I couldn't solve x^x^x=2, so I solved x^x^(x+1)=2 instead
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- Опубліковано 28 жов 2023
- We will solve a tetration of a tetration equation x^x^(x+1)=2, which becomes (x^x)^(x^x)=2. This equation has one real solution and it's solvable by using the Lambert W function. Check out how I used Newton's method to get the approximated root to x^x^x=2 • solving the tetration ...
What’s the Lambert W function? • Lambert W Function (do...
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Thank you all!
![[柴犬ASMR]曼玉Manyu&小白Bai 毛发护理Spa asmr](http://i.ytimg.com/vi/0TsXQ7z2Dh4/mqdefault.jpg)
so you want both solutions to 2^x-3x-1=0 (not regular algebra)
ua-cam.com/video/GJbzsmccFtw/v-deo.html
I recently came across your video on how to solve integrals.
Although I have completed my bachelor's in computer science and currently pursuing web3 development... I am always interested in Mathematics.
I want to know and learn mathematics from 0 to 100.
From the very basic to the very advanced.
I am ready to put in years into it. Like I'm not literally free in my life and want to learn this. I am currently 23 years old and wish to complete this mathematics in the coming 5 to 8 years. Or even maximum 20 years.
I want to know what resources I should use. What books I must use. And what should be proper roadmap for it.
My reason for learning maths this way is because I want to grasp all the topics 100% and take to the next level by doing a research in Physics. [I will also study Physics likewise]
But I need to learn maths first.
Can you guide me? (I know it's impossible to guide like this, but still can you sometimes make a post related to 0 to 100 maths roadmap)
@@Mr.Pro-aksh26I would also love that
I don’t fear the math, I fear the amount of marker boxes he has in the back
It can solve it numerically. The truth is "it cannot interpret the input" as it's written on your screenshot. If you explicitly write the expression using the parenthesis, it will solve it.
I was about to say the same thing. However I'm glad the typo created a new video :D
Yes the input: " given x^(x^(x+1))= 2 find x " returns x~=1.379970...
Wait but if I type "x^x^(x+1)=2" then it does interpret it correctly
Clickbait✅
@@elmarhoppieland actually, you're right. In my case, it also worked out, but I was thinking that it's due to the fact that I have a premium version of Wolframalfa.
6:04 "It is not because it's cool. It's supercool."
This video starts a bit fishy but ends up as yet another great catch by our favorite problem solver.
I know it will sound complex, but what if the fish is not real?
A-hahahahahahahahahahahahahaha! I SEE WHAT YOU DID THERE!
When you get the green marker out, things are getting serious. Because you have reached the scenario of the four colour map problem :)
I’m always looking forward to listen the usual: « And you take your fish back! » Thank you Coach Bprp!
😆
4:20 Lembedari function of the Lembedari function 😂
Nice video as always! Now I'll be glad if you try to solve this equation:
x^x^(x+2) = 2
It can not be solved by using LambertW function, but it is solvable in terms of superroot! So I hope this experience will be interesting for you!
what is the solution?
i tried looking into a general solution for x in terms of m and k
x^(x^(x+k) = m
this is the closest i got:
lnx (e^ ln(x + k) = lnm,
no clue how to transform x+k into x or vice-versa and then use lambert, then substituting the values for m and k
@Tritibellum Oh, there is no general solution to your equation because it requires different superroots depending on k parameter. Like comparing these 2 equations:
x^x^x = 2 and x^x^(x+1) = 2
Solution for the left equation is 3rd superroot of 2, while for other one is 2nd superroot of 2nd superroot of 2 (which is not equal to 4th superroot of 2!). See, it even has a various amount of superroots that can not be merged into a single one.
And, by the way, the first equation is not solvable in terms of LambertW function, while second is.
