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March 21[edit]

explicit rule[edit]

Can anyone give a step-by-step description on writing a rule in equation form of the following pattern?

2,6,24,120

Well, without doing all of the work for you, check out factorial. Let us know if you're still stuck after that. :) --Kinu ^t/_c 00:40, 21 March 2008 (UTC)[reply]

lol I can't believe I didn't get that (the factorial part)! Still, I would appreciate it greatly if you can solve it for me as my brain is working a little more slowly than usual today.

ah (n+1)! is the answer...is there any other way to express this?

No, not really, that's the answer. You can write "(n+1)n(n-1)(n-2) ... 3.2.1", but that's just defining the factorial notation. --Tango (talk) 01:10, 21 March 2008 (UTC)[reply]

recursive rule[edit]

Speaking of rules and equations, can anyone help me write a recursive rule for the following equation?

2,5,14,41,122

Calculate the differences between the terms (5-2, 14-5, etc) and see if you spot a pattern. --Tango (talk) 01:11, 21 March 2008 (UTC)[reply]

yes i see a pattern...a_n=a_n-1+3^n-1

Yes. You can also search integer sequences in OEIS. "2,5,14,41,122" gives a non-recursive formula (and more complicated rules leading to other continuations). PrimeHunter (talk) 05:05, 21 March 2008 (UTC)[reply]

Yes, but that's not really very useful. What's useful is knowing how to determine the pattern yourself, having a website do your homework for you is pointless. --Tango (talk) 15:42, 21 March 2008 (UTC)[reply]

One issue with this question is that there are quite many patterns that have these first five numbers; how do you decide which is `the' pattern? SmaleDuffin (talk) 17:23, 21 March 2008 (UTC)[reply]

It's an issue with all such questions - I think to be sure you need to know the context. It's presumably a homework question and the class will have been studying certain types of sequence, so it's going to one of those types. --Tango (talk) 18:13, 21 March 2008 (UTC)[reply]

Occam's razor. If there is not enough information to determine an object uniquely, the one with the simplest representation should be chosen.

{\frac {3^{n}+1}{2}}

is simpler than

n^{4}-4n^{3}+8n^{2}-2n+2\;\!

. -- Meni Rosenfeld (talk) 00:17, 24 March 2008 (UTC)[reply]

Only if it's absolutely essential that you come to a conclusion at all. If your life, livelihood, or peace of mind depends on deciding quickly, that's a good rule of thumb. If not, it's a good idea to slow down a bit, and accept that you just don't have enough information yet. Black Carrot (talk) 00:39, 25 March 2008 (UTC)[reply]

Exponential of an exponential[edit]

Can an equation of the form y = k(a + bc^x)^x be simplified? Neon Merlin 04:57, 21 March 2008 (UTC)[reply]

You could expand the xth power binomial, but I would say that's unsimplifying the above equation. A math-wiki (talk) 08:37, 21 March 2008 (UTC)[reply]

Peano Curve[edit]

Does anyone know where I could get hold of Peano's original paper on the curve, or a translation if it's in another language? Also, a question about the curve itself: Did Peano himself come up with these pictures of the curve, or did they come later when someone decided they wanted a concrete example? Black Carrot (talk) 05:50, 21 March 2008 (UTC)[reply]

Peano curve cites 'G. Peano, Sur une courbe, qui remplit toute une aire plane. Math. Ann 36 (1890), 157-160.', in case you missed that. The original text is available here. The article appears to contain a concrete example, but no pictures (at least in the pages I can get to work). I don't know about translations, but at least the article's in French, not a language of his own invention. Algebraist 11:04, 21 March 2008 (UTC)[reply]

Thanks! And apparently I can read it, which surprised the heck out of me. :) Black Carrot (talk) 07:37, 22 March 2008 (UTC)[reply]

Mathematical French is dead easy (possibly because we stole most of our terminology from them). Algebraist 13:49, 22 March 2008 (UTC)[reply]

Happy Easter![edit]

Another pagan festival approaches and all those wishing to receive a 'happy easter' message get one.

