How to get pinned: 曹老师, here's a math joke. Student: All right, what equation are we going to solve? Teacher Chow: 𝑥⁵+𝑝𝑥+𝑞=0. Just bring me some radicals, please. Student: What radicals do we need to solve 𝑥⁵+𝑝𝑥+𝑞=0 ? Teacher Chow: Bring radicals, please. Student: I couldn't find it...? Teacher Chow: Ok, the Bring radical is here! : ) Get the joke? If not, here's the Wikipedia page for this new inverse function: en.wikipedia.org/wiki/Bring_radical
@@blackpenredpen You're welcome! Except... this radical is complicated because it would look like the inverse of 𝑥⁵+𝑥, but it isn't. The graph depicted in the Wikipedia article about the Bring radical is really a reflection on the y-axis. However, if you go to the Russian version of the Wikipedia article concerning the Bring radical, it will give you a proof of the solution. Just scroll down and you should see Доказательство (transliterated as dokazatel'stvo), which means "proof", and on the right of Доказательство, click on [показать] (transliterated as [pokazat']) to show the proof. The only difference is that if you define the Bring radical as the inverse of 𝑥⁵+𝑥 instead of the Wikipedia definition (Yes, all versions with different languages about the Wikipedia article concerning the Bring radical agree.), the fraction inside the Bring radical in the solution for 𝑥⁵+𝑝𝑥+𝑞 becomes negative instead of positive. Hope that helps! By the way, your sqrt(i) video is about to become the 2nd video to get 1 million videos on your channel, after the "100 Integrals" video! So excited! : D #YAY
@@ashlok2003 Hi! First of all, I'm not Tejas. I'm Alex Eduardo Gonzalez. Also, are you talking about integrating √(sin(𝑥)) or integrating the integral of √(sin(𝑥))? Both of them require special functions, but the integral of √(sin(𝑥)) requires an elementary function. Thus, integrating the integral of √(sin(𝑥)) requires knowing how to integrate √(sin(𝑥)) itself. Here's why some integrals of elementary functions may not have an elementary antiderivative. en.wikipedia.org/wiki/Liouville%27s_theorem_(differential_algebra)
In 7:20 you can write 1/sqrt(2) as (1/2)^(1/2) and then when you take the ssrt on both sides you get 1/x=1/2 directly, so x=2. But then, you should somehow determine other solutions (like x=4, or the complex ones)
As I watched this video I thought it was just another math video I could geek over. Then I realized you were solving one of my childhood dream problems!!! X is the base and the power, which always confused me as a kid! 6th grade me can now rest in peace, thank you :)
If you look on a graph, 2^x is equal to x^2 at 3 points, being 2,4 (the 2 values findable via this method), but there is another solution at around -0.761
You should do a video on half-derivatives and non-integer derivatives, just like Payam did. I found the half-derivative of sin(x) and ln(x) just using taylor series. Not many people understand non-derivatives, so I would like to see more on them
Question 1: Do you have an idea how to integrate ssrt(x) and similar functions?; Question 2: Do you know why the value of x^-x numerical integration (from positive zero to infinity) has a value of 1,994.... and never reaches 2? Is it just another, specially defined, transcendental number, or can it be expressed as a function of e or pi, or something like this?
Ssrt is the inverse of x^x or tetration(x,2) in my humble notation for avoiding confusion of 2^x. But what is inverse of tetration(x,n) = x^x^x... n times? Is there a general case mega square root or hyper square root or oreo square root?
@@MarkMcDaniel I have been making videos for the Differential Equations course I am teaching this semester. I have ~80 on my channel if you are interested.
Ideas for new videos on super square root: - What is the taylor series of the super square root of x ? - Use super square root to calculate x^(x^...) - Use super square root to solve x*a^x=b where a and b are constant. - Use super square root to solve x+a^x=b where a and b are constant. - Use super square root to solve x^(x^a) where a is a constant.