@@fantisciousOh, if I show the solution before bprp finds it himself, I will just spoil it for him and the audience. But I can give hints on how to approach to solution ;)
To solve this equation, I used these 2 cool properties of tetration:
(ᵃx)^(ᵇx) = (ᵇ⁺¹x)^(ᵃ⁻¹x)
(ᵃx)^x = ¹⁺¹/ᵃ(ᵃx)
You can try and prove these properties by yourself, and I hope it will bring you same delight as for me, when I first time discovered and proved them.
Nice job👍 have never seen the terms ln and e being manipulated in that sort of way. I had also never seen the lambertw function before as something that is similar to a natural logarithm.
PROF UR videos extremely interesting.
That's a W right there.
With your videos I will eventually unlock the secrets of the numerical universe. But I'll never be able to answer the primordial question: why do happy-faced fish have horns?
Those are eyebrows 😆
I always enjoy your videos, but I think my favorites are the ones where you make stuff up (like the fish) to keep your viewers from getting confused by yet another variable. Placeholders for the win.
I appreciate that! Thanks.
Great problem and solution! Wish I could give more than 1 thumb up!
2:45 - that was a smart move O_O
Can we get more algebra like this
As a law student that just likes math I have no idea what just happened. I'm just happy for the fish
As a highschool student, I agree
Wow, this is impressive 👍
Excellent!
Hey blackpenredpen, can you tetrate i to the length of i? Like i with an exponent tower of i's that is i long.
I'm wondering how to solve an equation written as a^x=(x+b)(x+c) b≠c. Could you please make a video about it? Love from Turkey 🇹🇷
Nice
i think thats not possible with w function or smth
@@matheus-f yeah you're right, Such an equation can only have a numerical solution.
@@matheus-f For example, it is impossible to find the roots of a 5th degree polynomial by separating it into radicals. for example -b/2a +- sqrt(b^2 -4ac)/2a... I thought maybe there was a solution to this.
I have an idea for when one takes the Lambert function of a Lambert function: The Lamborghini function, noted by the picture of a sleek car instead of a letter like "W".
or just W²(x)
IDK, people might think that's like cos^2(x).@@jellymath
@@Mark16v15Oh I see, although it is a convention that fⁿ(x) is the composition of f with itself n times, so W²(x) = W(W(x)), by convention
sin²(x) is just a special situation where writing sin(x)² is more painful, and sin²(x) generally comes up somewhat often in math, like the famous pythagorean identity sin²(x) + cos²(x) = 1. And I think sin(sin(x)) is less so common
Anyone correct me if I'm accidentally spreading misinformation haha
@@jellymath I don't know whether or not this is common, but I've never seen fⁿ(x) used to represent composition, only (f^-1)(x) used to represent the inverse of a function.
So how start the mathematics ofline classes by you my dear Sir??
That moment when a mathematician has run out of letters, both latin and greek, to represent various obscure variables in math, and we resort to pictographs as variables. 😂
I don't think it's so much that he's run out of letters, it's just that we don't have a letter that conveys the idea that it could be a number or equation, but is probably an equation, and ANY equation.
Thus although if y = x^fish, that means dy/dx = fish*x^(fish-1), since we rarely run into the fish being an equation in this instance, we use letters like "a" which usually is for a number, especially since derivatives are taught to first year calc students, whereas the Lambert function is taught way further down the road.
LOL the fish is tradition at this point... 😂
@@Mark16v15 In other words, all the greek and latin letters are already largely taken as they usually represent various other things in math, so he's using a fish because nothing else uses a fish pictograph. Or to put it more simply, we've run out of letters and are resorting to pictographs
Beautiful 🔥🔥🔥🔥
To solve x^x^x=a, you need to invent a new function similar to the Lambert W function: such that X(ye^ye^y)=y
Bro I love you❤❤❤❤❤❤❤❤❤
Is there any difference between x^(x^(x+1)) = 2 and x^((x+1)^(x)) = 2?
purely curious, when will we have 100 differential equation or like 100 real analysis
How long before the few transforms of the W function are added to Wolfram Alpha so it can solve these equations?
