The famous exponential equation 2^x=2x (ALL solutions)
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- Опубликовано: 1 авг 2024
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0:00 Let's do some math for fun!
0:29 Review Lambert W function
2:28 Solve 2^x=2x
6:13 Why it looks like just one answer
9:27 Check out Brilliant
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We could actually get both answers from W(-ln(2)/2) by hand
See here: instagram.com/p/CSfIWPchpB1/?
I believe people don’t like it because it is clear that xe^x can not have well defined inverse because it is not biyective. Now sin(x) is not either, it’s just to properly define the argument and be sure one stays there.
Wow, that's an interesting way of going about it! Videos like yours inspire me to share my own maths tricks!
As I see you are a very good mathematician. I was working with a lot of equations back then! One of my favourite exercises like 10-15 years ago was the following. We have a and b where a,b e N! a^b+b^a = 423393 and a^a + b^b = 16780341. What is the value of a and b? Resolve it without just trying out numbers and hope we have luck!
Try solving this strange one(final version): lim d/dx -lg(2)/lg(1-1/x) as x is approaching infinity
No available 😢
I watched about 10 of your videos asking myself, "Why is this weirdo doing math while holding a PokeBall??" Then I finally saw one and realized it's your microphone with a cover on it.
Oh and i thought he has it just incase a random PI-kachu appears.
@@CrazyT2009😂😂😂
The pens are his wands and the pokeball is his pondering orb. This dude is an actual wizard of mathematics.
I don't understand much of any of this, but I really like your enthusiasm and way of teaching, the 10 minutes flew by before I even realised. Very entertaining channel
The Lambert W function is not an analytic function. Thus, one cannot present its formula using basic operations. (The sum and multiplication of analytic functions such as polynomials (The constant function is a special case of a polynomial.), exponential functions and trigonometric functions. (I may have omitted something.)
@@heinrich.hitzingermate why r u saying that here
@@egggames8059 Because he is secretly a genius. Real sigma males will understand.
Steps:
6:58
I like how you fun about with math. it opens your mind to lots of possibilities.
Your grammar made my brain divide by 0.
@@shen144 Maybe because not everyone is a native english speaker ?
You should also note that W0(- ln x/x) = -ln x for 0
Imagine bprp at the end of an epic video pulling out a green pen to finish it off!
Bprp: hmm new idea let's introduce rainbow pen too
@@gamin8ing Underrated comment😂
Noiceee
@@gamin8ingnoo, only straight education is needed...
X=1 and x=2 are easy solutions you can guess and then you can show that g(x)=2^x - 2x > 0 for x>2
So x=2 is the biggest solution
Then you can show that g(x)
I am a class 11th students and I just got introduced to calculus few days ago and it's super interesting!!! Am more fascinated by the way this teacher switches to different marker in seconds!!!!!😳👍🏻
Finally I was waiting for an explanation of the 2 branches!! Now I get it
An claer and simple explanation of the two branches ! Thank you !
Well, 9:27 is already in the video description so I have nothing to do this time 😂
❤️ I love your videos! Thank You so much!
Recently there was a table "Derivatives For You" on the wall and now there is a painting "The Scream" by Edvard Munch. How are we to understand this?
"Maths for Fun" - "The Scream".
Pretty obvious, no?
he stuck the derivatives on his clothes so he no longer has that table on the wall
I’m beginning to suspect the Lambert family is paying you every time you make a video mentioning the name.
😂
Here surveyors use the Lambert conformal conic projection. It's the same Lambert.
@@pierreabbat6157 man, I wish I could get in on some of that Lambert money.
For some reason no one wants to use the Winchell conformal tesseract mapping.
This is one beautiful problem that links the obvious 2 solutions of 2x=2^x and the 2 forms of the W function. I wonder if there is a possible generalization here beyond 2.
Okay okay I’ll subscribe already. Can’t believe you made math interesting
Finally, the branches. The only thing we missed is how you don't need wolfram|alpha to figure out that -W_(0)(-ln2/2)/ln2 = 1. You coulda just gone with -ln2/2 = -ln2 · 2^(-1) = -ln2e^(-ln2)
Nice
Any ideas to find the -1 branch?