Talkin' about integrals, we can actually, for elementary trig. And inverse trig fun. Express in terms of complex form and get result directly with just knowing the derivate if e^x and lnx This indeed relives us from the process of remembring lengthy proofs
When I was in middle school I came up with a fancy iteration that converges to the second super root if x was between (1/e)^(1/e) to e^e based on infinite power towers. When I was a bit smarter, I realized that the complicated thing I wrote simplifies to a_n=x^{1/a_{n-1}} which would of course eventually converge to an s such that s^s=x, if it converged at all. Was still fun to explore, learned a lot about dynamical systems, solutions to x^(1/x)=c, and sequences that have different converging subsequences
Oh the complicated thing I wrote in middle school was x^(1/x)^(1/x)^(1/x)... converges to such an s if x is in that range. If x>e^e, there exist a distinct a and b such that a^b=b^a=x, and this sequence will alternate between getting closer to a and getting closer to b. You can find these pairs by looking at levels of x^(1/x)
Here's a cool property of tetration you can find via simple exponent rules. (x^^a)^(x^^b) = (x^^b+1)^(x^^a-1) Which means you can solve for (x^x^x)^x = 2 because x^x^x = x^^3 and x = x^^1, so (x^x^x)^x = (x^x)^(x^x), which you can by using the super square root twice
I've been trying to find the integral from 0 to 1 of an infinite height power tower. I realized that taking 1 - the integral from 0 to 1 of the inverse of the infinite power tower (which is x^(1/x)) gives the same area, and so my new problem seemed more reasonable. However, I have yet to find a good way to integrate this, because unlike for x^x or x^-x, transforming this function into a taylor series expansion of e^x creates a divergent integral. Some help would be much appreciated!
BPRP, you write that *ssrt((1/x)^(1/x))=1/x* if so, however, couldn't you then not have continued by employing this to solve the equation of *1/x=ssrt(1/sqrt(2))* because the right hand side of that equation is equal to: *ssrt(1/sqrt(2))=ssrt(sqrt(1/2))=ssrt((1/2)^(1/2))=1/2* ? The last equation of this series following from the one I mentioned in this comment first (just insert 2 instead of x). . . and finally via multiplication by 2x we find *2=x* as a solution. edit: right hand side is also equal to: *ssrt(1/sqrt(2))=ssrt(1/sqrt(sqrt(2^2)))=ssrt(sqrt(1/sqrt(4)))=ssrt(sqrt(sqrt(1/4)))=ssrt( ((1/4)^(1/2))^(1/2) )=ssrt( (1/4)^(1/4) )=1/4* And then from 1/x=1/4 follows x=4.
SSRTs that treat unity as a pinhole expire in decrepitation before meeting Hairy Wau and using any squeeze play. Assuming the fish is chopped unity before artificial exultation.
W(x) is basically the inverse of the function x•e^x, so w(x)•e^w(x) and w(x•e^x) are both x. As for e, explaining what it is, that’s too complicated for me, but all you need to know to wrap your head around this concept is that e~2.71818, is irrational and is also transcendental.
If we're going to stick with the topic of unusual functions, maybe go over the Ackerman function next? Or maybe other functions like the hyperfactorial or pickermans hyperfactorial, or maybe even a video on the super logarithm
I, too, don't much care for the "sub-s" notation for "super-squareroot." How about this? Long ago, I was toying around with tetration (but not knowing it by name), and invented a symbol for the tetration root. It was like the radical, but instead of the "check-mark" at the left end, there was a little circle, which was an extension of the upward line at the start of the radical. And analogously to ordinary radicals, you could put a number inside the circle that designates the 'height' of tetration, and which when left blank, was taken as 2 by default. Sorta like this: x = ⁽²⁾√y = º√y; xˣ = y x = ⁽³⁾√y; x^(xˣ) = y etc. That isn't quite what the symbol looks like, but it's as close as I can get with the symbols at my disposal. Fred
Ahhh I see. I just learned about this recently and found it very interesting! Btw I read your marathon reply from the other comment. Thank you Fred for always being so supportive and I found it really cool that we have similar hobbies, too! I actually did a 11 mile run tonight too! I never liked running until a high school classmate invited me to run the LA marathon with him and another guy. Ever since then, I developed this weird love for it! Yea... I don't know but I just like it. So I have been running marathons on and off since then.
@@blackpenredpen Yep, you've got the bug!! I guess I never had a marathon "bug," but I had the distance running "bug." Maybe there's no real difference. In that vein, I recommend (if you haven't already discovered them) the writings of "the runner's philosopher," George Sheehan, a cardiologist who, in his 40's, quit the practice of medicine to take up running, including writing & speaking about it. He was a longtime columnist for _Runner's World,_ and wrote several books. His essays really spoke to the runner's soul. He died of prostate cancer in 1993, at almost 75. Cheers! Fred
I don't like how the ssrt() takes an integer input to the transcendentals twice before giving an integer solution. Is there an exact algebraic way? Ideally without guessing/verifying.