I just tried it and it answered correctly
Wolfram can solve this equation.
It just has trouble correctly interpreting the input without a few more clarification parentheses
Now I really need to decide which fish should I cook for dinner
then can you please do the general form? x^x^(x+a)=b !!!
So X=1/W after everything cancels out :)
Great !
It would be really helpful if you clearly and slowly pronounced "Lambert W function" once at the beginning. I was confused for quite a while, and the subtitles didn't help. Thanks for mentioning it in the description.
Horrible math teacher. Fish function?? WTF.
maybe just terrible student
What's Sigma(i=1,b) d^i/dx^i x^^(tetration)b
مرحبا سيدي كنت اريد ان اسألك هل نحن في تفاضلي الدالة نقوم بضرب dy/dx في dx ام ماذا ؟
وان كنا نقوم بذلك فكيف نقوم بضرب dy/dx في dx علي الرغم من ان dy/dx مجرد رمز وانه ليس عملية ؟
This is kind of a never ending problem! There are concepts like "differential forms" and "operator theory" in more advanced math which give different interpretations.
My advice: dy/dx is an operation which takes functions and gives you the derivative. When you multiply by dx, you are just doing a weird trick which saves time.
For example lets say you have dy/dx = y. You might have seen:
1/y dy = dx and then integration. But this is unnecessary! Instead you can do: 1/y dy/dx = 1 and then integrate both sides as int(1/y dy/dx * dx) = int(dx). The trick is that this is the same as last situation by substituting u=y(x), so du=dy/dx * dx. So it just saved time!
I am so happy I can’t even begin to understand this.
Next time, after solving a difficult solution, do the Dr Payem's "Chalk Throw"!
Nice!
Thanks!
x^x^x=2 is REALLY close to "let x=1.476684 what is x^x^x"
also WA doesn't give that ouput anymore. it gives an approx but it still times out trying to get the exact answer.
Geniel, hermoso, fabuloso, divertido, ...
its already been a long time to switch channels name to blackpenredpenbluepen
wow
I dont understand anything but I like watching this lol
Could you not plot x^x^x into a graphing calculator? I feel like i am missing something, could someone clarify this please.
Yes, the clarification is that he is seeking an exact solution in terms of known functions, rather than a numerical approximation to the solution. It is the difference between pi and 3.14... . Exact solutions are useful in pure math as the methods to obtain them lead to new ideas. If you are solving an engineering problem (e.g. building something in real life) then numerical solutions tend to be fine. Any single variable problem of the form "Find x in the interval [a,b] such that f(x) = 0" can be solved up to an error E (some small number) in a number of computational steps that scales as (b - a) / E.
I want more videos featuring the alpha fish 🐟
Double lambert W function
Imagine working with tetration series n^(n+1)^(n+2)^(...)
converge to infinity, because if the value of n < infinity, will have a (n+x)^y, that y will make the value of exponential positive, and if a value of the serie get positive, the serie converge to infinity. i think this without demonstration, i really no know if im right
note: if n=infinity, the serie will converge to infinity too
I love the fish ❤❤
W(W(ln 2) = world war ln 2
How old are the students you teach?
The Lambert W is always such a cop-out...
We ❤ the fish!
Is this the first video where he’s shown an ungodly amount of markers in the background
No. 😆
It’s a business expense
I love the alpha fish
I wouldn't be surprised if blackpenredpen genuinely is better at calculating math problems than WolframAlpha.
Woah, looks like I need to invent X(n), which is the inverse of ne^(ne^n), and if x^x^x = y, then x = e^X(ln(y))
Wait, how is 1.3 (+ other decimals) tetrated 3 times (and the fact that last power is 2.3...) equal to two tho?
