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January 10

Finding the area under a curve (calculus question)

I've been struggling with a problem. I need to find the area under the curve y = 27 - x^3 on the interval [1, 3].

area = lim [as n --> infinity] ∑ from [i = 1] to [n] of f(c_i)∆x
area = lim [as n --> infinity] ∑ from [i = 1] to [n] of (27 - [1 + i/n])(2/n)

I think I know the rules for how to manipulate sigma notation. Maybe I just keep making subtle algebra mistakes, but my friend and I can't figure it out. (The textbook and my class say the answer is 34.) Can anyone show me how to get this answer? Thanks! —anon.

The easiest way is with integration: $\int _{1}^{3}(27-x^{3})dx$, which is exactly what sums become as the partitions become infinitely small. Strad (talk) 00:21, 10 January 2008 (UTC)

Ah, but we haven't learned how to evaluate definite integrals yet, except with the formula I wrote above... :-( —anon

So you wanna do it the hard way. FIrst of all, your formula lost the cube. Secondly, $c_{i}$ should be the start of the interval plus $i/n$ times the length of the interval, which makes it $1+2i/n$. The sum should be

$\sum _{i=1}^{n}{\frac {2}{n}}[27-(1+2i/n)^{3}]$

Since n is not dependent on i, the 2/n can be taken out of the sum:

${\frac {2}{n}}\sum _{i=1}^{n}[27-(1+2i/n)^{3}]$

The cube has to be expanded as in $(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}$:

${\frac {2}{n}}\sum _{i=1}^{n}[27-(1+3{\frac {2i}{n}}+3{\frac {4i^{2}}{n^{2}}}+{\frac {8i^{3}}{n^{3}}})]$

Now the sum can be split into simpler sums and everything not dependent on i is moved outside the sums (like the 2/n earlier):

${\frac {2}{n}}\left[\sum _{i=1}^{n}26-{\frac {6}{n}}\sum _{i=1}^{n}i-{\frac {12}{n^{2}}}\sum _{i=1}^{n}i^{2}-{\frac {8}{n^{3}}}\sum _{i=1}^{n}i^{3}\right]$

Those simple sums can all be found at Summation#Identities.

${\frac {2}{n}}\left[26n-{\frac {6}{n}}\cdot {\frac {n(n+1)}{2}}-{\frac {12}{n^{2}}}\cdot \left({\frac {n^{3}}{3}}+{\frac {n^{2}}{2}}+{\frac {n}{6}}\right)-{\frac {8}{n^{3}}}\cdot \left({\frac {n^{4}}{4}}+{\frac {n^{3}}{2}}+{\frac {n^{2}}{4}}\right)\right]$

The rest should be easy:

${\frac {2}{n}}\left[26n-3(n+1)-\left(4n+6+{\frac {2}{n}}\right)-\left(2n+4+{\frac {2}{n}}\right)\right]$

${\frac {2}{n}}\left[26n-3n-3-4n-6-{\frac {2}{n}}-2n-4-{\frac {2}{n}}\right]$

$52-6-{\frac {6}{n}}-8-{\frac {12}{n}}-{\frac {4}{n^{2}}}-4-{\frac {8}{n}}-{\frac {4}{n^{2}}}$

$34-{\frac {26}{n}}-{\frac {8}{n^{2}}}$

And you can see what happens when n goes to infinity.

--tcsetattr (talk / contribs) 05:17, 10 January 2008 (UTC)

Do you know (or can you prove) that the "area" under the curve y = 27 - x³ is equal to [(the area under the curve y = 27) minus (the area under the curve y = x³)] ? This won't solve your underlying problem, but it may make the algebra easier. Tesseran (talk) 03:27, 10 January 2008 (UTC)

roman

were there any weaknesses in the roman numeral system? —Preceding unsigned comment added by Invisiblebug590 (talk • contribs) 07:39, 10 January 2008 (UTC)

It's very hard to write large numbers and it's hard to do calculations. -- Meni Rosenfeld (talk) 08:35, 10 January 2008 (UTC)

It was used essentially to number the legions, after all. PMajer (talk) 12:46, 10 January 2008 (UTC)

The biggest problem with the Roman system (along with many other ancient number systems) was the lack of a zero for place holding. This meant they had to continually invent new symbols as larger and larger numbers were required until by the Middle Ages virtually all of the alphabet had a numerical value. I don't entirely hold with the widespread opinion that it is hard to do calculations with roman numerals. In particular, it is often said that multiplication of large numbers and long division are difficult. I have practised both of these a bit just to see if this is true. I found that the difficulty was mainly my unfamiliarity with the system. Once this was overcome I found it actually easier in some respects: the symbols retain the same value wherever you write them so you do not need to worry about column values. SpinningSpark 16:39, 10 January 2008 (UTC)

