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February 12

Is there are transition point between the spirals in the image when applying a Möbius transformation to spirals?

Hello,

this question has been bugging me for some time now. Let us consider a spiral in the complex plane, starting in zero and moving "towards infinity". A Möbius transformation will be a curve, acting like a spirak initially, and then at infinity approaching another point in the complex point, again like a spiral (but this time, it's turning in the other direction).

But... is there a "transition point"? A point where the curve stops being "like the first spiral" and becomes "the second spiral"? I've been trying to pinpoint this point, I tried to come up with definitions or even ways to calculate it, but the more and more I'm experimenting, the more I'm wondering whether or not there really is a point like that? Can anyone help? Thanks!

Evilbu (talk) 11:48, 12 February 2008 (UTC)

The transition point would be a point of inflexion, wouldn't it? They aren't usually hard to find. --Tango (talk) 13:40, 12 February 2008 (UTC)

Yeah. I think the choice of transition point has mostly to do with personal preference, but the inflection point (the point where the spiral changes direction of rotation) is probably the most obvious one. To find it, first find the equation of the transformed spiral, and then find the point where the ratio of the first and second derivatives has an imaginary part of 0. This will guarantee that the ratio is real, thus the second derivative (acceleration) has the same direction as the first (velocity), thus there is no rotation component and this is an inflection point. -- Meni Rosenfeld (talk) 14:14, 12 February 2008 (UTC)

I'm not so sure. I tried taking $\Gamma :[0,\infty )\rightarrow \mathbb {C} :r\rightarrow re^{ir}$ and a Mobius transformation $M(z)={\dfrac {az+b}{z+d}}$ and came out with the only point where the equation of the transformed spiral is 0 is $r=i$ which lies outside the given range of r. -mattbuck (Talk) 23:24, 12 February 2008 (UTC)

I think you misunderstood. The transformed spiral probably won't go through 0, and the ratio of the second and first derivatives probably won't go through 0, but the ratio will cross the real axis. -- BenRG (talk) 02:21, 13 February 2008 (UTC)

I'll take this opportunity to mention that I wasn't completely accurate in my post. For a smooth curve, the criterion I have given is necessary, but not sufficient, for inflection. However, for the curve in question there will usually be only one point satisfying it, so it can be used to identify the point. -- Meni Rosenfeld (talk) 08:46, 13 February 2008 (UTC)

Oblique Asymptote

Is there a way to find the oblique asymptote of a polynomial function without using polynomial long division or calculus? Acceptable (talk) 23:52, 12 February 2008 (UTC)

Polynomials don't have asymptotes, do you mean a rational function? [1] has a good method - basically, you just take the limit of the function as x tends to infinity to find the gradient, and then take the limit of the function minus the bit you just found to find the constant term. --Tango (talk) 00:11, 13 February 2008 (UTC)

Sorry my bad, I indeed meant rational functions. Acceptable (talk) 01:23, 13 February 2008 (UTC)

There is actually a much simpler way. Just take the two highest terms of the numerator and divide them by the highest term of the denominator. For example, the asymptote of ${\frac {2x^{5}+3x^{3}-x^{2}+4x-5}{3x^{4}+8x^{3}-2x^{2}+5x-7}}$ is ${\frac {2x^{5}+0x^{4}}{3x^{4}}}={\tfrac {2}{3}}x$. -- Meni Rosenfeld (talk) 08:34, 13 February 2008 (UTC)

Does that always work? It looks like a rough method of taking limits, but I think it only works on that function because the degree of the numerator is one higher than that of the denominator. You would need to adjust that method to work for other degrees. You also still need to find the constant term. --Tango (talk) 14:52, 13 February 2008 (UTC)

That's true. I did assume that the function satisfies this condition, as this is the usual case where the term "oblique asymptote" would be used. For finding the asymptote in a more general case, you need to take the terms of the numerator from the degree of the denominator and upwards. This still avoids limits, derivatives, or polynomial long division. -- Meni Rosenfeld (talk) 15:06, 13 February 2008 (UTC)

Don't think it works if the difference in degree is >1. e.g. ${\frac {x^{3}+2x^{2}+5x+4}{x+3}}$ yields ${\frac {x^{3}+2x^{2}+5x}{x}}=x^{2}+3x+5$ by your method. However, a quick long division yields ${\frac {x^{3}+2x^{2}+5x+4}{x+3}}=x^{2}-x+8-{\frac {20}{x+3}}$, i.e. an asymptote of x^2-x+8. Long division is the easiest method in such cases, methinks. --PalaceGuard008 (Talk) 22:19, 13 February 2008 (UTC)

You are of course absolutely right. Not one of my best days, apparently. -- Meni Rosenfeld (talk) 22:43, 13 February 2008 (UTC)

February 13

THE PHYSICS MAGAZINE GUY,WITH AN ALGEBRA QUESTION FROM AN ALGEBRA MAGAZINE

Hey,guys I`ve got a sequence question the sequence for the first problem is 1,1 over two,one over three,one over four. I need the sequence completed.I also have a problem involving The Fibanocci Sequence. the numbers are 1,2,3,5,8,13.I also have to draw the next three figures in a sequence in which there are oval with one side,shaded black.Another,one with the top side shaded black.And,one last one with the right side black. Also,another one with a line through the circle.Another,with a cross in it,Another with several spokes. There,is a triangle with no lines One with a triangle going through it. And a last one with three across from the left. —Preceding unsigned comment added by Yeats30 (talk • contribs) 00:10, 13 February 2008 (UTC)

You want to know what comes next in the sequence $1,{\frac {1}{2}},{\frac {1}{3}},{\frac {1}{4}}$? I think you can work that out for yourself... You haven't actually said what the second question is. And picture questions aren't really going to work unless you can reproduce the picture, however they aren't usually that difficult - some general advice: try to break the sequence down into different things that are changing (the shape, the colour, the orientation, etc.) and consider each separately. --Tango (talk) 00:16, 13 February 2008 (UTC)

Bullshit this is from an algebra magazine. If you read an algebra magazine, you are capable of working out a simple sequence like this. Stop asking wikipedia users to do your homework for you. -mattbuck (Talk) 01:56, 13 February 2008 (UTC)

Mattbuck, please try to keep civil even when dealing with someone who isn't following the rules themselves (i.e. posting homework questions). Did anybody bother to notify the user on his talk page that he'd been outed? No, so I'm going to do so now. Confusing Manifestation(Say hi!) 05:38, 13 February 2008 (UTC)

Sorry, I kind of figured he'd have read the countless people who have outed him in his questions, since if he asks them, he presumably wants answers and thus reads the responses. -mattbuck (Talk) 10:27, 13 February 2008 (UTC)

I like how he uses the winning formula of "magazines". "HI HERE'S A HOMEWORK QUESTION FROM A HOMEWORK MAGAZINE" =) --PalaceGuard008 (Talk) 11:06, 13 February 2008 (UTC)

I particularly liked his question about how much someone weighs in a lift which climbs five floors. The carpet was made of broccoli. -mattbuck (Talk) 11:28, 13 February 2008 (UTC)

Predicate logic

I'm not 100% of the "rules" of manipulating predicates but I have this problem and I want to see if I've manipulated it correctly:

R, P, Q are all in the same domain. Use rules of inference and logical equivalence to prove that these premises together lead to a contradiction.

