Bprp, I have one more question. You know how we solved this problem of maximizing x^y w/ x+y=8 using derivatives? Suppose we restricted the domain of x,y to be natural numbers again. Could we then use discrete calculus to solve this problem. That is forward difference/backward differences are now analogies to the first derivative?
"I really love this question, let's... I really love this question, let's have a look. I really love this question, let's have a look." - blackpenredpen
While most of this went completely over my head, it was a fun watch, and I am quite proud of myself that my first thoughts were "Are they positive numbers, and are they whole numbers?"
The 2 most important parts of the video is recognizing that you have to take the derivative to find the min and max and then realizing that u can use the Lambert W function, everything else is following up on those 2 things
found this channel last night. i used to "teach"/tutor math in college and have always been passionate about teaching despite not being very good at it, so I love watching you explain intermediate math in very straightforward ways. but, i personally tapped out of math education in late calculus and linear algebra, and to see a video on more advanced subjects that I'M not even familiar with like the Lambert W... it made me feel very warm and cozy. This is a wonderful channel and I'm delighted to have found it.
Using the Lagrange multiplier will give a system of equations identical to solving using the methods of single-variable calculus, ie the solution x+xln(x)=8.
for a lot of people the concept of powers might not be that clear or straightforward. Younger students who may have an interest comes to mind. Of course however, most high-schoolers or college level students (who take math courses), will see it very clearly.@@xlr8_bs514
(even -ωₙ)^(even ωₙ+8) where n is arbitrary Because a negative when exponentiated to a positive even is positive, this produces a value greater than ωₙ₂ Where n⇐∞, the maximum approaches Ω.
I mean, if you had it so X could be negative, the max is theoretically infinity, but that min is theoretically negative infinity, which makes the question irrelevant. Knowing that is actually why I clicked on the video in the first place.
@@TheSkullkid16 Exactly. You end up with an equation of -(n-8)+n=8 and then have (-(n-8))^n. The limit would basically be negative infinity to the power of infinity plus 8.
@@FourOneNineOneFourOne I did learn some of it in high school and even at university, but I graduated from university 5 years ago, my maths class had finished 2-3 years prior, and I have been working in a completely unrelated area. And in my life the only areas of maths I use are arithmetics and trigonometry 😂 so, the other things get forgotten easily. I think we didn't reach this difficulty of calculus at high school, or, if we did, it's gone from my memory, because I understood about half of the video at most. I don't remember any of the functions, sadly.
@@FourOneNineOneFourOne At least in my case, Calc wasn't even offered unless you qualified to do College Courses while still in Highschool. I got through Calc 1 just fine, but most of what he was showing here is not stuff we really covered. This is the first time I ever heard of W(xe^x)=x, and I would have literally no idea how to parse something like that in a casual way. While I have worked with Log/ln/e a bit back then, it's been over a decade, and I was never fond of dealing with those at all, so I just lost it completely.
@@FourOneNineOneFourOne What if one has no aptitude for math? As a high school student, or today more than 30 years later, I could no more understand what he is explaining then I could run 400m in 48 seconds (I never broke the 50s barrier 50.7, and only passed algebra as a senior.) The ability to do high school math, beyond basic algebra, is a mental gift, and is no different than the ability to any number of other things. And no amount of hard work in the world will overcome a lack of ability. If you can do this stuff, congrats to you, but no one “should” be able to do anything you can do.
@@JonesyTerp1 "no amount of hard work in the world will overcome a lack of ability". I'm not sure I agree, particularly for all the activities you mentioned, but, putting that aside... The problem is, even if you have the "innate ability", for most (many?) things, you still have to do the "hard work" before you are proficient enough to be convinced you have the ability. So, to find out if you have the ability to do "a thing", you have to take a leap of faith and put in the hard work to find out. One would hope you enjoy "the thing" so if it turns out you can't do it, at least you had fun trying. It comes down to the point, does the hard work "develop" the aptitude, or does it "expose" it? I think there are some people in each group, along with a group who work hard and still can't, and another group who never seriously try.
@@TomJones-tx7pb but what I stopped with is an implicit function. Explicit solution presents the way to directly calculate the value for a given argument rather than to bruteforce it
Same for me. I also solved the problem before watching the video and got stuck with x ln(x) + x = 8 but I just type it into Wolframalpha and see that it has no "nice solution". In my opinion, there is no benefit in expressing this equation with the W function. The equation x ln(x) + x = 8 can already easily be solved by a computer and for a human the W function does not give you any additional information
@@adammizaushev Good point for classical math, if you have a pre-calculated W function, but modern way of solving is using a computer to get as accurate an answer as you desire with numerical analysis.
@@TomJones-tx7pb Yeah, I was only speaking in the matter of getting intuitive, natural understanding of a function to a human, which is got by its explicit expression, rather than implicit one
@@abrahamholleran4162 so like, you get that my skepticism came exclusively from the idea that e^{W(100e)-1} is a natural number right? Like, sqrt(2) is "close" to 1.4, but the fact that it's irrational is kind of a really important thing about it
@@frimi8593 They weren't replying to you, but rather to the original comment. I'm sure they would assume that you were doubting it being an integer. (It can be difficult to tell what people mean, since some use the word "equals" too loosely. You could say something like "exactly 24" or just explicitly doubt it being an integer.)
Yes, x=0 is a local minimum because to the left the value of P is unbounded. That means that I think the result we found is not a global max but a local max. Wolfram is mistaken 😮 because it calls it a global max. Or have we told Wolfram that x>=0 ?😊
Wolfram alpha probably ignores negatives entirely here because any noninteger negative x can't be raised to the power 8-x when restricted to the reals.
As soon as I saw the question I noticed there's no upper limit because we can have absurdly large negative numbers for X (or Y). Once he put the constraint in that they had to be positive values, I was sure the answer was x=y=4. I just assumed the answer was going to be integers. Don't know why I fell into that trap, it just felt intuitive.
@@KenFullman Exactly, I was thinking the same thing. Because, you can have the equation as such: y=n, and x= -(n-8) For example, you could have y=1,000,008. x would then equal -1,000,000 and the answer would still equal 8, and you'd be left with -(1,000,000)^1,000,008 which would be absolutely huge, and can still get bigger, into infinity. Because you'd have -(n-8)+n=8 and then (-(n-8))^n. As long as you can have y be 8 more than a negative value of x, you can get infinitely higher values for x and y.
@@KenFullman If the equation had been x*y instead of x^y, then that intuition would be correct, and I frequently encounter situations where it's useful to optimize x*y by setting x=y (maybe you do too). So maybe that explains how you fell into that trap. :)
It's always about local extremum, cause blackpenredpen used Fermat's theorem that states that *local extrema* in the interior of a domain must occur at points where the derivative equals zero or undefined.
honestly, I understood everything you said (despite this being the first time I've heard about lambert w function) and this made me so happy (also, the fish you draw are amazing)
hey @blackpenredpen, can you make a shirt that shows a white and green pencil and says "I really hate this question, let's not take a look ~whitepencilgreenpencil"? pweeeasee???
