all solutions to 2^x-3x-1=0 (transcendental equation)
Vložit
- čas přidán 22. 10. 2023
- Learn how to solve this transcendental equation 2^x-3x-1=0 with the Lambert W function! This is definitely not a regular math equation that you see in school!
Subscribe to @blackpenredpen so you don't miss the "fixed point iteration" video!
Lambert W function introduction: • Lambert W Function (do...
For another challenge, check out this video: • so you want a VERY HAR...
🛍 Shop my math t-shirt & hoodies: amzn.to/3qBeuw6
----------------------------------------
💪 Support the channel and get featured in the video description by becoming a patron: / blackpenredpen
AP-IP Ben Delo Marcelo Silva Ehud Ezra 3blue1brown Joseph DeStefano
Mark Mann Philippe Zivan Sussholz AlkanKondo89 Adam Quentin Colley
Gary Tugan Stephen Stofka Alex Dodge Gary Huntress Alison Hansel
Delton Ding Klemens Christopher Ursich buda Vincent Poirier Toma Kolev
Tibees Bob Maxell A.B.C Cristian Navarro Jan Bormans Galios Theorist
Robert Sundling Stuart Wurtman Nick S William O'Corrigan Ron Jensen
Patapom Daniel Kahn Lea Denise James Steven Ridgway Jason Bucata
Mirko Schultz xeioex Jean-Manuel Izaret Jason Clement robert huff
Julian Moik Hiu Fung Lam Ronald Bryant Jan Řehák Robert Toltowicz
Angel Marchev, Jr. Antonio Luiz Brandao SquadriWilliam Laderer Natasha Caron Yevonnael Andrew Angel Marchev Sam Padilla ScienceBro Ryan Bingham
Papa Fassi Hoang Nguyen Arun Iyengar Michael Miller Sandun Panthangi
Skorj Olafsen Riley Faison Rolf Waefler Andrew Jack Ingham P Dwag Jason Kevin Davis Franco Tejero Klasseh Khornate Richard Payne Witek Mozga Brandon Smith Jan Lukas Kiermeyer Ralph Sato Kischel Nair Carsten Milkau Keith Kevelson Christoph Hipp Witness Forest Roberts Abd-alijaleel Laraki Anthony Bruent-Bessette Samuel Gronwold Tyler Bennett christopher careta Troy R Katy Lap C Niltiac, Stealer of Souls Jon Daivd R meh Tom Noa Overloop Jude Khine R3factor. Jasmine Soni L wan na Marcelo Silva Samuel N Anthony Rogers Mark Madsen Robert Da Costa Nathan Kean Timothy Raymond Gregory Henzie Lauren Danielle Nadia Rahman Evangline McDonald Yuval Blatt Zahra Parhoun Hassan Alashoor Kaakaopuupod bbaa Joash Hall Andr3w11235 Cadentato Joe Wisniewski Eric Maximilian Mecke Jorge Casanova Alexis Villalobos Jm Law Siang Qi Tancredi Casoli Steven Sea Shanties Nick K Daniel Akheterov Roy Logan
----------------------------------------
Thank you all!
#algebra #math #blackpenredpen
so you want a VERY HARD math question?!
czcams.com/video/Rg3dBosfZ3Y/video.html
2:22 Learning is to eliminate enough unsuccessful paths. More of that.
Best math variables:
α ❌
Fish ✅
🐟
Fish ❎
X ✅
Hehe I am your sister 😅😅😅
0:38
"First off, we need to have a fish - I call this the _alpha_ fish!"
lmao
When you realise that x=0 is also "an" solution
Wdym “when you realise” it’s immediately obvious
@@zimzimal8547 not for everyone bastard
“a solution”, not “an solution”.
Baka mitai
@@zimzimal8547 no it isn't
It's clear from the testing of easy candidate values that one solution is 0 and that there's another between 3 and 4.
Interesting to be shown how to solve it properly, thank you.
Testing of easy candidate values is the best name I've heard for "Throwing numbers at the function and see what sticks" :))
@@flawnel
Great, that's something. Cheers :)
Plug n chug am I right
@@eldunari6676
I prefer that to suck it and see. ;)
I just ran into your video, I remember following your channel ages ago, being a student and trying to figure out how to solve calculus exercises.