Exact, and by the way another thing to recall is that an efficient number system was not so developed, just because there was not so a big need for it, although for instance Greek mathematics was extremely advanced. The point is that, at that time, a geometric formalism was indeed the most natural and satisfactory thing for both the theoretical and practical sake. Think about this: if you have to draw a project, and you have to compute a length, or an angle, you can do it directly by a geometric construction, like the ones described by Euclid (under this respect Euclid's Elements is the AutoCAD of the antiquity). On the contrary for example, measuring a length, then computing numerically the product by square root of 2, then measuring and drawing a segment of that length, is ridiculous compared to the efficiency of the drawing algorithm (make a square on your segment, take the diagonal). The numeric formalism become something really useful and efficient only after the invention of the mechanical printing by Gutenberg. Only then numerals became the perfect way to store mathematical information in a compact way, like in tables of logarithms, etc. If you wish, we had a similar situation with the binary system. It is not that Euler or Gauss ignored it, the fact is that it was of little use for them, and it had been so indeed, till the invention of the computers.PMajer (talk) 18:02, 10 January 2008 (UTC)

Confidence Ellipsoid, Least Squares

I'm working through a derivation of the confidence ellipsoid of a ordinary least squares problem. Several sources (Scheffe 1959, Wikipedia) all seem to give the following as the solution:

$({\boldsymbol {\beta }}-\mathbf {b} )^{\prime }\mathbf {X} ^{\prime }\mathbf {X} ({\boldsymbol {\beta }}-\mathbf {b} )\leq ps^{2}F(p,n-p,1-\alpha )$

where p is the number of parameters fit in the model. When I do the derivation, I get it should be n, the number of data points. In particular,

$({\boldsymbol {\beta }}-\mathbf {b} )^{\prime }\mathbf {X} ^{\prime }\mathbf {X} ({\boldsymbol {\beta }}-\mathbf {b} )$

=$(\mathbf {X} ({\boldsymbol {\beta }}-\mathbf {b} ))^{\prime }\mathbf {X} ({\boldsymbol {\beta }}-\mathbf {b} )$

And $\mathbf {X} ({\boldsymbol {\beta }}-\mathbf {b} )\sim \sigma N_{n}(0,I)$ (since X is n by p and $({\boldsymbol {\beta }}-\mathbf {b} )$ is p by 1, the dimension of the product will be n by 1.). Thus $({\boldsymbol {\beta }}-\mathbf {b} )^{\prime }\mathbf {X} ^{\prime }\mathbf {X} ({\boldsymbol {\beta }}-\mathbf {b} )$ should be $\chi _{n}^{2}$, not p degrees of freedom, so the ratio

${\frac {(({\boldsymbol {\beta }}-\mathbf {b} )^{\prime }\mathbf {X} ^{\prime }\mathbf {X} ({\boldsymbol {\beta }}-\mathbf {b} ))/n}{{\hat {\sigma }}^{2}}}$

should be distributed $F_{n,n-p}$, not $F_{p,n-p}$, which seems to be what every book I check says. Anyone spot my mistake? --TeaDrinker (talk) 20:52, 10 January 2008 (UTC)

The simplest thing to do is note that β-b is p-dimensional, so that while multiplying it by X gives an n-dimensional result, it is one that is always in a p-dimensional (or less) subspace of Rⁿ. Matrix multiplication can never increase the span of a linear space. Baccyak4H (Yak!) 04:49, 11 January 2008 (UTC)

I should add that the operative part here is that the df for the numerator chi squared comes from the rank of the covariance matrix of b's normal distribution, which is p for an identifiable model. The rank of the subsequent n-dimensional normal's covariance matrix is thus still p, even though the multivariate normal is n-dimensional. Baccyak4H (Yak!) 15:39, 11 January 2008 (UTC)

Thanks! I was pretty sure my proof fell apart somewhere, I was just having trouble tracking down where. I'm thinking, based on what you say, the proof falls apart on the independence of the numerator and denominator, needed for the F distribution (although I have not been able to fully convince myself of that). In any event, thanks! -TeaDrinker (talk) 21:23, 11 January 2008 (UTC)

