# Roots $(x,y)$ of the equations of the type $x^y=y^x=k, k>1$

Here I want to share a numerical method of approximating the roots of the equation $x^y$ = $y^x$ = $k$, where $k$ is a real number, $k>1$.

Interestingly I came upon it while playing with a scientific calculator two years ago, and later on I proved the algorithm, and wrote the code for it. I discovered this entirely by accident, I had not even wanted to do this. I just saw something and realized that 'this' might work for something like 'this'. To know what the two 'this' refer to, read ahead.

Here is my claim.

We define a sequence $a_{i}$ such that $a_{i}$=$k^{1/a_{i-1}}$ , and $a_{1}$=$k^{1/m}$ where m is any positive real number. Now we go on evaluating the terms of the sequence. Let us take the odd numbered terms to be a sub-sequence $M$, and the even numbered terms to be a sub-sequence $T$. Then my claim is that,

The sub-sequence $M$ is convergent, and so is the sub-sequence $T$ and if their limits are $z$ and $g$ respectively, then $(z,g)$ and $(g,z)$ are solutions to our equation.

And here goes the proof:

$P$ = $a_{1}$ = $k^{1/m}$, and $b_{j+1}$ = $k^{1/b_{j-1}}$ for all j in $N$ ( $N$ is the set of all natural numbers.) and m is any positive real number. (domain of m is $(0,∞)$ )

Let us take the two sub-sequences { $a_{1}$ , $a_{3}$, $a_{5}$ …… } and { $a_{2}$, $a_{4}$, $a_{6}$, ….. } that is the sub-sequences formed by the odd numbered terms and that formed by the even numbered terms respectively.

Let the sequence formed by the odd numbered terms be M and the sequence formed by the even numbered terms be T.

Now we express $a_{n+2}$ in terms of $a_{n}$ . By definition,

$a_{n+2}$ = $k^{1/a_{n+1}}$ =$k^{1/{k^{1/a_{n}}}}$ __ $(1)$

Our first observation is that the two sub-sequences that I have considered are both bounded below by $0$ and above by $k$ itself.

The lower bound is easy to observe because we know that the exponential function $a^x$ is positive for all real $x$ and positive $a$. And our sequence is of the exponential form only. So $0$ is indeed a lower bound.

Now we can establish the validity of $k$ as an upper bound by the method of contradiction.

Let us assume that the sequence $P$ contains a number which is greater than $k$ itself. i.e. $a_{h}$ = $v>k$ for some $h$ in $N$.

But $a_{h}$ = $k^{1/{a_{h-1}}}$ = $v$ necessary implies that $a_{h-1}$ is less than $1$ and greater than $0$ i.e. $1/{a_{h-1}}$ is greater than $1$, and $k^{1/{a_{h-1}}}$ is greater than $k$. All this holds if $a_{h-1}$ is less than $1$. But again

$a_{h-1}$ = $k^{1/{a_{h-2}}}$ . Now if we take logarithm to the base $k$ on both sides ,

then on L.H.S. we get something negative as $a_{h-1}$ belongs to $(0,1)$ by assumption and $k$ is greater than $1$ (domain of k is $(1,∞)$ ) but on R.H.S we have $1/{a_{h-2}}$ which is positive because $a_{h-2}$ is positive as the lower bound of the sequence {$a_{1}$, $a_{2}$, $a_{3}$, ….} is 0. Thus a negative number on the L.H.S is equal to a positive number on the R.H.S by assumption that $a_{h}>k$ for some natural number $h$.

So we have a direct contradiction if $a_{h}>k$ for some natural number $h$, therefore this assumption must be false. So the negation of this statement must be true. That is ‘$a_{h} for all natural numbers $h$.’ So we have established that $k$ itself is an upper bound.

Our second aim is to prove that these two sub-sequences are monotonic.

Let us consider the sub-sequence M={ $a_{1}$, $a_{3}$, $a_{5}$, .......} and we take $M_{1}$ = $a_{1}$ , $M_{2}$ = $a_{3}$, $M_{3}$ = $a_{5}$, and in general $M_{k}$ = $a_{2k-1}$ for all $k$ in $N$.