Now I'm graduated, but still enjoyed the video, and it took me back to that time, living off of coffee at the library, sharpening my pencil and the table full of the residual eraser lol
Welcome back! Hope all is well for you : )
A nifty general trick to learn is that once you get 2^x = 3x+1, you can use the substitution y = 3x+1 to get rid of that annoying constant 1. This gives you y = 2^((y-1)/3) = 2^(y/3) * 2^(-1/3) or 2^(-1/3) = y*2^(-y/3) = y*e^(-y(ln 2)/3). Then multiplying both sides by -(ln 2)/3 gives you the desired "fish" z = -y(ln 2)/3 on the right-hand side.
TBH, we have become computer assisted scientists and engineers with ever weakening math muscle. I once was tasked to setup a monte-carlo for the aviation industry to estimate the lifetime of a jet engine. I remember I replaced the whole monte-carlo simulation by solving an integral with maxima (compound Weibull). I wondered why no one in the R&D thought about it. Your video gives a clue.
This is one of the big reasons why I want to be good at math
Physics of photonic crystals is the same way. Every paper published is all numerical simulations or calculations with no fundamental equations to guide the reader. Just hand waving, ad-hoc arguments. Currently in a back in forth with Physical Review A because their "expert" referee did a simulation and it didn't agree with our analytical formulation. I hope your work landed you some nice job security. You have a valuable skill.
You were fortunate in having an amenable problem. Most problems are not like that.
Most transcendental equations are not analytically solvable, so there is little room for “math” math. As others have stated, the overwhelming majority of problems are only solvable by approximation.
I get what you mean by "math" muscle. What do you think is the best way to train it?(a high school student wishing to enter the number 1 or 2 best engineer school in my country and who is passionate of maths)
Glad to be able to get this one on my own, my working was nowhere near as neat and simplified as yours though
Nicely done!
Is Lambert W really a function? I see the concept is useful and it's worthwhile giving it a name, but is it a function that humans can calculate?
I keep seeing it in things I ask WolframAlpha. It's starting to feel like I'm going to have to learn how to calculate using it :/ so interested in any responses you get..
It doesn't have an elementary form, so it can't be calculated exactly. But the same is true of functions like ln or sine. And just like with those, there are ways to get closer and closer to the exact answer.
That said, just like with those functions, you're mostly expected to put W(x) into a calculator. It's just that most regular calculators don't have that function built in, so you have to use websites like Wolfram Alpha.
In short, the productlog or Lambert W function can be useful. There just isn't any way to solve xe^x = a without it. It can'e be broken down into any simpler functions.
If by calculate, you mean express in terms of a finite number of operations of arithmetic, integer powers, and roots, then there is no way to do it.
But if by calculate, you allow for techniques such as iteration and infinite series, such that you can calculate it accurately enough for your purposes, then there certainly is a way to calculate it. Steve has a video on it here:
czcams.com/video/Qb7JITsbyKs/video.html
I finally understand a video, im progressing thanks to you
keep up the good work :)
Productlog is so useful.I don't know why they don't teach it in high school.
It’s a bit useless in hs math for the most part, that and trying to understand it isn’t the most hs friendly when most people in highschool already struggle with other, more simple, concepts
most calculators dont have it (i think)
it also cant be computed by hand
@@askandpushpaltiwary8537 well you COULD always use newtons method of solving an equation but if you don’t have 20 hours of time, then only some can be solved by hand. An example would be 2*ln(2) which is equivalent to ln(2)*e^ln(2) but that’s a very specific case and wouldn’t be practical at all so yeah.
@@askandpushpaltiwary8537 thats a good point, but you dont need to calculate it. Its good to have the exact form not just the first three digits of an irrational (most likely) number
@@elquesohombre9931 Well good work takes time🤣🤣
I wonder, is there a way to get 0 as a solution using something similar to the first method of iteration? That would be interesting.
Taylor expansion around x=0, because 2^x expanse is converge in R so the 2^x-3x-1 also converge.
And when you expand the constant term is eliminated so the x can be sub out to equal to 0, but the remain infinitive polynomial is too complex to be reduced
In fact the general: a^x - bx - 1 = 0 always have a root x = 0.
Except if a=0
To the people confused by the fish: don't worry, fish doesn't have to be real.