A = $\forall x[(R(x)\lor (\exists yP(y)))\to (\neg Q(x))]$

B = $\exists x[(Q(x)\land R(x))\lor (Q(x)\land \forall yP(y))]$

Starting with A:

1. $\forall x[(R(x)\lor (\exists yP(y)))\to (\neg Q(x))]$ Premise

2. $\forall x[\neg (R(x)\lor (\exists yP(y)))\lor (\neg Q(x))]$ Implication

3. $\forall x[(\neg R(x)\land \neg (\exists yP(y)))\lor (\neg Q(x))]$ Negation

4. $\forall x[(\neg R(x)\land (\forall y\neg P(y)))\lor (\neg Q(x))]$ Negation of existential quant.

Now here comes the bit I'm not sure about:

5. $\forall x[(\neg Q(x)\lor \neg R(x))\land (\neg Q(x)\lor (\forall y\neg P(y)))]$ Distribution on 4, then commutative rearrangement.

Supposing I can distribute as demonstrated in 5., I now have A looking similar to B:

A = $\forall x[(\neg Q(x)\lor \neg R(x))\land (\neg Q(x)\lor (\forall y\neg P(y)))]$

B = $\exists x[(Q(x)\land R(x))\lor (Q(x)\land \forall yP(y))]$

If I have made it thus far with no mistakes, I would like to know where to take this next. I can "see" that these two statements could contradict but I'm unsure how to prove it. If there is an error in my working so far, please let me know! Damien Karras (talk) 08:51, 13 February 2008 (UTC)

All that predicate stuff is making my head spin (I only did the tiniest bit of it), but what happens if you use the definition $\exists xS(x):=\neg \forall x\neg S(x)$ on, say, B, and then bring the second negation throughout the expression? Confusing Manifestation(Say hi!) 11:27, 13 February 2008 (UTC)

Sorry, I know it's hard to follow but I'm trying to do it step by step so that you can see how I arrived at the conclusion. I hope I am right in assuming that if I rearrange B's existential quantifier $\exists x[...]$ to $\neg \forall x\neg [...]$ I will have B as:

$\neg \forall x[(\neg Q(x)\lor \neg R(x))\land (\neg Q(x)\lor \neg (\forall yP(y)))]$

Which is almost the negation of A save for the $\neg (\forall yP(y))$ part! Damien Karras (talk) 12:57, 13 February 2008 (UTC)

I'm not sure how rigorous this is, but it seems to work intuitively, at least, so might help. A implies:

$(\forall x[R(x)\to (\neg Q(x)])\land (\forall x[(\exists yP(y))\to (\neg Q(x))])$

You can then use those implications to substitute into B, getting:

$\exists x[(Q(x)\land (\neg Q(x)))\lor ((\neg (\exists yP(y)))\land \forall yP(y))]$

Which is clearly false. I'm fairly sure that's all true, but I may have missed out a few too many steps by just using intuition... --Tango (talk) 14:43, 13 February 2008 (UTC)

The step you're asking about is correct. You're almost there. You just have to pull the "not" out all the way to the left. When you do, you'll get A as equivalent to precisely notB. Or, to shorten the proof, you could have pulled it out to begin with and rearranged the inside separately. Black Carrot (talk) 18:51, 13 February 2008 (UTC)

If you want to look up the rules, Laws_of_classical_logic has a list of them. Black Carrot (talk) 18:53, 13 February 2008 (UTC)

Yes, when I said "clearly false" I was assuming people could perform that one last manipulation themselves. It's the splitting of the implication and the substitutions that I thought might need to be done in more steps to be completely rigorous. --Tango (talk) 19:32, 13 February 2008 (UTC)

I was talking to the OP. Your solution is good too, though. Black Carrot (talk) 07:17, 14 February 2008 (UTC)

Just to check, this is in reference to the distributive step 5 further back, right? I have to try and extract all those "nots" out of A and I'll get something that looks identical to "not" B? Damien Karras (talk) 08:20, 14 February 2008 (UTC)

That's right. And by definition, the statement "B and notB" is a contradiction. Black Carrot (talk) 07:44, 15 February 2008 (UTC)

Alrighty then, going back to A and B in the equivalent forms:

A = $\forall x[(\neg Q(x)\lor \neg R(x))\land (\neg Q(x)\lor (\forall y\neg P(y)))]$

B = $\exists x[(Q(x)\land R(x))\lor (Q(x)\land \forall yP(y))]$

Now, if I extract the "nots" in A to the left, I have:

A = $\forall x\neg [(Q(x)\land R(x))\lor (Q(x)\land \neg (\forall y\neg P(y)))]$

Going back to the previous point that $\exists x[...]$ is equivalent to $\neg \forall x\neg [...]$, I can write B as:

B = $\neg \forall x\neg [(Q(x)\land R(x))\lor (Q(x)\land (\forall yP(y)))]$

The two statements are almost a contradiction aside from the fact that $\neg (\forall y\neg P(y))$ is not equivalent to $(\forall yP(y))$, correct? Damien Karras (talk) 09:47, 15 February 2008 (UTC)

You're right, sorry. The existential bit doesn't work out quite right. I was thinking of

A = $\neg \exists x[(Q(x)\land R(x))\lor (Q(x)\land \exists yP(y))]$

B = $\exists x[(Q(x)\land R(x))\lor (Q(x)\land \forall yP(y))]$,

but I thought the ends would be the same. You can get a contradiction from this, but it's not as pretty as "B and notB". Since the problem is all the way on the inside, it'll have to be pulled out. Let's do it for B, starting from where I wound up.

B = $\exists x[(Q(x)\land R(x))\lor (Q(x)\land \forall yP(y))]$

B = $\exists x[(Q(x)\land R(x))]\lor \exists x[(Q(x)\land \forall yP(y))]$

B = $\exists x[(Q(x)\land R(x))]\lor [\exists xQ(x)\land \forall yP(y)]$

Then for A,

A = $\neg \exists x[(Q(x)\land R(x))\lor (Q(x)\land \exists yP(y))]$

A = $\neg \exists x[(Q(x)\land R(x))]\land \neg \exists x[(Q(x)\land \exists yP(y))]$

A = $\neg \exists x[(Q(x)\land R(x))]\land [\neg \exists xQ(x)\lor \neg \exists yP(y)]$

A = $\neg \exists x[(Q(x)\land R(x))]\land [\neg \exists xQ(x)\lor \forall y\neg P(y)]$.

From here, I think it's easiest to combine the two. Since the original assumption is A AND B, that's not hard to do. A AND B = C,

C = $\neg \exists x[(Q(x)\land R(x))]\land [\neg \exists xQ(x)\lor \forall y\neg P(y)]\land {\bigg (}\exists x[(Q(x)\land R(x))]\lor [\exists xQ(x)\land \forall yP(y)]{\bigg )}$.