We know by order property (Law of Trichotomy) ‘Given two reals $a$ and $b$, either $a=b$, or $a>b$, or $a.’ Here we observe that $M_{n}$ is not equal to $M_{n+1}$ for any natural number $n$, because of the relation between two consecutive terms of the sequence { $a_{1}$, $a_{3}$, $a_{5}$, …. } derived in $(1)$ above.

So $M_{1}>M_{2}$ or $M_{2}>M_{1}$:

Case I:

$M_{1}>M_{2}$ i.e. $a_{1}>a_{3}$ . Next we proceed by the principle of mathematical induction to show that $M_{n}>M_{n+1}$ for all natural numbers $n$. In fact we have already checked the first step for $n=1$.

Next we assume the validity of the statement to hold true for some $n=k$. i.e. $M_{k}>M_{k+1}$ or $a_{m}>a_{m+2}$ (where obviously $m={2k-1}$

Now

$M_{k}>M_{k+1}$

$a_{m}>a_{m+2}$

$1/{a_{m}}$ < $1/{a_{m+2}}$ [reciprocating the positive numbers reverses the sign of inequality]

$k^{1/{a_{m}}}$ < $k^{1/{a_{m+2}}}$

$1/{k^{1/{a_{m}}}}$ > $1/{k^{1/{a_{m+2}}}}$ [reciprocating the positive numbers reverses the sign of inequality]

$k^{1/{k^{1/{a_{m}}}}}$ > $k^{1/{k^{1/{a_{m}}}}}$

$a_{m+2}$>$a_{m+4}$ (by the relation derived in $(1)$ )

i.e.

$M_{k+1}$ > $M_{k+2}$

So we have $M_{1}$ > $M_{2}$ and $M_{k}$ > $M_{k+1}$ implies $M_{k+1}$ > $M_{k+2}$ for all natural numbers $k$.

So $M$ is a strictly decreasing monotone sequence.

Having established that we also see that $M_{1}$ > $M_{2}$ i.e. $a_{1}$ > $a_{3}$ implies that

$1/{a_{1}}$ < $1/{a_{3}}$

$k^{1/{a_{1}}}$ < $k^{1/{a_{3}}}$

$a_{2}$ < $a_{4}$

Therefore $a_{1}$>$a_{3}$ implies that $a_{2}$ < $a_{4}$ . If we consider the sub-sequence $T$ consisting of { $a_{2}$, $a_{4}$, $a_{6}$, ….. } then we have $a_{2}$ < $a_{4}$ i.e. $T_{1}$ < $T_{2}$ . By a similar inductive argument as above we can see that

$T$ is a strictly increasing monotone sequence.

It is to be noted that the conclusions drawn above are under the heading of [ Case I: ]

Case II:

$M_{1}$ < $M_{2}$ . Under this consideration we proceed exactly as in Case I: and prove that

$M$ is a strictly increasing monotone sequence.

And on the same lines as in Case I: we have that

$T$ is a strictly decreasing monotone sequence.

1. In both the cases which are mutually exhaustive we see that $M$ and $T$ are monotone sequences of the opposite nature, i.e. if one is increasing the other is decreasing.

2. Also $M$ and $T$ are sub-sequences of a sequence $P$ which is bounded as a whole with $0$ as a lower bound and $k$ as an upper bound.

3. So $M$ and $T$ themselves are also bounded with $0$ as a lower bound for both and $k$ as an upper bound for both.

So from 1, 2 and 3 we see that M and T are both monotonic and bounded.

By the Monotone-Convergence theorem, ‘Any monotonic bounded sequence has a finite limit.’

Since the sequences $M$ and $T$ meet the requirements of the theorem, they are convergent and have a finite limit. Let the limit of $M_{n}$ as $n$ tends to $∞$ be $z$ and the limit of $T_{n}$ as $n$ tends to $∞$ be $g$.

Now we note that as $n$ tends to infinity the terms of the sequence $P$ = { $a_{1}$, $a_{2}$, $a_{3}$,….. } take the limiting values $z$ and $g$ alternatively because these are the limits of the odd numbered sub-sequence and the even numbered sub-sequence respectively which have alternate terms occurring in $P$.