My question is - what simple formulas give a good 'egg' shape. Links please, and it's just for curiousities sake, so don't exert yourselfs too much.87.102.16.238 (talk) 12:08, 21 March 2008 (UTC)[reply]

This doesn’t quite answer the question, but both Ellipsoid and Oval have a section on the egg shape. GromXXVII (talk) 12:35, 21 March 2008 (UTC)[reply]

Look here for all your oval curve needs. — Kieff | Talk 13:02, 21 March 2008 (UTC)[reply]

Super!87.102.16.238 (talk) 13:53, 21 March 2008 (UTC)[reply]

simple question[edit]

can we finde,or do we have afunction where,limit[f(x)] approches to infinity when ,x,approches to zero,while,f(0)=known value?it sounds no big deal but i think,yes there is such afunction exists.Husseinshimaljasimdini (talk) 12:23, 21 March 2008 (UTC)[reply]

Define f by f(0)=0, f(x)=1/x elsewhere. Bo Jacoby (talk) 12:35, 21 March 2008 (UTC).[reply]

Well 1/|x| elsewhere I believe, else you’d have to talk about one handed limits. GromXXVII (talk) 12:39, 21 March 2008 (UTC)[reply]

I wonder something like the differential of the dirac delta function?87.102.16.238 (talk) 15:17, 21 March 2008 (UTC)[reply]

as such take a look at Dirac_delta#Representations_of_the_delta_function the differential of the limits (n to infinity) of these functions.

eg y=d/dx( lim (e^{-x^2n} ))

when x=0 y=(1 error) =0, when x=0+delta x y=big

?87.102.16.238 (talk) 15:21, 21 March 2008 (UTC)[reply]

Also maybe consider the differential of the function y=e^{1/x^2} .At x=0 dy/dx = 0?87.102.16.238 (talk) 18:06, 21 March 2008 (UTC)[reply]

e^{1/x^2} isn't defined at 0, so can't have a derivative there. Even if you define y(0)=k though, for some k, it won't be continuous (because the limit at that point is positive infinity), so there'll be no derivative. Black Carrot (talk) 06:05, 22 March 2008 (UTC).Well, i was not expecting all these answers.As amatter[reply]

of fact i was thinking of the numbers of sectors we can get by dividing acircle.Now if the function ,F(θ)=2pi\θ,describes numbers of sectors we get by dividing acircle where θ is the centeral angle,then how many sectors we will get if,θ=0?is it zero?Husseinshimaljasimdini (talk) 11:46, 22 March 2008 (UTC)[reply]

I'd say you get infinitely many. The angle is always positive, and the limit as it tends to 0 from above is positive infinity. --Tango (talk) 15:41, 22 March 2008 (UTC)[reply]

Of course if the sector angle is zero - the answer (in the real world) is 'doesn't work' ie impossible - hence the answer 'infinitely many'. It's possible that you are confusing the answer 'zero' with 'not possible' - in this case the two aren't the same. There are of course examples of things were the answer = 0 means - 'not possible' or 'doesn't happen'.87.102.16.238 (talk) 15:58, 22 March 2008 (UTC)[reply]

Algebra with probabilistic events[edit]

I'm a college freshmen, and I'm no further into my mathematical education than vector calculus, but recently I picked up a book on probability theory. Even if it's mostly out of my league, it sure looks interesting. Anyway, I've encountered a statement that seems to me intuitively correct but algebraically wrong. Surely I'm mistaken; the question is, how?

Since I don't know to what degree the notation that the author uses is standard, I'll briefly recapitulate it. An event $A_{i}$ is a set of one or more of the elementary events within a sample space. $\Omega$ is the set of all elementary events, and $O$ is the empty set. $A_{1}+A_{2}$ is defined as the union of $A_{1}$ and $A_{2}$ ; $A_{1}A_{2}$ is the intersection of $A_{1}$ and $A_{2}$ . As usual, addition and multiplication are associative and commutative, and addition is distributive over multiplication. Lastly, ${\overline {A_{1}}}$ , the complement of $A_{1}$ , is $\Omega -A_{1}$ .

The statement that puzzles me is:

${\overline {A_{1}+A_{2}}}={\overline {A_{1}}}\;{\overline {A_{2}}}$ .

