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February 26

Measure Theory

Let C be a semi-algebra on a given set $\Omega$ and $\mu :C\rightarrow [0,\infty ]$ such that $\mu (\emptyset )=0$ and $\mu$ is sigma additive.

We then define $\mu ^{*}(A)=\inf \sum _{k=1}^{\infty }\mu (D_{k})$ where $A\subset (\cup D_{k})$ and each $D_{k}\in C$ .

I am trying to prove that $\mu ^{*}(E)=\mu ^{*}(B\cap E)+\mu ^{*}(B^{c}\cap E)\,\forall B\in C$ and $\forall E\in P(\Omega )$ .

One inequality is easy to show because E is a subset of $(B\cap E)\cup (B^{C}\cap E)$ and $\mu ^{*}$ is monotonic.

But any hints as to how to show $\mu ^{*}(E)\geq \mu ^{*}(B\cap E)+\mu ^{*}(B^{c}\cap E)$ would be appreciated.

\sum \mu (D_{k})=\sum (\mu (D_{k}\cap B)+\mu (D_{k}\cap B^{c}))=\sum \mu (D_{k}\cap B)+\sum \mu (D_{k}\cap B^{c})

.

Furthermore,

(\cup D_{k})\cap X=\cup (D_{k}\cap X)

. Does this help? -- Meni Rosenfeld (talk) 00:41, 26 February 2008 (UTC)[reply]

This fact is necessary for proving the existence of the Caratheodory extension. The proof of it is very ugly and technical. See [1] starting from page 3. 128.139.226.37 (talk) 10:02, 28 February 2008 (UTC)[reply]

Are you sure this is relevant? There, they are starting with an outer measure and proving some things about it. Here, we start with a measure and construct from it what will later turn out to be an outer measure. Unless I am missing something, what I have written is one line away from a complete proof of the statement. -- Meni Rosenfeld (talk) 10:15, 28 February 2008 (UTC)[reply]

Grammars

Given {a, ba, bba, bbba} bⁿa|nEN
I calculated that S→aS → bSa → bbSa → bbbSa would be the grammar.
I wanted to make sure that I am on the correct path. NanohaA'sYuri^{Talk, My master} 00:04, 26 February 2008 (UTC)[reply]

I don't think that's right, because if "a" is in the set and "S→aS" is a rule, then you can get "aa", which isn't in the set. Am I misunderstanding the question? —Keenan Pepper 05:06, 26 February 2008 (UTC)[reply]

One way of reading the rule "S→aS" is the following:

If σ ∈ L(S), then also aσ ∈ L(S) – where L(S) here denotes the language produced using S as the start symbol.

So if bba ∈ L(S), as desired, then you would also get abba ∈ L(S), which would be wrong.

There is a context-free grammar for the language {a, ba, bba, bbba, ...} = {bⁿa|n∈N} with one nonterminal symbol and two production rules. I think the assignment calls for a context-free grammar. In a context-free grammar all rules have a single nonterminal symbol on the left-hand side. What you wrote, "S→aS → bSa → bbSa → bbbSa", does not look like (production rules of) a grammar. Production rules contain only one arrow. It looks more like the begin of a derivation. However, no context-free grammar could have a derivation step of the form aS → bSa. If at some stage of a derivation the string starts with a terminal symbol, like "a" here, then it will start with "a" in all following stages. --Lambiam 06:40, 26 February 2008 (UTC)[reply]

Completing the Square

I was just looking at the article on completing the square and saw in the derivation of the quadratic formula from the equation $ax^{2}+bx+c=0\,\!$ that $\displaystyle {x^{2}+{\frac {b}{a}}x+\left({\frac {b}{2a}}\right)^{2}=\left({\frac {b}{2a}}\right)^{2}-{\frac {c}{a}}\to \left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}-4ac}{4a^{2}}}}$ . What I am wondering (although it is probably very simple) is where does the ${\frac {b}{a}}x$ go from the first equation to the next. Much appreciated, Zrs 12 (talk) 02:00, 26 February 2008 (UTC)[reply]

Try expanding out

\left(x+{\frac {b}{2a}}\right)^{2}

. Remember, when you expand something like

(L+R)^{2}

, the result is not

L^{2}+R^{2}

, but

L^{2}+2LR+R^{2}

(the FOIL rule). —Bkell (talk) 02:34, 26 February 2008 (UTC)[reply]