Exceptional explanation of the unique "W" Function! However, I still need to exponentially repeat your Steps to the Solution (or repeat the Steps to the maximum number allowed) --- in keeping with the ultimate objective of this proposal (computing the maximum)! Alas, I am exhausted from this exhilarating mathematical exercise. Bravo! Thank you, sir blackpenredpen!
Maximizing x^y with constraints like x+y=8 was one of those math puzzles that used to leave me completely stumped, especially with concepts like the Lambert W function! What made a big difference for me was using tools that provide step-by-step explanations. SolutionInn’s AI study tool is fantastic for breaking down complex problems like this into manageable steps. If anyone’s feeling overwhelmed, tools like that can really help clarify things.
Hmm this video demonstrates the difference in thinking between an engineer and a mathematician. I paused the video when the question was posed, opened excel, used one cell as an input for X, defined another cell as Y by making it 8 - the X cell, then defined a 3rd cell as the X cell to the power of the Y cell. At this point I used the solver function in excel to maximise the value of the 3rd cell by altering the value of the X input cell. Got the answer in less than a minute by brute forcing it 😅. I will admit blackpenredpen's method was more eloquent.
I'm right there with you. I might have done a column for x, another for 8-x and one more for the function. Range 3-5 with 0.05 increments, maybe refine around the maximum and called it good at 3.51.
f(x) = x^(8-x) ln(f(x))' = f'(x)/f(x) the max is at f'(x) = 0, and f(x) can't be infinity so the max is at ln(f(x))' = - ln(x) + (8 - x)/x = 0 Ask a computer for an approximation
We just covered that a couple of months ago. Wouldn’t it be easier to just substitute y = kx and get kx/x = k => limit depends on the slope k, therefore it doesn’t exist?
How would you solve Lambert W by hand? This seems extremely hand wavey by just invoking an inverse of what feels arbitrary, and then asking wolfram alpha to actually solve it. How would this be any different than just asking wolfram to do it from the beginning? Does W have an alternative definition that allows this to actually be calculated reasonably?
Basically it's xe^x = 8e where x = lnx + 1. Solving xe^x isnt trivial, but solving lnx + 1 is. So yes Basically a calculator that could solve xe^x = number would be able to solve the original equation, you need like a TI89 to do it
This is precisely why it doesn't really get covered in any mathematics education. It's not particularly useful because it basically requires you to ask the computer for the numerical approximation, which it could have done without going through the manipulation of the equation. There are applications of it in various science fields, but not really understanding those fields, I don't really know what's gained from putting the answer in terms of the Lambert W instead of whatever form it was in before. Special functions are defined for all sorts of problems that lack answers in terms of elementary functions; for something as simply defined as the Lambert W, it shouldn't be too surprising that it pops up in the answer of all sorts of problems in science that lack closed form elementary solutions.
Another Chatgpt response when I asked it to solve using calculus: Certainly! To solve this problem using calculus, we first express \(y\) in terms of \(x\) using the constraint \(x + y = 8\), which gives us \(y = 8 - x\). Our function to maximize is therefore \(f(x) = x^{8 - x}\). To find the maximum value, we need to: 1. Take the derivative of \(f(x)\) with respect to \(x\). 2. Set this derivative equal to zero and solve for \(x\) to find critical points. 3. Check these critical points to determine which one gives the maximum value of \(f(x)\). The derivative of \(f(x)\) can be challenging because it involves both the power rule and the chain rule. Let's calculate it. from sympy import symbols, diff, solve # Define the symbol x x = symbols('x') # Define the function f(x) = x^(8 - x) f = x ** (8 - x) # Take the derivative of f with respect to x f_prime = diff(f, x) # Find the critical points by solving f'(x) = 0 critical_points = solve(f_prime, x) # Show the derivative and critical points f_prime, critical_points The critical point for maximizing the function \(f(x) = x^{8 - x}\) is \(x = \exp(-1 + \text{LambertW}(8e))\), and the maximum value of \(f(x)\) at this point is approximately 280.905. This result confirms our earlier graphical analysis, demonstrating that calculus provides a precise way to determine the maximum value of \(x^y\) given the constraints \(x, y \geq 0\) and \(x + y = 8\).
I got pretty far into the question before I eventually got stuck... Of course the Lambert W function was involved lmao, I still dont know how to properly use that thing
If you ask GPT-4 instead of GPT-3.5, you get the correct answer. ChatGPT will now write python programs to calculate precise results for questions like these.
chat gpt just take info from the internet/database and provide an interpretation. if the database doesnt have the answer, chatgpt wont have it. it doesn't "think" on its own. edit: actually it DOES learn, for it's an AI, but it specializes in predicting what the response to your question is base on the database of information it's accessing. But i dont think it can do math.
@@kingpetunfortunately chatgpt doesn't have the capabilities to do calculus but it might be able to do basic arithmetic, they could integrate something like Wolfram alpha into chatgpt one day
@@maddenbanh8033chatgpt has emergent problem solving as a result of understanding human language. Gpt-4 is able to use that problem solving to load arithmetic into premade calculators, like programming languages or Wolfram, in order to “solve” calculus and other math problems. It figures out what steps are necessary, and loads those steps into something that can actually get a confident result. Then it takes that result and moves on to the next step
Tried solving using desmos. Plotted x+y=8 first. Then plotted x^y = some constant b. The value of b where the two graphs only have one point of intersection is the maximum value of the function, and that came out to be approximately 280.902799
As others have pointed out, you could use the same method to get x=k/W(k*e) for the max of x^y when x+y=k. The case of k=8 is interesting though, because 4^4 and 3^5 are so close together. We were already pretty close with 3.5^4.5.
This can also be solved using binary search import math low = 0 high = 8 delta = 1e-6 while abs(high-low) > delta: mid = (low+high)/2 first = math.pow(mid-delta, 8-mid+delta) second = math.pow(mid, 8-mid) third = math.pow(mid+delta, 8-mid-delta) if firstthird: high = mid else: break print(math.pow(low, 8-low))
Here wrote a little python script don’t know if this will help max_num = 0; x = 1 y = 7 while (int(x+y) == 8): print(x+y) if (max_num < x**y): max_num = x**y; x+=0.0001 y-=0.0001 print(x+y) print(max_num)
It may be because I never studied it so I don't know much about it, but I don't really like the Lambert function. Since there's no way to determine its value without using wolpharm alpha or a software, it's not really explicitating or finding the value of X. I feel it's just writing an implicit equation in a different way that's still an implicit equation. It's like saying "there's no solution but here's one that isn't really one".
That's an inherent problem with inverses. Think of it like roots and logs where there's no other way to express those numbers. For example, √2 and ln(2) have no nice ways of expressing them, and you generally need a calculator to compute their values. Not much you can do about it when you have an irrational answer.