That is as $n$ tends to infinity the sequence $P$ takes limiting values $z$ and $g$ alternately i.e. $z$, $g$, $z$, $g$, $z$, $g$, $z$, $g$, $z$, $g$, ……….. But we have defined the function as { $a_{n}$ = $k^{1/{a_{n-1}}}$ }. So as $n$ tends to infinity we have successive values coming as $z$ = $k^{1/g}$ and the next term as $g$ = $k^{1/z}$.

From the above two equations we get $g^z=k$ and $z^g=k$. So we can say that $(z,g)$ is a solution to the equation $x^y$ = $y^x$ = $k$.

Hence Proved.

When I was playing about with this, the first value I got this on was 3. So I have displayed the code for k=3 below. Interestingly for $3$ the two convergence values $z$ and $g$ are so close by that they seem to be identical. As a result it also gives the result for $x^x=3$. In fact it actually gives two different values which are very close by, hence it seems that they are equal.Interesting, huh?

For those interested here is the code in Python and the output with it. You are welcome to copy paste it and test it with your own values.

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 print 'This program has been written to verify the conjecture devised by me, and then prove it by the methods of calculus. In the following lines, you will be prompted for 3 numbers' print 'The first number is \'a\' where we want to find the roots of the equation x^y=y^x=a, a>1.' print 'The second number is \'m\' where we start with the mth root of a.' print 'The third number is \'g\' where g is the number of iterations of the recursion we are calculating for approximating the limits of the two subsequences.' print 'Please make sure g>100.' print 'The two limits will be stored in p and q.' a=int(raw_input('a=:')) m=float(raw_input('m=:')) g=int(raw_input('g=:')) s=m for i in range(g): s=a**(1.0/s) print 'The',i+1,'th iteration yields ', s if i==g-2: p=s if i==g-1: q=s 

[and the output of the program is]