My idea as a notation for the super square root (maybe someone else posted it already idk): Just write the same square root notation but mirrored from left to right xD
How can you super square root a number if x is greater than (1/e)^(1/e), but less than 1 [ (1/e)^(1/e) < x < 1 ] ? As the x^x graph is concave up, then ssrt(x) would have two solutions on the interval 1/e < ssrt(x) < 1. How would the ssrt(x) work in this situation?
I checked to see, and the super square root of a number is ± the super square root of that number IF the number is even, which in the case with 256 is true, so really the answer is 4 or -4.
I know a bit about tetration, but I didn't know what a super-square root is; I didn't think it would be related to tetrations! What do you know, you learn something new everyday...I bet finding the super-cube root (x^x^x)/super-quartic root (x^x^x^x) would be very fun indeed...Actually, is it possible to find the super-cube root and super-quartic root?
If tetration is shown by a mirrored version of exponantiation, why don't we use a mirrored square root sign for the super square root? -----\ 2 \- Something like this
My initial answer is no. Any complex number added to itself must make the same "angle" in the complex plane, but multiplied with itself the angle doubles, so only for real numbers (in this case, only 2) could it work.
Blue Blue In most cases it's useful that you define 0^0 to be 1. But you can also argue that 0 to the power of any positive integer is 0 so why not to the power of 0. In math you define what is useful.
@@SeeTv., well, I am not a math teacher or something like that, but I think, you can´t argue, that you can define what is useful. It has to be consistent. Else you get into another Parallel-Math-Universe.
How to get pinned:
曹老师, here's a math joke.
Student: All right, what equation are we going to solve?
Teacher Chow: 𝑥⁵+𝑝𝑥+𝑞=0. Just bring me some radicals, please.
Student: What radicals do we need to solve 𝑥⁵+𝑝𝑥+𝑞=0 ?
Teacher Chow: Bring radicals, please.
Student: I couldn't find it...?
Teacher Chow: Ok, the Bring radical is here! : )
Get the joke?
If not, here's the Wikipedia page for this new inverse function: en.wikipedia.org/wiki/Bring_radical
WOW! I never knew about this! Thanks and yes, you will be pinned!
@@blackpenredpen You're welcome! Except... this radical is complicated because it would look like the inverse of 𝑥⁵+𝑥, but it isn't. The graph depicted in the Wikipedia article about the Bring radical is really a reflection on the y-axis.
However, if you go to the Russian version of the Wikipedia article concerning the Bring radical, it will give you a proof of the solution. Just scroll down and you should see Доказательство (transliterated as dokazatel'stvo), which means "proof", and on the right of Доказательство, click on [показать] (transliterated as [pokazat']) to show the proof.
The only difference is that if you define the Bring radical as the inverse of 𝑥⁵+𝑥 instead of the Wikipedia definition (Yes, all versions with different languages about the Wikipedia article concerning the Bring radical agree.), the fraction inside the Bring radical in the solution for 𝑥⁵+𝑝𝑥+𝑞 becomes negative instead of positive. Hope that helps!
By the way, your sqrt(i) video is about to become the 2nd video to get 1 million videos on your channel, after the "100 Integrals" video! So excited! : D
#YAY
Hey Tejas,
I have an question for you.
Why we can't integrate ∫ √sinx .
☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️
@@ashlok2003 Hi! First of all, I'm not Tejas. I'm Alex Eduardo Gonzalez.
Also, are you talking about integrating √(sin(𝑥)) or integrating the integral of √(sin(𝑥))? Both of them require special functions, but the integral of √(sin(𝑥)) requires an elementary function. Thus, integrating the integral of √(sin(𝑥)) requires knowing how to integrate √(sin(𝑥)) itself.