I would explain more, if I could, but well... it's complex.👌
This technique seems like it might be helpful for solving the collatz conjecture.
the fish function is my favorite
fixed point iteration method video plz
What is the backstory to Lambert choosing to call this function W?
7:37 Now I'm wondering if you can't shuffle the W(-ln2/...) contents with similar tricks
(And presumably it has something to do with i?)
Well i drew graphs of y=2^x and y=3x+1. The intersection point is somewhere between x=3 and x=4
in epsilon delta there should be simple epsilon/delta=f'(a)
✌️😁✌️( 2^x-3x-1) =0
Multiply both side with a 0 so, 0 = 0 ✌️😁✌️
x1=0, x2～=3.53767 by for loop
Isnt Lambert W function aproximated via fixed-point iteration? Or there's a way to find the solutions analytically. Also great video💯💯
Well it's easy to see that -ln2/(3cbrt2)=-ln2/3*e^(-ln2/3) so the zeroth branch can be found analytically in this case.
I love the fish gimmick
You may want to solve this (other then graphically): x*4^(1/x)) + (4^x)/x -12=0; real solutions are 0.5 and 2...
Hey where’s the sphere mic?
@blackpenredpen can you prove this : integral from 0 to infinity of (root x times ln(x)) / (1+x^2) dx = (pi)^2 / (2root2)
Using complex Analysis it is easy.
A square Root combined with a logarithm means that Share a branch cut Over the positive real axis.
Consider a keyhole contour with the complex function f(z) = (exp(log(z)/2)*log(z))/(1+z^2). The contour are composed by two straight lines and two centered circles.
It is easy to show that Over the two circles the modulus of f(z) are bounded by:
f(z)
Linear equation
What is the Integration of -cotx cosec²x? My book gives answer (cot²x)/2 (by u sub) but d/dx of (csc²x)/2 is also -cotx csc²x. Please reply
This is why the +C is important because (cot²x)/2 and (csc²x)/2 both have the required derivative [they just differ by ½ which you can show via trig identities]. Hope that helps! :)
(cosec^2 x)/2
= (1 + cot^2 x)/2
= 1/2 + (cot^2 x)/2
The two answers are said to be off by a constant (1/2 in this case) which are considered equivalent as far as indefinite integrals are concerned due to the +c
@@Ninja20704 thanks bro. This freaking formula, I totally forgot. As it was useless in our syllabus questions.
@@MathNerd1729 thanks bro for the help
at 7:36 is ln not log. idk if was an error or not because get the same answer
one fish, two fish, red fish, blue fish, alpha fish, betta fish.
4:22 we there isn't base for the function as in the normal log
Log base x
W base x??
If A = 2023(10^n) + 1, where n is a positive integral, then can A be the square of an integral?
I love math
On the left side
W((-ln2)/(3(2^1/3)) can be rewritten as
W((-1/3)ln2*e^((-1/3)ln2)) which evaluates to
(-1/3)ln2
This gives the zero solution
yup!
0:02 me: what’s loy?
0:03 me: oh that’s just log
What if (for example) n=2? Does the W-Lambert function not also give a new value than? Or does it only give a value we already got? Or does it give no value at all?
Little late here, but only the 0 and -1 branches can give real solutions so if n=2 you would get a complex one.
me after bprp says: "Lambert W function", I know video'll be a banger
دمت گرم خدایی❤
Hello, great solution, but I didn't understand what does n=0, n=-1 etc. mean. Could someone explain?
What happens if n = 1, or 2, or a fraction, or an irrational number, or a complex number?
n=0 is the only branch that stays real for its entire domain, while n=-1 can give real answers within a certain range. The other branches will always give out complex results.
The branches are distinct and don't have in between numbers. Long story short it's all about how complex multiplication works - those quantized 2pi rotations.
Innit like (2^x-3x-1)^x=0
Instead of writing it as -1/(3cbrt(2)), couldn't you have written it like (-1/3)2^(-1/3)? Then using the ln2 and a base change in the next step it would turn into (-ln2/3)e^(-ln2/3), which means the Lambert W would reduce it down to simply -x-1/3 = -ln2/3
You get -x-1/3 = -1/3 if you do that, in which case you only get x = 0 because lambert W identity works like this: W_(0)(xe^x) = x when x>= -1, W_(-1)(xe^x) = x when x
Can you try solving x^2 + 2^x = 0?