@@sethb124 but at least with square roots you can have a general idea of their value, you can at least figure out a min/max borders, like, you know that square root of 2 is somewhere between 1 and 2, you can, like, try out 1,1 * 1,1 then 1,2 * 1,2 and so on and figure out a few decimals this way. None of that with the Lambert function. Either you leave it at that or use a software. At least as far as I know, there's no way to even figure out its value, better off go back to the implicit expression and try out values there.
@AcaciaAvenue Yeah, that's true. It's not super easy to estimate, but for problems like these, it's the most "exact" answer you'll get. It's not very intuitive, but I'd argue neither are lnx and e^x. It's still a valid answer because there are ways of estimating it (or else computers couldn't do it), it's just not an intuitive function.
@@sethb124That's the point I wanted to make. It's not an "exact" answer to me if I have no means of giving even an estimate numerical value. It's just picking something with no answer and writing it differently, but still with no answer. If there were some means to calculate an estimate value of the lambert function without feeding it to a software and tell it "do it for me", then I will change my mind but I looked it up but there seems to be none. Therefore I don't see the utility of the Lambert function.
Totally agree. Just seems like a fix to me. And calling it exact is a stretch too. I solved this by implicit differentiation and Newton-Raphson. Didn't take long at all.
Yeah. I majored in math and in four years of college, I never met Lambert. In some ways, I have always felt that Va. Tech let me down. (This was early 70's.)
Yeah, it's rare to see it in math curriculum because it doesn't really have deeper theoretical value AFAIK. Perhaps more likely to see it in an applied math course.
@@lukeg378 Agreed, master's or even PhD in math probably would not touch this topic because it tackles theory over application. Applied math or engineering seems far more likely to cover this. That said, any self-respecting mathatician with a graduate degree should be well prepared to do their own research and learn about this after finding out about its existence.
I just want to add to the chorus of people with Math master degrees that didn't encounter this until UA-cam. This guy has had some really interesting stuff that you don't see in school. I can see why this doesn't get covered, because it's generally a way of solving equations for a numeric value; it doesn't really have much use in doing anything more fundamental. You could theoretically use this method to invert any function that has a good range of monotonicity to it if inverting that function could be useful in simplifying an equation. You'd then have to figure out how to numerically calculate the values of the inverse function. While computers do that for the Lambert W and a bunch of other functions already, they can also just solve the original equation numerically instead of trying to find a closed form solution in terms of a nonelementary function.
With a TI-84 or the like you could do the integral definition and use it. It wouldn't be super quick but it would work. Could write a program on it to do it for you.
If you are going to program a calculator, you might as well skip W and go straight to x^(8-x). Much more direct. You won't get the W but still get the answer.
6:10 "Fishes" is correct here. It's a weird corner of English: it's "fishes" if they're easily countable, "fish" otherwise. No stranger than infinities, I guess.
Fishes is 'multiple fish of different species', if there are multiple fish of the same species, then they're fish. 6 fish, all salmon, vs 3 fishes, salmon, trout, mackerel.
@@Azmodon If you order fish for dinner, you might get 2 fishes on your plate. It's not just types of fish, it's simply countable vs uncountable. At least, per my OED Style Guide.
It's one of those funky words that is really dependent on the article used in front of it like the fishes vs fish like you were saying on it denoting a type but nothing more it seems
You have a lovely collection of white board pens, but at 7:00 when you write over the black X with the red pen it made me feel a little bit ill inside. You then cured this illness with a lovely explanation of a powerful tool I'm sure I'll use again at some point.
Great video. I also arrived at x+xlnx=8 using Lagrange, but afer reaching that equality, i cannot make any progress. Still cannot understand why should it be raised to the second power timestamp 12:30
I went through my YT feed and marked all the politics, sports, pop culture, and crime stories as "not interested", and now finally I'm seeing interesting content like this in my feed. What took me so long?😀
@@richatlarge462I get not being interested in pop culture, sports and stuff, but why did you get rid of politics? Politics are highly important. Its what governs us and bad politics will have bad consequences in society and in people's lifes. When people aren't interested in politics, you eventually get leadership like Russia. So please take an interest in it. I'm sure you make use of your right to vote.
I haven’t done the calculation, but could you not have set F(x,y) = x^y - lambda * ( x + y ) and then minimized this function by eliminating the lambda parameter?
if you mean if there are any direct methods (such as division etc.) the answer is no. there are other methods available though like the one I mentioned or halley's method (not as straight forward) can research more on ur free time if ure interested!
0:41 Imma try to calculate that mentally. (-4)^12 is the same as 4^12 (since the power is an even number). 4^12 = (2^2)^12 = 2^24 = 2^20 × 2^4 (ik from memory that 2^20 = 1048576) = 1048576 × 2 × 2 × 2 × 2 Ok imma stop now. But that's certainly gonna be equal to over 16 million.
You can actually eyeball this. Notice how 3^5 and 4^4 are very similar ? So similar, one might think it's 2 sides of a parabola curve (Almost same Y value with a different X value, with the Y going up then reaching a peak then going down, exactly like a normal distribution curve). Since the 2 numbers are nearly identical, one could assume that the peak is the number in the middle between the two. Therefore 3.5. And since the number 4 yields higher number than the number 3, it's reasonable to assume the correct number is closer to 4 than 3. (Think of it like this : Since the number for 4 is higher, it means there has been less downwards trend from the peak to 4, than upwards trend from 3 to the peak. Therefore the peak is closer to 4 than 3. From there, if you keep following that logic, i bet you can narrow it down to 3.59.) I didn't understand anything from the math equations but just figured i could share my view on this. Also, for some people that don't know why this could be useful, having an idea of where the answer should be helps with verification process. It lets you know if your math is wrong. Example : let's say you're doing a physics question about the speed of a plane, and your answer turns out to be -250km/h. You already know your math is wrong because the speed is supposed to be positive. You're doing a math question about how much does John weigh based on information about his diet, activity and the fact that he is 203cm tall. Your answer from your math is 25kg. But you already know that a human that size can't weigh 25kg so you already know you need to check your math again. Edit : obviously this is assuming X,Y > 0. Just mentioning this since it's not present on thumbnail.
Thank you for this comment. I didn't understand what he was doing until your comment made me realize that if you graphed it as a parabola, you could use the derivative to find the inflection points.
i used desmos for non whole numbers, and i used this equation: y = x^(8-x) I got 3.535^(8 - 3.535), which is approximately 280, which is greater than 4^4, or 256.
Wait if x + y = 8 if we make x a number like -100 and y = 108 cant we create infinitely large numbers with x^y since it is a negative that will become a positive due to multiplying it an even amount of times. Like cant x be an almost infinitely small number and y the positive of x + 8 and well have infinitely large values for x ^ y? Also haven't watched the video yet im in class rn so please don't hate me.
