Wikipedia:Reference desk/Archives/Mathematics/2012 May 21

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May 21

Kernel

Hi everyone.
Let $g:\mathbb {R} \rightarrow [0,\infty )$ continuous with $\int _{\mathbb {R} }g(x)dx=1$. Show that, for all bounded $f\in C(\mathbb {R} )$, $\forall x\in \mathbb {R} \int _{\mathbb {R} }f(x-y)tg(ty)dy\rightarrow f(x)$ as $t\rightarrow \infty$ and that this convergence is uniform if $f$ is uniformly continuous. Widener (talk) 15:16, 21 May 2012 (UTC)

I thought about integrating by parts but that would require differentiability not just continuity. It sort of suggests this approach since $\int tg(ty)dy=\int g(u)du=1$ (where u = ty). Widener (talk) 15:23, 21 May 2012 (UTC)

I just had a thought. If you make the substitution $u=ty$ right from the beginning, you get $\int _{\mathbb {R} }f(x-{\frac {u}{t}})g(u)du$. Is it rigorous to claim that this approaches $\int _{\mathbb {R} }f(x)g(u)du$ as t approaches infinity (by continuity of f and g perhaps?) Widener (talk) 15:29, 21 May 2012 (UTC)

These are standard facts about the approximation of the identity via convolution. A friendly reference is e.g. Wheeden - Zygmund book, Measure and Integral: An Introduction to Real Analysis. If you can't find it, let me know. --pma 22:15, 21 May 2012 (UTC)

I'm afraid I can't find this book. Widener (talk) 23:16, 21 May 2012 (UTC)

Is what I've done so far correct? Or entirely a wrong approach? Widener (talk) 00:05, 22 May 2012 (UTC)

How about this argument:

$\int _{\mathbb {R} }f(x-y)tg(ty)dy$

$=\int _{\mathbb {R} }f(x-{\frac {u}{t}})g(u)du$

$={\frac {d}{du}}\int _{\mathbb {R} }\int _{\mathbb {R} }f(x-{\frac {u}{t}})g(u)dudu$

$={\frac {d}{du}}\left(\left(\int _{\mathbb {R} }f(x-{\frac {u}{t}})du\right)\left(\int _{\mathbb {R} }g(u)du\right)\right)$

$={\frac {d}{du}}\int _{\mathbb {R} }f(x-{\frac {u}{t}})du$

$=f(x-{\frac {u}{t}})\rightarrow f(x)$
as $t\rightarrow \infty$ since $f$ is continuous. Is this a valid argument? Still not sure about the uniform continuity part either. Widener (talk) 02:40, 22 May 2012 (UTC)

Actually, this doesn't work :( Is there a standard proof for this math problem? If so, what is it? Widener (talk) 02:52, 22 May 2012 (UTC)

I will add some lines. What you wrote is a bit strange :) especially in the $dudu$. Also note that in the initial statement, one has to assume in addition either f integrable, or g with compact support; otherwise the integral $\int _{\mathbb {R} }f(x-y)g(y)dy$ may be undefined.(wrong rmk) --pma 05:41, 22 May 2012 (UTC)

Proving that the Mandelbrot set has a bound of 2.

It seems to be pretty well known among those who are familiar with the Mandebrot set ${\mathcal {M}}$ that $\left(c\in \mathbb {C} \land |c|>2\right)\Rightarrow c\not \in {\mathcal {M}}$ (or, in words, that the Mandelbrot set does not contain any points outside of the closed circle of radius 2 centered at 0 in the complex plane). Where can I find a relatively simple proof of this fact? — Trevor K. — 17:21, 21 May 2012 (UTC) — Preceding unsigned comment added by Yakeyglee (talk • contribs)