Commute and distribute, to get

C = $[\neg \exists xQ(x)\lor \forall y\neg P(y)]\land {\Bigg (}{\bigg (}\neg \exists x[(Q(x)\land R(x))]\land \exists x[(Q(x)\land R(x))]{\bigg )}\lor {\bigg (}\neg \exists x[(Q(x)\land R(x))]\land [\exists xQ(x)\land \forall yP(y)]{\bigg )}{\Bigg )}$

Since that bit towards the middle is a contradiction in its proper form, it's automatically equal to False, and since it's one side of an OR, the other side absorbs it.

C = $[\neg \exists xQ(x)\lor \forall y\neg P(y)]\land {\bigg (}\neg \exists x[(Q(x)\land R(x))]\land [\exists xQ(x)\land \forall yP(y)]{\bigg )}$

C = $[\neg \exists xQ(x)\lor \forall y\neg P(y)]\land \neg \exists x[(Q(x)\land R(x))]\land \exists xQ(x)\land \forall yP(y)$

I commute and distribute the exists-Q bit the same as the previous bit, and get

C = $\exists xQ(x)\land \forall y\neg P(y)\land \neg \exists x[(Q(x)\land R(x))]\land \forall yP(y)$

Commute,

C = ${\bigg (}\exists xQ(x)\land \neg \exists x[(Q(x)\land R(x))]{\bigg )}\land {\bigg (}\forall yP(y)\land \forall y\neg P(y){\bigg )}$

As long as there's a y in the universe to act on, the right side is false, so the entire thing is false, and equivalent to a contradiction. That's not quite the same as saying it is a contradiction, since there's nothing to stop y from ranging over an empty set, but they probably want that case excluded. Black Carrot (talk) 04:13, 16 February 2008 (UTC)

On the other hand, if the universe is empty, then the left side is false instead. So I guess it works. I can't figure out how to get an actual contradiction-type contradiction, though, and I'm not certain it's possible with formal manipulations. Black Carrot (talk) 04:56, 16 February 2008 (UTC)

February 14

Three calculus problems (studying for a test)

1. I need to evaluate the indefinite integral ∫x²√(1-x)dx. I tried using u-substitution:
u = 1 - x
x = 1 - u
du = -dx
dx = -du

∫(1 - u)²u^1/2•-du + C
-∫(1 - 2u + u²)u^1/2du + C
-(∫u^1/2du - ∫2u^3/2du + ∫u^5/2du) + C
-∫u^1/2du + ∫2u^3/2du - ∫u^5/2du + C
-2u^3/2/3 + 4u^5/2/5 - 2u^7/2/7 + C
-2(1 - x)^3/2/3 + 4(1 - x)^5/2/5 - 2(1 - x)^7/2/7 + C

The textbook says the answer is -2/105(1 - x)^3/2(15x² + 12x + 8) + C. Is this just an equivalent form that I'm not recognizing?

2. This time it's a definite integral: ∫₁⁹ 1/[(√x)(1 + √x)²]dx. Again I tried u-substitution:
u = 1 + √x
du = (1/2)x^-1/2dx = [1/(2√x)]dx

2∫_u(1)^u(9) u^-2du
2∫₂⁴ u^-2du
2•(u^-3/-3)]₂⁴
2[4^-3/-3 - 2^-3/-3] -(2/3)(1/64 - 1/8)

But the book claims that the answer is 1/2! (That exclamation mark denotes my frustration, not a factorial.)

3. I need to find the value of c guaranteed by the mean-value theorem for integrals for the function f(x) = 2sec² x over the interval [-π/4, π/4].
f(c) = [1/(b - a)]∫_a^bf(x)dx
f(c) = [1/(π/4 - [-π/4])]∫_-π/4^π/4(2sec²) x)dx
f(c) = [1/(π/2)]2 • (tan x)]_-π/4^π/4
f(c) = (4/π)[tan (π/4) - tan (-π/4)]
f(c) = (4/π)[1 - (-1)]
f(c) = 8/π

Yet the official answer given is ±arccos (√π)/2 ≈ ±0.4817 ... what am I doing wrong??

Any help is much appreciated. Thanks so much. --anon

Hello. Yes, the first is an equivalent form - just factor out the (1-x)^1.5 and it should be obvious. In the second, the substitution is fine, but you integrated u^-2 to give u^-3/-3, when the correct integral is -u^-1 since you add one to the power, then divide through by it. As for question 3, your derivation is correct, but you missed the final step. You say f(c) = 8/pi, but you're asked to give c, not f(c).

Hope this helps. -mattbuck (Talk) 01:29, 14 February 2008 (UTC)

Dude, I love you. --anon —Preceding unsigned comment added by 141.155.31.240 (talk) 02:24, 14 February 2008 (UTC)

Surd Troubles

I've tried and i thought i did it correctly but as it seems i did not =[. Some help would be appreciated.

• Problem 1 ${\sqrt {63}}-{\sqrt {28}}={\sqrt {x}}$
  $={\sqrt {9}}\times {\sqrt {7}}-{\sqrt {4}}\times {\sqrt {7}}$
  $=3{\sqrt {7}}-2{\sqrt {7}}$
  $=1{\sqrt {7}}$
  $={\sqrt {7}}$
  but the textbook says $=5{\sqrt {7}}$
• Problem 2 - Simplify $4{\sqrt {7}}\times (-2{\sqrt {7}})$

$=4{\sqrt {\times }}(-2{\sqrt {)}}=(-8{\sqrt {)}}$
$={\sqrt {7}}\times {\sqrt {7}}={\sqrt {49}}$
$=-8{\sqrt {49}}$

• Problem 3 - Simplify $2{\sqrt {3}}(4{\sqrt {3}}-{\sqrt {6}})$

$=8{\sqrt {9}}-{\sqrt {18}}$
$=8\times 3\times 3-9\times 3$
$=24{\sqrt {3}}-3{\sqrt {3}}$
$=21{\sqrt {3}}$
textbook says $=24-6{\sqrt {2}}$
Thanks for the help =] !. —Preceding unsigned comment added by 121.219.227.217 (talk) 07:52, 14 February 2008 (UTC)

In problem 1, it's probably supposed to be a plus sign. In problem 2, since 49 is a square you can simplify it further. In problem 3, I think you should just do it more carefully. Black Carrot (talk) 08:02, 14 February 2008 (UTC)

Ah i figured out problem 2 and possibly 1( before i noticed your comment that'll teach me to refresh eh) [Note:In problem 1 a minus sign is written in the text book]

• Problem 2 - Simplify $4{\sqrt {7}}\times (-2{\sqrt {7}})$

$={\sqrt {7}}\times {\sqrt {7}}={\sqrt {7}}^{2}=7$
$=4{\sqrt {\times }}(-2{\sqrt {)}}=(-8{\sqrt {)}}$
$=-8\times 7$
$=(-56)$ (this is the answer in the text book)

• Problem 1 ${\sqrt {63}}-{\sqrt {28}}={\sqrt {x}}$

$={\sqrt {9}}\times {\sqrt {7}}-{\sqrt {4}}\times {\sqrt {7}}$
$=9{\sqrt {7}}-4{\sqrt {7}}$ ( i know it isnt simplifying it but its all i could think of)
$=5{\sqrt {7}}$

And i got futher on problem three but still hit a snag..