When you get P(x) = e^((8-x)lnx), can't you assume that max P => max (8-x)lnx and derive that instead? Then you don't have to deal with the exponential.
If I understand this right, this means we now have a generic formula for any number x + y has to add up to right? Because you can just replace the 8 with whatever the sum is?
Can you do an all-in-one calc question again? But please do a d/dx this time and put integrals inside Edit: and i know he did it 4wk ago but i want one with a derivative
What course teaches the Lambert W function? I'm pretty sure I didn't see it in my high school calculus class (although it was 16 years ago so maybe I forgot)
It's not taught in standard Mathematics education at all. It has some fringe applications in some sciences as a simpler method of talking about solutions to some problems, but it's really just a hack that has no theoretically implication on any deeper mathematics. That is, it provides a way to give you a (semi-)closed form solution to some problems that otherwise lack them. Some people might claim leaving your answer in terms of other functions like sin, e^ and ln are similar; you can't calculate those exact forms without a calculator either, and in some sense they're right, but there's nothing that stops you from creating a special function that exactly solves whatever equation you want to solve and then figuring out how to numerically calculate the special function for certain values that you have given to you from your data. This means there's nothing really special about the function.
It’s a common method of solving continuous restricted optimization problems (seeking for maxima/minima of functions of several variables (f(x, y) = x^y) with constraints on those variables (x + y = 8)). To understand the method, you might need to spend about a couple of hours😃
Much the same - I don't know if it is easier - but from the differential... -ln x -1 +8/x = 0 ln x + 1 = 8/x e.x = e^(8/x) e.8 = (8/x)e^(8/x) W(e.8)=2.26282674=8/x x=3.535401 x^(8-x)=280.9045567
Well, do we assume x and y positive? x=-50 y=58 still produces an x+y=8, meaning any arbitrarily big and even negative x will solve to a greater arbitrarily big value of x^y.
all solutions to 2^x-3x-1=0 (transcendental equation)
ua-cam.com/video/GJbzsmccFtw/v-deo.html
Bprp, I have one more question. You know how we solved this problem of maximizing x^y w/ x+y=8 using derivatives? Suppose we restricted the domain of x,y to be natural numbers again. Could we then use discrete calculus to solve this problem. That is forward difference/backward differences are now analogies to the first derivative?
🎉🎉🎉
The finest value of X I’ve found is 3.53540103 but I have no clue how to express the limit of x.
I think the essence of this problem is to get an approximation without a calculator. The video never got to it at all.
(-4)¹⁶=16,777,216
I literally thought I was glitching with how many times I heard him say 'I really like this question" lmao
Yeah, I really like this question
he really likes the question
7:56 you are
I thought he was glitching.
i wonder if he likes the question
"I really love this question, let's...
I really love this question, let's have a look.
I really love this question, let's have a look."
- blackpenredpen
It’s a dude that holds a little lapel mic. I don’t think he’s too concerned with edit quality.
I don't know about editing, but I think he really loves this question
where are his other videos with that start lol
I love this question too!
He really loves this question
lambert W function be like :
input→fishy fish
output→fish
😂😂😂
I see what you did there
More like fish*e^fish 😂
BPRP’s ‘lil fish drawings are so cute. Fish*e^fish return fish for life.
lambert w function: defishifier
While most of this went completely over my head, it was a fun watch, and I am quite proud of myself that my first thoughts were "Are they positive numbers, and are they whole numbers?"
The 2 most important parts of the video is recognizing that you have to take the derivative to find the min and max and then realizing that u can use the Lambert W function, everything else is following up on those 2 things
@@sfglim5341 I understood it all till we hit that lambert W thing. I have never seen it before
@@jack002tuberits a very handy tool in college/uni algebra and calculus.
@@jack002tuber It's just a function, you don't need to delve into it too much
found this channel last night. i used to "teach"/tutor math in college and have always been passionate about teaching despite not being very good at it, so I love watching you explain intermediate math in very straightforward ways. but, i personally tapped out of math education in late calculus and linear algebra, and to see a video on more advanced subjects that I'M not even familiar with like the Lambert W... it made me feel very warm and cozy. This is a wonderful channel and I'm delighted to have found it.
Using Lagrange multipliers should solve the problem.
x=1/lambert_w(1) and y=8-1/lambert_w(1)
for the ignorant, forgetful, and lazy like myself:
en.wikipedia.org/wiki/Lagrange_multiplier
Using the Lagrange multiplier will give a system of equations identical to solving using the methods of single-variable calculus, ie the solution x+xln(x)=8.
I thought of this method too!
At the very start for (-4)^12 it is the same as (-2)^24, for anyone wondering just how much bigger it is.
Isn't that common sense lol
Or more directly, (1024)^2.4.
Which is slightly more than 16 million, because 2⁴ = 16, and 2²⁰ is 1024² > 1,000²
So yeah, I'm not arguing with "bigger" 😉
for a lot of people the concept of powers might not be that clear or straightforward. Younger students who may have an interest comes to mind. Of course however, most high-schoolers or college level students (who take math courses), will see it very clearly.@@xlr8_bs514
@@xlr8_bs514ikr
This channel has helped me see the usefulness of the product log in some obscure situations, but it comes up frequently enough to remember it.
See, the mistake at the end is that you asked ChatGPT 3.5, you should have asked ChatGPT 8/W(8e).
Elite comment. Simply beautiful
Underrated comment
W COMMENT THIS DESERVES TO BE POPULAR
THE creativity omg
🎉
This is so underrated
I did not notice that x and y should be nonnegative, else you could have (-2k)^{8+2k} which goes to infinity.
(even -ωₙ)^(even ωₙ+8)
where n is arbitrary
Because a negative when exponentiated to a positive even is positive, this produces a value greater than ωₙ₂
Where n⇐∞, the maximum approaches Ω.
@@Enlightenment0172!??
I mean, if you had it so X could be negative, the max is theoretically infinity, but that min is theoretically negative infinity, which makes the question irrelevant. Knowing that is actually why I clicked on the video in the first place.
I was thinking just that!
@@TheSkullkid16 Exactly. You end up with an equation of -(n-8)+n=8 and then have (-(n-8))^n. The limit would basically be negative infinity to the power of infinity plus 8.
As a non-mathematician I understood nothing but the way you explain is so cool that it made me more enthusiastic for maths😊
as non-mathematician you should still know highschool maths, unless you're still in school, then I guess you'll still learn about this.
@@FourOneNineOneFourOne I did learn some of it in high school and even at university, but I graduated from university 5 years ago, my maths class had finished 2-3 years prior, and I have been working in a completely unrelated area. And in my life the only areas of maths I use are arithmetics and trigonometry 😂 so, the other things get forgotten easily. I think we didn't reach this difficulty of calculus at high school, or, if we did, it's gone from my memory, because I understood about half of the video at most. I don't remember any of the functions, sadly.
