important notation note: when one writes, say, x^x^x^x without any parentheses, by default the computation is performed from the top down, so x^x^x^x = x^(x^(x^x)) and not ((x^x)^x)^x. if you launch the iteration considering the latter (that is, [z_0 = i, z_{n+1} = z_{n} ^ i] instead of the correct [z_0 = i, z_{n+1} = i^z_{n}]) you don't converge to that value you've found in the video, but fall into the 3-cycle "0.20..., 6.12e-17-i, 4.81...)"
@@katakana1 not sure I got their point about "local maximum of the amount of solutions". the equation i^z = z has infinite amount of solutions as ln(i) = i * pi / 2 + 2*pi*n with n integers. the simulation converges to the principal branch
careful with parentheses when writing power towers! the expression at 1:04 doesnt represent the spiral. its not even a tower, it collapses down with each parathesis. writing these is annoying for this reason, since we dont have a convenient notation for the anti-log operation
I came to the comments to say the same thing: the expression at the beginning is not a power tower, it's iterated exponentiation in the wrong direction.
There is a candidate for convenient antilog... 'exp' with a subscript of the base. It's not terribly commonly used, but I have occasionally seen it to represent exponentiation when the important argument is meant to be the exponent. exp₂(x) = 2^x
If you substitute z=i^i^(…) like in the video as follows: i^i^z=z you get multiple solutions. Adding another i gives you even more solutions. Repeating the pattern, you reach a local maximum of the amounts of solutions after which the amount of solutions decreases again. Maybe worth a follow-up video!
Somewhere in between Wolfram Alpha and tabulating everything on an electric abacus is one of my fall-back methods: Excel. When I don't know how to find a value within a known range, I can get Excel to do the calculations over 10000+ values that are, say, 0.00001 apart. I can have a column check an answer in two columns and report greater than or less than, and I can just look for where that flips. If It's not in the list, I change the starting value to try to get it in range. If I find a flip point, the lower value (assuming I'm counting up from the starting value) before the flip is my new starting value, and I add some post-decimal-point zeros to the increment value. In a short time I find a value to 15 decimal places. But then I always investigate what better, more accepted means of finding that value is.
TI-36X Pro Engineering/Scientific Calculator: I was able to easily program the algorithm of _Newton's method_ into the calculator itself such that the only thing I had to do was choose an initial value and then hit the [ENTER] key a bunch of times, until 'convergence.' In less than 2 seconds, I could get a zero with the precision of 12 correct decimal places. Everyone else was doing this 'by hand,' so to speak, whilst I busted out the entire *_hw_* assignment in less tim than it took everyone else to do a _single problem;_ the TI-36X Pro has feature that takes the derivative of a function at x=a; it also has an "OP" function, where every time you hit {ENTER}, it would run the function. If one made the 'OP" function to f(a) -> x (store the value of the function evaluated at x=a into the variable x), then one would have a recursive function
The real question: why are there three "branches" in that graph. It looks suspiciously like the tour a point goes on when inside the mandelbrot set. There's probably one lurking somewhere inside here...
for anyone who also got confused at 8:06 about where w*e^w=-i*pi/2 comes from, it comes from the W(z)e^(W(z))=z equation that was on screen before it. If you plug -i*pi/2 into that equation and let w=W(-i*pi/2), you get w*e^w=-i*pi/2.
While i know it’s prob too long and technical for the video, how would we prove that the sequence converges in the first place? Cause people always say we must make sure the infinite recursive sequence/pattern converges before we can do our substitutions, but i have no idea how to show that.
you can say the result is the limit of a sequence of maps (if it exists) z_{n+1} = f(z_n) where f(z) = i^z with the initial term z_1 = i. If the limit exists then it must be a fixed point of the map f. To show that the limit exists you just have to show that the initial term z_1 = i lies within the basin of attraction for one of the fixed points of the map.
@@eastonrocket兀 yeah so idk what im thinking when making this 1 year old comment, i think im referencing someone who make √-1 as i but yea ima change it so that my comment isn't false
@5:39 you have implemented log(i) = ln|i|+arg(i) over ln(i)... you will need to divide log(i) with log(e) as well to get to ln(i) so ln(i) will actually be equals to (iπ)/(2*log(e)) or you can say ln(z)/z = iπ/(2*log(e))
im kinda interested in knowing what would be i-th tetration of i. (or to be more exactly, i = e^i(pi)/2, since i know there are infinitely many roots for i, but i want the "principal" root) the only problem would be to define what would be the i-th tetration of some number x... maybe in form of integral, series, product, etc etc...