• Problem 3 - Simplify $2{\sqrt {3}}(4{\sqrt {3}}-{\sqrt {6}})$

$=8{\sqrt {9}}-{\sqrt {18}}$
$={\sqrt {8}}\times {\sqrt {3}}\times {\sqrt {3}}-{\sqrt {9}}\times {\sqrt {2}}$
$=24-3{\sqrt {2}}$
(Why isnt this 6!) textbook says $=24-6{\sqrt {2}}$
—Preceding unsigned comment added by 121.219.227.217 (talk) 08:25, 14 February 2008 (UTC)

For problem 1 - just because the textbook made a mistake, doesn't mean you have to, too. ${\sqrt {63}}-{\sqrt {28}}$ is equal to ${\sqrt {7}}$. If the textbook says something else then it is wrong, and you shouldn't invent new math rules to cover up for it.

For problem 3 - take a closer look at your very first step. -- Meni Rosenfeld (talk) 08:33, 14 February 2008 (UTC)

Ah i see the problem i didn't know you had to do that, well i did learn something new now.

$=8{\sqrt {9}}-2{\sqrt {18}}$
$={\sqrt {8}}\times {\sqrt {3}}\times {\sqrt {3}}-{\sqrt {2}}\times {\sqrt {9}}\times {\sqrt {2}}$
$=24-6{\sqrt {2}}$

Thanks Guys =] you've been a big help! —Preceding unsigned comment added by 121.219.227.217 (talk) 08:49, 14 February 2008 (UTC)

Of the three expressions above, the first and last are equal, but the middle one is different. There are too many root signs there.  --Lambiam 18:37, 14 February 2008 (UTC)

Something else seems wrong, problem 1 says: ${\sqrt {63}}-{\sqrt {28}}={\sqrt {x}}$, do they want you to simplify the left side or solve for x on the right (they're very different problems). If the book answer is to be believed then I suspect they asked you to simplify ${\sqrt {63}}-{\sqrt {28}}$. A math-wiki (talk) 02:17, 15 February 2008 (UTC)

I suspect this was indeed the question, but ${\sqrt {63}}-{\sqrt {28}}$ is not equal to $5{\sqrt {7}}$. -- Meni Rosenfeld (talk) 09:21, 15 February 2008 (UTC)

You seem to be having trouble formatting your equations. Do you need any help? Black Carrot (talk) 07:57, 15 February 2008 (UTC) 07:48, 15 February 2008 (UTC)

relation between arithmatic mean, median and mode

what is the relation between arithmatic mean,median and mode? what are its limitations?please explain in detail how the relation is derivedKasiraoj (talk) 14:18, 14 February 2008 (UTC)

Mode (statistics), Arithmetic mean, Median explains what each is (in the first paragraph).. They are different measures of an 'average' - there is no arithmetic relationship between them - that depends on what you are measuring - sometimes they can all be the same, sometimes widely different.. 'Arithmetic mean' is the one most closely similar to someones concept of an average. The 'mode' can be used on non-numerical sets of data - such as popular names.

Do a search outside wikipedia should give less technical descriptions of these things and their uses..87.102.114.215 (talk) 14:56, 14 February 2008 (UTC)

You can find out something about the skewness of a distribution by comparing the mean, median and mode - I can't remember the details, but that article gives some explanation of it (most of it rather technical, though). --Tango (talk) 17:06, 14 February 2008 (UTC)

Mean, median and mode are central tendency measures: They show where data concentration occur.

The mean is the most frequently used of the three concepts. The fact of its taking into account every point of information is its great advantage, but it also could be a drawback, especially whenever outliers are present. In such cases, usually the median is used as a central tendency measure.

As Tango points out, mean, median and mode relate to skewness. Right-skewed distributions have a mean greater in value than the median, and viceversa for negative-skewed distributions. Usually the mode is located in such a way that the median stands between mode and mean. The article on skewness has a section that deals with Pearson skewness coefficients, measures which elaborate on this fact. Pallida  Mors 18:11, 14 February 2008 (UTC)

The article Mode (statistics) also gives a comparison of mean, median and mode.  --Lambiam 18:41, 14 February 2008 (UTC)

On Average

How many different measures of the "average" of a set of data are there? I know about the mean, median, and mode, are there any others? --Carnildo (talk) 21:23, 14 February 2008 (UTC)

Average has a list of measures of central tendency. --PalaceGuard008 (Talk) 21:28, 14 February 2008 (UTC)

February 15

Simple question on exponents

With an expression like 4^3^2, does one consider this as (4^3)^2 or 4^(3^2)? Myles325a (talk) 00:32, 15 February 2008 (UTC)

In what context? TeX doesn't seem to like it at all: <math>x^y^z</math> yields an error message (Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. TeX parse error: Double exponent: use braces to clarify"): {\displaystyle x^{y}^{z}} ).

There's a good argument for treating it as right-associative: if you wanted (x^y)^z you'd be better off writing it as x^(y*z) but if you want x^(y^z) there's no alternative. That's how it's interpreted in some programming languages, Perl for example, although the operator is spelled ** because ^ means something else. The double-asterisk comes from FORTRAN, which also treats it as right-associative. --tcsetattr (talk / contribs) 01:11, 15 February 2008 (UTC)

In TeX you typeset it as $x^{y^{z}}$, which makes it obvious what is intended when you look at the source. But you have to know the convention (which, as you say, is quite standardly right-associative; there's no ambiguity whatsoever) to interpret the output. --Trovatore (talk) 01:25, 15 February 2008 (UTC)

Proof by contradiction

In the article about mathematical proofs, there is a proof that ${\sqrt {2}}$ is irrational, this proof assumes that if a² is even then a must also be even. My question is : can you prove that if (a² is even ) then (a must also be even ) even if ${\sqrt {2}}$ is rational ? --George (talk) 02:51, 15 February 2008 (UTC)

Sure, that's easy to prove. Assume a is odd. Then it is of the form $2n+1$, for some integer n. (I take this as the definition of an odd number; if you use a different definition the proof will be slightly different.) Then $a^{2}=(2n+1)^{2}=4n^{2}+4n+1=2(2n^{2}+2n)+1$. If n is an integer, then $k=2n^{2}+2n$ is also an integer, so $a^{2}=2k+1$ is odd by definition. But this contradicts the fact that $a^{2}$ is even, so the assumption that a was odd must have been false. —Keenan Pepper 03:58, 15 February 2008 (UTC)

I changed the question , can you take another look at it ? Thanks--George (talk) 04:01, 15 February 2008 (UTC)

Um, it seems like you just removed the phrase "for any integer a" from the hypothesis. This makes the statement false, because as far as I know even and odd apply only to integers. If $a={\sqrt {2}}$, then $a^{2}$ is the even integer 2, but a itself is neither even nor odd because it is not an integer. Am I missing something? —Keenan Pepper 04:11, 15 February 2008 (UTC)

