2:26 _Obviously this just means k to the power of 1 so when you multiply powers of the same base you add the exponents 1 plus 1 over k minus one is gonna be k minus 1 plus 1 over k minus 1 which is equal to k over k minus one_
It isn't necessary for n=1 in order for x and y to be rational, but it is necessary for n to divide m, so m=qn, where q is an integer. It just so happens that when n=1, the values of x and y become the same as if you made the substitution.
THX I think of this equation when I was in 8th grade , At that time I only think 2 kind of solutions x=y x=2 and y=4 Now I know how this equation is so popular from many long years ago
This is pretty cool but the rational solutions are gonna be pretty elusive, no? Having the exponent 1/k-1 be rational is hard to prove necessary. What about k=pi? the irrationality of pi^pi/(pi -1) i think is only conjectured, but if not, let pi be an arbitrary transcendental number. Rational numbers raised to irrationals are not necessary irrational either so a claim like y and x can't both be rational for irrational k doesn't seem easy to prove either.
We treated x as the only independent variable. I was curious to know what will the solution(s) be if we have both x and y as independent variables. On plotting the curves x^y and y^x we see they tangentially touch each other at point (1,1) which is of course apparent from the equation but i was wondering if i could get more solutions. As it happens we don't, so yes making x the only independent variable makes the problem interesting. Thank you so much! 😊💐👍
2 and 4 are the only integer solutions because (m+1)/m is only an integer when m is 1. All other consecutive numbers are coprime to one another, meaning their quotient can't be integers
Why would you consider y to be a linear function of x? Not even a complete linear function at that. Obviously this choice gives some solution, but how do you know we don't have more solutions?
It's interesting to extend this to z^w = w^z, w, z ∈ ℂ. Then the solutions lie on a hypersurface (4-D). With Mathematica and ContourPlot3D, keeping one of the components of z or w fixed and equating, say, the absolute values, you can take a 3D immersion of the solution hypersurfaces. They look pretty weird and I couldn't begin to guess whether they're connected, compact or any of the other topological criteria. I strongly doubt they're simply connected, though. Actually, now that I come to think of it, they'll be a mess because of branch cuts and so forth.
I worked all this out myself about four years ago (though I used α instead of k because k implies an integer). You get some awesome solutions for k=φ (φ^φ, φ^φ²)*, k=φ² (φ^(2/φ),φ^(2φ²)), and k=φ³ (φ^(3/2φ),φ²^(3φ²/2). (Those were inspired by the k-1 nature of the denominator and the properties of powers of φ). The solutions for k=5 and k=9 are also pretty intriguing (∜5, 5 ∜5) and ∜3, 9∜3). OBTW, the limit of the given equation as k approaches but is not equal to 1 is e, Euler's constant†. I recently gave this as a puzzle on a whiteboard at work in my ongoing efforts to help my co-workers exercise their math skills - In my case, I gave out x,y pairs and asked for the equation with the topic of this video as a follow up. I am sending the link to this video to those who tried to work it. My co-worker's investigations turned up the fact that xy - x - y - 2 = 0 is a surprisingly good fit for the curve, which may prove useful as a starting point to anyone who wants to iteratively calculate the y for a given x. * I actually came up with this a few days before Trump posted about news covfefe, and I only barely restrained myself from posting about a logical function cov(x,y) which was true if xʸ = yˣ, and noting this solution in "fe" and "fefe". Until now. † I couldn't come up with any other transcendental solutions that didn't give away the form of the solutions; if you do, please let me know!
I dont know if im alowed to do this but this is my way : Let x = a.y And we can change the equation to become x^(1/x) = y^(1/y) Substitut x with a.y and we get Y= a^(1/(a-1)) X=a^(a/(a-1)) (a) is a parameter !!!!
@@azzteke is that if a = 1, the denominator of the exponents of the powers would be zero, which cannot happen. now if a = 0, when substituting in the expression that determines the value of x, then x = 0^0, which is a mathematical indeterminacy. hence, that a cannot be 0 or 1.
good place to stop! this brings us to the end of the video, i hope you enjoyed it, thanks for watching, i will se you tomorrow with another video, be safe take care and BYE BYE!!!!!!!! *please don't get offended sybermath :(
Thanks for posting this video. I've often wondered what the solution to this would be and never had the math skills to do it. But now the question is, what is the parameter for the solion x = y? Because that leads to K^(1/(K-1)) = K^(K/(K-1)). I'm out of my league again. Help.