@@theuserings What do you mean?
I see, but what about -1 branch?
Thanks for the video
The fact that him making confused faces like he's geniuenly confused for teaching purposes is so hilarious 😂
I'm more confused at trying to decipher this in to English.. 😂
Where does the -1 in the "parameter" to the W function come from? What do the other values (not 0, 1) of that parameter represent when they are used? (Are they the complex roots of the original equation?)
Since the Lambert w function is an inverse function and it's not bijective you have to choose the branch. It happens to be that -1,1,0 are the easiest branches to work with. The intervals which the branches are are not consistent and the solutions are countably infinite. I suggest you look at the graph of the function and maybe that will clear up why.
5:44 "And that's a good place to stop."
This question *_*exists*_*
Logarithm:- *Did anyone summon me?*
I'm so glad you don't do sound effects anymore
Plz explain zeta function and riemann hypothesis 🙄
the last person who wanted to prove this in an open environment already died in January 2019 🙄
@@ymj5161 proving something and explaining what it is and what is states are two very different things...
@@anshumanagrawal346 lololol
@Castlier how are you calculating it I mean how did you know that it will converge at π²/6, is there any formula...
please kindly make videos on vector calculus.
Your videos are very interesting
iˣ=2 then x=?
X = 2^i (i guess, i don't really know)
x=log(i)(2)
=ln(2)/ln(i)
=ln(2)/(πi/2 + 2πni), n is an integer
So the principle value is ln(2)/(πi/2), which is the same as ln(4)/πi
That's what I think.
@K.SRIKANTH REDDY MATHEMATICS yes, but that's exactly what I said, just slightly rearranged.
Wait, sorry. I am an idiot. My bad
log_i(2)
Because:
log_i(i^x)=log_i(2)
so log_i cancels out the i in i^x
We can generalize to a ** (x - y) = x ** z. , where y >= 1, a and z > 0.
The equation to study is f(x) = ln(x) / (x - y). If a > 1, there always two distinct solutions. If a 1, there is only one solution.
Thanks sir 🙏
How to enter indices in W, in the Wolfiman calculator in the Lamberte formula?
Amazing!
I liked the graph of lambert W(x) function.
nice video, i liked it
This reminds me of how encapsulated funk takes place in real life and industries of skateboards.
Great video as always. What's Edvard Munch's The Scream doing in the background, tho?
😆
Bprp can u plz bring more content related to Recurance relations I'll appreciate it (at high school level) 😃😊
It's not the fact that this has two answers that surprises me. It's that the answer can produce integers, but have no analytic way to reduce it. Is there really no way to take your answer in the box and show those answers are 1 and 2 without approximating the W() function?
See the pinned comment
So ja great Video its so interesging. I'm finished my Abitur last Month but i Like to See thos Videos furthmore💅🤪🤖✨
Hello blackpenredpen, how are you? Im sorry but I would like to program the lambert w function, can you help me? Is there a site to visit that could help me. Thanks so mucho for the content by the way, you are so smart! Salute you!
We can also log 2^x/2=x
X.lg2--lg2=lgx
Lg2(x--1)=lgx now remove log
2x--2=x
X=2
I have a question about complex numbers :
If I have, m = a + bi & n = c + di , where a, b, c, d are real numbers and (i^2) = -1, then is,
n < m or, n > m?
The way to determine the `size` of complex numbers is to take their magnitude
M>N if |M|>|N|
|M| = sqrt(a^2+b^2)
|N| = sqrt(c^2+d^2)
You broke my mind when you multiplied both sides by -ln
great!
This was really fun!
Thanks bprp!
That lambert guy must have been a genius
Thanks for your hard work 😸 i wish you good luck , greetings from Ukraine ))
Could u tell me, why we take n tends to infinity in limit where is infinity already undefined.
because we want to see what happens to the function as it gets closer to infinity
very Nice.
