Wikipedia:Reference desk/Archives/Mathematics/2009 July 29

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July 29

Sin Graph

Im just stuck on one tiny bit..

The graph of y = Sin x undergoes the following transformations

            *The period is halved [[this bit is the trouble]]
            *The amplitude is doubled
            *The function is translated π/6 units to the right

I managed to come up with y = 2 sin (x - π/6)

Now im not sure but would... y = 2 sin 2(x - π/6) do it.. i think it might but im not entirely sure why..well its either 2 or 1/2.. outside the bracket somewhere or inside.. (Can it only go outside the brackets?)

Cheers, 124.180.244.92 (talk) 10:31, 29 July 2009 (UTC)

You're quite right, that doubles the frequency and halves the period. I always write sin with lower case otherwise I think it must be the problem of Sin. I don't know why programs like Mathematica insist on capitalising it. Dmcq (talk) 11:14, 29 July 2009 (UTC)

The names of Mathematica built-in functions are not designed to correspond with how they are usually written, and they all begin with a capital letter. It does allow you to enter input in TraditionalForm, which accepts lowercase sin. -- Meni Rosenfeld (talk) 19:01, 29 July 2009 (UTC)

To do those transformations in the order you stated (for clarity, giving names f1,f2... to the intermediate steps):

1. start with f₁(x) = sin x
2. halve the period, i.e. double the frequency => f₂(x) = f₁(2x) = sin 2x
3. double the amplitude => f₃(x) = 2 sin f₂(x) = 2 sin 2x
4. translate π/6 units to the right => f₄(x) = f₃(x-π/6) = 2 sin 2(x-π/6)

Of course you can check this visually with a graphing calculator. —Preceding unsigned comment added by 67.117.147.249 (talk) 16:33, 29 July 2009 (UTC)

Calculating a limit explicitly.

Hi, I'm fairly sure (for example)

$a=\lim _{x\to \infty }{{1} \over {\sqrt {x+1}}}=0$

But how does one come to this conclusion "solidly", so to speak? I can see that with each increase in n, a decreases, but this is just by observing the behaviour of say, n = 1, n = 10, n = 100 and so on. Is there a way to resolve the limit algebraically, and can all reasonable limits (say, the limits of rational expressions) be calculated explicitly? Thanks. 81.149.255.225 (talk) 11:28, 29 July 2009 (UTC)

For the formal definition of the limit, check our article on (ε,_δ)-definition_of_limit. For the limits of rational functions at infinity, it depends entirely on the degrees involved. If the degree of the numerator exceeds that of the denominator, the limit will be infinite (of positive or negative sign). If the degree of the denominator exceeds that of the numerator, then the limit will be 0. If the degrees equal, then the limit will be the ratio of the coefficients of the leading terms. In general, there is no easy way to determine the limit of an arbitrary function at infinity, and much energy has been expended in the finding of limits for various types of functions; this is one of the major topics in differential calculus. Ray^Talk 11:48, 29 July 2009 (UTC)

A general method of calculating the limits of rational functions at infinity (which then leads to the rules that Ray has outlined above) is to factor the largest terms out of the numerator and denominator. In this case, the largest term in the numerator is 1, and the largest term of the denominator is ${\sqrt {x}}$, which can be factored and calculated as follows:

$\lim _{x\to \infty }{\frac {1}{\sqrt {x+1}}}=\lim _{x\to \infty }{\frac {1}{{\sqrt {x}}{\sqrt {1+{\frac {1}{x}}}}}}=(\lim _{x\to \infty }{\frac {1}{\sqrt {x}}})(\lim _{x\to \infty }{\frac {1}{\sqrt {1+{\frac {1}{x}}}}})=(0)(1)=0$

It may not look pretty in this case, but in general, the right-hand expression, which is the part that can get ugly, always evaluates to a limit of 1 (if you do it right), so it's not really such a big deal. --COVIZAPIBETEFOKY (talk) 12:22, 29 July 2009 (UTC)

(e/c) If you take the value of $\lim _{x\to \infty }{\frac {1}{\sqrt {x}}}$ for granted, it's much easier to use the squeeze theorem:

$0\leq \lim _{x\to \infty }{\frac {1}{\sqrt {x+1}}}\leq \lim _{x\to \infty }{\frac {1}{\sqrt {x}}}=0.$

— Emil J. 12:35, 29 July 2009 (UTC)

Also note that in this particular case, it is a trivial fact that ${\sqrt {x}}$ diverges to infinity as x goes to infinity, and therefore ${\sqrt {x+1}}$ also diverges to infinity as x goes to infinity, and therefore ${\frac {1}{\sqrt {x+1}}}$ goes to 0 as x goes to infinity. --COVIZAPIBETEFOKY (talk) 12:31, 29 July 2009 (UTC)