Forgive me my lame English. Some years ago, I tried to prove, that it is only one integer pair as a solution and proved it by follows way. 1. It is evident, that either x and y can not be 0, therefore we can power both sides by (1/x)(1/y) => x^(1/x) = y^(1/y) and to analyze only 1 part (e.g. left) 2. By brief examination of we can say: 2.1 It has one maximum at = x=e and this can be found by derivative An this is only one such specific point 2.2 On interval (1,inf) it goes from 1 up until x reaches e as max and then goes down from this max to 1 2.3 The interval (0,1] isn't interesting - the expression value is below 1 2.4 So, each x on (1,e) wil have one and only one "friend" on (e,inf.) such, that has the same value of expression 2.5 In (1, e) ()we have only one integer and therefore we can not have more integers pairs (2,4 and 4,2 I recognize as 1 pair) 3. Then I tried find the rational pairs. Your suggestion to use kx as y helped me. If k is like 1+ 1/a it allows to get rational pairs
Excellent tutorial, fantastic perspective! Just based off of your accent I think you are Turkish aswell right ? Just an assumption you don't have to answer it haha.
Solution x=y........both side by exponential 1/x....both side by exponential 1/Y......result...x exponential (1/x) = Y exponential (1/y).....which means all X=Y....can be a solution for the equation.
Why did you say that n = 1? If the base (m + n)/m is a perfect square, n can equal 2, right? And if the base is a perfect cube, n = 3, and so on... is there any reason why (m + n) / m can't be a perfect square?
There are no two perfect squares that differ by 2; so, (m+n)/m being the square of a rational number would mean either that n isn't 2 or that m+2 and m have a common factor (although, it will turn out that this still doesn't allow squares). The first case would mean that you're taking a root other than the square root; and the second would mean that m is even, which makes the solution equivalent to one for m'=m/2 and n'=1. A similar argument will work for any other root just by noting that any two nth powers of positive integers differ by more than n.
@@haricharanbalasundaram3124 If m and n are both negative, the sign cancels and the analysis remains unchanged. So, we only have to consider the case where their signs differ. If (m+n)/m > 0, the only thing that changes is that the numerator of this fraction is smaller than the denominator. They still differ by too little to both be nth power. If (m+n)/m < 0, even roots will be imaginary; so, we only have to think about odd n. In this case, the argument changes slightly as -1 will be an nth power. However, other than the difference of 2 between -1 and 1, all other difference between nth powers will still be larger than n. And, the 2 doesn't help, since n is odd. (Hopefully it pretty much goes without saying why we don't need to think about m=0 or m+n = 0.)
@@Phantoms3709 Yes! But answers like that is too obvious. He said that we could ignore x and y to be the same. Then it would be more interesting answers.
@@adipy8912 you’re right, interestingly enough if you graph this function there are infinite answers but I suppose the only 2 different integer solutions are 2 and 4
You have made an assumption that x and y have linear relationship. Just like we can't we can't assume that sin(x)=k cos(x). You got a solution, but not the only solution.
a^x=(b^n)^x is equivalent to: - a=b^n if a>= 0 and b^n>=0 (for the latter to be true we have b>=0 or n is even (n=2k ...English is not my first language)) - -a=b^n if a0 - a=-b^n if a>0 and b^n
x^y=y^x=1/x^y x^y=1/x^y implies x^y*x^y=1 implies x^2y=1 x^2y=1 implies x^2y=x^0 becouse power of zero to any number is 1 x^2y=x^0 implies 2y=0 implies y=0 and x=0 is the correct answer
2:26
_Obviously this just means k to the power of 1 so when you multiply powers of the same base you add the exponents 1 plus 1 over k minus one is gonna be k minus 1 plus 1 over k minus 1 which is equal to k over k minus one_
I've been looking for something like this for ages! Thanks!
It isn't necessary for n=1 in order for x and y to be rational, but it is necessary for n to divide m, so m=qn, where q is an integer. It just so happens that when n=1, the values of x and y become the same as if you made the substitution.
That's right!
Almost at 9k! Your channel has been growing exponentially. Challenge for the readers: find the equation of Sybermath's channel in subscribers
Going out on a limb here and predicting logistic growth. :)
@@emanuellandeholm5657 Well, that's pretty much exponential growth until it starts going slowly XD
@@diogenissiganos5036 Exp growth constrained by finite carrying capacity.
THX I think of this equation when I was in 8th grade , At that time I only think 2 kind of solutions
x=y
x=2 and y=4
Now I know how this equation is so popular from many long years ago
THANK YOU VERY MUCH SIR, YOUR CHANNEL HAS REALLY INSPIRED ME ON HOW TO GO ABOUT MY CHANNEL.
totally agree with this
You're welcome! Good luck!