Thanks Bkell, I understand it a little better now. ~~Expanding it gives $x^{2}+{\frac {b}{a}}x+{\frac {b}{a}}$ , right? Where does ${\frac {b}{a}}x$ go on the right side? Many thanks, Zrs 12 (talk) 03:04, 26 February 2008 (UTC)~~ I got it. When expanded

\left(x+{\frac {b}{2a}}\right)^{2}=x^{2}+{\frac {b}{a}}x+{\frac {b^{2}}{4a^{2}}}

. That was easy enough. I just had to think and write it out. Thanks! Zrs 12 (talk) 03:25, 26 February 2008 (UTC)[reply]

Glad to help. —Bkell (talk) 04:23, 26 February 2008 (UTC)[reply]

Geometry Help

In a triangular prism with the width 3, and each diagonal side 3 and a length of 7, how would you find the surface area and volume?

--Devol4 (talk) 07:41, 26 February 2008 (UTC)[reply]

Triangular prism helps with the volume. Zain Ebrahim (talk) 11:09, 26 February 2008 (UTC)[reply]

Well, if I understand the question correctly, your prism can be interpreted geometrically as two equilateral triangles joined by rectangles. To find the volume, just multiply cross-sectional area (the triangle) by the length, and to find surface area, just find the area of all components and sum them. -mattbuck (Talk) 11:10, 26 February 2008 (UTC)[reply]

What I don't understand in Devol4's question is what exactly the width, the diagonal side, and the length means in such a prism. – b_jonas 10:52, 29 February 2008 (UTC)[reply]

egytian method

divide 16 by 9 —Preceding unsigned comment added by 41.205.189.9 (talk) 15:27, 26 February 2008 (UTC)[reply]

16/9. =) –King Bee ^{(τ • γ)} 15:35, 26 February 2008 (UTC)[reply]

I assume the OP means the "Egyptian method". Here's one explanation of the process. — Lomn 15:49, 26 February 2008 (UTC)[reply]

We do have an article on this, check out Egyptian fraction. After having done the calculations, you can check your work using these calculators. --NorwegianBlue^talk 18:37, 26 February 2008 (UTC)[reply]

maths trick i just found out

think of a number and put it in your calculator [make sure you remember it]
add 5
multiply your answer by 6
divide your answer by 2
subtract your answer by 15
divide your answer by the number you first thought of

answer is always = 3
check source at [doxadeocollege.co.za] —Preceding unsigned comment added by 196.207.47.60 (talk) 16:15, 26 February 2008 (UTC)[reply]

That is true. It's not really a trick so much as obfuscation - simply writing it out on a piece of paper and you can see the following steps:

$S_{2}:x\rightarrow x+5$
$S_{3}:x+5\rightarrow 6(x+5)$
$S_{4}:6(x+5)\rightarrow {\frac {6(x+5)}{2}}=3(x+5)$
$S_{5}:3(x+5)\rightarrow 3(x+5)-15=3x+15-15=3x$
$S_{6}:3x\rightarrow {\frac {3x}{x}}=3$

-mattbuck (Talk) 16:30, 26 February 2008 (UTC)[reply]

And what if the number I choose is 0? –King Bee ^{(τ • γ)} 16:36, 26 February 2008 (UTC)[reply]

Define 0/0:=3. That's afterall one reason 0/0 is indeterminant, because processes that result in 0/0 could in fact by any real number, or infinity (by appropiate choice of the process). 130.127.186.122 (talk) 17:15, 26 February 2008 (UTC)[reply]

How is "obfuscation" not a more specific word for "trick"? In fact, most actual magic tricks use nothing but clever obfuscation. It's so hard to make the hand quicker than the eye, but so easy to find gullible spectators. Take the classic "cup and ball" routine, for instance. You keep the mark's eyes on the cup and ball, and get caught clumsily trying to remove the ball, so that his attention will be on you when your friend picks his pocket. Black Carrot (talk) 17:32, 26 February 2008 (UTC)[reply]

