Mathematics desk
< August 7	<< Jul \| August \| Sep >>	August 9 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

August 8[edit]

Tetration[edit]

Do there exist two different positive real numbers x and y such that $x^{x}=y^{y}$ ? Using tetration, this can rewritten as ${^{2}x}={^{2}y}$ . GeoffreyT2000 (talk) 04:33, 8 August 2015 (UTC)[reply]

Just consider the graph of the function x^x on the interval (0,1). --JBL (talk) 04:40, 8 August 2015 (UTC)[reply]

If 0 ≤ x < e⁻¹ ≃ 0.367879 then 0.692201 ≃ e^−e⁻¹ < x^x ≤ 1 and then there is a number y = f(x) such that e⁻¹ < y ≤ 1 and x^x = y^y. Note that two numbers are always different, otherwise they are only one number. Note also that positive numbers are always real. Bo Jacoby (talk) 06:57, 8 August 2015 (UTC).[reply]

It is not true either that "two different numbers" or "positive real numbers" is redundant, as you seem to suggest. Actually I'm kind of amazed that anyone would take the time to criticize these standard turns of phrase. The first usage indicates that the answer "x = 2, y = 2" is not welcome, which would have been ambiguous without the word "different"; the second sets the domain of solutions and disambiguates from "positive rational numbers," "positive surreal numbers," etc. Certainly neither of these choices of wording are as problematic as your unnecessary introduction of the symbol f without definition.--JBL (talk) 15:37, 8 August 2015 (UTC)[reply]

I'm also confused by your criticism of the questioner's wording. A standard Google search on the quoted phrase "two different positive real numbers" yields only ten results, but four are mathematics text books and one a journal article. If we substitute "different" with the more preferred "distinct", a search on "two distinct positive real numbers" yields numerous results, with five of the first ten being text books. -- ToE 16:25, 8 August 2015 (UTC)[reply]

I was not critisizing - merely pointing out an innoscent redundancy. Sorry. The function f points out that y depends on x. Bo Jacoby (talk) 18:21, 8 August 2015 (UTC).[reply]

As JBL suggested, I looked a graph of f(x) = x^x, and there is a minimum when e⁻¹ -- see WolframAlpha: minimum of x^x. Bo Jacoby pointed out the entire range of matching numbers; that range includes 0^0 = 1 = 1^1 and also (1/2)^(1/2) = 1/sqrt(2) = (1/4)^(1/4). Are there any other pairs with simple representations? --DavidCary (talk) 17:28, 13 August 2015 (UTC)[reply]

Decimal / binary integral[edit]

Define f(x) for 0 <= x <= 1 as the decimal interpretation of the binary expansion of x (for example, f(1/2) = f((0.1)₂) = 1/10). Numerically the integral from 0 to 1 seems to be equal to 1/18. How can it be proven that the integral exists? 24.255.17.182 (talk) 18:56, 8 August 2015 (UTC)[reply]

Half of 0 to 1 has 1 in the first place which sums to 1/20. Half of 0 to 1 has 1 in the 2nd decimal place making 1/200 etc so the sum is 1/20+1/200+1/2000... = 1/20*(1/(1-1/10))= 1/18. So yes the integral is 1/18. Dmcq (talk) 20:52, 8 August 2015 (UTC)[reply]

Another way: Let

a_{n}=\int _{0}^{2^{-n}}f(x)\ dx

. Show that

a_{n}=2a_{n+1}+20^{-n-1}

and solve the recursion to find

a_{n}=20^{-n}/18

and

a_{0}=1/18

.

Also, this is to find the value of the integral; to merely prove that it exists, it suffices to see that f is nondecreasing and bounded in the interval. -- Meni Rosenfeld (talk) 21:21, 8 August 2015 (UTC)[reply]

I suppose the question was to show the integral exists. It is Lesbegue integrable since the sum can be got by adding a sequence of little boxes - one of height 1/10 from 0.5 to 1, one of height 1/100 from 0.25 to 0.5 and another from 0.75 to 1, then 4 of height 1/1000 - 0.125 to 0.25, 0.375 to 0.5, 0.625 to 0.75 and 0.875 to 1. This is a countable sum with the limit of 1/18 as shown above and equals the given function except at a countable number of ponts. And even the countable number of differences can be got rid of by having the boxes have one end open as for instance the first being of height 1/10 on the range closed at 0.5 but open at 1. Dmcq (talk) 21:39, 8 August 2015 (UTC)[reply]

It follows from Meni Rosenfeld's comments that f is even Riemann integrable. —Kusma (t·c) 05:16, 9 August 2015 (UTC)[reply]