Caps lock abuse
Nice
Another great explanation, SyberMath! I have solved the x and y values in whole numbers.
Excellent!
This is pretty cool but the rational solutions are gonna be pretty elusive, no? Having the exponent 1/k-1 be rational is hard to prove necessary. What about k=pi? the irrationality of pi^pi/(pi -1) i think is only conjectured, but if not, let pi be an arbitrary transcendental number. Rational numbers raised to irrationals are not necessary irrational either so a claim like y and x can't both be rational for irrational k doesn't seem easy to prove either.
We treated x as the only independent variable. I was curious to know what will the solution(s) be if we have both x and y as independent variables. On plotting the curves x^y and y^x we see they tangentially touch each other at point (1,1) which is of course apparent from the equation but i was wondering if i could get more solutions. As it happens we don't, so yes making x the only independent variable makes the problem interesting. Thank you so much! 😊💐👍
There is also an parametric solution using Lambert W function
2 and 4 are the only integer solutions because (m+1)/m is only an integer when m is 1. All other consecutive numbers are coprime to one another, meaning their quotient can't be integers
Why would you consider y to be a linear function of x? Not even a complete linear function at that. Obviously this choice gives some solution, but how do you know we don't have more solutions?
It's interesting to extend this to z^w = w^z, w, z ∈ ℂ. Then the solutions lie on a hypersurface (4-D). With Mathematica and ContourPlot3D, keeping one of the components of z or w fixed and equating, say, the absolute values, you can take a 3D immersion of the solution hypersurfaces. They look pretty weird and I couldn't begin to guess whether they're connected, compact or any of the other topological criteria. I strongly doubt they're simply connected, though. Actually, now that I come to think of it, they'll be a mess because of branch cuts and so forth.
Very interesting and very deep!
whats looks like? chaotic?
Thanks...its really a very good problem...and you provide it's satisfactory answer 👍👍👍☺️
Can't there be for n=2 an m such that m+2/m is a square? in that way you would get more rational solutions, and maybe more integer solutions as well
One of your best videos yet.
Love from France 😁
Wow, thank you! 💖
I worked all this out myself about four years ago (though I used α instead of k because k implies an integer). You get some awesome solutions for k=φ (φ^φ, φ^φ²)*, k=φ² (φ^(2/φ),φ^(2φ²)), and k=φ³ (φ^(3/2φ),φ²^(3φ²/2). (Those were inspired by the k-1 nature of the denominator and the properties of powers of φ). The solutions for k=5 and k=9 are also pretty intriguing (∜5, 5 ∜5) and ∜3, 9∜3). OBTW, the limit of the given equation as k approaches but is not equal to 1 is e, Euler's constant†. I recently gave this as a puzzle on a whiteboard at work in my ongoing efforts to help my co-workers exercise their math skills - In my case, I gave out x,y pairs and asked for the equation with the topic of this video as a follow up. I am sending the link to this video to those who tried to work it.
My co-worker's investigations turned up the fact that xy - x - y - 2 = 0 is a surprisingly good fit for the curve, which may prove useful as a starting point to anyone who wants to iteratively calculate the y for a given x.
* I actually came up with this a few days before Trump posted about news covfefe, and I only barely restrained myself from posting about a logical function cov(x,y) which was true if xʸ = yˣ, and noting this solution in "fe" and "fefe". Until now.
† I couldn't come up with any other transcendental solutions that didn't give away the form of the solutions; if you do, please let me know!
Unfortunately, φ isn't transcendental - only irrational.
@@-wx-78- I meant other than e. I found plenty of algebraic solutions that weren't obvious.
@@SlidellRobotics Algebraic numbers can't be transcendental, by definition of latter.
@@-wx-78- To be more explicit: (e,e) was the only transcendental solution I found that didn't give away the mechanics of finding it.
I find out a general solution to the problem. Unfortunately, there is no enough space for me to write down here.
Hehe! This is nice! 😂
I dont know if im alowed to do this but this is my way :
Let x = a.y
And we can change the equation to become x^(1/x) = y^(1/y)
Substitut x with a.y and we get
Y= a^(1/(a-1))
X=a^(a/(a-1))
(a) is a parameter !!!!
(a) can not be 0 or 1 beacaus undifine
@@tonyhaddad1394 "beacaus undifine" is not English.
@@azzteke is that if a = 1, the denominator of the exponents of the powers would be zero, which cannot happen.
now if a = 0, when substituting in the expression that determines the value of x, then x = 0^0, which is a mathematical indeterminacy.
hence, that a cannot be 0 or 1.