Here's why some integrals of elementary functions may not have an elementary antiderivative.
en.wikipedia.org/wiki/Liouville%27s_theorem_(differential_algebra)
@@einsteingonzalez4336 math joke
Prove that 1=2
Method 1
Let WW1 and WW2 are equal
WW1=WW2
WW gets cancelled
1=2
Hence proved
Method 2
1+1=1+1
1(binary math)=2(maths)
1=2
Hence proved
Prove that 0=1
1=2(proven above)
1=1+1
1-1=1
0=1
Hence proved
I think writing it as ssrt(x) looks cleaner than writing it as √x_s
HitzCritz I think so too
obama clip
Like this video or I will give the fish to Oreo!
Things get weird when I read the pinned post before watching the video.
Is it just me or is the comment 8 hours before the video...
blackpenredpen, well, give it to oreo. I don‘t mind. 😉
Liked anyway!
do super log
@@brewbearz1136 its just you
Oreo likes the fish!
Love your channel...
Can you make videos on how to differentiate and integrate the nth operator function of x with respect to many variables at once?
If anyone is thinking who is oreo, he is his bunny 🐇🐰😂.
Chirayu Jain, a bunny likes fish?
Solution: You don‘t have to eat it if you like it...
blackpenredpen, proof it! 😉
@@chirayu_jain
That Oreo is Peyam's bunny.
The Oreo that I mentioned is my gf's cat.
In 7:20 you can write 1/sqrt(2) as (1/2)^(1/2) and then when you take the ssrt on both sides you get 1/x=1/2 directly, so x=2.
But then, you should somehow determine other solutions (like x=4, or the complex ones)
What can you do next? Keep ypour promises!
1. Sing the quadratic formula song!
2. Solve an easy integral in German language! :-)
1. you probably don't want to hear me sing, trust me. That's why I haven't done that.
2. I need Peyam's help.
: )))))
i smell math
As I watched this video I thought it was just another math video I could geek over.
Then I realized you were solving one of my childhood dream problems!!! X is the base and the power, which always confused me as a kid! 6th grade me can now rest in peace, thank you :)
Respect to Professor Terrence Tao 😂
For sure!
At x^1/x=√2
x^1/x=2^1/2
Wait x just equal to 2?
If you look on a graph, 2^x is equal to x^2 at 3 points, being 2,4 (the 2 values findable via this method), but there is another solution at around -0.761
OMG I never heard of super square root until today! This is so awesome and it's always fun to learn new stuff from you man! YAY!
Only 7 % people know about ssrt, actually I also didn't knew about it before watching the video 😅.
Know*
@@mr.nobody5242 15% of all statistics are made up on the spot
You should do a video on half-derivatives and non-integer derivatives, just like Payam did. I found the half-derivative of sin(x) and ln(x) just using taylor series. Not many people understand non-derivatives, so I would like to see more on them
Question 1: Do you have an idea how to integrate ssrt(x) and similar functions?; Question 2: Do you know why the value of x^-x numerical integration (from positive zero to infinity) has a value of 1,994.... and never reaches 2? Is it just another, specially defined, transcendental number, or can it be expressed as a function of e or pi, or something like this?
I believe the integral of 1/x^x does not have a closed form, but it can be written as an infinite sum. Look up the sophomore's dream function
@@soupisfornoobs4081 Well, Bernoulli integrated from 0 to 1, while I would like to integrate from 0 to infinity...
@@MrMatthewliver there's a general form for integrating from 0 to x
Ssrt is the inverse of x^x or tetration(x,2) in my humble notation for avoiding confusion of 2^x.
But what is inverse of tetration(x,n) = x^x^x... n times?
Is there a general case mega square root or hyper square root or oreo square root?
That would just be super roots. For example, the inverse of x^x^x (otherwise known as x^^3) would be the super cube root.
"Super nth roots" to be exact.
We miss you professor✌🏻💯
This is so cool. I'm actually starting to understand
hey, do you have stuff on TRANSFORMS? and their INVERSES? these topics are coming up in my course.
Me too, currently crunching through Diff. Eq.
@@MarkMcDaniel I have been making videos for the Differential Equations course I am teaching this semester. I have ~80 on my channel if you are interested.
He has tons of videos on the Laplace transform and it's inverse transform if that's what you mean
@@technoguyxYeah, that is exactly what I meant!
Ideas for new videos on super square root:
- What is the taylor series of the super square root of x ?
- Use super square root to calculate x^(x^...)
- Use super square root to solve x*a^x=b where a and b are constant.