I know already that the second iteration of the polynomial $P_{c}:\mathbb {C} \rightarrow \mathbb {C}$ where $P_{c}(z)=z^{2}+c$ for some $c\in \mathbb {C} \land |c|>2$ will have a modulus greater than $c$ by the Triangle Inequality. Since we can define the Mandelbrot set by ${\mathcal {M}}{\overset {\text{defd}}{=}}\left\{c\in \mathbb {C} :\exists s\in \mathbb {R} ,\forall n\in \mathbb {N} ,|P_{c}^{n}(0)|\leq s\right\}$, or, in other words, that for all $c\in \mathbb {C}$, the sequence $P_{c}(0),P_{c}^{2}(0),P_{c}^{3}(0),\ldots ,P_{c}^{n}(0),\ldots$ must be bounded, it is clear that the sequence diverges when we have $|c|>2$. The simple proof by the Triangle Inequality is as follows: $|P_{c}^{2}(0)|=|(P_{c}(0))^{2}+c|\geq |P_{c}(0)|^{2}-|c|\leq |c|=|P_{c}(0)|$. However, this tactic cannot be used to show that the subsequent terms increase, and even if they did, it would not be sufficient to show that the sequence ultimately diverges as $n\rightarrow \infty$. What would be the next step in proving this notion? — Trevor K. — 04:17, 22 May 2012 (UTC) — Preceding unsigned comment added by Yakeyglee (talk • contribs)

I think a proof can be found here. Basically you show that for all |z|>R: |f(z)| > |z|. Ssscienccce (talk) 17:49, 24 May 2012 (UTC)

boys and girls

Jack has two children, and at least one of them is a girl. What is the probability that both children are girls? The answer is 1/3 because the possible combination is GG GB BG BB. This is base on http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Archives/Mathematics/2009_September_23.

My question is base on this scenario and you dont know the gender of the kids, is it right to assume that the probabilty of one kid being a girl is 3/4? If yes so that means the probability of a kid being a girl or a boy is affected by other kids? I know im missing something. 203.112.82.129 (talk) 18:39, 21 May 2012 (UTC)

No, the probability is 1/2, because the options are the second kid is a boy, or a girl, 50% chance of being a girl and therefore 50% both girls.--Gilderien Chat|List of good deeds 18:42, 21 May 2012 (UTC)

What you are missing is conditional probability. This concept takes a little getting used to, and our article is probably not that helpful. So I'll try to explain. First, let's be clear and ignore actual biological effects of birth order on sex (there are some, but that is not the spirit of the problem). Your question can be phrased as "what is the probability of having two girls, conditioned on the fact that you have one girl. And the answer is 1/3, instead of the 1/4 you would get as the probability of having two girls with no conditioning. However, if we ask "what is the probability of having a girl as your second child, conditioned on the fact that you have one girl, the answer is still 1/2, because these events are independent. You may be interested in the Lewis Carroll's "pillow problem" [1], or the Monty Hall problem, which both illustrate how conditional probabilities can lead to counter-intuitive answers. SemanticMantis (talk) 20:05, 21 May 2012 (UTC)

(edit conflict)First, given 2 children with at least 1 girl, then there are three possibilities GG, GB, and BG. So yes, the probability is 1/3 that both are girls. It is also true that the probability that at least 1 child is a girl is 3/4, however, this does not mean what you seem to think it does. The probability that at least 1 child is a girl is not the same as the probability that a given child is a girl. For example, there are two cases where the first child is a girl, so the probability that the first child is a girl is 1/2, as expected. In short, you are confusing "at least one of these is X" with "this one is X"; think of it like this: it's very likely that someone in a casino will come out ahead by a good margin, it is not very likely that it will be you. Phoenixia1177 (talk) 20:09, 21 May 2012 (UTC)

So base on what i understand on your explanation, given that you have two kids, you have a 3/4 chance that one of them is a girl? 203.112.82.129 (talk) 21:01, 21 May 2012 (UTC)

Nevermind my question, didnt notice you already answered it. So in this problem, the knowledge of what gender is older or younger affects the probability right? 203.112.82.1 (talk) 21:15, 21 May 2012 (UTC)

try this. you select a 2 child family with at least one girl who was born on tuesday the 13 of may. whats the chance the other child is a girl. The solution is seriously counter intuitive. — Preceding unsigned comment added by 86.148.182.74 (talk) 20:31, 21 May 2012 (UTC)

May i just point out that this depends on how you select Jack. Did you randomly pick a family, which happened to have one girl-in which case the probability of the second child being a girl and a boy is equal. Or did you specifically select a family with at least one girl, in which case it is a 1/3:2/3 split. — Preceding unsigned comment added by 86.148.182.74 (talk) 20:29, 21 May 2012 (UTC)

Oh now i understand, its like if i want a dice to give me number 4, ill have more chance if i throw 20 dice at once. Thanks guys. 203.112.82.129 (talk) 22:06, 21 May 2012 (UTC)