I don't know what's wrong, but it doesn't seem right. I'll think about it again, maybe I am missing something.--George (talk) 04:32, 15 February 2008 (UTC)

As User:Keenan Pepper writes, we assume you must be talking about integers. And note that any integer is either of the form 2b for some b or else of the form 2c+1 for some c, call them type-0 and type-1 integers, respectively, for what gets added in either case to the multiple of 2. Any type-1 integer 2c+1 will have square equal to 4c²+4c+1. But if we define d to be 2c²+2c, then we can see that the type-1 integer has a square of the form 2d+1, so the square, too, is type-1. So, because all type-1 integers have type-1 squares, no type-1 integers have type-0 squares. Thus if a² is type-0—and therefore not type-1—neither can a itself be type-1; the only possible type for a is thus 0. (P.S. - feel free to replace type-0 with even and type-1 with odd. Furthermore, the proof made no mention at all of ${\sqrt {2}}$)—PaulTanenbaum (talk) 04:34, 15 February 2008 (UTC)

Urgent integration question!

This is for school. I need to find the area bounded by the graphs of f(y) = -y² + 2y and g(y) = -y. I equated them and got the intersection points (0,0) and (3,-3). I'm supposed to find the area by integrating in terms of y, so based on what the graphs look like, I guess it should be ∫_-3⁰(-y² + 2y - [-y]) = ∫_-3⁰(-y² + 3y), which would then give me
(-y³/3 + 3y²/2)]_-3⁰. The upper limit yields only zero values, so this should be equal to -(-(-3)³/3 + 3(-3)²/2) = -(9 + 27/2) = -22.5. Aside from the disconcerting fact that this "area" value is negative (though I know that can happen), it seems far too large to be reasonable. What am I doing wrong? My utmost thanks for any help, anon.

The problem is with the integration limits. Maybe you got confused because the independent variable is called "y" here? Given your intersection points (y,f(y))=(0,0) and (3,-3), think again about what range of y you should integrate over. --mglg_(talk) 06:16, 15 February 2008 (UTC)

Also, make sure you are subtracting the small function from the large one - if you do the opposite you will get a negative result. -- Meni Rosenfeld (talk) 09:15, 15 February 2008 (UTC)

It helps to draw a rough graph with these kinds of questions. Then you can see how to use the intersection points and what to integrate. --Tango (talk) 13:43, 15 February 2008 (UTC)

Number of solutions for a system of linear equations over finite fields.

How can we find out the number of solutions for a system of linear equations over finite fields. Are there any results giving the number of solutions in terms of the number of unknowns, "rank", etc.? 210.212.228.94 (talk) 05:58, 15 February 2008 (UTC)

I think Gaussian elimination still works. Black Carrot (talk) 07:56, 15 February 2008 (UTC)

Yeah, according to the article Gaussian elimination can be performed over any field. Black Carrot (talk) 08:16, 15 February 2008 (UTC)

Yes; but what about the number of solutions? Is there any result giving any bounds for it? ( in terms of the number of variables, rank, q and n if our field is GF(qⁿ). 210.212.228.94 (talk) 10:23, 15 February 2008 (UTC)

Exactly the same results hold as for infinite fields (that's the beauty of linear algebra). If we have m equations in n unknowns, then this is the same as a matrix equation Ax=b where A is an m x n matrix, of rank r say and nullity n-r. If b is outside the image of A then there are no solutions, otherwise there is an (n-r)-dimensional space of them. If the ground field has q elements, then the solution space has q^n-r. Algebraist 12:22, 15 February 2008 (UTC)

Sobolev Spaces

Okay, so in my graduate PDE class we are working with the Sobolev Space denoted by $H^{1}(0,1)$ which consists of all square integrable functions (over the interval (0,1)) such that their first derivatives are also square integrable over (0,1). Since this is our first introduction to Sobolev Spaces, our teacher gave us a few functions to determine if they are in $H^{1}$ or not. My question is that, is it just enough to show that the first derivative of the function is square integrable or do we have to show that the function itself is also square integrable? Because I think that there is some theorem in analysis which says that if a function's derivative is square integrable, then the function itself is also square integrable. If not, then does anyone have any example of a function (could be multi-variable) whose derivative is square integrable but the function itself is not?A Real Kaiser (talk) 08:22, 15 February 2008 (UTC)

How about ${\frac {1}{\sqrt {x}}}$? 134.173.92.17 (talk) 08:45, 15 February 2008 (UTC)

Erm, you probably wanted a bounded domain. Scratch that. 134.173.92.17 (talk) 08:55, 15 February 2008 (UTC)

You want what is called the Poincare inequality: $\int _{0}^{1}|f(t)-a|^{2}~dt\leq C\int _{0}^{1}|f'(t)|^{2}~dt$, where $a$ is the average value of f on [0,1], and C is some constant independent of f.

Basically this says that for functions in a sobolev space on a bounded domain (finite volume), the sobolev norm up to the first derivative is equivalent to the norm of the first derivative. In an unbounded domain this is not true, as the previous example 1/sqrt(x) on [1,oo) shows. JackSchmidt (talk) 18:21, 15 February 2008 (UTC)

In case you actually needed a proof. If g is a continuous function on [0,1], then m(t)=g(0)+t*(g(1)-g(0)) has the property that f=g-m is square integrable if and only if g is, f'=g'-(g(1)-g(0)) is square integrable if and only if g' is, and f(0)=f(1)=0. Then $|f(x)|\leq \int _{0}^{x}|f'(t)|\leq \left(\int _{0}^{x}|f'(t)|^{2}~dt\right)^{\tfrac {1}{2}}\cdot {\sqrt {x}}$ where the first inequality is the fundamental theorem of calculus, and the second is cauchy-schwarz applied to f'*1 (you can use Hoelder inequality in general), so $|f(x)|^{2}\leq \int _{0}^{1}|f(t)|^{2}~dt$ after squaring both sides and replacing the factors on the right with their maxes. Integrating over x=0..1 gives the result:

$\int _{0}^{1}|f(x)|^{2}~dx\leq \int _{0}^{1}|f'(x)|^{2}~dx$

Note that one used that the volume of the domain was 1 really quite a few times, so for a more general domain there will be some constants. For the complete proof you might need some density arguments to say that it is ok to only have considered C^1 functions, and you need to handle switching from f to g if you want an explicit inequality. On wikipedia, I think this version is called the Friedrichs' inequality, but it and several others were called Poincare in our analysis class. JackSchmidt (talk) 18:59, 15 February 2008 (UTC)

Wow, thanks! This is exactly what I was looking for.A Real Kaiser (talk) 06:18, 16 February 2008 (UTC)

relation between A.M, Median and Mode

in acadamic books it is given that the relation between arithmatic mean, median and mode is mode = 3MEDIAN - 2 A.M Please explain this and explain any limitations for it because for some uni-model group of values this formula is not working out. F0r example