Power of parametric equations. 💪
way to go Syber, excellent tutorial
Glad you liked it!
i have a question why you didn’t consider x = -2 and y=-4 in integer solution it will also satisfy the condition right
You're right!
fasinating as always!
Thank you!
8:27
unique outros are so cool
good place to stop! this brings us to the end of the video, i hope you enjoyed it, thanks for watching, i will se you tomorrow with another video, be safe take care and BYE BYE!!!!!!!!
*please don't get offended sybermath :(
Why would I? 🙂
Good place to stop
@@shreyan1362 no he wont cuz it gives us potential to come again tmrw for the next premeir ...isn't it?
Thanks for posting this video. I've often wondered what the solution to this would be and never had the math skills to do it. But now the question is, what is the parameter for the solion x = y? Because that leads to K^(1/(K-1)) = K^(K/(K-1)). I'm out of my league again.
Help.
Very intresting
Never expected the end 😻
Hello,
What is wrong with x=1, y=1, and x=2, y=2; what is the rule?
Regards,
Second other than SyberMath lol
very interesting video again, I think I see a trend with these lmao
I don't count! 😁
Haha math elite!!!! I have received more hearts than you from sybermath ha ha , zzz,, z😎😎😎😎😎💯💯❤️❤️🤑🤑🤪💓👍👍🔍🔍👍🤪🤑🤑🤑🤑🤑🤑🤑🤑🤪🤪🤪💓💓💓👍👍🔍🔍📌📌📌😎📌🔍👍💓🤪💯💯❤️❤️❤️🤑🤑🤑🤑
I think the equation works for all integers where x=y and y=x
For any positive x except for 1/e, 1 , e. there exists a unique y not equal x such that x^y=y^x.......
Also -2 and -4 works perfectly and they are integers!
yes!
It's almost like a horizon you crossing after that it's prime number factorization
-2,-4 is also a solution set
(-2)^-4=1/16
(-4)^-2=1/16
Hmm that's nice! I never thought about this 😁
complexe solution please
nice solution sir thank u
Most welcome
Sir You take y=Kx..
means y related by x linearly
How u can take..?
It's like exploring the solutions.
Why there're missing the simplest? X=1 , Y=1 ?
Too trivial! 😂
Forgive me my lame English. Some years ago, I tried to prove, that it is only one integer pair as a solution and proved it by follows way.
1. It is evident, that either x and y can not be 0, therefore we can power both sides by (1/x)(1/y) => x^(1/x) = y^(1/y) and to analyze only 1 part (e.g. left)
2. By brief examination of we can say:
2.1 It has one maximum at = x=e and this can be found by derivative An this is only one such specific point
2.2 On interval (1,inf) it goes from 1 up until x reaches e as max and then goes down from this max to 1
2.3 The interval (0,1] isn't interesting - the expression value is below 1
2.4 So, each x on (1,e) wil have one and only one "friend" on (e,inf.) such, that has the same value of expression
2.5 In (1, e) ()we have only one integer and therefore we can not have more integers pairs (2,4 and 4,2 I recognize as 1 pair)
3. Then I tried find the rational pairs. Your suggestion to use kx as y helped me. If k is like 1+ 1/a it allows to get rational pairs
Nice. Your English is fine!
Excellent tutorial, fantastic perspective! Just based off of your accent I think you are Turkish aswell right ? Just an assumption you don't have to answer it haha.
What about if X=y? All numbers are true to this equation!
yes but we are also interested in other solutions
X=Y is obvious solution
Cool!
Superb 👍
Thanks 🤗
Can u pls do trigonometry or DE or integrals next
DEs are too hard! 😁
The next video has already been planned. It will be on complex numbers. Let me think of a good integral
@@SyberMath x^x, jk
Btw, x=y=-1 are also the solutions... :)
wowww ....... you are unstopabe, syber.Love from Bangladesh
💖
Solution x=y........both side by exponential 1/x....both side by exponential 1/Y......result...x exponential (1/x) = Y exponential (1/y).....which means all X=Y....can be a solution for the equation.
Why did you say that n = 1? If the base (m + n)/m is a perfect square, n can equal 2, right? And if the base is a perfect cube, n = 3, and so on... is there any reason why (m + n) / m can't be a perfect square?
There are no two perfect squares that differ by 2; so, (m+n)/m being the square of a rational number would mean either that n isn't 2 or that m+2 and m have a common factor (although, it will turn out that this still doesn't allow squares). The first case would mean that you're taking a root other than the square root; and the second would mean that m is even, which makes the solution equivalent to one for m'=m/2 and n'=1.