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 Python 2.7.4 (default, Apr 6 2013, 19:54:46) [MSC v.1500 32 bit (Intel)] on win32 Type "copyright", "credits" or "license()" for more information. >>> ================================ RESTART ================================ >>> This program has been written to verify the conjecture devised by me, and then prove it by the methods of calculus. In the following lines, you will be prompted for 3 numbers The first number is 'a' where we want to find the roots of the equation x^y=y^x=a, a>1. The second number is 'm' where we start with the mth root of a. The third number is 'g' where g is the number of iterations of the recursion we are calculating for approximating the limits of the two subsequences. Please make sure g>100. The two limits will be stored in p and q. a=:59 m=:12 g=:120 The 1 th iteration yields 1.40465930662 The 2 th iteration yields 18.2263033042 The 3 th iteration yields 1.25071725987 The 4 th iteration yields 26.053685862 The 5 th iteration yields 1.16941684899 The 6 th iteration yields 32.6816132681 The 7 th iteration yields 1.13288274786 The 8 th iteration yields 36.5710934875 The 9 th iteration yields 1.11794948021 The 10 th iteration yields 38.3723079533 The 11 th iteration yields 1.11211377009 The 12 th iteration yields 39.1137902945 The 13 th iteration yields 1.10987575602 The 14 th iteration yields 39.40404115 The 15 th iteration yields 1.10902381716 The 16 th iteration yields 39.5154051688 The 17 th iteration yields 1.10870043713 The 18 th iteration yields 39.5578041471 The 19 th iteration yields 1.10857782176 The 20 th iteration yields 39.5738988513 The 21 th iteration yields 1.10853134915 The 22 th iteration yields 39.580001567 The 23 th iteration yields 1.10851373827 The 24 th iteration yields 39.5823145817 The 25 th iteration yields 1.108507065 The 26 th iteration yields 39.5831911051 The 27 th iteration yields 1.10850453636 The 28 th iteration yields 39.5835232458 The 29 th iteration yields 1.10850357821 The 30 th iteration yields 39.5836491008 The 31 th iteration yields 1.10850321515 The 32 th iteration yields 39.5836967895 The 33 th iteration yields 1.10850307758 The 34 th iteration yields 39.5837148595 The 35 th iteration yields 1.10850302546 The 36 th iteration yields 39.5837217065 The 37 th iteration yields 1.1085030057 The 38 th iteration yields 39.583724301 The 39 th iteration yields 1.10850299822 The 40 th iteration yields 39.5837252841 The 41 th iteration yields 1.10850299538 The 42 th iteration yields 39.5837256566 The 43 th iteration yields 1.10850299431 The 44 th iteration yields 39.5837257977 The 45 th iteration yields 1.1085029939 The 46 th iteration yields 39.5837258512 The 47 th iteration yields 1.10850299375 The 48 th iteration yields 39.5837258715 The 49 th iteration yields 1.10850299369 The 50 th iteration yields 39.5837258791 The 51 th iteration yields 1.10850299367 The 52 th iteration yields 39.583725882 The 53 th iteration yields 1.10850299366 The 54 th iteration yields 39.5837258831 The 55 th iteration yields 1.10850299366 The 56 th iteration yields 39.5837258836 The 57 th iteration yields 1.10850299365 The 58 th iteration yields 39.5837258837 The 59 th iteration yields 1.10850299365 The 60 th iteration yields 39.5837258838 The 61 th iteration yields 1.10850299365 The 62 th iteration yields 39.5837258838 The 63 th iteration yields 1.10850299365 The 64 th iteration yields 39.5837258838 The 65 th iteration yields 1.10850299365 The 66 th iteration yields 39.5837258838 The 67 th iteration yields 1.10850299365 The 68 th iteration yields 39.5837258838 The 69 th iteration yields 1.10850299365 The 70 th iteration yields 39.5837258838 The 71 th iteration yields 1.10850299365 The 72 th iteration yields 39.5837258838 The 73 th iteration yields 1.10850299365 The 74 th iteration yields 39.5837258838 The 75 th iteration yields 1.10850299365 The 76 th iteration yields 39.5837258838 The 77 th iteration yields 1.10850299365 The 78 th iteration yields 39.5837258838 The 79 th iteration yields 1.10850299365 The 80 th iteration yields 39.5837258838 The 81 th iteration yields 1.10850299365 The 82 th iteration yields 39.5837258838 The 83 th iteration yields 1.10850299365 The 84 th iteration yields 39.5837258838 The 85 th iteration yields 1.10850299365 The 86 th iteration yields 39.5837258838 The 87 th iteration yields 1.10850299365 The 88 th iteration yields 39.5837258838 The 89 th iteration yields 1.10850299365 The 90 th iteration yields 39.5837258838 The 91 th iteration yields 1.10850299365 The 92 th iteration yields 39.5837258838 The 93 th iteration yields 1.10850299365 The 94 th iteration yields 39.5837258838 The 95 th iteration yields 1.10850299365 The 96 th iteration yields 39.5837258838 The 97 th iteration yields 1.10850299365 The 98 th iteration yields 39.5837258838 The 99 th iteration yields 1.10850299365 The 100 th iteration yields 39.5837258838 The 101 th iteration yields 1.10850299365 The 102 th iteration yields 39.5837258838 The 103 th iteration yields 1.10850299365 The 104 th iteration yields 39.5837258838 The 105 th iteration yields 1.10850299365 The 106 th iteration yields 39.5837258838 The 107 th iteration yields 1.10850299365 The 108 th iteration yields 39.5837258838 The 109 th iteration yields 1.10850299365 The 110 th iteration yields 39.5837258838 The 111 th iteration yields 1.10850299365 The 112 th iteration yields 39.5837258838 The 113 th iteration yields 1.10850299365 The 114 th iteration yields 39.5837258838 The 115 th iteration yields 1.10850299365 The 116 th iteration yields 39.5837258838 The 117 th iteration yields 1.10850299365 The 118 th iteration yields 39.5837258838 The 119 th iteration yields 1.10850299365 The 120 th iteration yields 39.5837258838 >>> g**z 58.999999999999986 >>>z**g 59.00000000000006 

Also I have started with the 12th root of 59 here just to demonstrate that it does not matter what value of $m$ we are choosing at the beginning.