@@FourOneNineOneFourOne At least in my case, Calc wasn't even offered unless you qualified to do College Courses while still in Highschool. I got through Calc 1 just fine, but most of what he was showing here is not stuff we really covered. This is the first time I ever heard of W(xe^x)=x, and I would have literally no idea how to parse something like that in a casual way.
While I have worked with Log/ln/e a bit back then, it's been over a decade, and I was never fond of dealing with those at all, so I just lost it completely.
@@FourOneNineOneFourOne What if one has no aptitude for math? As a high school student, or today more than 30 years later, I could no more understand what he is explaining then I could run 400m in 48 seconds (I never broke the 50s barrier 50.7, and only passed algebra as a senior.) The ability to do high school math, beyond basic algebra, is a mental gift, and is no different than the ability to any number of other things. And no amount of hard work in the world will overcome a lack of ability. If you can do this stuff, congrats to you, but no one “should” be able to do anything you can do.
@@JonesyTerp1 "no amount of hard work in the world will overcome a lack of ability". I'm not sure I agree, particularly for all the activities you mentioned, but, putting that aside...
The problem is, even if you have the "innate ability", for most (many?) things, you still have to do the "hard work" before you are proficient enough to be convinced you have the ability.
So, to find out if you have the ability to do "a thing", you have to take a leap of faith and put in the hard work to find out. One would hope you enjoy "the thing" so if it turns out you can't do it, at least you had fun trying.
It comes down to the point, does the hard work "develop" the aptitude, or does it "expose" it? I think there are some people in each group, along with a group who work hard and still can't, and another group who never seriously try.
Considering x >= 0, y >= 0 on my own, I got stuck with x(ln(x) + 1) = 8
Thank you for introducing the W function
yeah but you already had a function, so rephrasing it as another function is not really solving anything.
@@TomJones-tx7pb but what I stopped with is an implicit function. Explicit solution presents the way to directly calculate the value for a given argument rather than to bruteforce it
Same for me. I also solved the problem before watching the video and got stuck with x ln(x) + x = 8 but I just type it into Wolframalpha and see that it has no "nice solution". In my opinion, there is no benefit in expressing this equation with the W function. The equation x ln(x) + x = 8 can already easily be solved by a computer and for a human the W function does not give you any additional information
@@adammizaushev Good point for classical math, if you have a pre-calculated W function, but modern way of solving is using a computer to get as accurate an answer as you desire with numerical analysis.
@@TomJones-tx7pb
Yeah, I was only speaking in the matter of getting intuitive, natural understanding of a function to a human, which is got by its explicit expression, rather than implicit one
10:03 Keep in mind rationalizing the denominator. That case is not needed to put W() in the denominator.
Great explanation. But what's up with your fish having eyebrows?
Fishes also deserve to have personality 😢
these are the most mischievous fish i've ever seen
The most fascinating about this, is if you do this with X + Y = 100, the maximum will be 24^76, which is totally asymmetrical
You mean to tell me that e^{W(100e)-1} is equal to 24?
Close - the x is still e^(productlog(100e)-1) or 23.947
@@abrahamholleran4162 so like, you get that my skepticism came exclusively from the idea that e^{W(100e)-1} is a natural number right? Like, sqrt(2) is "close" to 1.4, but the fact that it's irrational is kind of a really important thing about it
W(e)=1 so it's a shame there aren't identities for cases a constant precedes e.
@@frimi8593 They weren't replying to you, but rather to the original comment. I'm sure they would assume that you were doubting it being an integer.
(It can be difficult to tell what people mean, since some use the word "equals" too loosely. You could say something like "exactly 24" or just explicitly doubt it being an integer.)
didnt understand a word of this but i am now holding a W and a bunch of fish
pretty straight forward
Yes, x=0 is a local minimum because to the left the value of P is unbounded.
That means that I think the result we found is not a global max but a local max. Wolfram is mistaken 😮 because it calls it a global max.
Or have we told Wolfram that x>=0 ?😊
Wolfram alpha probably ignores negatives entirely here because any noninteger negative x can't be raised to the power 8-x when restricted to the reals.
As soon as I saw the question I noticed there's no upper limit because we can have absurdly large negative numbers for X (or Y). Once he put the constraint in that they had to be positive values, I was sure the answer was x=y=4. I just assumed the answer was going to be integers. Don't know why I fell into that trap, it just felt intuitive.
@@KenFullman Exactly, I was thinking the same thing. Because, you can have the equation as such: y=n, and x= -(n-8) For example, you could have y=1,000,008. x would then equal -1,000,000 and the answer would still equal 8, and you'd be left with -(1,000,000)^1,000,008 which would be absolutely huge, and can still get bigger, into infinity. Because you'd have -(n-8)+n=8 and then (-(n-8))^n. As long as you can have y be 8 more than a negative value of x, you can get infinitely higher values for x and y.
@@KenFullman If the equation had been x*y instead of x^y, then that intuition would be correct, and I frequently encounter situations where it's useful to optimize x*y by setting x=y (maybe you do too). So maybe that explains how you fell into that trap. :)
It's always about local extremum, cause blackpenredpen used Fermat's theorem that states that *local extrema* in the interior of a domain must occur at points where the derivative equals zero or undefined.
honestly, I understood everything you said (despite this being the first time I've heard about lambert w function) and this made me so happy
(also, the fish you draw are amazing)
I often find your videos a bit beyond my skill level but I found this one really approachable and fun. thanks for this!
yeah this guy is using calc 1 and maybe calc 2
This is a great video and you are a wonderful mathematical communicator! Thank you for creating this!
I really hate this question, let’s not take a look.
evil blackpenredpen be like:
@@yesiamrussian whitepencilgreenpencil
hey @blackpenredpen, can you make a shirt that shows a white and green pencil and says
"I really hate this question, let's not take a look
~whitepencilgreenpencil"? pweeeasee???
video ends
I'm happy to say that this is the first problem from your videos that I have been able to solve !🎉
This is super fun to watch. I had no idea about the W function.
Nice problem.
Here's the Sagemath code (using Newton's method):
sage: reset()
sage: f(x)=x^(8-x)
sage: g(x)=diff(f(x),x)
sage: n(x)=x-g(x)/diff(g(x),x)
sage: x=4.0
sage: x=n(x);x
3.35700838162158
sage: x=n(x);x
3.53287101659463
sage: x=n(x);x
3.53539951720032
sage: x=n(x);x
3.53540103555960
sage: x=n(x);x
3.53540103556015
sage: x=n(x);x
3.53540103556015
sage: f(x)
280.904556712676
Note: I picked 4.0 as the starting value because it is between 0 and 8.
十分感謝老師解答😊😊
Exceptional explanation of the unique "W" Function! However, I still need to exponentially repeat your Steps to the Solution (or repeat the Steps to the maximum number allowed) --- in keeping with the ultimate objective of this proposal (computing the maximum)! Alas, I am exhausted from this exhilarating mathematical exercise. Bravo! Thank you, sir blackpenredpen!