See Boy or Girl paradox.—Wavelength (talk) 23:56, 21 May 2012 (UTC)

Closure of set of functions

Let $(X,d)$ be a compact metric space, and $F$ an equicontinuous family of functions from $X$ to itself. Suppose $g:X\rightarrow \mathbb {R}$ is continuous. Define $G=\{g\circ f:f\in F\}\subset C(X,\mathbb {R} )$. Show that the closure of G is compact.
First of all, what is the closure of G? What definition of compactness is the correct one to use? Widener (talk) 20:55, 21 May 2012 (UTC)

It's just a consequence of the Arzelà-Ascoli theorem: since $F$ is equicontinuous, it is a relatively compact subset. The set $G$ is just the image of $F$ via the continuous map $f\mapsto g\circ f$, so it is relatively compact too. Which definition of compactness? In a metric space all definitions of compactness coincide, so no matter.As to the closure of $G$: by continuity, it contains the set $\{g\circ f\ :\ f\in {\overline {F}}\}$, but the latter set is closed, because it is compact, therefore it is the closure of $G.$ --pma 21:48, 21 May 2012 (UTC)

I think Widener was asking what the question meant, not how to prove it — specifically, what topology to put on the family of functions from X to X. I'm guessing that's the topology of pointwise convergence, but I don't have time to think about it now to see if that makes sense. --Trovatore (talk) 22:05, 21 May 2012 (UTC)

OK, then the topology should be the topology of uniform convergence; but in fact on equicontinuous families it coincides with the topology of pointwise convergence (the one induced by the product topology; incidentally, this fact yields to a proof of AA theorem, via Tychonoff's theorem)--pma 22:23, 21 May 2012 (UTC)

Fruits

Another question on the link in boys and girls section. A man bought fruit at the rate of 16 for $24 and sold them at the rate of 8 for $18. What is the percent profit? Answer is 40%.

My question is why? On the thread, someone answered its 50% (which i think is the obvious answer) but i have a feeling that 50% is incorrect since i think that it has a conter intuitive solution as well same as the boys and girls question.203.112.82.128 (talk) 23:04, 21 May 2012 (UTC)

I don't know why, he makes a profit of $0.75 per fruit, and he payed $1.50 per fruit. Assuming that he sold the same number of fruit as what he purchased, the total profit would be 50%. If he did not sell all his fruit, then the total profit may be 40% of the total cost. We need to know how many fruit he purchased and how many he sold. Plasmic Physics (talk) 02:41, 22 May 2012 (UTC)

Now that I've thought about it, what is the profit compared with? Plasmic Physics (talk) 02:50, 22 May 2012 (UTC)

Here are few equations:

z = $18/8 × y − $24/16 × x = $2.25 × y - $1.50 × x (where z is total profit, x is total fruit bought, y is total fruit sold)

z_{% of bought} = 16 × 100%/$24 × z/x = 200%/$3 × z/x

z_{% of sold} = 8 × 100%/$18 × z/y = 400%/$9 × z/y

Take your pick, we need to know the values of x and y, and whether you're trying to find z_{% of bought} or z_{% of sol}. Plasmic Physics (talk) 03:12, 22 May 2012 (UTC)

I think the percent profit should be 33 -- that's the percentage of revenue that is profit. The return on investment, on the other hand, is 50%. I can't see how a value of 40% can arise. Looie496 (talk) 18:21, 22 May 2012 (UTC)

Where does 33% come from? You seem to be comparing $18 with $24 - but that's comparing apples and oranges if you'll pardon the pun. The only valid comparison is 8 for $12 (purchase price) vs. 8 for $18 (sale price). That's a 50% profit. That assumes he sold everything he bought, which is a reasonable assumption. Any assumption that he sold less than he bought is not a reasonable assumption. -- ♬ Jack of Oz ♬ ^{[your turn]} 02:09, 23 May 2012 (UTC)

Seems that the relevant definitions are: Markup = (Sale price / Cost) - 1 and Profit Margin = (Sale price - Cost) / Sale price. You want the profit margin so: (9/4 - 6/4) / (9/4) = (3/4) * (4/9)=1/3=33% Ssscienccce (talk) 18:03, 24 May 2012 (UTC)