value usually is _not_ really the solution, but only an approximation to the solution? That is _not_ what "solution to an equation" actually means!

[@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

[@dlevi67]

@@meateaw Small correction - W(x) is the inverse of x(e^x)

[@atscxyw61qupim7]

how do we input W function on a scientific calculator?

[@asparkdeity8717]

And to those complaining, we got a near identical question in our Cambridge maths entrance exam, the very paper I sat had a question with the lambert-W function. Don’t believe me, look up STEP II 2021 Q4. Not something I had ever learnt in school or heard of at the time, but given its introduction I was still able to do the question. It’s not about solving the question for an exact answer using a calculator, but it’s about understanding and applying new techniques to gain an analytic closed form solution to an unseen problem. It actually tests your true mathematical ability.

[@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!

[@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?

[@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!_

[@rickdesper]

I got a PhD in math without ever hearing about it It's not terribly important. But now that it's built-in to mathematical software a bunch of people think it's fair game for math puzzles. But really, there are countless functions that have inverses that we cannot put in closed form. How interesting is this particular one? I guess it depends on how often you want the inverse of a specific function. It's nice that Woflram-Alpha apparently has decided to hard-code this, but for the most part we don't want to work with functions that are not in a closed form of combinations of simple computations. Existential proofs that certain functions have inverses aren't very interesting, in general. There are infinitely many (uncountably many!) 1-1 functions and they're all invertible. I don't see what the appeal is here.

[@dlevi67]

@@rickdesper It has significant amount of use in physics, chemistry and biosciences.

[@bjornfeuerbacher5514]

I got a PhD in physics without ever hearing about it. Only in the last about 5 years, I keep seeing KZbin videos about it... :D But as others already have mentioned: It apparently has lots of applications in physics.

[@cguy96]

I think people are missing the fact that the Lambert W function is not just some arbitrary inverse, otherwise Presh could have just said P(2^x+x) = 5 and stopped there. The Lambert W function has been extensively researched, has a lot of properties, and identities, and is quite useful. This is why Presh went to the trouble to reformulate the problem into the product-log form.

[@Tim3.14]

Yeah, plus there are math programs (like Wolfram Alpha / Mathematica) that have a predefined W function for you to use.

[@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.

[@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

[@Qermaq]

Presh, this was a really well-paced and thorough explanation of the Lambert W function. Great job! Would you do a sequel looking at the sort of calculus needed to derive the approximate value?

[@TitanOfClash]

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

[@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.

[@Rhesa-jc3on]

@@danmerget EXCELLENT!! That is just 1 of the many reasons that I love math - that there's more than just 1 way!!

[@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?

[@rickdesper]

ln() is considered a function in closed form. W() is not. ln x has been computed with a hand-held calculator for a very long time. W() is not easily computable. The Taylor series for ln x is easily written with coefficients in a closed form. The same is not true for W().

[@bjornfeuerbacher5514]

@@rickdesper "ln() is considered a function in closed form" What is that supposed to mean? I never heard about a "function in closed form". "The Taylor series for ln x is easily written with coefficients in a closed form. The same is not true for W()." W has a rather simple Taylor series, what are you talking about?!?

[@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)

[@bjornfeuerbacher5514]

That does not give the actual solution, but only an approximation to the solution.

[@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!

[@asparkdeity8717]

Another name for it is the “Product Log function”

[@octobermathematics]

Thankyou Presh for explaining it so nicely.

[@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.

[@ahojg]

The Qs and tasks are not the hardest ones, but I like the way you treat them when providing other related info, context, connections.

[@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.)

[@sonicbreaker00]

first try x=1 and x=2 to see that the solution must be closer to 2 than 1. now assume x = 2 - y and use 1st-order Taylor for exponential [note: 2^x = exp(x.ln2)]. then you get a linear equation in y with solution y = 1/(1+4.ln2) = 1/(1+4*0.69) = 0.265 (surely every one remembers ln(2)=0.69 ... think about half-life of exponential decays like in radioactivity). this then gives x = 2 - y = 1.735 without use of any special functions or a calculator ... all paper and pencil. and x = 1.735 is pretty close to the actual answer of 1.7156.

[@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

[@dlevi67]

@@lucabastianello9830 Actually, the two expressions are NOT equivalent. The second one is an expression representing a lower bound for W in the original solution. It is an operator - or better, a multi-branched function. Neither more nor less so than the 'normal' logarithm.

[@leo-um3pj]

seen too many bprp videos and i immediately knew that lambert w function would be the key to solving

[@Musterkartoffel]

Same lmao. Quite suprised he showed prime newtons videos instead of his, even tho both are very good

[@paulkolodner2445]

I always like to solve these problems iteratively. In this case, start with a guess for x0 between 0 and 5. The next iteration x1 comes from setting 2^x1 +x0 = 5 or x1 = (ln(5-x0))/ln(2). The next iteration gives x2 = (ln(5-x1))/ln(2), and so on. This converges rapidly. If you try it the other way, ie, 2^x0 +x1 = 5 or x1 = 5 - 2^x0, it doesn't converge.

