if x+y=8, find the max of x^y (Lambert W function)
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- Опубліковано 15 вер 2024
- If x+y=8 for nonnegative x and y, then find the maximum value of x^y. This question seems like a typical calculus optimization problem, but to find the exact answer, we will need to use the Lambert W function. • Lambert W Function (do...
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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
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
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
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
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!
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.
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.
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.
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
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.)
Great explanation. But what's up with your fish having eyebrows?
Fishes also deserve to have personality 😢
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!
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!
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.
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
pretty straight forward
This is super fun to watch. I had no idea about the W function.
I really hate this question, let’s not take a look.
I'm happy to say that this is the first problem from your videos that I have been able to solve !🎉
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
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.
I'm a computer scientist who hasn't done math in a couple of years, it felt good to get this on my own 😌
didnt understand a word of this but i am now holding a W and a bunch of fish
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.
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)
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))
This guy really loves this question
My approach before watching the video:
x^y is not defined for x0.
Since x^y < 1 for x>0 and y
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..😊
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
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)
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
I don't know, I just have the feeling...but I'm pretty sure he loves the question
十分感謝老師解答😊😊
Fun fact: if x+y=2e, the max of x^y is registered when x=y. In all other cases: if x^yy, if x^y>2e, max of x^y is reached when x
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...
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.
amazing explanation.❤
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
I appreciate that you don't waste time while giving presentation. 👌🏻
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.
I like how he shows that ChatGPT is unable to find the solution at the end.
Sure, it can spit out an explanation for how to find the solution, because those sorts of explanations are in its training data, but it can't process the solutions for itself because it doesn't "understand" what the explanation means. It can't create a mathematical model, much less perform the operations to process it, because it has no understanding or procedure for that. It only recognizes that this sort of problem resembles the maxima/minima problems from its training data, and regurgitates a procedure.
It's useful to recognize the limitations of the language model.
I don't pay for v4 but v3.5 has always been awful at calculus and has only gotten worse as they have limited the product. I love learning how to solve problems (hence why I am here) but I have yet to find a good AI solution.
Please DO NOT let UA-cam sabotage your otherwise excellent videos with nested ads. They throw off the viewers' concentration.
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.
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!
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
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
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
He might really love this question
Calling it out, the answer is going to be pi isnt it?
Spoiler : It wasn't. But I am an engineer so good enough
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.
Solved it with Lagrange multipliers. Glad I still remember Calculus III.
I think he really likes this question
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
A math symbol I didn't know existed: 🐟(e^🐟)
or if you're me and don't know calc but you know python and are a fan of bruteforcing, you write this lol
def max(iteration_amt, sum_amt):
i, x, y, max_found = 0, 0, 0, 0
while i < sum_amt:
x = sum_amt - i
y = sum_amt - x
if x**y > max_found:
max_found = x**y
best_nums = [x, y]
i += iteration_amt
return max_found, best_nums
print(max(0.001, 8))
(decrease iteration amount for higher decimal precision)
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.
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.
ChatGPT can’t even exceed calculus teacher
That’s what I like to see❤
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 \(x + y = 8\).
Ahh, Lambert W. I had to use it to solve an optimal control problem for landing a spaceship once.
Make it so, Ensign Cainghorn.
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.
Brillant explanation, I was able to follow along completely!
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.
Quick education on fish or fishes: a fish is one singular fish, the fish are multiple fish from the same species of fish, the fishes are multiple fish from multiple species of fish :)
the question gets pretty easy using Lagrange multipliers, but you'll end with a x(lnx + 1) = 8.... and then i got stuck, thanks god to W(.) function
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
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..........
Two iterations of Python got me the answer: "for f in [x * .01 for x in range(340,370)]: print(f, f**(8.0-f))"
duh:
without excluding negative numbers: (-infinity)^(infinity+8)
because both are even, it gives a positive answer which is infinitely large, thus providing a max that has no end
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
Quite interesting. I didn't know the Lambert W function and ended up with (8-x)/x - ln x = 0.
I found the approximation using my almost 40y old pocket calculator by transforming this to x = 8/(ln x + 1) and iterating with the initial value x0=4.
Convergence isn't very fast, but at the end of the day you have to solve this by using some form of iteration.
Nevertheless a really interesting function.
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
looked for math, found numerical methods 😢
I have absolutely no ideia why UA-cam recommend this video. Everything went WAAAAY over my head!
My youngest son hit his high school math teacher with a similar answer as there was no criteria on a range for the problem, as with her and must all be positive.
Notice that going from X^(8-x) to e^(8-x)lnx is valid if and only if x is > 0 ( can't be = to 0)
if X + Y = 8 then Y = - x +8. That looks like slope intercept form for a line that goes to infinity. So the answer is infinity^ infinity for max X ^Y.
All of this, just to get .2 higher value.
But hey! Progress is progress! XD
Keep up the content! You've earned a sub!
I'm glad that I was able to solve it by myself. I found the maximum value of x by using newton's method directly as opposed to the lambert w function though (I vaguely recall that you find the values of W using newton's method anyway right?) .
My first thought was x=-1000, y=1008 works so there's no bounds to the ultimate max value of x^^y.
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.
This is FASCINATING. You are a Master my friend 👏👏👏👏👏👏👏👏
Thank you,! I wasted some time trying to solve this only with derivates and found only (0, 8) and (8,0) 😞.
I don't even remember the Lambert W from school. Very nice!
Instructions unclear. I now have 1,024 fishes.
This really reminded me how I hate this weird way to write 1, so confusing. And using chained fractions is ugly and unnecessarily confusing. And writing e^1 instead of e is curious.
Rather than maximize f(x) = x^(8-x) I look the natural log first. Easier to take the derivative of g(x) = (8-x) ln x.
The W function didn't impress me, in the end we need a computer to spit out an approximation.
Before watching the video and spoiling the answer, I think the answer is that there is no max for x^y, in other words, it can be as high as we want (infinite). Here's how:
let 's take y to be equal to some number 2*n (any large even number). Now calculate x = 8 - y. This means that x+y = 8. x^y will be a huge positive number. We can make it as large as we want by choosing a big enough n.
I love how all his fish are evil
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.
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.
there is no maximum: for any pair number y, we have (8-y)^y as big as possible.
example y=1008 x=-1000 , -1000^1008 = 10^3024...
with the constraints given where x can be a negative integer, the highest is incalculable ( infinity )
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. 😅
Over compiling 4+4. Variables are exactly that and anything can be done with them.