Wow, that was an interesting way of going about it! Videos like yours inspire me to share my own maths tricks!
the equation 2ˣ= 2x can be solved in a simpler, graphical way: we plot y =2ˣ and y= 2x, after which we look at the intersections of the data with the graph and these points will be solutions to this equation.
therefore, x=1; x=2
i just looked at it b4 he did the math and found 1 & 2 as solutions. After he did the math I had a mental breakdown
An interesting generalization: a^x=a*x, 1=ax a^(-x)=ax e^(-x ln(a)),-ln(a)/a = -ln(a) x e^(-x ln(a)), so you have W(-ln(a)/a) in general. This means you have no real solution if -ln(a)/a0), one real solution if ln(a)/a=-1/e, two real solutions if -1/e
Is it possible to get an integral of 1-((x-1)/x)^x dx?
WolframAlpha just says it doesn't know.
This is what I missed by not majoring in math in college? Chuck in a W. Chuck in a e. Chuck in a Log or a ln. 1 can be anything, 2 has no meaning. Then out of left field, tan, then sin of theta, the sec. Obvious.
You are in love with lambert function🤩😍
Who isn’t? 😆
hi bprp! is there a W-1 = f (W0)? in other words, is it possible to find W-1 having found Wo?
Do you mean:
Is there an f(x), such as f(Wₒ(t)) = W₋₁(t)?
In other words:
Is it possible to express W₋₁(t) using Wₒ(t)?
@@lukandrate9866 exactly
That I am not sure. Unless we have the vertical distance as what I pointed out in the video.
Fun fact tho, W1(-1/e)=W0(-1/e)=1
Math is the thing where when you’re learning something knew, if you look away for a second, you will be lost.
I used to watch your videos in high school and couldnt understand a damn thing, now im in college studying cc and everything is clear now, mostly your calculus videos
I want an approximation of Lambert W function with respect to other existing functions qwq
Assuming integers ... 1 & 2
Took about 2 seconds to work out in my head
@7:00
OK, so if W0 for the solution gives X = 1, that means that W0(-ln(2)/2) = W(ln(1/sqrt(2))) = ln 2
This is the first time I think I've ever seen you put the result of the W function into something that is not just a Wolfram numerical answer
Is there an analytical way to come up with that result?
You missed a minus sign, its actually -ln(2). Now the reason is, technically, you can rewrite -ln(2)/2 as -ln(2)*e^(-ln(2)), now see that this is in the form of xe^x, hence, W(-ln(2)/2)=-ln2. And also notice, if you multiply and divide by 2, we get -2ln(2)/4, which is -ln(4)e^(-ln(4)), hence W(-ln(2)/2)=-ln(4)=-2ln(2) if you restrict the range of W(x) to y
Hey I have a pretty interesting question.can you solve this equation? "Logx(base a)=a^x”
Your awesome
You make me love highschool maths, especially while I'm high
I find it fascinating that such an innocent looking function as x(e^x) has a nose bleedingly crazy integral for its inverse.
What is ur hand sir
Sir I've been watching your videos and it really helped me develop interest in mathematics...earlier I scored 17/50 marks in previous maths test and now it's been 3 months the last test I got 48/50 and I'm the topper of my class.
Thank you Sir......
The man on the painting shows his confusion 😂
When I do productlog equations I don't convert the number to base e first. I do it in the original base and convert to base e or whatever afterwards using this change of base formula.
W[base b](x)=W(x ln(b))/ln(b)
I think its much simpler
Gigachads: Graph the equations and find the common points
If reported that the original scream painting goes missing, we know who we'll be seeing😁
When I saw the title
My mind: x=2
We should have different bases for the W function like how we can have different bases for logs. The one issue I could think of is notation because W has multiple real branches (ln has multiple branches but it only has on real branch).
Мне непонятно вот это уравнение:
W(x)*e^(W(x))=x
Откуда оно взялось?
~~~
I don't understand this equation:
W(x)*e^(W(x))=x
Where did it come from?
1.20 why the second one is true?
Got exp(-W(-ln(2)/2) /2 when I did it. Results in the same results when evaluating in wolfram alpha so guess correct. But no clue how to reduce it to bprp solution without just doing his derivation 😞
How do you compute values of W_(-1)(x) by hand?
Use newtons method but pick x₁= some negative number
@@lukandrate9866 That's only an approximation, I wanna compute the actual precise values by hand
@@theimmux3034 It is like computing precise value of ln(5) by hand. It is impossible, the only thing you can do is to make a very accurate approximation
@@lukandrate9866 The precise value in the case of ln5 would be ln5 and it would be what I was looking for
@@theimmux3034 Ok so why you don't like the precise value of W₋₁(-0.23) as W₋₁(-0.23)?