First off, it sounds better, and to me, a mathematical trick is some way of simplifying a result or theory. For instance, an easy way to calculate the multiplicity of a pole is to find the multiplicity of the zero at that point of the function's reciprocal - that I would consider a mathematical trick. -mattbuck (Talk) 18:38, 26 February 2008 (UTC)[reply]

Trig equation

(Note: this isn't homework, but I'd appreciate a hint rather than being told the answer) I'm having trouble with solving $\!\sin 2x+\sin ^{2}x=1$ . We've not covered this in class, so the only things I can see ( $\!\sin 2x=2\sin x\cos x=2\sin x\sin(x+{\frac {\pi }{2}})$ or $\!\sin 2x-\cos ^{2}+1$ or some combination thereof) don't seem to be awfully helpful. Any pointers – is there an identity I need that I don't have? Angus Lepper^{(T, C, D)} 20:58, 26 February 2008 (UTC)[reply]

Combine the double-angle identity

\sin 2x=2\sin x\cos x\,\!

with the Pythagorean identity

\sin ^{2}x+\cos ^{2}x=1\,.

--Lambiam 21:09, 26 February 2008 (UTC)[reply]

To 'idiot proof' the above... (almost full solution commented out by Meni Rosenfeld) AirdishStraus (talk) 22:28, 26 February 2008 (UTC)[reply]

Why would you give a full solution when the OP explicitly asked you not to? --Tango (talk) 22:41, 26 February 2008 (UTC)[reply]

Or call them an idiot? Black Carrot (talk) 23:04, 26 February 2008 (UTC)[reply]

Tango is right. I have commented out the solution. -- Meni Rosenfeld (talk) 10:15, 27 February 2008 (UTC)[reply]

I agree Tango is right. I got carried away doing the near full solution and had forgotten the 'hint only' request. My apologies. On a seperate note... 'idiot-proofing' something is a mere colloquialism for unambiguous clarification; I never used the word idiot to directly describe anyone and wouldn't ever use it in ANY case. Maybe Black Carrott hasn't come across this term before? AirdishStraus (talk) 11:15, 27 February 2008 (UTC)[reply]

I have no doubt that you didn't intend to insult the OP, but using the phrase "idiot-proof" in that way could very easily be misinterpreted, especially if someone isn't a native English speaker, so it's probably best to avoid it. --Tango (talk) 13:06, 27 February 2008 (UTC)[reply]

Triangle

The angle bisector of alpha w_α, the median of s_b and the altitude of c h_c of a acute-angled triangle ABC cross in one point, if w_α, the side BC and a circle around the point H_c, which goes through the corner A, have also a common point together.

How can one proof this statement? --85.178.22.17 (talk) 23:18, 26 February 2008 (UTC)[reply]

You need to do it one step at a time

1. First find the intersection of the "angle bisector of alpha w_α and the median of s_b" as a function of the parameters of a generalised triangle eg (0,0) (a,b) (c,d) as the points of the triangle.

2. Then find the coodrinates of the crossing point of "the median of s_b and the altitude of c h_c of a acute-angled triangle ABC"

Next you need to show that the two equations abtained above give the same point if the third premise is true.. ie that "w_α, the side BC and a circle around the point H_c, which goes through the corner A, have also a common point together."

3. I'm not sure what H_c is here - but if you do you should be able to generate some sort of parametric equation relating the vertexs of the triangle. {{ Sound's like Hc is the centre of the circle that goes through A and also passes through the intersection of (w_α and the side BC ) - this makes somewhere Hc along a line then}}

Then you need to show that if for instance the equation obtained in 1 is true at the same time as the equation obtained in 3. then equation 2 is also true.87.102.93.245 (talk) 11:43, 27 February 2008 (UTC)[reply]

HINT for the "angle bisector of alpha w_α, the median of s_b and the altitude of c h_c" to cross at one point the triangle must be at least iscocoles since we have three lines intersecting at one point and two of them "angle bisector of alpha w_α" "and the altitude of c h_c" already cross at A.. Because the median cannot pass through A, the crossing of the "angle bisector of alpha w_α" "and the altitude of c h_c" must give a line (ie they are the same line) and not just a single point. therefor ac=ab I hope that is enough to get you finished.87.102.93.245 (talk) 12:26, 27 February 2008 (UTC)[reply]