P.S. Please leave your comments and feedback for improvement.

Note by Soumava Pal
4 years ago

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## Comments

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@Otto Bretscher Please help me out here.

This algorithm relies on the fact that $x \mapsto k^{k^{\frac{-1}{x}}}\$ is a contraction for k > 1

How do I show this (in an alternate way)?

- 3 years, 12 months ago

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- 3 years, 12 months ago

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Magnificent work!

- 3 years, 12 months ago

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Thanks. XD

- 3 years, 12 months ago

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@Soumava Pal I made a small change to fix the formatting. Can you check the code to see if there is something you want to edit out?

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@Agnishom Chattopadhyay Thanks for pointing it out, I did not even notice. I have pasted the code for a wrong hypothesis here. I am modifying it immediately. This is the last code I have written. Do tell me if it is all right now.

- 4 years ago

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Haha, glad that I could help. I did not notice anything wrong at first, either. I just wanted to fix the formatting (which I'll do again).

By the way, thanks for sharing your work. This is Brilliant. (Pun intended with genuine appreciation)

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Haha. Pun accepted. Thanks for the appreciation.

P.S. Could you please tell me how to include code in replies and solutions? So far all I have done is to give ideone links. Because everytime I include code, the indentations don't come properly, till I have used a new line between every other line.

 1 print "Hello World!" 

- 4 years ago

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I included a line in your comment for reference. You could click the edit and check it out.

Here is the full reference

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Thanks a lot!

- 4 years ago

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@Soumava Pal that is great! ... BTW how did you exactly get this idea?

- 4 years ago

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@Abhinav Raichur As I said, I was playing with the calculator, precisely I was doing the following steps over and over again:

Type a number k. Then type the symbol ^.

Then type 1/m where m is any integer you like.

Then you get some result.

Then you type k again, followed by ^, and then press the button marked 'Ans' on your calculator.

Then go on pressing '=' again and again.

I just found this pressing of '=' repeatedly quite fun from my childhood.

Incidentally I found that this value stopped deviating after a while.

I tried it with more numbers.

In this way I got to this observation. Though at first I thought it would be the solution to $x^x=k$ because I was taking small values of k, the convergence values were very close. But as I went on taking larger values they alternated. I had to modify my observation. I only found a proof to this last year, and have discussed it with many important people before publishing it here.

P.S. I also have another observation of mine that I got by this same game of mine, pressing a particular button again and again on the calculator. But that observation was said to be wrong by a famous professor. I need to check it again, but it seems to work every time. That is material for another note.

- 4 years ago

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That is great use of the calculator ... all I used to do is force the calculator to stack error :p .. +1 for patiently developing a rigorous proof .. and yes!, waiting for your next note :)

- 4 years ago

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@Abhinav Raichur Good to know. It will be some time before I can publish my next note because my board exams for the 12th are from March 1st, and I need to get decent marks in Chemistry and Physics. Also I need to consult some more before publishing that note. Meanwhile, Thanks for your support.

- 4 years ago

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just to share my experience, I used to press a lot of nested sines ( i.e. sin(sin(sin(...)) ) and the answer was always near to zero... till my tenth grade I would remember sine as the function that returned zero if you'd press it lot times .... then came trigonometry :)

- 4 years ago

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@Abhinav Raichur Well then I would suggest you to proceed along those lines further, because the button I used to press repeatedly on my calculator for my other observation that I mentioned has something to do with Trigonometry. Go ahead with it.

- 4 years ago

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Well there is already a theorem in ergodic theory for that .. and I also convinced that fact myself by drawing graphs

- 4 years ago

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I too convinced myself by drawing graphs. I don't know what ergodic theory is, but I have seen a proof for the sin(sin(sin.... (x))))..)) and that of cos(cos(cos(.....(x)))...)) having the property that they both converge on repeated composition. However my function differs a little from these two in the fact that it uses multiple trigonometric functions and it's convergence value is also the value given by a trigonometric function.

- 4 years ago

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Interesting ... looking forward for that note ... bye :)

- 4 years ago

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Thanks, bye. :)

- 4 years ago

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