You can tell if you wanna just express the lambert function without using the lambert function. Not just saying "I wanna a precise value". But I think W(x) is better than an infinite sum expansion or some other non-elementary functions
it do has a simple way to solve it right.
A different solution of this equation can be seen on my new channel called L+M=N at ruclips.net/video/CC-L-OP71CM/видео.html
I wish I could double subscribe to you. You SUCH A GOOD TEACHER!!!
Thanks 😃
nice ;D
Question suggestion: x^2 - y^3 = 1, x and y are all integers, what are x and y? Note that there is only one answer for x and y, and you probably already found out x = 3 and y = 2
How does that work? (3)(2) - (2)(3) = 1???
@@weirdassbird I mean 3^2 - 2^3 = 1
multiply both sides by x
x * 2^x = 2 * x^2
at this point.. idk lol
Boom! I don't know Lambert W function has subscript, like wow that's how you define hidden number
what about n+n = n*n = n^n = n^^n = ... = n^...(infinite times)^n
Blue pen black pen red pen YAAAAAY!
Yes 😂👌👌
The function f(x) = ln(x) / (x-1) , x > 0 , with f(1) = 1 is strictly decreasing and range all positive numbers.
Solutions for a**x = a*x , a > 0 . x = 1 is always a solution, if a > 1 , f(x) = ln(a) is the only one else.
I am feeling really stupid. How can x be 1 in that function?
@@stevendeans4211 Do you mean f(x)? If so, because the limit of f when x approaches 1 is also 1.
@@stevendeans4211 He specified f(1) = 1 at the discontinuity. He isn’t putting x = 1 in the function. For a > 1, f(a) = ln(a)/(a - 1) , f(2) = ln(2) but I don’t see the point. You can graph this function in Desmos in three parts: 0 < x < 1, x = 1, and x > 1. The discontinuity at x = 1 is removed by specifying f(1) = 1. (It is also true that f(a) = ln(a)/(a - 1) for 0 < a < 1.) At x = 1 the limit of the function from right and left has the form 0/0 so L’Hôpital’s rule applies. (The righthand limit is -1 as is the lefthand limit. If this is confusing, it is the fault of the terminology. A good reference is Olmsted’s Advanced Calculus)
@@honortruth5227 I get it. I misread the nomenclature. Thanks
I had a similar problem in school once
X^2 = 2^x
X = π
In RieMann Geometry, π can be solved ☺️ 🤣 into Whole Numbers...
If you believe 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... = 2.
You can use this Concept to solve Pi 😜☺️ = 2 in RieMann Geometry Mathematics...
Can someone tell me wether my approach also works:
2^x = 2x | :x
x^(-1)*2^x = 2
e^(-lnx)*e^xln2 = 2 | ln(…)
-lnx+xln2 = ln2 | :ln2
-lnx:ln2 + x = 1
-log_2(x) + x = 1 | +log_2(x), -1
x-1 = log_2(x) | (…)^2
x^2 -2x +1 = x | -x
x^2-3x+1 = 0,
and solving this is just a quadratic.
Would that be a valid solution?
Nvm, it isn‘t but where is the mistake?
1 and 2 are the real answers that I got
Maestro
Well its exponential vs linear so u can just plug numbers till it stops working, 0 doesnt work, 1 works, 2 works, 3 doesnt and any number further wont either, hence answer is 1 and 2
Please solve x²=2^x ❤
How to solve a slightly more difficult case 2^x=2x+5?
Does this mean you can get infinitely many answers with any n?
Yes if you allow complex solutions.
x = 2^(x-1)
I got x = (W[ln{2}2x^2])/ln{2}. Can you tell me what I did wrong
When I hear lambert function I instantly change video ,why? Because if I were alone I wouldn't be able to find the answer to lambert function by heart , so If at the end I have to search for a plot why on earth couldn't I plot the original functions and just see where they meet at? It's like we do a lot of effort to do it formally but we end up doing what we were avoiding at the beginning. ....so....