[@dandeliondesign6155]

An economy university student once taught me to make a graphic instead of trying to solve it mathematical. And you can also try out numbers with decimals to get a good rounded result. It took me 4 attempts to get to 1.8. I tried 1 -> 3, 2 -> 6, 1.5 -> 3.75, 1.8 -> 5.04. 1 was too small and 2 too big. Looking at the results it must between 1.5 and 2, but closer to 2. Hence 1.8 was chosen as the next input. If I continue for more decimals then 1.79 -> 4.9941, 1.791 -> 4.998681, etc. The time to do this with a calculator beats the mathematical solve, which you also have to round up or down.

[@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...

[@brucemapaya0000]

By using trial and error one can show the x lies between 2 and 1...and by choosing the mid section of this range, such that x=3/2....we find that the answer is much closer to 5....so the the range is between (3/2 , 2) By minimizing the range : (3/2 + 1/5 , 2 - 1/5)... ,one can get an approximate answer

[@bjornfeuerbacher5514]

As he explained in the video (2:45 to 2:55), there are cases in which you want to have an exact solution, not only an approximate one.

[@johnpollard9828]

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

[@donnybrooklads]

Luckily for me as a terrible math student AI solved it for me in 1.25 seconds.

[@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

[@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!

[@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.

[@yonatan2009]

One thing I realized in math is that devision comes befoure multiplication and why is no one talking about it.

[@radscorpion8]

I'm glad I stopped trying to solve it after a while on my own. All I know are basic log rules. At some point you just realize there are only so many ways to rewrite the equation and you need some help :P. I have never heard of the Lambert W function before, but it sure was interesting to learn about it, especially with copilot's help. So I assume the lambert W function is in our calculators somewhere? It better be or I have no idea how the harvard students are doing this exam! I assume its all still paper and pencil

[@varathan3558]

The best I could aproximately think is that: easilly we see that 1.5

[@aram5642]

I feel so greatly indifferent and unchanged after this.

[@locheyoutube5252]

No wonder this is new to me. I was already out of school before this was even being taught! 😮

[@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.

[@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?"

[@asparkdeity8717]

We literally got a question like this in our STEP exam for Cambridge maths, despite having never learnt it in school. It’s about how you well and quickly you are able to understand and apply totally new concepts

[@renatofernandez175]

"You just need to solve this equation"

[@brianmcguinness9642]

Now I just need to find an algorithm for calculating W.

[@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 !

[@cguy96]

I am ecstatic that I even remembered there WAS a Lambert W function

[@carly09et]

graph 2^x & 5-x the intersection is the solution! Da

[@DrigoToes]

So, I'm a PhD in maths and I've never ever heard about this W thing in my life!

[@artzyalicia5546]

Early squad! Love your videos, man!

[@asparkdeity8717]

Nice simple rearrangement problem, in fact several KZbin videos cover the general case of a^x + bx = c 2^x = 5-x ==> 1 = 2^(-x) (5-x) ==> 32 = (5-x) 2^(5-x) ==> 32 = (5-x) exp(ln(2^(5-x)) ==> 32ln(2) = ln(2)(5-x) exp((5-x)ln(2)) ==> W(32ln(2)) = ln(2) (5-x) ==> x = 5 - W(32ln(2)) / ln(2)

[@CasualTS]

I don't have an exact solution, but with just a basic calculator and guess and test methodology I got to the approximation of x=1.715 in about 2 minutes.

[@rahulgoel3120]

If we are allowed a calculator, newton rhapson method of approximation works great for these kind of problems

[@HemantPandey123]

Try graphs. y= 2^x and y = 5-x.

[@StevenMRSenior]

Admittedly I never did the lambert W function at school or university but on investigation it has been around for 250 years.

[@prestonhensinger598]

Ive lost my trust in youtubers. I’d love to say i learned something but now that u didn’t explain the inner workings of the W function I’m going to need to watch another video to learn about that. Thanks

[@dlevi67]

There are no more "inner workings" to it than the definition, which Presh has spent the first half of the video in explaining. What "inner workings" are there to the square root of something?

[@barackfrans6777]

Prime Newton and mind yr decision u guys makes my dad I'm always excited for yr videos God u guys❤

[@ir2001]

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

[@HighMojo]

Now if I could only find the LambertW function key on my calculator.

[@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. 🤓

[@benjamingross3384]

I used Newton's method by hand one time and got 1.794 but there may have been arithmetic errors. No one said anything about an exact solution so I'm satisfied.

[@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.

[@dlevi67]

Probably something to do with waveguide design?

[@TheChamp1971]

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

[@rickdesper]

There is no "Harvard University Entrance Exam." My answer would be "it's a number between 1 and 2. We can use a computer to approximate it." My other answers would be "why do you want this?" and "You know there's no closed form for the inverse to 2^x +x, right?"

[@bjornfeuerbacher5514]

But the video shows that there _is_ a closed form - if one uses the W function. Or what exactly do you _mean_ with a "closed form"?