- Use super square root to solve x+a^x=b where a and b are constant.
- Use super square root to solve x^(x^a) where a is a constant.
It is much more fashionable to use the Prada log ;). (Especially when using the Chen Lu!!)
ssrt(exp(-π/2)) = i
a = exp(-π/2) < exp(-1/e)
Are there any rules for the ssrt(xy) or ssrt(x/y) or similar? Could you do a video on this?
Why is ssrt(fish^fish)=fish?
Well, I see, it is the inverse function...
For a new proof: give us an idea of Tree(3)...
Nah, let's go for the full TREE(3)... :D
@@erikkonstas 1st guy: Feferman-Schütte!
2nd guy: Bless you!
I wish I could a great teacher like you
I remember when the name of this video was: "Super Square Root & Solving x^2=2^x" Ahhh, great times
There are 3 answers to x^2=2^x. How do we use the super square root to conclude this?
That's right. wolframalpha shows x = {-0.766665, 2, 4}.
Well in the last step we can again use that fish formula and get
ssrt((1/2)^(1/2)) = 1/2
And so X=2
But why did this only give a single solution?
I left it like that bc ssrt is a multi valued function, just like the lamber w function.
01:50 super logarithm?
That sounds very interesting.
Please, make (whenever you want, of course) a video about super logarithm :)
Talkin' about integrals, we can actually, for elementary trig. And inverse trig fun. Express in terms of complex form and get result directly with just knowing the derivate if e^x and lnx
This indeed relives us from the process of remembring lengthy proofs
When I was in middle school I came up with a fancy iteration that converges to the second super root if x was between (1/e)^(1/e) to e^e based on infinite power towers. When I was a bit smarter, I realized that the complicated thing I wrote simplifies to a_n=x^{1/a_{n-1}} which would of course eventually converge to an s such that s^s=x, if it converged at all.
Was still fun to explore, learned a lot about dynamical systems, solutions to x^(1/x)=c, and sequences that have different converging subsequences
Oh the complicated thing I wrote in middle school was x^(1/x)^(1/x)^(1/x)... converges to such an s if x is in that range. If x>e^e, there exist a distinct a and b such that a^b=b^a=x, and this sequence will alternate between getting closer to a and getting closer to b. You can find these pairs by looking at levels of x^(1/x)
Superexponent(n)=Tetration(n)=2^2^2^...^2 with n 2's in base 2
Supersqrt(n^n)=n
Now introducing...
Superlog(Tetration(n))=Tetration(Superlog(n))=n
I ask Wolframalpha:
True or false: productlog(e^(e+1))=e
Then it said false…
…
Here's a cool property of tetration you can find via simple exponent rules.
(x^^a)^(x^^b) = (x^^b+1)^(x^^a-1)
Which means you can solve for (x^x^x)^x = 2
because x^x^x = x^^3 and x = x^^1,
so (x^x^x)^x = (x^x)^(x^x), which you can by using the super square root twice
Thats beauty of blackpenredpen...
3:04 what if you're not allowed to use Wolfram Alpha ?
You have to work out the productlog. (idk how)
Edit: this might help
en.m.wikipedia.org/wiki/Lambert_W_function
I'm guessing that we will need a "super" imaginary number if super-powers are widely useful
I've been trying to find the integral from 0 to 1 of an infinite height power tower. I realized that taking 1 - the integral from 0 to 1 of the inverse of the infinite power tower (which is x^(1/x)) gives the same area, and so my new problem seemed more reasonable. However, I have yet to find a good way to integrate this, because unlike for x^x or x^-x, transforming this function into a taylor series expansion of e^x creates a divergent integral. Some help would be much appreciated!
I was wondering how could you solve this one : 2^n > (n+1)^2.
It is a lot more harder than it looks, good luck !
I tried (x+1)^x=x^(x+1) maybe this video will help us
Is the answer 2^n/2 -1 > n
Cool! 😎 Maybe some more difficult equations with ssrt, like: ssrt(f(x)) = g(x)?
Wow, Marvel really has outdone DC with this one
*slamming fists on table*
**Differentiate it!**
**Differentiate it!**
**Differentiate it!**
Is this not his first video on ssrt(x)?
Awsome Awsome as always
This is cool (as always), but is there real world examples for the ssrt?