Please do let me know if i am being plainly dumb, or if i am latching on to something😅
peace
I know this is pretty late but
in tetration you start by calculating the power tower from the top so first you take 1.37 raised to 2.37 which is approximately 2.11 then you take 1.37 raised to 2.11 which is approximately 2
@@AttyPatty3 thx 😄
Nice!.wolfram can solve it (at least the numerical value) if you help him and write : x^(x+1)=ln2/lnx
you can "solve" any single variable problem numerically. "Find x in the interval [a,b] such that f(x) = 0" can be solved up to an error E (some small number) in a number of computational steps that scales as (b - a) / E. (just make a grid with that many points and evaluate f(x) at each of the points). The computational cost of numerical methods scales exponentially with the number of variables, and many real-world problems have a large number of variables (e.g. simulating the price of 1000 items in an economy, or simulating a piece of cloth made of a million points that each have a position in 3D space, etc). This is one reason that clever methods for exact solutions are useful.
Sir why don't you tell us about your degrees
fish for life
If you switch to "math input" wolfram alpha can solve it just fine.
With the premium membership plan , it's possible 👀
I see!
It says: cannot interpret the input
On Wolfram Alpha:
x^(x^(x+1)) = 2
Numerical solution
x ≈ 1.37997039661106...
3^x+x=30, Solve the value of x thank you
3
I forgot to mention, I want you to prove that x is 3, I'm curious how to find out
Sir, please solve x^(x+1)=2
The original problem is somewhere around 1.4766843
Me watching the whole video without knowing what is the lmperdabliu function
If x^x^x=2, x=1,41421356231... which approximately equals square root of 2. Now, it is widely known that infinite tetrate od sqrt (2) is 2, but I have not supposed it becomes so close even with sqrt(2)^^3.
5:30 I suppose you didn't prove that identity.
Can we solve x^x^(x+2)=2?
ua-cam.com/video/ef-TSTg-2sI/v-deo.htmlsi=fHtkI67xFtStYwa3
You are going to help me reenter college and earn my degree while I'm in the army. Thanks
Sir I am from India
I don't understand how to pass from the second line to the 3rd line.
a^(bc) = (a^b)^c = (a^c)^b, a, b, c > 0
We have a = x, b = x^x and c = x with x > 0*. You apply this property in the second line and you have the third line.
* x = 0 cannot be a solution since x^(x + 1) = 0^(0 + 1) = 0¹ = 0 and then we have x^(x^(x + 1)) = 0⁰, which isn't defined.
x < 0 cannot be a solution since
- for x rational or irrational, we get complex values instead of real ones (the equation is equaled to 2 and 2 is real).
- for x integer, we have
• x = -1 implies (-1)^((-1)^(-1 + 1)) = (-1)^((-1)⁰) = (-1)¹ = -1 < 0 and 2 > 0.
• x = -2 implies (-2)^((-2)^(-2 + 1)) = (-2)^((-2)^(-1)) = (-2)^(1/(-2)¹) = (-2)^(-½) = 1/(-2)½, which is not real and 2 is real.
• x = -3 implies (-3)^((-3)^(-3 + 1)) = (-3)^((-3)^(-2)) = (-3)^(-⅑) = 1/(-3)⅑ < 0 and 2 > 0.
We see that odd negative integers give negative values (and 2 is positive) and even negative ones give complex values (and 2 is real). Therefore, if x isn't zero nor negative, then it must be positive (x > 0).
Fun fact: W(ln(x))=ln(ssrt(x))
x=W(W(ln2))=1,37997....
❤
can you solve this [ 2^x+x²=0 ] ???
Haters will ask why you cannot cancel the Ws at the end.
What is W ?
The Lambert W function. You can Google it for a good answer.
Sorry its wrong. Doesn't for x=5 orso. Then it becomes x exp 125. But alright i guess with either 2 or 1 it works.