0, 0, 0, 2, 2, 3, 18

its mean is 12.5, meadian is 2, mode is 0 which do not satisfy the above relation: Please explain in detailKasiraoj (talk) 08:22, 15 February 2008 (UTC)

It is definitely not true in general; and I have never heard of it. What book is this and what exactly does it say? --Spoon! (talk) 09:06, 15 February 2008 (UTC)

You should have read the links provided to you earlier (and followed up there instead of posting a new question). This relation is just an empirical rule of thumb that has been found to approximately hold in some distributions commonly encountered in practice - as I understand, in particular those that resemble the normal distribution (and yet have some asymmetry). -- Meni Rosenfeld (talk) 09:12, 15 February 2008 (UTC)

Arithmetic mean = Sum of all items divided by number of items
Median = The number with the middle value
Mode = The number that appears most in the list
(I think you know that^)
The book should give the limitations for the equation. If not, the book should be wrong. Visit me at Ftbhrygvn (Talk|Contribs|Log|Userboxes) 03:51, 16 February 2008 (UTC)

Percentage Calculator

I'm looking for a calculator online that can perform this simple function: I have two (or more) values. I want the calculator to add up the values and tell me what the percentage of each value is compared to the total value. Basically, I'm trying to calculate vote totals from election websites that list vote amounts but not the percentages, and I know there's got to be an easier way than using a calculator and adding up the values and dividing manually. — Preceding unsigned comment added by Rc251 (talk • contribs)

You could try using a spreadsheet (Excel or similar). --Tango (talk) 16:01, 15 February 2008 (UTC)

This online spreadsheet is easy to learn and to use. --hydnjo talk 04:10, 16 February 2008 (UTC)

Or use google and just type in the calculation. For example '50/95*100' in google will return a 'google calculator' answer (http://www.google.co.uk/search?hl=en&q=50%2F95*100&btnG=Search&meta=) not a search result. ny156uk (talk) 10:30, 16 February 2008 (UTC)

which number is higher?

hi there, which number or fractions is higher? 3.5%(7/2), 3.4%(17/5), 4.3%(43/10), 3.1%(31/10), 1.8%(9/5), or 5.4%(27/5)? —Preceding unsigned comment added by Don Mustafa (talk • contribs) 15:41, 15 February 2008 (UTC)

You've already got them in decimal form, so you can easily which is the bigger number. Start by looking at the whole part (ie. the bit before the decimal point), and see which is larger. If there is one with a whole part larger than all the others, that number is larger. If there are two (or more) the same, then you look at the fractional parts (ie. the bits after the decimal points) of them and see which of them is larger, and then that number is the larger (if they have the same number of digits after the point, it's just like with whole numbers, if they have different numbers of digits, add 0's to the end of the shorter to make them the same length, and then it's just like whole numbers). Alternatively, you can write all the numbers in a column making sure to line up the decimal points above each other (and make each digit the same size so they all line up - squared paper might help), then start with the column of digits furthest left, and see which is larger (if a number doesn't have a digit in that column, you know it's smaller), if two have the same digit, compare the next digits until you find ones that are different, and the number with the large digit is the larger number. Hope that helps. --Tango (talk) 15:59, 15 February 2008 (UTC)

You got to be kidding me right?. Which number is larger? 3.5 , 3.4 , 4.3 , 3.1 , 1.8 or 5.4 . This seems to be such an easy question. 122.107.129.141 (talk) 23:05, 15 February 2008 (UTC)

Hi. Please don't make any assumptions about the OP's (what does OP mean anyway) age or mathematical abilities. Thanks. ~AH1^(TCU) 02:58, 16 February 2008 (UTC)

It means Original Poster, and it's necessary to make assumptions to even provide an answer. Black Carrot (talk) 04:19, 16 February 2008 (UTC)

I got the numbers from [[2]] [[3]] and [[4]]. Please, have a look the deep red colored ones and calculate the percentages into numbers. —Preceding unsigned comment added by Don Mustafa (talk • contribs) 01:46, 18 February 2008 (UTC)

which number is higher? 2

Which number is higher? 23.5%, 26.9% or 20.1%? —Preceding unsigned comment added by Don Mustafa (talk • contribs) 15:49, 15 February 2008 (UTC)

Ignore the percent sign and just treat them as you would any other numbers (see above). --Tango (talk) 15:59, 15 February 2008 (UTC)

Is this a joke? Surely someone who can type with capitals and know that "2" is bigger than "1" can read numbers... --PalaceGuard008 (Talk) 02:50, 16 February 2008 (UTC)

February 16

Coordinate system

Hello friends. I have a cartesian coordinate system, where four points are given: (0|0), (0|1), (1|0), (1|1). You have now to move these points so that at the end these points are at (0|0), (1|3)(1|1), (3|0), (2|-1). But there's one rule given: One point can only be moved, if there is another point between its old place and its new place. Now my question: Is it possible to get this latter layout from the first one? --85.178.63.105 (talk) 10:30, 16 February 2008 (UTC)

Sounds similar to Hi-Q, but without removals. Here's a solution. I don't guarantee it's minimal. It was easy enough (working backwards from the goal to the starting position) that I wonder if there is some other constraint you forgot to mention. Each of my moves jumps over a point that is exactly halfway to the destination, which you didn't even ask for, so call that a bonus.

(0,1) -> (2,-1) -- the (1,0) peg is in the middle of the jump.
(1,1) -> (1,-1) -- the (1,0) peg is in the middle.
(1,0) -> (1,-2) -- the (1,-1) peg is in the middle.
(1,-2) -> (3,0) -- the (2,-1) peg is in the middle.
(1,-1) -> (5,1) -- the (3,0) peg is in the middle.
(5,1) -> (-1,-3) -- the (2,-1) peg is in the middle.
(-1,-3) -> (1,3) -- the (0,0) peg is in the middle.

--tcsetattr (talk / contribs) 10:57, 16 February 2008 (UTC)

Oh, I'm so much sorry I do not mean (1|3) but (1|1) at the end. Can you write the solution again for this way please? --85.178.63.105 (talk) 11:10, 16 February 2008 (UTC)

In that case, assuming the middle peg has to be halfway between the starting and ending positions, it's not hard to show that there's no solution. I'll withhold the details for the moment since this looks like the kind of problem that might be assigned as math homework. Try starting with the ending position and seeing where you can go from there to get a feel for what's going on. -- BenRG (talk) 19:09, 16 February 2008 (UTC)

No, this is not a homework or something like that. My friend told me if I want to puzzle over something and gave me this exercise for the weekend. Now I don't want to tell him that I failed to do this [and I want to get a beer for free]. So please can you give me please the solution? --85.178.32.123 (talk) 23:22, 16 February 2008 (UTC)

You want me to help you defraud your friend out of a beer? I'll give you a hint. Take a checkerboard with red and white squares and put some tokens on red squares. Now move the tokens around according to the rules of this game. You will never be able to move a token onto a white square (do you see why?). This means if your friend had told you to go from, say, (0|1), (0|-1), (1|0), (-1|0) to (0|0), (0|1), (1|0), (1|1), you would know it was impossible, because the former positions are all on the same color and the latter positions aren't. It's also impossible to do it in the other direction, because all of the moves in this game are reversible. That coloring argument doesn't work for the problem he actually gave you, but a slightly different coloring does work. To figure out that coloring, start with the pieces in the ending position of your friend's problem and try moving them around until you can see which squares are reachable and which aren't. -- BenRG (talk) 00:22, 17 February 2008 (UTC)