[@kytnCars]

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

[@SJrad]

idk as a student Ive never liked functions such as the lambert w function or logarithms because they feel too black box-y. i don’t really understand the underlying process of how it works, just that it does the thing i want to do

[@mauriziograndi1750]

Ln10/2=1.151292 1.151292x1/2=0.575646 1.151292+0.575646= 1.726938 2^1.726938=3.310245 3.310245+1.726938= 5.0 X = 1.726938

[@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.

[@sjeremy]

hehe, I did it but I havent simplify to 32 from e^5ln(2)

[@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!

[@dlevi67]

It's a standard non-elementary function with 250+ years of study behind it. Not a "so called function" that someone has made up just now.

[@sumanjangid1250]

@dlevi67 I'm sorry dear.. By "so called", I didn't mean to detract such a useful function. With little knowledge about the function on my part ,I just said this...but l really like your sincere approbation towards Mathematics.😊

[@dlevi67]

@@sumanjangid1250 No need to apologise - it was just meant as a clarification that Lambert's W isn't something made up by Presh on the hoof (which seems to be a rather pervasive idea in the comments). Sorry if it came across as too abrupt!

[@valentinoromitti6005]

Are really people asking why you can create "magical functions" like that? Could you know the answer to ln(78.4) without a calculator? The W function is a well-known function that can be approximated just like sin, cos, or ln. If you think of it like a magical function it's just because you haven't learned it in school.

[@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

[@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.

[@Kounomura]

A joke comes to mind about the Lambert W function..... Someone ran over a horse at the corner of a street with an unpronounceable and hard-to-describe name. Two policemen are dispatched to find out what happened. The two police officers discuss for some time how to write down or announce the name of the street over the phone. One of them suddenly comes up with a brilliant idea: "Hey Jack, let's pull the horse over to Forget-me-not Street around the corner."

[@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

[@ForcefieldDown]

So if this function isn't commonly taught in schools, why is Harvard expecting an answer involving it? Or would they accept an approximation of the result arrived at through different means?

[@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.

[@somersetcace1]

The mathematical equivalency of unraveling a Rubik's cube.

[@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.

[@dlevi67]

@@TheJaguar1983 Yes, that's exactly what it is - more than complicated, I'd say awkward, though, at least for x>0. Unlike sin or log that have 'nice' series expansions, where a pattern is apparent (x/1 -x^2/2 + x^3/3 ...; x/1 -x^3/3! + x^5/5! ...), W's expansion is quite gnarly. The first 5 terms read: x - x² +3x³/2 - 8x⁴/3 + 125x⁵/24 - ... Then there is the "complication" that W (like the complex logarithm) is multi-branched, and the above expansion only holds for values near 0... so perhaps complicated _is_ the right term! 😁 EDIT - screwed up copying the expansion formula - sorry.

[@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?

[@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

[@dlevi67]

Fair point. I don't think this question was ever put as a "written answer, get it right or wrong" type of question to a high school student trying to get into Harvard (or any other university). It was either used for post-grad admission, or as a 'discussion' question during an interview.

[@madcow3417]

I now remember why I quit math after differential equations.

[@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

[@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.

[@asparkdeity8717]

Yeah but you don’t get a calculator in your Harvard Interview

[@ikocheratcr]

So, I am not alone of not knowing about Wlambert function in math class.

[@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

[@dlevi67]

So the answer of e^x = 3 x = ln(3) is not an answer because it rewrites the equation in terms of other predefined functions?

[@looldrole]

Is that not a circular argument? Aka, as we cannot solve numerical, let’s create w(x) and then solve numerical? Or is it that w(x) is actually a closed form expression?

[@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.

[@djdoc06]

Can interpolate pretty fast, graphically.

[@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.

[@TheInterestingInformer]

How do you interpret the lambert W function once you get the answer though?

[@divijsharma5610]

Binary search in to get approximated answer as answer lies between 1 and 2 and function is increasing.

[@QuovatisPS]

I left the video disappointed that W wasn't really explained. How does a computer evaluate it?

[@KaivalyaChess]

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

[@exoplanet11]

Hey...solve the problem w/ Desmos. Great idea! In fact, that's the idea I suggested in the comments of the original video! Thanks @MindYourDecisions !

[@patrickfeuillet6878]

Most trancendental equations such as this one do not have a clean analytical solution.

[@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.

[@user255]

If we would input that to Wolfram Alpha, then we would have done it in "2^x + x = 5" phase...

[@dlevi67]

That depends on what you are asking Wolframalpha to do - if you are saying "this is too difficult for me, I need help in solving the whole equation", then fair enough. But if you solve the equation in terms of W, and then want to have a numerical approximation for that, Wolframalpha is not a bad choice.

[@user255]

@@dlevi67 Yeah, I agree. I was only half serious, but I felt disappointed that there weren't some trick to do the approximation by hand.

[@dlevi67]

​@@user255 It's quite hard. The coefficients of the series expansion around zero (which are nice and easy in say trigonometric functions or exp/log) are gnarly, and the radius of convergence is small.

[@AbhijeetSharma-w6e]

√3

[@dlevi67]

Close, but no cigar.