2😎🤙
@@JoaoPedro-cv7hn4+4=0? 🤨
this video is similar: czcams.com/video/ndA0sF_0Rwk/video.htmlsi=YfV8rBOIopB6OK3F
Isn't a zero also an answer? :]
oh dayum
You must have missed it because the video does say so.
@@rogerkearns8094 ah, right, it really does in 7:37! I missed that. Also it's not written at the end as a solution, so that got me confused. ;)
yes, and it is in the video
@@Lohikaarme1984it technically still is in the final answer. When we wrote the final answer with the lambert W function, with an arbitary branch n, that is not one solution but an entire set of solutions, which does contain 0.
At a glance you know the answer is between 3 and 4. If 3 significant figures is OK create a spreadsheet and vary X from 3 to 4 in 0.001 steps. How is that different, really, from asking Wolfram Alpha to solve it? But it was a fun ride. 😅
What is W?
what is w?
Him saying alphafish
Me a chess player:😂😂
Great👍
No.
No, I do not want that.
I don't know why the algorithm decided I *did* , but I don't.
I want other people who want the answer to be able to find it. Congratulations! Well, done! Now never speak to me again Mathman.
3^x+x=30, Solve the value of x thank you
Solving with graphics
or the fixed point iteration! : )
I think this problem can be solved using calculus. Is it a correct method?
i like it!
was terminating it with log_2[1 + 3] intentional?
but a graph of this in wolfram alpha....
0 = -v + 1 + 3 log_2[v]
Me after I realise that 0 is also an answer
👁️👄👁️
👁👄👁
I've been thinking about this problem for long, could you make a video about it?
Solve for a, b, c and d:
a=bc+bd+cd
b=ac+ad+cd
c=ab+ad+bd
d=ab+ac+bc
I suppose to get a start on where solutions might be, I'd let a=b=c=d (since it's clear there exist solutions where this is true), so a=3a^2. 3a^2-a=0. a(3a-1)=0, a=0 or 1/3.
Then loosen it a bit, a=b, c=d. so a=ac+ac+cc=2ac+c^2, c=aa+ac+ac.= 2ac+a^2. difference in those is a-c = c^2-a^2 = (c-a)(c+a), divide by (a-c) (only valid if a not equal c) to get 1=-(c+a), c = -a-1. back into the earlier equation, a = 2a(-a-1) + (-a-1)^2. becomes a^2+a-1 = 0, solutions are the golden ratio, a=-phi c=1/phi and then the other way around.
The same kind of cancellation to get relationships works with all 4 variables (c-d = ad+bd-ac-bc = (a+b)(d-c)), it's just progressively messier to work back to quadratic equations. You'd also want to check the intermediate step a=b, c not equal d, and a=b=c, to make sure you find all solutions.
❤❤❤
You already made a video of creating a formula for a^x+bx+c=0 1month ahead just use that 🐟🐟🐟
can't you just take the x root of 2^x, and the x root of 0=0.
so x =-0.5
why is't it x=0?
Constant
x=0 is a sol but there's another one a bit more than 3
In minute 4:00 Why didn't you write the right side of the equation as (-1/3).2^(-1/3) ????!!!!!! So it made the solution easier?!?!
just remove the x?
where's the x?
Wat da fish doin’
CAN YOU PLS BRING HARD GEOMETRY FOR ONCE????????
honestly i think trying random numbers would be easier
-1 is also a valid answer
1/2 + 2 = 0 ?
Just draw the plot of the function to get approximate solutions and then iterate nu merically. No need for Lambert W or Mr Wolfram's nonsense.
ladies and gentlemen, isn't it???
Why did you shave your beard 😭😭
A challenge for you
Prove that log2=0.3010
Are you a strict maths teacher
✌️😁✌️2+2= , my answer is you know ✌️😁✌️
If god is all powerful, can he divide 1 by zero?
FEESH
hi
🐟
That sounds fish-e 😝
Is it possible? Is it even a question 😂😂. If youre watching bprp you alredy know its possible.
hahaha!
First comment pin
rohadj meg youtube a reklámaiddal együtt
geek
touch grass
You mean you did all of that work for nothing?!
?
@@blackpenredpen You did a lot of calculation to end with the answer being zero.
Calculators can’t explain the math, only people can…