Trimming my nose hair. Ask me anything.

[HeavyArtillery]

I have the answer.

[vashti]

What's that picture in your avatar?

[HeavyArtillery]

some big boobies!

[boobaa]

Suppose xy = yx, where x and y are positive real numbers, with x < y. Show that x = 2, y = 4 is the only integer solution. Are there further rational solutions? (That is, with x and y rational.) For what values of x do real solutions exist?

[vashti]

Why are men so obsessed with fat deposits on the chest when they are so phobic about fat anywhere else?

[HeavyArtillery]

Suppose xy = yx, where x and y are positive real numbers, with x < y. Show that x = 2, y = 4 is the only integer solution. Are there further rational solutions? (That is, with x and y rational.) For what values of x do real solutions exist?

We have xy = yx, with 0 < x < y.

Let y = rx, where r > 1.
Then xrx = (rx)x.
Taking the natural logarithm of both sides, rx ln x = x (ln r + ln x).
Dividing by x, and rearranging, (r−1) ln x = ln r.
(Notice that r = 1 would be a solution, from which y = x.)
With r > 1, ln x = (ln r) / (r−1).
Taking exponentials, x = e(ln r) / (r−1) = (eln r)1/(r−1).
Therefore we have the parametric solution, x = r1/(r−1), y = rx = rr/(r−1).

If we set r = 1 + 1/k, where k > 0, we have x = (1 + 1/k)k, y = (1 + 1/k)k+1.
It is a well known result that, as k tends to infinity (equivalently, r tends to 1), the expression for x tends to e from below, and that for y tends to e from above.

Since r > 1, 1/(r−1) is positive, and so x > 1, and therefore y > 1.
(Notice that, if we allow 0 < r < 1, 1/(r−1) and r/(r−1) are negative, and so again x and y are greater than 1. Of course, this must be the case, because r = a and r = 1/a yield essentially the same solution, with x and y swapped.)

If r is slightly greater than 1, we get solutions with x just less than e, and y just greater than e. On the other hand, if r is large, we get x slightly greater than 1, and y slightly greater than r.

Rational solutions

Seeking solutions in rational x, y, note that r = y/x must also be rational. If we set r = 1 + 1/k, then k must be rational.
If k is rational, then x = (1 + 1/k)k is rational if, and only if, k is an integer. A proof follows.

Let k = a/b, a fraction in its lowest terms; that is, a and b are relatively prime.
Then (1 + 1/k)k = [(a + b)/a]a/b, which is clearly rational if b = 1.

If b > 1, then, since a and b are relatively prime, a + b and a are also relatively prime.
Therefore, for [(a + b)/a]a/b to be rational, both a + b and a must be perfect bth powers of integers.
However, this is impossible.
For example, if b = 2, a + 2 and a cannot both be perfect squares, for the difference between two positive perfect squares is at least 3.

More generally, if u, v, and c are positive integers, with c > 1, using the binomial theorem, (u + v)c − uc = cuc−1v + ... + vc > c.
Hence two distinct perfect bth powers cannot differ by b.
Therefore (1 + 1/k)k = [(a + b)/a]a/b cannot be rational if b > 1.

So (1 + 1/k)k is rational if, and only if, k is a positive integer.
Therefore all rational solutions are of the form x = (1 + 1/n)n, y = (1 + 1/n)n+1, where n = 1, 2, ... .

Notice that, for any rational solution, 2 less than or equal to x < e < y less than or equal to 4. There is a countable infinity of rational solutions, and an uncountable infinity of real solutions.

We have shown that, for a rational solution, k is an integer; that is x = [(a + 1)/a]a.
If a > 1, a and a + 1 are relatively prime, and so [(a + 1)/a]a is a fraction in its lowest terms, with denominator > 1, and therefore not an integer.
So the only integer solution is x = 2, y = 4.

Notice that nowhere in the derivation of the parametric solution, above, did we assume that x, y, and r are real.
Setting r = −1, we get perhaps the simplest complex solution: x = −i, y = i.
Euler's formula states that eit = cos t + i sin t, where t is any real number.
Setting t = pi/2, we have ei*pi/2 = i.
Hence the principal value of i−i = (ei*pi/2)−i = e−i*i*pi/2 = epi/2 is approximately equal to 4.8104773809653516554730356667.

Robin Hankin wrote to point out that this problem is solved formally by Lambert's W function, which will yield complex solutions. See the references below for further details.

Taking the natural logarithm of both sides of the equation, we get y ln x = x ln y. Using implicit differentiation:

y/x + y' ln x = ln y + xy'/y.
Grouping terms containing y', we have y' (x/y − ln x) = y/x − ln y, from which we get the pleasantly symmetrical formula:
y' = (y/x − ln y) / (x/y − ln x).

Multiplying numerator and denominator by xy, we have y' = (y2 − y ln yx) / (x2 − x ln xy).
Since xy = yx, we get an alternative formula y' = y2(1 − ln x) / x2(1 − ln y).

When x = y = a, we get y' = 1, except when a = e, in which case the derivative is undefined. This reflects the fact there is no unique tangent line at the point (e, e). In fact, the derivative exists at every point on the graph except for (e, e).
The slope of the tangent at the point (2, 4), for example, is 4 (1 − ln 2) / (1 − 2 ln 2) is approximately equal to −3.1774.
When x not equal to y, we have x < e and y > e, or x > e and y < e.
Therefore, 1 − ln x and 1 − ln y have opposite signs, and so the slope is always negative.

All rational solutions to the equation are given by x = A000169(n+1)/A000312(n), y = A000312(n+1)/A007778(n), where n = 1, 2, 3, ... . For instance, n = 2 yields the solution (9/4)27/8 = (27/8)9/4.

[HeavyArtillery]

Why are men so obsessed with fat deposits on the chest when they are so phobic about fat anywhere else?

Because they feel good.

[boobaa]

What's the geometrical meaning of the central extension of the algebra of diffeomorphisms of the circle?

[HeavyArtillery]

What's the geometrical meaning of the central extension of the algebra of diffeomorphisms of the circle?

Let f(x) be a function defined on the positive integers such that:

f(x) = x/2 if x is even
f(x) = (3*x+1)/2 if x is odd

Then the conjecture is: iterates of f(x) will eventually reach 1 for any
initial value of x.

OR

Pi

[boobaa]

Actually you didn't answer the geometrical meaning part.

What is the answer to ultimate question?

[HeavyArtillery]

Actually you didn't answer the geometrical meaning part.

What is the answer to ultimate question?

yes i did. im done trimming now.

question time is over.

[Mish]

What's the meaning of life?


You realize now any unanswered question will be interpreted as you not knowing the answer

[HeavyArtillery]

What's the meaning of life?


You realize now any unanswered question will be interpreted as you not knowing the answer

The meaning of life is that one thing. What that is, is up to you.

[1averagejoe]

This guy is going to be DM's new butt buddy when it comes to math.

[Junket]

This guy is going to be DM's new butt buddy when it comes to math.

Yes, assuming he didn't simply control-C, control-V into Google.