That makes sense to me insofar as the event that neither $A_{1}$ nor $A_{2}$ happens is the same as the combined event that $A_{1}$ doesn't happen and $A_{2}$ doesn't happen. But:

${\begin{aligned}\Omega -A_{1}-A_{2}&=(\Omega -A_{1})(\Omega -A_{2})\\\Omega -A_{1}-A_{2}&=\Omega ^{2}-\Omega A_{1}-\Omega A_{2}+A_{1}A_{2}\\\Omega -A_{1}-A_{2}&=\Omega -A_{1}-A_{2}+A_{1}A_{2}\\O&=A_{1}A_{2}\\\end{aligned}}$

which clearly isn't necessarily true. —Saric (Talk) 17:26, 21 March 2008 (UTC)[reply]

Your deduction of the second line from the first is flawed, as is your deduction of the last from the third. You have been misled by the author's non-standard and ghastly notation into thinking that the things he has called +, - and * behave as you expect things with those names to behave: in the first case, you've assumed - interacts properly with * (it doesn't), and in the second you've used a cancellation law which doesn't hold in this setting. I recommend finding a better book. Algebraist 17:43, 21 March 2008 (UTC)[reply]

Oh, and FYI the statement in question is (one of) De Morgan's laws. Algebraist 17:47, 21 March 2008 (UTC)[reply]

(ec)I think the problem you're having is because set difference (ie. complement) and union aren't inverses of each other.

A=(A\cup B)-B

is only true if

A\cap B=\emptyset

. You need to be careful when using "+" and "-" that you don't assume it works in the way as addition and subtraction of numbers. --Tango (talk) 17:51, 21 March 2008 (UTC)[reply]

The usual notation for this set subtraction, known as relative complement, is not A − B but A \ B. Using that notation helps one to avoid the pit you fell in. The intersection of two such complements satisfies the identity

(P \ A) ∩ (Q \ B) = ((P ∩ Q) \ A) \ B.

There is no such thing here as "minus times minus equals plus". --Lambiam 22:22, 21 March 2008 (UTC)[reply]

So these set operations don't work quite like the numerical equivalents, after all. I think I can guess why the author of the book liked this notation— so long as $A_{1}A_{2}=O$ , the probability of $A_{1}+A_{2}$ is the probability of $A_{1}$ plus the probability of $A_{2}$ , and likewise for multiplication. But if the analogy isn't perfect, I guess the notation is more harmful than helpful. Yeah, I'll probably want to get a different book. Thanks for helping me out here.

(Me, "a college freshmen", indeed! Obviously my graduation from high school hasn't deterred the typo fairy.) —Saric (Talk) 18:55, 22 March 2008 (UTC)[reply]

Likewise for multiplication? If by "

O

" in "

A_{1}A_{2}=O

" you mean the empty set (so that A₁ and A₂ are mutually exclusive), it is not true in general that this implies P(A₁A₂) = P(A₁)P(A₂). For example, take A₁ = heads and A₂ = tails in a fair coin flip. --Lambiam 21:03, 22 March 2008 (UTC)[reply]

Yeah, you need mutually exclusive and independent. If you're only dealing with such simple probability, then the notation works fine, but it rapidly falls apart once you do anything real. --Tango (talk) 22:28, 23 March 2008 (UTC)[reply]

Solving "almost" superincreasing knapsacks problem[edit]

I know that it is really easy to solve the knapsack problem if all elements are super increasing, however, what if all of the elements were "almost super increasing". By almost super increasing, I mean that instead of having each successively larger integer being greater than the sum of all of the smaller integers. Each successively larger integer will be larger than, say, one-half of the sum all of the smaller integers. For example a valid set of "~1/2 almost super increasing integers" would be {8,11,17,26,38,58,86,130,195,292,438,657,985}; Would this instance of knapsack be as difficult to solve as the general knapsack problem? Or maybe somewhere between the difficulty of general knapsack and super increasing knapsack? 24.250.129.216 (talk) 19:16, 21 March 2008 (UTC)mathnoob[reply]

We don't know for sure whether the general knapsack problem is difficult to solve, so a proof that this is "somewhere between" easy and hard, but neither of the two, would imply a proof of P ≠ NP.

My gut feeling is that this is still NP-hard, but I don't readily see an applicable reduction like that of Exact cover to the knapsack problem. --Lambiam 22:39, 21 March 2008 (UTC)[reply]

The only difference between a super increasing sequence and this sequence is that in this sequence each term T(n) is larger than the sum of all terms from T(0) to T(n-3); while a super increasing sequence has each term T(n) larger than the sum of all terms T(0) to T(n-1). A super increasing sequence has a polynomial time solution.