Never have I seen a man so excited about getting fish
Hey bro could you make a video on graph of "continuous everywhere differentiation nowhere" function.
I can’t graph it by hand tho
@@rleim weierstrass function also
@@rleim"takagi"🤨 heard 1st time
@@blackpenredpen ohh i mean a video on it. And thanks for replying
Weierstrauss function, dirichlet function,,,,,,,,they are just a family of a huge group of functions☺☺☺
BPRP, you write that
*ssrt((1/x)^(1/x))=1/x*
if so, however, couldn't you then not have continued by employing this to solve the equation of *1/x=ssrt(1/sqrt(2))* because the right hand side of that equation is equal to:
*ssrt(1/sqrt(2))=ssrt(sqrt(1/2))=ssrt((1/2)^(1/2))=1/2* ?
The last equation of this series following from the one I mentioned in this comment first (just insert 2 instead of x). . . and finally via multiplication by 2x we find *2=x* as a solution.
edit:
right hand side is also equal to:
*ssrt(1/sqrt(2))=ssrt(1/sqrt(sqrt(2^2)))=ssrt(sqrt(1/sqrt(4)))=ssrt(sqrt(sqrt(1/4)))=ssrt( ((1/4)^(1/2))^(1/2) )=ssrt( (1/4)^(1/4) )=1/4*
And then from 1/x=1/4 follows x=4.
SSRTs that treat unity as a pinhole expire in decrepitation before meeting Hairy Wau and using any squeeze play. Assuming the fish is chopped unity before artificial exultation.
I did not know that.
Hey Tejas,
I have an question for you.
Why we can't integrate ∫ √sinx .
☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️
Ashlok Chaudhary who says we can’t? Check my video!
Ragtime at the end 😁
I have a lot of question, I havent learned calculus or anything and I dont know what's w and e
W(x) is basically the inverse of the function x•e^x, so w(x)•e^w(x) and w(x•e^x) are both x. As for e, explaining what it is, that’s too complicated for me, but all you need to know to wrap your head around this concept is that e~2.71818, is irrational and is also transcendental.
Why not write the square root symbol backwards to denote ssqrt?
Next video: super square root of a pentation, hexation, septation, octation and the generalized fomular for tower math :D
Integral ln(x) /cos(x)^2
I didn't understand that function W(x) ... nevermind that lesson was still truly interesting !
The next video should be about integral of ssrt(x)
The answer is 2
:)
What about in polynomial equation form?
If we're going to stick with the topic of unusual functions, maybe go over the Ackerman function next? Or maybe other functions like the hyperfactorial or pickermans hyperfactorial, or maybe even a video on the super logarithm
"now to factor this expression we will use the Super Factorizer 2000-operation!"
Why dont you do more videos on tetration and ssrt? Like the olden days :D
I don't have new ideas of those topics. But once I do, I will. : )
I like Super stuff. Probably my favorite word ever. :D
Eres un crack :D
Why not just use Knuth's up-arrow notation?
If I understand well, division by 2 is a lower-square-root function.
I, too, don't much care for the "sub-s" notation for "super-squareroot." How about this?
Long ago, I was toying around with tetration (but not knowing it by name), and invented a symbol for the tetration root. It was like the radical, but instead of the "check-mark" at the left end, there was a little circle, which was an extension of the upward line at the start of the radical.
And analogously to ordinary radicals, you could put a number inside the circle that designates the 'height' of tetration, and which when left blank, was taken as 2 by default.
Sorta like this:
x = ⁽²⁾√y = º√y; xˣ = y
x = ⁽³⁾√y; x^(xˣ) = y
etc.
That isn't quite what the symbol looks like, but it's as close as I can get with the symbols at my disposal.
Fred
Ahhh I see. I just learned about this recently and found it very interesting!
Btw I read your marathon reply from the other comment. Thank you Fred for always being so supportive and I found it really cool that we have similar hobbies, too! I actually did a 11 mile run tonight too! I never liked running until a high school classmate invited me to run the LA marathon with him and another guy. Ever since then, I developed this weird love for it! Yea... I don't know but I just like it. So I have been running marathons on and off since then.
@@blackpenredpen Yep, you've got the bug!! I guess I never had a marathon "bug," but I had the distance running "bug." Maybe there's no real difference.
