Got stuck at a point -2

While working with algebra, I got a very interesting equation to solve which is as follows

\[a^y=y\] \[\Rightarrow a^{a^{y}}=y\] You may continue like this and you will get \[\boxed{\red{a^{a^{a^{a^{a...}}}}=y}}\] Where \(a\) is known, \(a∈(0,e^{\frac{1}{e}}]\), can you solve for \(y\) such that \(y∈(0,e]\)?

Any solution will be appreciated

Note by Zakir Husain
3 months, 1 week ago

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This property or in fact this question is known as tetration.

Interestingly if we let

aaaaa=2a^{a^{a^{a^{a^{}}}}} = 2 a2=2    a=2a^2 = 2 \implies a = \sqrt{2}

And again

bbbbb=4b^{b^{b^{b^{b^{}}}}} = 4 b4=4    b=2b^4 = 4 \implies b = \sqrt{2}

Now, since a=ba=b,     aaaaa=bbbbb    2=4\implies a^{a^{a^{a^{a^{}}}}} = b^{b^{b^{b^{b^{}}}}} \implies 2=4 ?

Does this mean 2=42 = 4?

Mahdi Raza - 3 months, 1 week ago

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Aren't those two two different equations. So, you can't equate both of them, ain't it?

Aryan Sanghi - 3 months, 1 week ago

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Rethink, i have edited my comment a bit

Mahdi Raza - 3 months, 1 week ago

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@Zakir Husain

Mahdi Raza - 3 months, 1 week ago

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well all values of aa where a(1,e1e)a∈(1,e^\frac{1}{e}) follow this behaviour

Zakir Husain - 3 months, 1 week ago

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Yes

Mahdi Raza - 3 months, 1 week ago

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@Mahdi Raza - It is also a math failure, a potential Math also fails! -3. Well, what could be it's explanation?

Zakir Husain - 3 months, 1 week ago

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I remember seeing this question somewhere. The answer said that it is limited for some values of aa or bb, but I don't remember much.

Vinayak Srivastava - 3 months, 1 week ago

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This might help :) The number of x’s are finite, yet the graph starts to approach a straight line already.

Jeff Giff - 3 months, 1 week ago

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The straight line at the end is x=e1e1.444667861009766x = e^{\frac{1}{e}} \approx 1.444667861009766

Mahdi Raza - 3 months, 1 week ago

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@Mahdi Raza Oh! That is interesting :)

Jeff Giff - 3 months, 1 week ago

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@Jeff Giff, For infinite xx you should graph y=xyy=x^y instead of this poor approximation, because this will give you the exact graph.

Zakir Husain - 3 months, 1 week ago

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@Zakir Husain Oh, I never thought of that! Silly me :)

Jeff Giff - 3 months, 1 week ago

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@Jeff Giff, how you put this image in the comment?

Zakir Husain - 3 months, 1 week ago

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@Zakir Husain First try posting the image in a question. Then copy the link form. Finally paste it onto the comment :)

Jeff Giff - 3 months, 1 week ago

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@Zakir Husain What i do is go to a previous solution, insert media, copy the location code, and paste it in the comments. It is tedious i know

Mahdi Raza - 3 months, 1 week ago

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@Jeff Giff, see the graph below:

I only want a formula which relates all the xx and yy of the points of xy=yx^y=y in the rectangle formed by the xaxis,yaxis,y=e,x=e1ex-axis,y-axis,y=e,x=e^\frac{1}{e}. As I said in my comment

Zakir Husain - 3 months, 1 week ago

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@Zakir Husain Oh. I’ll try to help:)

Jeff Giff - 3 months, 1 week ago

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@Zakir Husain Maybe it’s related to [(x+a)n+b]×c[(x+a)^n+b]\times c

Jeff Giff - 3 months, 1 week ago

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@Mahdi Raza- Explanation to the contradiction

Let x=yyy...x=yxx=y^{y^{y^{.^{.^{.}}}}} \Rightarrow x=y^x assuming x0x ≠ 0 y=x1xy=x^\frac{1}{x}

Now the maxima of x1xx^\frac{1}{x} is e1ee^\frac{1}{e} and for any xx(e,)x | x \in (e,∞), y(1,e1e)y \in (1,e^\frac{1}{e}).

And for any xx(1,e)x | x \in (1,e), y(1,e1e)y \in (1,e^\frac{1}{e})

So for any two variables x1x_1 and x2x_2 x1(1,),x2(1,)| x_1\in (1,∞),x_2 \in (1,∞) and x11x1=x21x2x1=x2x_1^\frac{1}{x_1}=x_2^\frac{1}{x_2} \nRightarrow x_1=x_2

And in your case, applying above statement a=212a=2^\frac{1}{2} and b=414a=b2=4b=4^\frac{1}{4} \Rightarrow a=b \nRightarrow 2=4

Zakir Husain - 3 months ago

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  • If you haven't read the note above and the comment above yet, then kindly read them first.
@Chew-Seong Cheong, @Justin Travers , @jordi curto , @Yajat Shamji , @Mahdi Raza , @Kumudesh Ghosh , @Alak Bhattacharya , @Vinayak Srivastava , @Aryan Sanghi , @Jeff Giff , @Marvin Kalngan .