If you will use an iterative method it would have been easier to approach directly i^i^i^i^i^i^i... since the very beginning instead of partially calculating it and then iterating
The great Russian mathematician Lobachevsky said: "Mathematics is gymnastics of the mind" It is certainly useful to periodically do gymnastics of the mind. But still!!!! What practical advice do you have?? Where do these infinite power series apply in practice???? .
This’ll be a fun problem to give my son. I would really like to see a good proof for why the limit as x goes to zero (from the right) of the nth tetration of x is zero if n is odd and one if n is even…
I have a subtle problem with this. z is not i^z coz the z in the power has one less i. This is my problem with proofs that 0.9 recurring is 1 and harmonic number diverges to infinity rather than log(infinity). Harmonic sum reaches infinity if done upto e^infinity. If infinity is 1/0 then the number line is scalable and probably 1 is a simple infinity with al properties except scale .
The reason why you can say z = i^z is not rigorously explained in the video. You are basically trying to find the fixed point of the function f(z) = i^z. If the expression i^i^i^i... converges to a finite value, then it must follow that this value is a fixed point of the function f(z), i.e. applying any more iterations will not change its value. As such, this value must be the solution to the equation f(z) = z, or i^z = z. Your intuition that the harmonic series tends to infinity logarithmically is totally correct, but we have other ways to express it rigorously. log(infinity) is still infinity. However, if you would take the limit as x goes to infinity of the harmonic series divided by log(x), you will see that the result is 1 www.wolframalpha.com/input?i=lim+x+-%3E+inf+of+%28sum+k%3D1+to+x+of+1%2Fk%29+%2F+ln%28x%29, meaning both functions approach infinity at the same rate. This is one way your idea can be formalized, by framing it as a comparison of growth rates between different functions. Regarding 0.99... recurring, there are many, many different ways to prove that no real number exists between that number and 1, as such, that number must be 1, since there are always infinite real numbers between two distinct real numbers. Your notion that 0.99... is lacking some infinitesimal amount to reach 1 can be formalized by using other number systems, such as the surreal numbers. But this notion does not exist within the real numbers.
I’m bored as heck this summer, since I sit around doing nothing but imaginary and complex math problems in my notebook. I’ll add this to the list of fun stuff to do on my break! I (the reader) will do this exercise first then come back to the video. Then I might hara$$ my Professor about it afterwards LMAO. EDIT: Ok, I tried it, and I believe I found out that it was impossible for me to do given the current tools I have. I’d have to know what the lambert function is and what arg(z) is, and I didn’t, so whomp whomp.
Hey... At the beginning when you plotted the points i think the parantheses are doing something else than you want them to I think you were plotting i^(i+i+i+...) Because of the product property of exponents This is a common mistake while coding. You can easily fix it by changing your loop statement Ig mathematicians are just more prone because it can't be bug fixed
I believe the simulation is correct but he just wrote it down wrongly. What he wrote down is something like the sequence a[n] = i^(i^n) which has no limit.
You start with a power tower at 0:21, but then you don’t calculate it as one at 1:11. Power towers are evaluated right-to-left, as you yourself explained in another video. I.e., 3^3^3 is 3^(3^3), not (3^3)^3 as in the vid. Does it matter with i? I don’t know.
i is the imaginary unit (the square root of -1) and a^b means a multiplied by itself b times So i^i is i multiplied by itself i times, which on the surface makes no sense, but if you do the math, it gives a real number, the result being approximately 0.20
@@alexterra2626 The exact number is e^(-pi/2). Remember, e^(ipi)+1=0. Subtract 1 from both sides, square root both sides, then raise each side to i to get e^(-pi/2). Simple! Easy! FUN 🙃
I don't think all the math you did after deriving z = i^z does anything. The solution to the problem is simply the solution(s) to this equation. You could have just solved z=i^z numerically using your favorite method. It's not becoming more "correct" because you use some fancy functions to derive another more contrived equation that can also only be solved numerically.