A similar argument will work for any other root just by noting that any two nth powers of positive integers differ by more than n.
@@SSGranor Thanks a lot... what about negative values of m or n? That should work, right?
@@haricharanbalasundaram3124 If m and n are both negative, the sign cancels and the analysis remains unchanged. So, we only have to consider the case where their signs differ.
If (m+n)/m > 0, the only thing that changes is that the numerator of this fraction is smaller than the denominator. They still differ by too little to both be nth power.
If (m+n)/m < 0, even roots will be imaginary; so, we only have to think about odd n. In this case, the argument changes slightly as -1 will be an nth power. However, other than the difference of 2 between -1 and 1, all other difference between nth powers will still be larger than n. And, the 2 doesn't help, since n is odd.
(Hopefully it pretty much goes without saying why we don't need to think about m=0 or m+n = 0.)
@@SSGranor Thanks a lot, much appreciated!
Good video
Thanks
Yes , we assume x =4, y= 2.. 4^2= 2^4=16 (proved)..
Plzzz do some on DE
Hmmm Can You Teach Integral??
Probably! 😁
Check this playlist:
Calculus: ruclips.net/p/PLvPOIUdohGFAmPjF08A7llzsa4qtIAa_K
what about x=2 and y=2 isnt that another solution?
x and y cannot be the same
@@adipy8912 there’s literally a function in where x = y
@@adipy8912 we can also say x=3 y=3
@@Phantoms3709 Yes! But answers like that is too obvious. He said that we could ignore x and y to be the same. Then it would be more interesting answers.
@@adipy8912 you’re right, interestingly enough if you graph this function there are infinite answers but I suppose the only 2 different integer solutions are 2 and 4
The most basic solution is x = 1 and y = 1
How about x=n and y=n?
That's also right
Good jobb !!!!! Challenging problem!!!
Thank you!!!
You have made an assumption that x and y have linear relationship. Just like we can't we can't assume that sin(x)=k cos(x). You got a solution, but not the only solution.
Just changing variables here
William, you have made an assumption that K is a constant 😮
Don't forget the trivial (x = 0, y = 0). ;)
😁
If this equation was a RIDDLE... ; then the answer will be dead easy.... ;
ie. X=Y. 🤔🤪
😁
(-2;-4) should also be a solution
How?
(-2)^(-4) =((-2)^2)^(-2)=(4)^(-2)=(-4)^(-2)
a^x=(b^n)^x is equivalent to:
- a=b^n if a>= 0 and b^n>=0 (for the latter to be true we have b>=0 or n is even (n=2k ...English is not my first language))
- -a=b^n if a0
- a=-b^n if a>0 and b^n
(a^x)^(1/x) is only possible if a^x>=0, because of the log function implicitedly used ... I think..i'm not certain myself about the last part
x=y=e is the bifurcation point
bifurcation point?
@@SyberMath where trivial and nontrivial solutions cross.
X=1, Y=1.
Is it?
From x^(kx) = (kx)^x
Plot a graph
X=4, y =2
Only that?
good vid... obviously...
x^y=y^x=1/x^y
x^y=1/x^y implies x^y*x^y=1 implies x^2y=1
x^2y=1 implies x^2y=x^0 becouse power of zero to any number is 1 x^2y=x^0 implies 2y=0 implies y=0 and x=0 is the correct answer
my computer can find some solutions but not in a closed form. interesting. haha
Humans are smarter than computers!
How x^k = kx
(4,2)
I did enjoy the video until the last:
_Unfortunately, or maybe fortunately..._
😅
😁
X=1 AND Y=1
That's it?
Wowww❤️❤️🥰love from india🙏
Thank you! 💖
2,4
isso é genial
]1,infinity[
x=y=1
That's it?
I think i am third 😍
Good job! You got Bronze Medal! 😁
@@SyberMath oh i am glad to recieve the medal.😁😁
Damn you're so good....
Thank you! Am I?
@@SyberMath yes u are!
@@SyberMath damn you're so good
Jajaja khe?
Nedense türkmüs gibi geliyor
For u solve this problem
1) u must be an African
Y=X?? Hahaha
Just that? 😄
@@SyberMath is that correct?? Then.. ha! 😎😎😎
first! take that math elite !
😁
rip :(((
@@MathElite and here i am missing premiere for straight two days
@@akshatjangra4167 yeah me too☹️☹️
@@vivekbhutada3049 i am sort of stuck in a wedding :(
Total bs.