In that vein, I recommend (if you haven't already discovered them) the writings of "the runner's philosopher," George Sheehan, a cardiologist who, in his 40's, quit the practice of medicine to take up running, including writing & speaking about it. He was a longtime columnist for _Runner's World,_ and wrote several books. His essays really spoke to the runner's soul.
He died of prostate cancer in 1993, at almost 75.
Cheers!
Fred
I don't like how the ssrt() takes an integer input to the transcendentals twice before giving an integer solution. Is there an exact algebraic way? Ideally without guessing/verifying.
W(Ln(x))=Ln(super raiz de indice 2 de x).
What is the w(x) function?
Please do the integral of w(x)
I’d have the inverse tetration be a backwards square root like how tetration looks like a backwards exponent
A great video!
My idea as a notation for the super square root (maybe someone else posted it already idk):
Just write the same square root notation but mirrored from left to right xD
I realy love you tshirt! I need this 🤩
How can you super square root a number if x is greater than (1/e)^(1/e), but less than 1 [ (1/e)^(1/e) < x < 1 ] ? As the x^x graph is concave up, then ssrt(x) would have two solutions on the interval 1/e < ssrt(x) < 1. How would the ssrt(x) work in this situation?
What is the super square root of i?
I know the square root of Onion, this is an advance.
Is 1/ssrt(1/sqrt2) the same as ssrt(sqrt2)?
How about ssrt(sqrt(i)) !?
i wish the s was under the hook for the square root, like under the curve...hopefully you get what i mean
below where a 3 would go for cube root
How about differentiating√(x+√(x+√(x...))) Upto x times
12 years of learning math and this is the first time I've heard super square root
(I'm Asian btw)
Can you solve integral of sqrt(sin(x)) / cos(x) ?
maybe u should do the sin(10°) with the cubic formula
Could be here some supercomlex numbers like ssrt(-1) or ssrt(0)? Or it is boring complex numbers?
The complex numbers are algebraically closed, you can't leave them through any kind of arithmetical methods.
I checked to see, and the super square root of a number is ± the super square root of that number IF the number is even, which in the case with 256 is true, so really the answer is 4 or -4.
I know a bit about tetration, but I didn't know what a super-square root is; I didn't think it would be related to tetrations! What do you know, you learn something new everyday...I bet finding the super-cube root (x^x^x)/super-quartic root (x^x^x^x) would be very fun indeed...Actually, is it possible to find the super-cube root and super-quartic root?
Do fractional tetration.
First 3 seconds: 6:00
Btw what is the super square root of i, where i is sqrt(-1)🤔🤔🤔
That would be a nice video
Hey people I made a video on it
You forgot 2nd part - not only x^(1/x)=√2 but also (-x)^(1/x)=√2. Then we get
x=-1/ssrt(√2)
What about - 2/ln(2) * LambertW(ln(2)/2)? Is that one of your solutions?
slog next?
If tetration is shown by a mirrored version of exponantiation, why don't we use a mirrored square root sign for the super square root?
-----\
2 \-
Something like this
If you have
x+x = x*x = x^x
then the only solutions I could find were 2 and 0 if you define 0^0 as 0.
Are there any other solutions?
Maybe complex?
SeeTv, don‘t define 0^0 as 0! Clearly undefined, isn‘t it?
Definitely wouldn't say that 0^0=0 especially as the limit of the x^x function approaching 0 is actually 1.
My initial answer is no. Any complex number added to itself must make the same "angle" in the complex plane, but multiplied with itself the angle doubles, so only for real numbers (in this case, only 2) could it work.
Blue Blue In most cases it's useful that you define 0^0 to be 1.
But you can also argue that 0 to the power of any positive integer is 0 so why not to the power of 0.
In math you define what is useful.
@@SeeTv., well, I am not a math teacher or something like that, but I think, you can´t argue, that you can define what is useful. It has to be consistent. Else you get into another Parallel-Math-Universe.
Gauss was multiplication.
Euler was exponentiation.
Terence Tao is tetration!
Lambert function of Lambert function
Maybe you could write the super square root flipped (the same way as th super 2nd power)
what about the half derivative of the super square root?
Perhaps solutions to tetration towers and the inverses that have the value i?
If 3 (left superscript) 3 is 3^3^3, what is 3 (left superscript) 3 (left superscript) 3 ?