I'm a computer scientist who hasn't done math in a couple of years, it felt good to get this on my own 😌
Maximizing x^y with constraints like x+y=8 was one of those math puzzles that used to leave me completely stumped, especially with concepts like the Lambert W function! What made a big difference for me was using tools that provide step-by-step explanations. SolutionInn’s AI study tool is fantastic for breaking down complex problems like this into manageable steps. If anyone’s feeling overwhelmed, tools like that can really help clarify things.
Always fun mentally juggling x^y | x+y = c among the integers; enjoyed seeing the solution for the reals with a refresher on the W function
6:13 I didn't understood how did fish e^ fish is fish. And you said it is in calculus 1,but I have not heard of it
Are there solutions if X and Y are complex numbers?
No, because you cannot compare the size of imaginary numbers. You would have to rephrase to find the maximum of |x^y| for instance.
Complex numbers arent ordered bud
This guy really loves this question
Hmm this video demonstrates the difference in thinking between an engineer and a mathematician. I paused the video when the question was posed, opened excel, used one cell as an input for X, defined another cell as Y by making it 8 - the X cell, then defined a 3rd cell as the X cell to the power of the Y cell. At this point I used the solver function in excel to maximise the value of the 3rd cell by altering the value of the X input cell. Got the answer in less than a minute by brute forcing it 😅. I will admit blackpenredpen's method was more eloquent.
I'm right there with you. I might have done a column for x, another for 8-x and one more for the function. Range 3-5 with 0.05 increments, maybe refine around the maximum and called it good at 3.51.
f(x) = x^(8-x)
ln(f(x))' = f'(x)/f(x)
the max is at f'(x) = 0, and f(x) can't be infinity so the max is at ln(f(x))' = - ln(x) + (8 - x)/x = 0
Ask a computer for an approximation
We just covered that a couple of months ago. Wouldn’t it be easier to just substitute y = kx and get kx/x = k => limit depends on the slope k, therefore it doesn’t exist?
I think you commented on a wrong video?
@@blackpenredpen Oh wow. That's weird. Yeah, something was wrong with the API, I guess.
How would you solve Lambert W by hand?
This seems extremely hand wavey by just invoking an inverse of what feels arbitrary, and then asking wolfram alpha to actually solve it. How would this be any different than just asking wolfram to do it from the beginning?
Does W have an alternative definition that allows this to actually be calculated reasonably?
No easy way, that's why it was not used till 1980 despite being found in 1780
Basically it's xe^x = 8e where x = lnx + 1. Solving xe^x isnt trivial, but solving lnx + 1 is. So yes Basically a calculator that could solve xe^x = number would be able to solve the original equation, you need like a TI89 to do it
This is precisely why it doesn't really get covered in any mathematics education. It's not particularly useful because it basically requires you to ask the computer for the numerical approximation, which it could have done without going through the manipulation of the equation. There are applications of it in various science fields, but not really understanding those fields, I don't really know what's gained from putting the answer in terms of the Lambert W instead of whatever form it was in before. Special functions are defined for all sorts of problems that lack answers in terms of elementary functions; for something as simply defined as the Lambert W, it shouldn't be too surprising that it pops up in the answer of all sorts of problems in science that lack closed form elementary solutions.
He might really love this question
amazing explanation.❤
Another Chatgpt response when I asked it to solve using calculus:
Certainly! To solve this problem using calculus, we first express \(y\) in terms of \(x\) using the constraint \(x + y = 8\), which gives us \(y = 8 - x\). Our function to maximize is therefore \(f(x) = x^{8 - x}\).
To find the maximum value, we need to:
1. Take the derivative of \(f(x)\) with respect to \(x\).
2. Set this derivative equal to zero and solve for \(x\) to find critical points.
3. Check these critical points to determine which one gives the maximum value of \(f(x)\).
The derivative of \(f(x)\) can be challenging because it involves both the power rule and the chain rule. Let's calculate it.
from sympy import symbols, diff, solve
# Define the symbol x
x = symbols('x')
# Define the function f(x) = x^(8 - x)
f = x ** (8 - x)
# Take the derivative of f with respect to x
f_prime = diff(f, x)
# Find the critical points by solving f'(x) = 0
critical_points = solve(f_prime, x)
# Show the derivative and critical points
f_prime, critical_points
The critical point for maximizing the function \(f(x) = x^{8 - x}\) is \(x = \exp(-1 + \text{LambertW}(8e))\), and the maximum value of \(f(x)\) at this point is approximately 280.905.
This result confirms our earlier graphical analysis, demonstrating that calculus provides a precise way to determine the maximum value of \(x^y\) given the constraints \(x, y \geq 0\) and \(x + y = 8\).
I think if you wanted another number other than 8, all you need to do is replace 8 in the solutions with that number so get the answer
I got pretty far into the question before I eventually got stuck... Of course the Lambert W function was involved lmao, I still dont know how to properly use that thing
It's right there on his shirt.
(i'm kidding)
If you ask GPT-4 instead of GPT-3.5, you get the correct answer. ChatGPT will now write python programs to calculate precise results for questions like these.
Thanks!
I really like this video.
I really like this video, let's have a look.
I really like this video, let's have a look!
Much more honest than other youtubers who cut every 2 seconds...
I appreciate that you don't waste time while giving presentation. 👌🏻
It's out of the scope of ChatGPT's algorithms LMAO
chat gpt just take info from the internet/database and provide an interpretation. if the database doesnt have the answer, chatgpt wont have it. it doesn't "think" on its own.
edit: actually it DOES learn, for it's an AI, but it specializes in predicting what the response to your question is base on the database of information it's accessing. But i dont think it can do math.
@@kingpetunfortunately chatgpt doesn't have the capabilities to do calculus but it might be able to do basic arithmetic, they could integrate something like Wolfram alpha into chatgpt one day
@@maddenbanh8033chatgpt has emergent problem solving as a result of understanding human language. Gpt-4 is able to use that problem solving to load arithmetic into premade calculators, like programming languages or Wolfram, in order to “solve” calculus and other math problems. It figures out what steps are necessary, and loads those steps into something that can actually get a confident result. Then it takes that result and moves on to the next step
Tried solving using desmos. Plotted x+y=8 first. Then plotted x^y = some constant b. The value of b where the two graphs only have one point of intersection is the maximum value of the function, and that came out to be approximately 280.902799
Solved it with Lagrange multipliers. Glad I still remember Calculus III.
As others have pointed out, you could use the same method to get x=k/W(k*e) for the max of x^y when x+y=k. The case of k=8 is interesting though, because 4^4 and 3^5 are so close together. We were already pretty close with 3.5^4.5.