Now a new problem arises, if x11x1=x21x2;x1x2x_1^\frac{1}{x_1}=x_2^\frac{1}{x_2}; x_1≠x_2 then what is the relation between x1x_1 and x2x_2 ??

Zakir Husain - 3 months ago

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@Zakir Husain This might help :)

Jeff Giff - 3 months ago

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@Zakir Husain x1,x2x_1,x_2 are the xx values of the intersection between x1xx^{\frac 1x} and y=x11x1y=x_1^{\frac {1}{x_1}}.

Jeff Giff - 3 months ago

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@Jeff Giff The graphs are intersecting at (1,1)(1,1), that is when x1=1,x2=1x_1=1,x_2=1, in that case x1=x2x_1=x_2. I'm talking about the case
When x11x1=x1x2x_1^\frac{1}{x_1}=x^\frac{1}{x_2} but x1x2x_1 \neq x_2. For example when x1=4;x2=2x_1 = 4; x_2=2; then 212=2=4142^\frac{1}{2}=\sqrt{2}=4^\frac{1}{4} but 242 \neq 4

Zakir Husain - 3 months ago

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@Zakir Husain But the graphs in the pic aren’t the ones that I mentioned:)

Jeff Giff - 3 months ago

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@Zakir Husain I just meant to say that the graphs in the picture are symmetrical along x=y.x=y.

Jeff Giff - 3 months ago

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@Zakir Husain Also, x1x_1 is on the interval [0,e)[0,e) while x2x_2 is on the interval (e,](e,\infty] for x1x2x_1\neq x_2 :)

Jeff Giff - 3 months ago

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@Zakir Husain e.g. the graphs intersect at (2,2)(2,\sqrt2) & (4,2)(4,\sqrt2) so 212=414=2.2^{\frac12}=4^{\frac14}=\sqrt2.

Jeff Giff - 3 months ago

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@Zakir Husain Two diferents values has same image :

other ejemple : f(x) = x^2

x= -2 and x= 2 => f(x) = 4

Relation is f inverse f(x) = x\sqrt{x}

for x= 4 => f(x) = + 2 and -2

jordi curto - 3 months ago

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@Jeff Giff- See here, I have already written that

Zakir Husain - 3 months ago

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Solution will exist only when a(0,1)a \in (0, 1). Now, you can draw graphs for y = axa^x and y = x and find the solution.

Aryan Sanghi - 3 months, 1 week ago

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solution exists if a(0,e1e)a∈(0,e^{\frac{1}{e}}), I recalculated it just now

Zakir Husain - 3 months, 1 week ago

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Anything wrong in this graph?

Vinayak Srivastava - 3 months, 1 week ago

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@Vinayak Srivastava - Nothing is wrong with the graph but, it is only for the case a=1.4a=1.4 I want a general formula for the solution for any a(0,e1e)a∈(0,e^\frac{1}{e})

Zakir Husain - 3 months, 1 week ago

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@Aryan Sanghi- I am searching for a formula which gives value of yy for every value of aa where a(0,e1e)a∈(0,e^\frac{1}{e})

Zakir Husain - 3 months, 1 week ago

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a(1ee,e1e)a \in (\frac{1}{e^e}, e^{\frac{1}{e}})

Mahdi Raza - 3 months, 1 week ago

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Well the e1ee^\frac{1}{e} comes by taking the maxima of the function x1/xx^{1/x}, but the minima is not 1/ee1/e^e. If you know calculus and wanted proof for the given argument, then please comment.

Zakir Husain - 3 months, 1 week ago

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@Zakir Husain Sorry I haven't learned calculus

Mahdi Raza - 3 months, 1 week ago

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@Zakir Husain Yes, you are right. I got same value as yours.

Aryan Sanghi - 3 months, 1 week ago

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@Chew-Seong Cheong, @Justin Travers, @jordi curto, @Yajat Shamji, @Mahdi Raza, @Kumudesh Ghosh, @Alak Bhattacharya, @Vinayak Srivastava, @Aryan Sanghi, @Jeff Giff, @Marvin Kalngan. For every a(1,e1e)a∈(1,e^\frac{1}{e}) there exists 22 solutions for each aa, one of them will be yy such that y(1,e)y∈(1,e). I'm only considering that as solution for 1<a<e1e1<a<e^\frac{1}{e}, for 0<a<10<a<1 there is only one solution for each aa. Can anyone find any formula for these solutions where (0<y<e\red{0<y<e}). I have also edited the note a bit, because I don't need y>ey>e

Zakir Husain - 3 months, 1 week ago

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Why mention me? @Zakir Husain. I cannot do this stuff - please remove my mentions...

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