Something I'm not clear on: is this confined to the integer lattice? If not, it can still be solved. Black Carrot (talk) 02:37, 18 February 2008 (UTC)

February 17

Functions with different arguments

Is there a standard notation to distinguish different arguments/formulations of a given function that is usually expressed as a single argument? For example:

${\begin{aligned}\phi &=\quad \phi (\beta )=\arctan(\sec(o\!\varepsilon )\tan(\beta )),\\&=\quad \phi ({\widehat {\sigma }})=\arctan(\sec({\widehat {\alpha }})\tan({\widehat {\sigma }}));\end{aligned}}\,\!$

If you are given $\phi (37^{\circ }){}_{\color {white}|}\,\!$, it could be either $\phi (\beta ){}_{\color {white}|}\,\!$ or $\phi ({\widehat {\sigma }}){}_{\color {white}|}\,\!$! Would you incorporate the constant as a coargument——e.g., $\phi (o\!\varepsilon ;37^{\circ }){}_{\color {white}|}\,\!$ or $\phi ({\widehat {\alpha }}\backslash {37}^{\circ }){}_{\color {white}|}\,\!$, or something else?
In the case where the function, itself, is an argument of another function, expressed either singularly or as a bivariable, it would seem perfectly proper to express it likewise:

${\begin{aligned}\phi &=\phi ({\widehat {\mathrm {A} }},{\widehat {\sigma }})=\arcsin(\cos({\widehat {\mathrm {A} }})\sin({\widehat {\sigma }}));\\M&=\!\!\quad {M}(\phi )=M({\widehat {\mathrm {A} }},{\widehat {\sigma }});\end{aligned}}\,\!$

 ~Kaimbridge~00:44, 17 February 2008 (UTC)

I really can't understand what this question is asking. If $a=b$, then $f(a)=f(b)$. That's the substitution property of equality. Introducing a system of notation that breaks this rule is an awful idea. If you want $\phi (\beta )$ to have one definition and $\phi ({\widehat {\sigma }})$ to have a different definition, then you're really talking about two different functions, and you need to give them two different names. —Keenan Pepper 01:19, 17 February 2008 (UTC)

I think Kaimbridge is working with some already awful notation commonly used in some subject, and trying to modify it to make more sense. If so, it would help to see a reference where this notation is used. -- Meni Rosenfeld (talk) 15:18, 17 February 2008 (UTC)

Using functions like that is an abuse of notation. It's quite common in science where the difference between a physical value and the function for calculating that value is rather blurred (hence the common shorthand for "position is determined by time" of "r=r(t)", the first r is a physical value, the second r is a function). The idea is that it's obvious from context what you mean, if it isn't, you need a better notation. In formal mathematics, you should give different functions different names, so you should have a $\phi _{1}$ and a $\phi _{2}$ or something. Two functions are equal if and only if they give the same result for any given input, which your two forms of $\phi$ clearly don't, so they aren't the same function, so they shouldn't have the same name. (You can get away with using the same name if you can distinguish the functions by number of variables, since that removes the ambiguity, but it's often best to distinguish them more clearly.) --Tango (talk) 15:51, 17 February 2008 (UTC)

Okay, let me give you the actual scenerio.
For Earth, $\phi \,\!$ is geographic latitude and $\beta \,\!$ is the reduced latitude. At the same time, ${\widehat {\alpha }}\,\!$ is the spherical/geographical azimuth and ${\widehat {\sigma }}\,\!$ is the transverse geographical colatitude (likewise, ${\tilde {\alpha }}\,\!$ and ${\tilde {\sigma }}\,\!$ are the elliptical equivalent, based on $\beta \,\!$). My particular question had to do with derivative. With respect to latitude relationship,

${\begin{aligned}\beta (\phi )&=\arctan(\cos(o\!\varepsilon )\tan(\phi ));\\\beta '(\phi )&={\sqrt {m'(\phi )n'(\phi )}}={\frac {\cos(o\!\varepsilon )}{1-(\sin(\phi )\sin(o\!\varepsilon ))^{2}}};\end{aligned}}\,\!$

But in terms of $\sigma \,\!$ to latitude conversion,

$\phi '({\widehat {\sigma }})=\cos({\widehat {\alpha }});\quad \beta '({\tilde {\sigma }})=\cos({\tilde {\alpha }});\,\!$

While the functions equate, the derivatives obviously don't! In the particular situation I am dealing with, the elliptical values are localized, with respect to ${\widehat {\sigma }}\,\!$:

${\tilde {\sigma }}({\widehat {\sigma }})=\lim _{\Delta {\widehat {\sigma }}{\to }0}{\tilde {\sigma }};\quad {\tilde {\alpha }}({\widehat {\sigma }})=\lim _{\Delta {\widehat {\sigma }}{\to }0}{\tilde {\alpha }};\,\!$

So, $\beta =\beta ({\tilde {\sigma }})=\beta (\phi ({\widehat {\sigma }}))=\beta ({\tilde {\sigma }}({\widehat {\sigma }}))\,\!$.

The ultimate use and purpose for all of this is this equation:

$o'(\phi ({\widehat {\sigma }}))=C'{\Big (}\beta ({\tilde {\sigma }}({\widehat {\sigma }})){\Big )}{\tilde {\sigma }}'({\widehat {\sigma }})=C'{\Big (}\beta (\phi ({\widehat {\sigma }})){\Big )}\beta '(\phi ({\widehat {\sigma }})){\frac {\phi '({\widehat {\sigma }})}{\beta '({\tilde {\sigma }}({\widehat {\sigma }}))}};\,\!$

or

$C'{\Big (}\beta ({\tilde {\sigma }}({\widehat {\sigma }})){\Big )}\beta '({\tilde {\sigma }}({\widehat {\sigma }})){\tilde {\sigma }}'({\widehat {\sigma }})=C'{\Big (}\beta (\phi ({\widehat {\sigma }})){\Big )}\beta '(\phi ({\widehat {\sigma }}))\phi '({\widehat {\sigma }});\,\!$

Thus $\beta '({\tilde {\sigma }}({\widehat {\sigma }})){\color {white}.}\,\!$ is azimuth in nature while $\beta '(\phi ({\widehat {\sigma }})){\color {white}.}\,\!$ is elliptical. Is it just the author's responsibility to effectively distinguish the two via some indexing——e.g., $\beta '_{1}({\tilde {\sigma }}({\widehat {\sigma }})){\mbox{ and }}\beta '_{2}(\phi ({\widehat {\sigma }})){\color {white}.}\,\!$ or $\beta '({\tilde {\sigma }}({\widehat {\sigma }}))_{1}{\mbox{ and }}\beta '(\phi ({\widehat {\sigma }}))_{2}{\color {white}.}\,\!$——or is there some established special notation?  ~Kaimbridge~16:22, 17 February 2008 (UTC)

