Roots (x,y)(x,y) of the equations of the type xy=yx=k,k>1x^y=y^x=k, k>1

Here I want to share a numerical method of approximating the roots of the equation xyx^y = yxy^x = kk, where kk is a real number, k>1k>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 aia_{i} such that aia_{i}=k1/ai1k^{1/a_{i-1}} , and a1a_{1}=k1/mk^{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 MM, and the even numbered terms to be a sub-sequence TT. Then my claim is that,

The sub-sequence MM is convergent, and so is the sub-sequence TT and if their limits are zz and gg respectively, then (z,g)(z,g) and (g,z)(g,z) are solutions to our equation.

And here goes the proof:

PP = a1a_{1} = k1/mk^{1/m}, and bj+1b_{j+1} = k1/bj1k^{1/b_{j-1}} for all j in NN ( NN is the set of all natural numbers.) and m is any positive real number. (domain of m is (0,)(0,∞) )

Let us take the two sub-sequences { a1a_{1} , a3a_{3}, a5a_{5} …… } and { a2a_{2}, a4a_{4}, a6a_{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 an+2a_{n+2} in terms of ana_{n} . By definition,

an+2a_{n+2} = k1/an+1k^{1/a_{n+1}} =k1/k1/ank^{1/{k^{1/a_{n}}}} __ (1)(1)

Our first observation is that the two sub-sequences that I have considered are both bounded below by 00 and above by kk itself.

The lower bound is easy to observe because we know that the exponential function axa^x is positive for all real xx and positive aa. And our sequence is of the exponential form only. So 00 is indeed a lower bound.

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

Let us assume that the sequence PP contains a number which is greater than kk itself. i.e. aha_{h} = v>kv>k for some hh in NN.

But aha_{h} = k1/ah1k^{1/{a_{h-1}}} = vv necessary implies that ah1a_{h-1} is less than 11 and greater than 00 i.e. 1/ah11/{a_{h-1}} is greater than 11, and k1/ah1k^{1/{a_{h-1}}} is greater than kk. All this holds if ah1a_{h-1} is less than 11. But again

ah1a_{h-1} = k1/ah2k^{1/{a_{h-2}}} . Now if we take logarithm to the base kk on both sides ,

then on L.H.S. we get something negative as ah1a_{h-1} belongs to (0,1)(0,1) by assumption and kk is greater than 11 (domain of k is (1,)(1,∞) ) but on R.H.S we have 1/ah21/{a_{h-2}} which is positive because ah2a_{h-2} is positive as the lower bound of the sequence {a1a_{1}, a2a_{2}, a3a_{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 ah>ka_{h}>k for some natural number hh.

So we have a direct contradiction if ah>ka_{h}>k for some natural number hh, therefore this assumption must be false. So the negation of this statement must be true. That is ‘ah<ka_{h}<k for all natural numbers hh.’ So we have established that kk 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={ a1a_{1}, a3a_{3}, a5a_{5}, .......} and we take M1M_{1} = a1a_{1} , M2M_{2} = a3a_{3}, M3M_{3} = a5a_{5}, and in general MkM_{k} = a2k1a_{2k-1} for all kk in NN.

We know by order property (Law of Trichotomy) ‘Given two reals aa and bb, either a=ba=b, or a>ba>b, or a<ba<b.’ Here we observe that MnM_{n} is not equal to Mn+1M_{n+1} for any natural number nn, because of the relation between two consecutive terms of the sequence { a1a_{1}, a3a_{3}, a5a_{5}, …. } derived in (1)(1) above.

So M1>M2M_{1}>M_{2} or M2>M1M_{2}>M_{1}:

Case I:

M1>M2M_{1}>M_{2} i.e. a1>a3a_{1}>a_{3} . Next we proceed by the principle of mathematical induction to show that Mn>Mn+1M_{n}>M_{n+1} for all natural numbers nn. In fact we have already checked the first step for n=1n=1.

Next we assume the validity of the statement to hold true for some n=kn=k. i.e. Mk>Mk+1M_{k}>M_{k+1} or am>am+2a_{m}>a_{m+2} (where obviously m=2k1m={2k-1}

Now

Mk>Mk+1M_{k}>M_{k+1}

am>am+2a_{m}>a_{m+2}

1/am1/{a_{m}} < 1/am+21/{a_{m+2}} [reciprocating the positive numbers reverses the sign of inequality]

k1/amk^{1/{a_{m}}} < k1/am+2k^{1/{a_{m+2}}}

1/k1/am1/{k^{1/{a_{m}}}} > 1/k1/am+21/{k^{1/{a_{m+2}}}} [reciprocating the positive numbers reverses the sign of inequality]

k1/k1/amk^{1/{k^{1/{a_{m}}}}} > k1/k1/amk^{1/{k^{1/{a_{m}}}}}

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

i.e.