This can also be solved using binary search
import math
low = 0
high = 8
delta = 1e-6
while abs(high-low) > delta:
mid = (low+high)/2
first = math.pow(mid-delta, 8-mid+delta)
second = math.pow(mid, 8-mid)
third = math.pow(mid+delta, 8-mid-delta)
if firstthird:
high = mid
else:
break
print(math.pow(low, 8-low))
Here wrote a little python script don’t know if this will help
max_num = 0;
x = 1
y = 7
while (int(x+y) == 8):
print(x+y)
if (max_num < x**y):
max_num = x**y;
x+=0.0001
y-=0.0001
print(x+y)
print(max_num)
It may be because I never studied it so I don't know much about it, but I don't really like the Lambert function. Since there's no way to determine its value without using wolpharm alpha or a software, it's not really explicitating or finding the value of X. I feel it's just writing an implicit equation in a different way that's still an implicit equation. It's like saying "there's no solution but here's one that isn't really one".
That's an inherent problem with inverses. Think of it like roots and logs where there's no other way to express those numbers. For example, √2 and ln(2) have no nice ways of expressing them, and you generally need a calculator to compute their values. Not much you can do about it when you have an irrational answer.
@@sethb124 but at least with square roots you can have a general idea of their value, you can at least figure out a min/max borders, like, you know that square root of 2 is somewhere between 1 and 2, you can, like, try out 1,1 * 1,1 then 1,2 * 1,2 and so on and figure out a few decimals this way.
None of that with the Lambert function. Either you leave it at that or use a software. At least as far as I know, there's no way to even figure out its value, better off go back to the implicit expression and try out values there.
@AcaciaAvenue Yeah, that's true. It's not super easy to estimate, but for problems like these, it's the most "exact" answer you'll get. It's not very intuitive, but I'd argue neither are lnx and e^x. It's still a valid answer because there are ways of estimating it (or else computers couldn't do it), it's just not an intuitive function.
@@sethb124That's the point I wanted to make. It's not an "exact" answer to me if I have no means of giving even an estimate numerical value. It's just picking something with no answer and writing it differently, but still with no answer.
If there were some means to calculate an estimate value of the lambert function without feeding it to a software and tell it "do it for me", then I will change my mind but I looked it up but there seems to be none. Therefore I don't see the utility of the Lambert function.
Totally agree. Just seems like a fix to me. And calling it exact is a stretch too. I solved this by implicit differentiation and Newton-Raphson. Didn't take long at all.
I never saw the Lambert function and I have a Master's in math. At what level is it usually taught?
Yeah. I majored in math and in four years of college, I never met Lambert. In some ways, I have always felt that Va. Tech let me down. (This was early 70's.)
Yeah, it's rare to see it in math curriculum because it doesn't really have deeper theoretical value AFAIK. Perhaps more likely to see it in an applied math course.
@@lukeg378 Agreed, master's or even PhD in math probably would not touch this topic because it tackles theory over application. Applied math or engineering seems far more likely to cover this.
That said, any self-respecting mathatician with a graduate degree should be well prepared to do their own research and learn about this after finding out about its existence.
I definitely saw it in my undergrad textbook, I don't think it came up in the course material, though.
I just want to add to the chorus of people with Math master degrees that didn't encounter this until UA-cam. This guy has had some really interesting stuff that you don't see in school. I can see why this doesn't get covered, because it's generally a way of solving equations for a numeric value; it doesn't really have much use in doing anything more fundamental. You could theoretically use this method to invert any function that has a good range of monotonicity to it if inverting that function could be useful in simplifying an equation. You'd then have to figure out how to numerically calculate the values of the inverse function. While computers do that for the Lambert W and a bunch of other functions already, they can also just solve the original equation numerically instead of trying to find a closed form solution in terms of a nonelementary function.
x+y=8, which also means x+(-y)= 8 (where x>y) or -x+y = 8 (where x
yes I know x&y should be >= 0...save that message..you're welcome..😊
x+y is no the same as x+(-y)
is it possible to calculate W(n) with a regular calculator? or do you need one with a specific button for the lambert w function?
With a TI-84 or the like you could do the integral definition and use it. It wouldn't be super quick but it would work. Could write a program on it to do it for you.
@@ryanburkett949 but nothing with an old calculator like a casio fx-82ex they make me use at school?
Just graph xe^x = 8e and you solve it. Since once you get x then set that = ln x + 1 which you can solve by hand
If you are going to program a calculator, you might as well skip W and go straight to x^(8-x). Much more direct. You won't get the W but still get the answer.
6:10 "Fishes" is correct here. It's a weird corner of English: it's "fishes" if they're easily countable, "fish" otherwise. No stranger than infinities, I guess.
Fishes is 'multiple fish of different species', if there are multiple fish of the same species, then they're fish. 6 fish, all salmon, vs 3 fishes, salmon, trout, mackerel.
@@Azmodon If you order fish for dinner, you might get 2 fishes on your plate. It's not just types of fish, it's simply countable vs uncountable. At least, per my OED Style Guide.
It's one of those funky words that is really dependent on the article used in front of it like the fishes vs fish like you were saying on it denoting a type but nothing more it seems
You have a lovely collection of white board pens, but at 7:00 when you write over the black X with the red pen it made me feel a little bit ill inside. You then cured this illness with a lovely explanation of a powerful tool I'm sure I'll use again at some point.
Literally the video I watched before this was a video on the W function on your shirt..
And now I see why.. this video is also about the function.. algorithm got me
Great video. I also arrived at x+xlnx=8 using Lagrange, but afer reaching that equality, i cannot make any progress.
Still cannot understand why should it be raised to the second power timestamp 12:30
Just multiplying the powers
like
(2³)⁴ = 2^(3×4)
you have (z+1) x (z-1) which is (Z^2) -1 where z is W(8e)
Yes, thanks. But Am I missing something, because the two expressions differ in signs. One is addition, the other subtraction
They are both minus signs
Oh I got it, both are minus signs. 😅
I don't know why YT recommended me this and I'm not a science major, but this intrigued me so much lol.
I went through my YT feed and marked all the politics, sports, pop culture, and crime stories as "not interested", and now finally I'm seeing interesting content like this in my feed. What took me so long?😀
@@richatlarge462I get not being interested in pop culture, sports and stuff, but why did you get rid of politics? Politics are highly important. Its what governs us and bad politics will have bad consequences in society and in people's lifes. When people aren't interested in politics, you eventually get leadership like Russia. So please take an interest in it. I'm sure you make use of your right to vote.
x+y=8
y=-x+8
Therefore, x^y=x^(-x+8)
Take the first derivative and set it equal to zero
... that's as far as I got 😅
Then apply the Newton-Raphson method to solve for x
@@vanshjain3428 no. The lambert function 🤦🏾♂️
I haven’t done the calculation, but could you not have set F(x,y) = x^y - lambda * ( x + y ) and then minimized this function by eliminating the lambda parameter?
My approach before watching the video:
x^y is not defined for x0.
Since x^y < 1 for x>0 and y
Please DO NOT let UA-cam sabotage your otherwise excellent videos with nested ads. They throw off the viewers' concentration.
this was insanely epic, like for real, this was amazing and so fricking cool :>
If x can be on minus there is more posibble way, because x=-200 and y=208 there we have -200+208=8, so -200^208 is big positive number. Am I wrong?