Simpler way to solve is to notice
x^x^(x+1) = (x^x)^(x^x)
Which reduces the problem to 2 iterations of solving x^x = c
can you explain how x^x^(x+1)=(x^x)^(x^x) works please?
look at the video@@Zdv0rz
I have a video idea
Is there a closed form solution to
∞
Σ (n^(-n-1))
n=1
or
∞
Σ (n^{1-n))
n=1
Wolfram alpha can only give a decimal aoriximation
There's most definitely an exact solution to both, what you might be asking if whether there's a closed form of that solution.
@@erikkonstas yeah I just didn't know that term
what is the solution
@@fantiscious I can't solve it and neither can wolfram alpha, but bprp is good at math obviously so maybe he can figure it out
Easy... x = scbrt(2)
That’s some complex “fun maths” there
So, I discovered something and want to call it "Euler's Reciprocal Loop".
e^(i•pi) = (e^pi)^i because of power law, lets say e^pi is any number, lets call it x.
x^i = i-th root of x^(i•i), i•i is equal to -1, so we have i-th root of x^-1, which is the reciprocal of x.
(i-th root of 1)/(i-th root of x). Any root of 1 (even complex numbers) is just 1 itself. So we have 1/(i-th root of e^pi) because we have x=e^pi earlier.
For the root-power law (n-th root of x^m is equal to x^(m/n)), we have 1/e^(pi/i).
1 raised to any number (even complex numbers) is just 1 itself, we can use this to have (1^(pi/i))/e^(pi/i).
Due to having the same power (pi/i), we can have (1/e) ^(pi/i).
Now we dont need to worry about 1/e, we just need to get pi/i = i•pi because we need it to have 1/e^(i•pi). Here comes the a bit hard part.
Pi is basically 3,so we can use it to have the square root of pi squared, or square root of 9.i is just square root of -1.
Due to root-fraction law ((square root of x)/(square root of y) = square root of (x/y)) , we can use it to have square root of 9/(-1) which is just -9. the square root of -9 is i•sq(9) or i•pi!
Now we finally have (1/e) ^(i•pi), simplify it out and we have 1/e^(i•pi)!
For the proof, we gonna check the results.
e^i•pi is just -1, so 1/(-1) is just -1, or e^i•pi.
We can do this forever and we still have -1.
Be careful when manipulating complex exponents. Some exponent laws don't hold for complex numbers, for instance a^b^(1/c) ≠ a^(1/c)^b
Sorry, but you violated a lot of rules related to exponents and roots. When dealing with complex numbers, the rules of exponents do not apply. Similarly you cannot combine 2 square roots into a single one when negative numbers are involved. It's a conceptual error.
I wonder why you never explained, what W() is, how it works.
I have. Check this out ua-cam.com/video/Qb7JITsbyKs/v-deo.htmlsi=05xAbHwlaj9ChD_W
Wolfram alpha does give a numerical solution, x = 1.3799..
Sir , by any chance u could solve the equation. D/dx ( d/dx ( intergal where the max is 7 the minimum is -2 , ln(e^x-1)
Don't conjure the math man!
What the hell is this fish business?!
headache..
🌹🌹🌹
Wish you didn't have a fish!
is this a joke? the 3rd line is not the same as the 2nd line
x^(x*x^x) != (x^x)^(x^x)
not always ...
could you integrate (sin^-1(x))* e^2x
Im going to have to mark down points for no absolute value on the input for ln.
wait where should there be an absolute value?
@@RunstarHomer There shouldn’t, x as well as all the expressions in it may well be complex, so the natural log, if we choose the principal branch as is customary, is taken to be a function from C-{0} to C. On the contrary, it makes no sense to require an absolute value in the argument of ln, as for example ln(-1)=iπ ≠ ln(1)=0.
You cannot just add absolute values where ever you like because it can completely change the domain. For example ln(x^3)≠3ln|x| because the domains are no longer the same.
Absolute values are to be added only when it is required to keep the domain consistent. Such as ln(x^2)=2ln|x| or integrating 1/x.
x^x is always positive over its domain x>0. So there is completely no point in adding it.
You have double double-u now, do it once more to make www lol.