This is what I feel like when I watch a math video: beginning of the video: Yeah im smart this is easy stuff middle of the video: uhhh yea that make sense.... right? end of the video: *help my brain melted*
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Hi
important notation note: when one writes, say, x^x^x^x without any parentheses, by default the computation is performed from the top down, so x^x^x^x = x^(x^(x^x)) and not ((x^x)^x)^x. if you launch the iteration considering the latter (that is, [z_0 = i, z_{n+1} = z_{n} ^ i] instead of the correct [z_0 = i, z_{n+1} = i^z_{n}]) you don't converge to that value you've found in the video, but fall into the 3-cycle "0.20..., 6.12e-17-i, 4.81...)"
This might be why @jaysn1683 seems to find multiple solutions depending on the amount of i's specified in the tower!
@@katakana1 not sure I got their point about "local maximum of the amount of solutions". the equation i^z = z has infinite amount of solutions as ln(i) = i * pi / 2 + 2*pi*n with n integers. the simulation converges to the principal branch
An _i_ for an _i_ for an _i_ ... leaves the whole world plotted on an graph.
careful with parentheses when writing power towers! the expression at 1:04 doesnt represent the spiral. its not even a tower, it collapses down with each parathesis. writing these is annoying for this reason, since we dont have a convenient notation for the anti-log operation
i picked up on this too
I came to the comments to say the same thing: the expression at the beginning is not a power tower, it's iterated exponentiation in the wrong direction.
I was also gonna say this looks like some other caught it
What do you mean we don't have a convenient notation, what about:
i^i^i^i^i^...^i?
There is a candidate for convenient antilog... 'exp' with a subscript of the base. It's not terribly commonly used, but I have occasionally seen it to represent exponentiation when the important argument is meant to be the exponent.
exp₂(x) = 2^x
If you substitute z=i^i^(…) like in the video as follows:
i^i^z=z
you get multiple solutions. Adding another i gives you even more solutions. Repeating the pattern, you reach a local maximum of the amounts of solutions after which the amount of solutions decreases again.
Maybe worth a follow-up video!
Holy fuck what should I look up to read about this
I don’t no, I found out about it a few weeks ago when experimenting with WolframAlpha
But only the solutions in common between the different substitutions are actual solutions to the original question
Could you please elaborate further or link a post explaining the situation?
@@jaysn1683this same channel has a video on i^i itself being a real number
At this point change your name to Brilliant the Math Guy
4:18
@@James22104:17
Somewhere in between Wolfram Alpha and tabulating everything on an electric abacus is one of my fall-back methods: Excel. When I don't know how to find a value within a known range, I can get Excel to do the calculations over 10000+ values that are, say, 0.00001 apart. I can have a column check an answer in two columns and report greater than or less than, and I can just look for where that flips. If It's not in the list, I change the starting value to try to get it in range. If I find a flip point, the lower value (assuming I'm counting up from the starting value) before the flip is my new starting value, and I add some post-decimal-point zeros to the increment value. In a short time I find a value to 15 decimal places. But then I always investigate what better, more accepted means of finding that value is.
There is a better thing called coding
@@samueljehannoYeah but I'd have to learn that.
Fortunately imaginary unit is not called z. Otherwise this infinite tower would equal to a sleeping man.
But z is usually used for complex number notation, like z = a+bi lol
@@swordofjustice7444Zᶻᶻᶻ
0:54 minor correction, the parentheses are the wrong way around. Still, great video - I love infinite power towers!
I also noticed that and wasn't sure what he was doing
TI-36X Pro Engineering/Scientific Calculator: I was able to easily program the algorithm of _Newton's method_ into the calculator itself such that the only thing I had to do was choose an initial value and then hit the [ENTER] key a bunch of times, until 'convergence.' In less than 2 seconds, I could get a zero with the precision of 12 correct decimal places.
Everyone else was doing this 'by hand,' so to speak, whilst I busted out the entire *_hw_* assignment in less tim than it took everyone else to do a _single problem;_ the TI-36X Pro has feature that takes the derivative of a function at x=a; it also has an "OP" function, where every time you hit {ENTER}, it would run the function. If one made the 'OP" function to f(a) -> x (store the value of the function evaluated at x=a into the variable x), then one would have a recursive function
9:25 you wrote W(-π/2) and not W(-iπ/2) in the top right corner of the screen
The real question: why are there three "branches" in that graph. It looks suspiciously like the tour a point goes on when inside the mandelbrot set. There's probably one lurking somewhere inside here...
The points that converge do indeed form a fractal.
Eon?
for anyone who also got confused at 8:06 about where w*e^w=-i*pi/2 comes from, it comes from the W(z)e^(W(z))=z equation that was on screen before it.