So is impossible to say max of x^y
Can you solve Lambert W function without using calculator?
can use Newton Raphson method to approximate the function. Usually it's very accurate and good enough since the function converges very fast
if you mean if there are any direct methods (such as division etc.) the answer is no. there are other methods available though like the one I mentioned or halley's method (not as straight forward) can research more on ur free time if ure interested!
After watching so many of your videos, I could easily find the answer by myself 😂 The lambert w function is like a normal function to me now.
Before watching:
The problem is they did not account for x
Why comment before watching, lmfao
@@w.o.jackson8432 to attempt to answer and give their initial thoughts before the teacher explains the solution
Is it possible to solve that without Lambert W function?
It is if you only count non-negative integers
A math symbol I didn't know existed: 🐟(e^🐟)
Brillant explanation, I was able to follow along completely!
Would the lambert function work with a dolphin? Because strictly speaking a dolphin is a mammal, not a fish.
Mammals are not a problem. It even works with unicorns, hippogriffs, and other animals with imaginary components.
I really liked your explanation. Even tho I am out of maths for years now, with some thinking I was at least able to understand your steps!
that was just another beautiful solution by my fav youtuber.
0:41 Imma try to calculate that mentally. (-4)^12 is the same as 4^12 (since the power is an even number).
4^12 = (2^2)^12
= 2^24
= 2^20 × 2^4 (ik from memory that 2^20 = 1048576)
= 1048576 × 2 × 2 × 2 × 2
Ok imma stop now. But that's certainly gonna be equal to over 16 million.
All of this, just to get .2 higher value.
But hey! Progress is progress! XD
Keep up the content! You've earned a sub!
Can you please solve for a b and c the equation 4×a^2+4×b^2+3=4c+4 sqrt of a+b-c . Is for a friend.
You can actually eyeball this.
Notice how 3^5 and 4^4 are very similar ? So similar, one might think it's 2 sides of a parabola curve (Almost same Y value with a different X value, with the Y going up then reaching a peak then going down, exactly like a normal distribution curve).
Since the 2 numbers are nearly identical, one could assume that the peak is the number in the middle between the two. Therefore 3.5.
And since the number 4 yields higher number than the number 3, it's reasonable to assume the correct number is closer to 4 than 3. (Think of it like this : Since the number for 4 is higher, it means there has been less downwards trend from the peak to 4, than upwards trend from 3 to the peak. Therefore the peak is closer to 4 than 3. From there, if you keep following that logic, i bet you can narrow it down to 3.59.)
I didn't understand anything from the math equations but just figured i could share my view on this.
Also, for some people that don't know why this could be useful, having an idea of where the answer should be helps with verification process. It lets you know if your math is wrong.
Example : let's say you're doing a physics question about the speed of a plane, and your answer turns out to be -250km/h. You already know your math is wrong because the speed is supposed to be positive.
You're doing a math question about how much does John weigh based on information about his diet, activity and the fact that he is 203cm tall. Your answer from your math is 25kg. But you already know that a human that size can't weigh 25kg so you already know you need to check your math again.
Edit : obviously this is assuming X,Y > 0. Just mentioning this since it's not present on thumbnail.
Thank you for this comment. I didn't understand what he was doing until your comment made me realize that if you graphed it as a parabola, you could use the derivative to find the inflection points.
OK, I did not expect the word FISH to be stuck in my head after a maths video.
i used desmos for non whole numbers, and i used this equation:
y = x^(8-x)
I got 3.535^(8 - 3.535), which is approximately 280, which is greater than 4^4, or 256.
how do you get 3.535 from e^(W(8e)-1). I was with you up until that point
I used WolframAlpha. W is “productlog”
@@blackpenredpen I get that, but how does it get the answer. I just look at it but I don't know what to do.
5:12 the function x^(8-x) or x^-x also has a zero at x->inf but in that case y is negative.
That's interesting, how did you think of that?
Wait if x + y = 8 if we make x a number like -100 and y = 108 cant we create infinitely large numbers with x^y since it is a negative that will become a positive due to multiplying it an even amount of times. Like cant x be an almost infinitely small number and y the positive of x + 8 and well have infinitely large values for x ^ y?
Also haven't watched the video yet im in class rn so please don't hate me.
One fish - two fish, one sheep - two sheep, one buffalo - two buffalo
This is FASCINATING. You are a Master my friend 👏👏👏👏👏👏👏👏
When you get P(x) = e^((8-x)lnx), can't you assume that max P => max (8-x)lnx and derive that instead? Then you don't have to deal with the exponential.
If I understand this right, this means we now have a generic formula for any number x + y has to add up to right? Because you can just replace the 8 with whatever the sum is?
Thanks to you I discovered the Lambert W function.
Can you do an all-in-one calc question again? But please do a d/dx this time and put integrals inside
Edit: and i know he did it 4wk ago but i want one with a derivative
I don't know, I just have the feeling...but I'm pretty sure he loves the question
What course teaches the Lambert W function? I'm pretty sure I didn't see it in my high school calculus class (although it was 16 years ago so maybe I forgot)
It's not taught in standard Mathematics education at all. It has some fringe applications in some sciences as a simpler method of talking about solutions to some problems, but it's really just a hack that has no theoretically implication on any deeper mathematics. That is, it provides a way to give you a (semi-)closed form solution to some problems that otherwise lack them. Some people might claim leaving your answer in terms of other functions like sin, e^ and ln are similar; you can't calculate those exact forms without a calculator either, and in some sense they're right, but there's nothing that stops you from creating a special function that exactly solves whatever equation you want to solve and then figuring out how to numerically calculate the special function for certain values that you have given to you from your data. This means there's nothing really special about the function.
I have absolutely no ideia why UA-cam recommend this video. Everything went WAAAAY over my head!
Interesting, ChatGTP gave me a method that uses Lagrange-multipliers. Can you say something about this?
It’s a common method of solving continuous restricted optimization problems (seeking for maxima/minima of functions of several variables (f(x, y) = x^y) with constraints on those variables (x + y = 8)).
To understand the method, you might need to spend about a couple of hours😃
he really wanted to make sure we are aware of his appreciation for the question
Much the same - I don't know if it is easier - but from the differential...
-ln x -1 +8/x = 0
ln x + 1 = 8/x
e.x = e^(8/x)
e.8 = (8/x)e^(8/x)
W(e.8)=2.26282674=8/x
x=3.535401
x^(8-x)=280.9045567
a^x=a^y+2^z here(a,x,z,y are positive intigre.) now find the sum of all solution (ayz/(x-1)^2.) Please solve this sir..........
Well, do we assume x and y positive?
x=-50 y=58 still produces an x+y=8, meaning any arbitrarily big and even negative x will solve to a greater arbitrarily big value of x^y.