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

So we have M1M_{1} > M2M_{2} and MkM_{k} > Mk+1M_{k+1} implies Mk+1M_{k+1} > Mk+2M_{k+2} for all natural numbers kk.

So MM is a strictly decreasing monotone sequence.

Having established that we also see that M1M_{1} > M2M_{2} i.e. a1a_{1} > a3a_{3} implies that

1/a11/{a_{1}} < 1/a31/{a_{3}}

k1/a1k^{1/{a_{1}}} < k1/a3k^{1/{a_{3}}}

a2a_{2} < a4a_{4}

Therefore a1a_{1}>a3a_{3} implies that a2a_{2} < a4a_{4} . If we consider the sub-sequence TT consisting of { a2a_{2}, a4a_{4}, a6a_{6}, ….. } then we have a2a_{2} < a4a_{4} i.e. T1T_{1} < T2T_{2} . By a similar inductive argument as above we can see that

TT 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:

M1M_{1} < M2M_{2} . Under this consideration we proceed exactly as in Case I: and prove that

MM is a strictly increasing monotone sequence.

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

TT is a strictly decreasing monotone sequence.

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

  2. Also MM and TT are sub-sequences of a sequence PP which is bounded as a whole with 00 as a lower bound and kk as an upper bound.

  3. So MM and TT themselves are also bounded with 00 as a lower bound for both and kk 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 MM and TT meet the requirements of the theorem, they are convergent and have a finite limit. Let the limit of MnM_{n} as nn tends to be zz and the limit of TnT_{n} as nn tends to be gg.

Now we note that as nn tends to infinity the terms of the sequence PP = { a1a_{1}, a2a_{2}, a3a_{3},….. } take the limiting values zz and gg 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 PP.

That is as nn tends to infinity the sequence PP takes limiting values zz and gg alternately i.e. zz, gg, zz, gg, zz, gg, zz, gg, zz, gg, ……….. But we have defined the function as { ana_{n} = k1/an1k^{1/{a_{n-1}}} }. So as nn tends to infinity we have successive values coming as zz = k1/gk^{1/g} and the next term as gg = k1/zk^{1/z}.

From the above two equations we get gz=kg^z=k and zg=kz^g=k. So we can say that (z,g)(z,g) is a solution to the equation xyx^y = yxy^x = kk.

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 33 the two convergence values zz and gg are so close by that they seem to be identical. As a result it also gives the result for xx=3x^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.

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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]

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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 mm we are choosing at the beginning.

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

Note by Soumava Pal
3 years, 8 months ago

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1 vote

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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?

Agnishom Chattopadhyay Staff - 3 years, 8 months ago

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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.

Soumava Pal - 3 years, 8 months 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)

Agnishom Chattopadhyay Staff - 3 years, 8 months ago

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@Agnishom Chattopadhyay 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!"

Soumava Pal - 3 years, 8 months ago

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

Here is the full reference

Agnishom Chattopadhyay Staff - 3 years, 8 months ago

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@Agnishom Chattopadhyay Thanks a lot!

Soumava Pal - 3 years, 8 months ago

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

Abhinav Raichur - 3 years, 8 months 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 xx=kx^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.

Soumava Pal - 3 years, 8 months 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 :)

Abhinav Raichur - 3 years, 8 months ago

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@Abhinav Raichur 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 :)

Abhinav Raichur - 3 years, 8 months ago

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@Abhinav Raichur @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.

Soumava Pal - 3 years, 8 months ago

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

Abhinav Raichur - 3 years, 8 months ago

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@Abhinav Raichur 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.

Soumava Pal - 3 years, 8 months ago

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@Soumava Pal Interesting ... looking forward for that note ... bye :)

Abhinav Raichur - 3 years, 8 months ago

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@Abhinav Raichur Thanks, bye. :)

Soumava Pal - 3 years, 8 months ago

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@Abhinav Raichur @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.

Soumava Pal - 3 years, 8 months ago

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Magnificent work!

Nihar Mahajan - 3 years, 7 months ago

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Thanks. XD

Soumava Pal - 3 years, 7 months ago

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@Otto Bretscher Please help me out here.

This algorithm relies on the fact that xkk1x$x \mapsto k^{k^{\frac{-1}{x}}}\$ is a contraction for k > 1

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

Agnishom Chattopadhyay Staff - 3 years, 7 months ago

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@Otto Bretscher

Agnishom Chattopadhyay Staff - 3 years, 7 months ago

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