If you plug -i*pi/2 into that equation and let w=W(-i*pi/2), you get w*e^w=-i*pi/2.
Thx bro! I really appreciate that you posted this
While i know it’s prob too long and technical for the video, how would we prove that the sequence converges in the first place?
Cause people always say we must make sure the infinite recursive sequence/pattern converges before we can do our substitutions, but i have no idea how to show that.
you can say the result is the limit of a sequence of maps (if it exists) z_{n+1} = f(z_n) where f(z) = i^z with the initial term z_1 = i. If the limit exists then it must be a fixed point of the map f. To show that the limit exists you just have to show that the initial term z_1 = i lies within the basin of attraction for one of the fixed points of the map.
Always annoyed me that there's no closed form for i^i^i^i...
Does the sequence of the tetration of i touch the real axis ever again (other than i^i)?
7:06 You used -W(x) = W(-x), but W() is not an odd function. What's going on here?
That spiral looks so cool
It would be amazing to see the spirals filled in continuously via functional roots of f(x)=i^x. I wonder if those would simplify..?
Ain't no way somebody made the definition for √-1 as i and accidentally create a new branch of math
He did not make this.
@@eastonrocket兀 yeah so idk what im thinking when making this 1 year old comment, i think im referencing someone who make √-1 as i but yea ima change it so that my comment isn't false
@5:39 you have implemented log(i) = ln|i|+arg(i) over ln(i)... you will need to divide log(i) with log(e) as well to get to ln(i) so ln(i) will actually be equals to (iπ)/(2*log(e)) or you can say ln(z)/z = iπ/(2*log(e))
im kinda interested in knowing what would be i-th tetration of i. (or to be more exactly, i = e^i(pi)/2, since i know there are infinitely many roots for i, but i want the "principal" root)
the only problem would be to define what would be the i-th tetration of some number x... maybe in form of integral, series, product, etc etc...
believe it or not, i↑↑i is solvable
If you will use an iterative method it would have been easier to approach directly i^i^i^i^i^i^i... since the very beginning instead of partially calculating it and then iterating
The great Russian mathematician Lobachevsky said: "Mathematics is gymnastics of the mind"
It is certainly useful to periodically do gymnastics of the mind.
But still!!!!
What practical advice do you have??
Where do these infinite power series apply in practice???? .
This’ll be a fun problem to give my son. I would really like to see a good proof for why the limit as x goes to zero (from the right) of the nth tetration of x is zero if n is odd and one if n is even…
Thanks Bri the Guy Math!
Love it! Thank you!
An equivalent solution is (i*2/pi)*Lambert(-i*pi/2). Numerically the same although I don't really know how to reduce the solutions into each other.
Power towers, imaginary numbers, and pi... these are a few of my favourite things.
I have a subtle problem with this. z is not i^z coz the z in the power has one less i. This is my problem with proofs that 0.9 recurring is 1 and harmonic number diverges to infinity rather than log(infinity). Harmonic sum reaches infinity if done upto e^infinity. If infinity is 1/0 then the number line is scalable and probably 1 is a simple infinity with al properties except scale .
But one less than infinity is infinity, and log(infinity) is infinity
The reason why you can say z = i^z is not rigorously explained in the video. You are basically trying to find the fixed point of the function f(z) = i^z. If the expression i^i^i^i... converges to a finite value, then it must follow that this value is a fixed point of the function f(z), i.e. applying any more iterations will not change its value. As such, this value must be the solution to the equation f(z) = z, or i^z = z.
Your intuition that the harmonic series tends to infinity logarithmically is totally correct, but we have other ways to express it rigorously. log(infinity) is still infinity. However, if you would take the limit as x goes to infinity of the harmonic series divided by log(x), you will see that the result is 1 www.wolframalpha.com/input?i=lim+x+-%3E+inf+of+%28sum+k%3D1+to+x+of+1%2Fk%29+%2F+ln%28x%29, meaning both functions approach infinity at the same rate. This is one way your idea can be formalized, by framing it as a comparison of growth rates between different functions.
Regarding 0.99... recurring, there are many, many different ways to prove that no real number exists between that number and 1, as such, that number must be 1, since there are always infinite real numbers between two distinct real numbers. Your notion that 0.99... is lacking some infinitesimal amount to reach 1 can be formalized by using other number systems, such as the surreal numbers. But this notion does not exist within the real numbers.
