April 20

Television signals are radio waves which travel at the speed of light. The first commercial TV broadcasts began about 1950. How far have these radio waves travelled in light years? —Preceding unsigned comment added by 66.222.179.213 (talk) 01:48, 20 April 2011 (UTC)[reply]

• The wording of this really sounds like a homework question. Hint: how far does light travel in a year? How many years have elapsed since 1950? --Kinu ^t/_c 02:05, 20 April 2011 (UTC)[reply]

Also asked at the science desk. I recommend answerers answer over there. OP, please do not ask the same question at multiple desks- it's too confusing for answerers. Staecker (talk) 02:08, 20 April 2011 (UTC)[reply]

If it was 61 years ago then they've traveled 61 light-years, and if you don't see that then you probably don't understand what a light-year is. Michael Hardy (talk) 18:04, 21 April 2011 (UTC)[reply]

Suppose we have two functions f(t) and g(t). How would one going about solving for the functions if given the following expression: f(t) ≠ g(t) ≠ f(g(t)) ≠ g(f(t)) ≠ f(f(t)) ≠ g(g(t)) ≠ f(f(f(t))) ≠ f(f(g(t))) ≠ f(g(f(t))) ≠ f(g(g(t))) ≠ g(f(f(t))) ≠ g(f(g(t))) ≠ g(g(f(t))) ≠ g(g(g(t)))

If it is easier to solve at a particular point, I'd be happy with that result as well. I'm also interested in possibly extending this for permutations of greater sizes. 199.111.182.147 (talk) 03:53, 20 April 2011 (UTC)[reply]

Your headerline seems to indicate that you want the 14 values {f(t), g(t), f(g(t)), g(f(t)), f(f(t)), g(g(t)), f(f(f(t))), f(f(g(t))), f(g(f(t))), f(g(g(t))), g(f(f(t))), g(f(g(t))), g(g(f(t))), g(g(g(t)))} to be pairwise distinct for all values of t, while in the text you do not seem to object against f(t)=f(g(t)) when only f(t)≠g(t) and g(t)≠f(g(t)). Consider the case f=g=1_Ø. Then both functions have empty domains, and no value f(t) or g(t) is defined. Then your requirement is fulfilled for all values of t in the domain. Bo Jacoby (talk) 10:54, 20 April 2011 (UTC).[reply]

I meant the former case. 199.111.165.206 (talk) 14:53, 20 April 2011 (UTC)[reply]

Try ${\displaystyle f(t)=t^{2))$ and ${\displaystyle g(t)=2^{t))$ for special values of t.Sławomir Biały (talk) 11:05, 20 April 2011 (UTC)[reply]

... or ${\displaystyle f(t)=t+1\,,\,g(t)=t^{2))$ (and whole family of similar pairs of functions) ? Gandalf61 (talk) 11:17, 20 April 2011 (UTC)[reply]

Would f(f(t))=g(t) for t=2? I'm missing something, presumably. Grandiose (me, talk, contribs) 16:43, 20 April 2011 (UTC)[reply]

Yes, but they are not the same function of t, which is the point I think. Sławomir Biały (talk) 18:17, 20 April 2011 (UTC)[reply]

You can do this for any number N of functions and t>0 by simply having f_k(t)=Nt+k with k from 0 to N-1. Here with N=2 that just gives 2t and 2t+1. Try it with N=10 and see what you get :) Dmcq (talk) 11:24, 20 April 2011 (UTC)[reply]

Interesting. I still have trouble grasping why this is true; how do we know there isn't some strange combination that will equal another? 199.111.165.206 (talk) 14:53, 20 April 2011 (UTC)[reply]

Try it for all permutations of, say, 3 fs and gs. You can show that the result of a sequence of n fs and gs is a function of the form ${\displaystyle 2^{n}t+m}$ where ${\displaystyle 0\leq m<2^{n))$ and there is a simple one-to-one correspondence between each of these 2ⁿ sequences and the value of m. Gandalf61 (talk) 15:37, 20 April 2011 (UTC)[reply]

Follow-up question: I want to find f and g such that the set of all iterates involving f and g (of arbitrary length) is dense in L^2. Count Iblis (talk) 21:02, 20 April 2011 (UTC)[reply]