Empty set

One of my fellow ask me a simple question about Empty Set.As we call set that thing if there are some thing in it,eg people,number etc but an empty set is call a "SET" becoz there are nothing in it according to text base definition and i am not mathematician so plz simplefy me. —Preceding unsigned comment added by Usmanzia1 (talk • contribs) 02:26, 17 February 2008 (UTC)

Yes, an empty set is a set with nothing in it. It is a collection of objects, which makes it a set - just that this collection of objects is empty. x42bn6 Talk Mess 13:55, 17 February 2008 (UTC)

Try reading our article, Empty set, it should help. It's a rather strange concept to get used to, but it's common throughout maths - the lack of something is an example of a something (The empty product equals 1, the empty sum equals 0, the empty union is the empty set, the empty intersection is the universe, etc. etc.) --Tango (talk) 14:37, 17 February 2008 (UTC)

$\pm$ and $\mp$

What is the difference between $\mp \,\!$ and $\pm \,\!$? Thanks, Zrs 12 (talk) 03:21, 17 February 2008 (UTC)

When they are separate, there is no difference. However, if they are used in the same equation, such as "$a\pm b\mp c$", it often denotes that if you use + in the first instance, you match it with - in the second. The previous example would generally be used to indicate two possibilities: "$a+b-c$" and "$a-b+c$". 134.173.92.17 (talk) 03:53, 17 February 2008 (UTC)

Also see Plus-minus sign#Minus-plus sign. --hydnjo talk 03:58, 17 February 2008 (UTC)

Thanks, Zrs 12 (talk) 05:17, 17 February 2008 (UTC)

Stochastic matrix Question 4

$\mathbf {p} _{k}=\mathbf {v} P^{k}\,.$

We want to find the probability that the system is in a given state after a given number of time steps. The set of probabilities for each state after k time steps is given by the probability vector p_k. The purpose of the formula is that it gives an expression for the probability vector after k time steps in terms of the initial state vector v and the stochastic matric P - so if we know v and P we can find the probability vector at any subsequent time. The "mathematical induction" part just means that we can derive the general formula for p_k by looking at the formulae for p₁, p₂ etc. and then generalising the pattern that we see to k time steps. Can you see where the formulae that I give above for p₁, p₂ come from ? Can you see how they lead to a general formula for p_k ? Gandalf61 (talk) 09:35, 12 February 2008 (UTC)

I am assuming the formulae $\mathbf {p} _{k}=\mathbf {v} P^{k}\,$ that you gave above for p₁, p₂ came from Summation? If this is true then can the formula be put in Sigma notation format $\sum _{i=m}^{n}x_{i}=x_{m}+x_{m+1}+x_{m+2}+\cdots +x_{n-1}+x_{n}.$ ? --Obsolete.fax (talk) 05:28, 17 February 2008 (UTC)

I can't see any connection with summation at all. Gandalf61 (talk) 15:02, 21 February 2008 (UTC)

Then could you answer "Can you see where the formulae that I give above for p₁, p₂ come from ? Can you see how they lead to a general formula for p_k?" I don't know. Please help, would really appreciate. --Obsolete.fax (talk) 18:23, 23 February 2008 (UTC)

Then go back and carefully read the explanations that Lambian gave here and that I gave here. They show how the general formula for p_k is derived. If there is a step that you don't understand, then say which step you get stuck on. It is not possible to help you any further unless you say which part of the explanation you don't understand. Gandalf61 (talk) 11:09, 24 February 2008 (UTC)

Indefinite integral of the Gamma function

Is there any way to integrate (doesn't have to be closed form) Gamma(x)?

--wj32 ^t/c 10:34, 17 February 2008 (UTC)

Well, gamma is continuous, so yes, it has an antiderivative. The Integrator gives nothing, so I suspect the antiderivative can't be expressed in terms of functions anyone's bothered naming, but I could be wrong, and don't know anything like enough differential Galois theory to prove such a result. Algebraist 14:50, 17 February 2008 (UTC)

Factorials

How do you prove, or whats the mathematical proof of O!=1? I've asked this to several math teachers and they always answer someting like: "You just take it for granted", "It's just a rule", etc... —Preceding unsigned comment added by 201.167.101.193 (talk) 21:58, 17 February 2008 (UTC)

See Factorial#Definition. Basically, it's because it turns out that whenever you come across 0! in some setting, you want it to be equal to 1. For example, n! is the number of ways you can arrange n different objects in a line. How many ways can you arrange zero objects in a line? Well, one—there is one way to arrange zero objects, but that's it. —Bkell (talk) 22:12, 17 February 2008 (UTC)

It's not something that can be proved, it's simply part of the definition of the function. -mattbuck (Talk) 22:50, 17 February 2008 (UTC)

Sure, but the reason it's defined that way is because it's the "right" definition to make. Defining 0! to be anything but 1 would break a lot of elegant formulas and identities. —Bkell (talk) 23:16, 17 February 2008 (UTC)

A better question than "why does 0! equal 1" is "why does 0! exist in the first place", and for that matter, "why does factorial exist in the first place?" It exists, by that I mean it is given its own name and used all over the place, because it is a convenient way of arranging information. 0! is included because in many applications the input winds up as zero in some special cases, and those special cases might as well be dealt with officially. It equals 1 because that's what you'd put as the output in each of those individual cases anyway. (Contrast this with division by zero, where the most convenient answer varies and no single definition would do the job.) At each step, the point is to summarize things as efficiently as possible. To do that, you introduce a structure, and make refinements of it. Only once you have a structure to describe can you start proving things about it. Black Carrot (talk) 23:26, 17 February 2008 (UTC)

When you're working with natural numbers, it's best to think of the factorial function as counting permutations. That is, n! is the number of distinct ways of arranging n different objects in a line. If you have zero objects, there's only one way to arrange them (don't do anything), so 0! = 1. More formally, if n! is the number of bijections from an n-element set to itself, then 0! = 1 because there is precisely one bijection from the empty set to itself. Michael Slone (talk) 23:39, 17 February 2008 (UTC)

Another way to think of it is by observing that n!=(n+1)!/(n+1) for all n>1, so extend that for n=0, and you get 0!=1.--Tango (talk) 23:45, 17 February 2008 (UTC)

And there's the fact that the Gamma function, a beautiful extension of the factorial, has a value of 1 at 1 (Gamma(t+1)=t! for t non-negative integer). Pallida  Mors 03:10, 18 February 2008 (UTC)

February 18

Cubic function

If you go here and enter as the equation ax³ + bx² + cx + d = 0 and tell it to solve for x, you get a long expression as the solution. Can someone show algebraically how you can solve ax³ + bx² + cx + d = 0 for x? Or link to a page that does? Thanks. 70.162.25.53 (talk) 03:39, 18 February 2008 (UTC)

See cubic function, specifically the Cardano's method section. Be warned, it's not quite as nice as the quadratic formula... 134.173.92.17 (talk)