Very nice❕Thanks Bri.
So basically the infinith tetration of i
I’m bored as heck this summer, since I sit around doing nothing but imaginary and complex math problems in my notebook.
I’ll add this to the list of fun stuff to do on my break!
I (the reader) will do this exercise first then come back to the video. Then I might hara$$ my Professor about it afterwards LMAO.
EDIT: Ok, I tried it, and I believe I found out that it was impossible for me to do given the current tools I have. I’d have to know what the lambert function is and what arg(z) is, and I didn’t, so whomp whomp.
Yep, i^i=exp(pi/2)!
Awesome video
how do you evaluate i^i?
Is there any other reasons why we don't use the other values for the complex logarithm other than making it easier to work with?
please continue your real analysis playlist
Your visualization from 0:45 shows the formula as
(((i^i)^i)^i)^...
But what we are interested in, is
i^(i^(i^(...)))
Hey... At the beginning when you plotted the points i think the parantheses are doing something else than you want them to
I think you were plotting i^(i+i+i+...) Because of the product property of exponents
This is a common mistake while coding. You can easily fix it by changing your loop statement
Ig mathematicians are just more prone because it can't be bug fixed
Day 2 of please continue your real analysis playlist
1:00 youre not doing it correctly
The exponent shpuld be evaluated top down
I believe the simulation is correct but he just wrote it down wrongly. What he wrote down is something like the sequence a[n] = i^(i^n) which has no limit.
Nice video !
You start with a power tower at 0:21, but then you don’t calculate it as one at 1:11. Power towers are evaluated right-to-left, as you yourself explained in another video. I.e., 3^3^3 is 3^(3^3), not (3^3)^3 as in the vid. Does it matter with i? I don’t know.
A friendly reminder that complex exponentiation and logarithm are multi-valued functions
And also Newton's method doesn't always work
Let
y_0 = i = exp(i * pi/2)
y_1 = (y_0)^i = i^i = (exp(i * pi/2))^i = exp(-pi/2)
y_2 = (y_1)^i = i^i^i = (exp(-pi/2))^i = exp(-i * pi/2) = -i
y_3 = (y_2)^i = i^i^i^i = (exp(-i * pi/2))^i = exp(pi/2)
y_4 = (y_3)^i = i^i^i^i^i = (exp(pi/2))^i = exp(i * pi/2) =y_0
The derivation manifests a cyclic relation for i^i^i^.......
Therefore, i^i^i^......is non-convergent.
Letting z=i^i^i^i... and then solving for it presupposes that z actually exists. So you still need to prove that the limit exists!
lemme guess: -1.
edit: nope i guess not.
I tattooed this 😅
What does i^i mean?
are you stupid or stupid
i is the imaginary unit (the square root of -1)
and a^b means a multiplied by itself b times
So i^i is i multiplied by itself i times, which on the surface makes no sense, but if you do the math, it gives a real number, the result being approximately 0.20
@@alexterra2626
The exact number is e^(-pi/2).
Remember, e^(ipi)+1=0.
Subtract 1 from both sides, square root both sides, then raise each side to i to get e^(-pi/2).
Simple! Easy! FUN 🙃
Z=i^z take z√ i=√z z=-1
Now solve it downwards instead of upwards :)
(i^i^i...) is i because i is 1, but just tilted 90 degrees, and (1^1^1...) is 1
no, it`s not "just rotated 1".
in the first minute you didnt do the infinite i tower, you did (i^i)^i... instead you should have done i^(i^i)...
0:53 your parenthesis are wrong
4:20
Bro talks like CGP grey
BRILLIANT IS ACTUALLY HELPFUL but i stayed in neon league for 8 days now im raging
Tetration on ♾️...
also written as i{2}ω
no
yes
I always thought your name was pronounced "bree" not "brai" 💀
d
I don't think all the math you did after deriving z = i^z does anything. The solution to the problem is simply the solution(s) to this equation. You could have just solved z=i^z numerically using your favorite method. It's not becoming more "correct" because you use some fancy functions to derive another more contrived equation that can also only be solved numerically.
100th comment
This is what I feel like when I watch a math video:
beginning of the video: Yeah im smart this is easy stuff
middle of the video: uhhh yea that make sense.... right?
end of the video: *help my brain melted*
(0.4000 + 0.0392) + (0.4000 - 0.0392) * i
Just something I noticed.
At this point change your name to Brilliant the Math Guy