This limit can be proven directly from the definition. Let $\epsilon >0$. Take $S={\frac {1}{\epsilon ^{2}}}-1$. If $x>S$ then $0<{\frac {1}{\sqrt {x+1}}}<\epsilon$ and so $\left|{\frac {1}{\sqrt {x+1}}}-0\right|<\epsilon$, thus $\lim _{x\to \infty }{\frac {1}{\sqrt {x+1}}}=0$. -- Meni Rosenfeld (talk) 18:52, 29 July 2009 (UTC)

You mean $S=\epsilon ^{-2}-1$, I think. Algebraist 19:00, 29 July 2009 (UTC)

Thanks, fixed. -- Meni Rosenfeld (talk) 19:10, 29 July 2009 (UTC)

Product measure article/question

The article Product measure says

In fact, for every measurable set E,

$(\mu _{1}\times \mu _{2})(E)=\int _{X_{2}}\mu _{1}(E^{y})\,d\mu _{2}(y)=\int _{X_{1}}\mu _{2}(E_{x})\,d\mu _{1}(x),$

where E_x = {y∈X₂|(x,y)∈E}, and E^y = {x∈X₁|(x,y)∈E}, which are both measurable sets.

Is this true? It doesn't say explicitly, E_x is measurable for each x, but it seems to be saying that. I asked a question elsewhere and mentioned this article and the person suggested perhaps what is meant here is E_x is measurable for each x except possibly some collection of x of measure 0??? Would this be better? StatisticsMan (talk) 14:05, 29 July 2009 (UTC)

According to the definition, the σ-algebra of measurable subsets of the product measure is generated by sets of the form A₁ × A₂, where A_i is μ_i-measurable. It follows easily that if E is measurable, then E_x is measurable for each x. — Emil J. 14:31, 29 July 2009 (UTC)

Very well, so I am understanding this a bit better, I believe. In Royden, he does things differently. He starts with the algebra generated by the measurable rectangles and then uses the Caratheodory extension process to create a sigma-algebra CONTAINING the sigma-algebra generated by measurable rectangles. And, in this case, it's not always true that E_x is measurable when E is. Is this correct? StatisticsMan (talk) 14:47, 29 July 2009 (UTC)

I see. The Carathéodory construction always yields a complete measure, whereas the product measure is not necessarily complete (for example, the product of two one-dimensional Lebesgue measures is not complete), so this indeed makes a difference. I do believe that the Carathéodory construction will yield the completion of the product measure in this case, which I think implies that E_x is measurable in the completion of μ₂ for all x except for a set of measure 0 in the completion of μ₁. — Emil J. 15:02, 29 July 2009 (UTC)

Okay, it's making a lot more sense now. Problem is I have Royden and I also have Measure and Integration Theory by de Barra and he does it the way the article does it so the conclusions are different on this one specific issue and I was very confused. Thanks for the help. The reason I started looking into this specifically is, of course, a qual problem. I think I understand how it works now. Thanks! StatisticsMan (talk) 15:13, 29 July 2009 (UTC)

Warning: as usual, non sigma-finite measure situations provide annoying exceptions. Recall the well-known counter-example to the formula you wrote (and to Tonelli's theorem in measure theory): if X₁=X₂=[0,1]; E=X₁ x X₂ the diagonal; μ₀ is the counting measure and μ₁ is the Lebesgue measure, then the two integrals give 0 and 1. --pma (talk) 16:11, 29 July 2009 (UTC)

E = X₁ x X₂ gives infinity both ways. You might mean E = the diagonal, I suppose; however I do not see why it should be measurable in the product measure. — Emil J. 16:31, 29 July 2009 (UTC)

Yes you're right, corrected, it was the diagonal. The two σ-algebras are: the Lebesgue σ-algebra on X₁, and the parts of X₂. If I'm not wrong any set E is measurable in the product σ-algebra (= the one generated by the products of measurable subsets) if and only if the set of its sections {E^y: y in X ₂} is a (finite or) countable family of Lebesgue measurable sets. So the product σ-algebra is not complete. Also, the diagonal is not measurable in the product σ-algebra. But it is measurable in the Carathéodory extension, with infinite measure. Disclaimer: I did not check everything... --pma (talk) 16:39, 29 July 2009 (UTC)

Gompertz function right asymptote

Greetings! In Gompertz function would you please check the citation needed tags about whether the right asymptote is horizontal. I think that it is not but now I can't remember and need to go to work. Thank you! 99.35.128.24 (talk) 16:56, 29 July 2009 (UTC)

Just going by the formula on the page, the limit as t goes to infinity is a, as the article says. $e^{ct}$ goes to 0, so $e^{be^{ct}}$ goes to 1. Rckrone (talk) 19:51, 29 July 2009 (UTC)