Actually, I did not see where you dealt with the independence or not in an incorrect way. It is simply that the rank of the covariance matrix of X (β-b) is p, not n. So its squared norm is a χ² with p degrees of freedom, not n. Baccyak4H (Yak!) 04:44, 12 January 2008 (UTC)

Hmm, that is odd then. If I understand you, you're saying $X({\hat {\beta }}-\beta )$ is not $N(0,I\sigma ^{2})$, since that would be a projection of a p dimensional span (betas) into n dimensions and getting an n dimensional span. However,

${\hat {\beta }}\sim N_{p}(\beta ,(X^{\prime }X)^{-1}\sigma ^{2})$

${\hat {\beta }}-\beta \sim N_{p}(0,(X^{\prime }X)^{-1}\sigma ^{2})$

$X({\hat {\beta }}-\beta )\sim XN_{p}(0,(X^{\prime }X)^{-1}\sigma ^{2})$

$=N_{n}(0,X(X^{\prime }X)^{-1}X^{\prime }\sigma ^{2})$

$=N_{n}(0,\sigma ^{2}I)\,\!$

Perhaps I am making a simple mistake somewhere. Many thanks, your comments are very much appreciated! --TeaDrinker (talk) 18:04, 12 January 2008 (UTC)

The only mistake I see there is that on the last step, the projection or hat matrix (the big mess of X things) does not simplify to the n-dimensional identity. Rather it is the projection matrix for X, and n-dimensional matrix of rank p (again, starting with rank p matrix X and involving only multiplication and inversion cannot increase the rank past p). It is the rank of the covariance matrix that is important -- it becomes the degrees of freedom.

If you wanted to, you could come up with a matrix to premultiply X (β-b) which would give you an N_p(0, I), but which would cancel itself out in the middle of the crossproduct if you reckoned that multiplication first. And what is the squared norm of N_p(0, I)? You bet. Baccyak4H (Yak!) 05:40, 13 January 2008 (UTC)

Ahh, yes, I see it. I had failed to recognize the hat matrix, instead multiplied by $(XX')(XX')^{-1}$ (which then very nicely simplifies), the latter part, as I think about it, need not exist. Thanks! -TeaDrinker (talk) 06:19, 13 January 2008 (UTC)

January 11

Does an integer solution exist for this equation?

For the Diophantine equation: 10*X/Y+X^2/Y^2=4, are there integer values for both X and Y such that this equation is true. If there are solutions what method do I use to solve equation in this form[ A*X/Y+(X/Y)^2=B ; where A and B are constant integers and X and Y are variables]? 24.250.129.216 (talk) 00:28, 11 January 2008 (UTC)1337haxor

You can transform this equation to (X/Y + A/2)^2 = B + A^2/4. So X/Y = ±sqrt(B + A^2/4) - A/2 = (±sqrt(4*B + A^2) - A)/2 and there are solutions only if (4*B + A^2) is a perfect square, because else X/Y would have to be irrational or complex. Icek (talk) 01:12, 11 January 2008 (UTC)

What kind of function/shape of curve...?

Hi all - a question (more office work than homework, BTW).

Say you had a large number of boxes (n), each containing the same number (x) of items. At random, you remove items from boxes one at a time, without replacement. What kind of function would you have if you graphed number of completely emptied boxes against number of items removed? My guess it would be some form of sigmoid shape, but beyond that I'm at a loss to know what kind of function you'd get. Any ideas? Thanks in advance, Grutness...wha? 02:05, 11 January 2008 (UTC)

When you say 'at random', do you mean that you pick a box at random, then pick a random item from it, or that you somehow pick at item at random from all those remaining? Algebraist 02:23, 11 January 2008 (UTC)

Ah. Good point. I meant at random overall (e.g., if you had 1000 numbered marbles in one bag you were drawing from, and wanted to know at what rate "decades" of numbers would be completely removed). Mind you, knowing the solution to the other randomisation would also be interesting... Grutness...wha? 05:59, 11 January 2008 (UTC)

Then, the problem you are thinking of is related to the multi-hypergeometric distribution. This doesn't mean that the article can give you an easy formula for solving this issue, I'm afraid.

The other possible formulation of the problem is (I guess) much more complex.

Eventually, you could feed a spreadsheet or a CAS with a model of your problem and simulate for estimates. Pallida  Mors 07:14, 11 January 2008 (UTC)

The number of empty boxes is the sum, taken over all boxes, of the following quantity, which is a random variable:

1 if box #i is empty, 0 otherwise.

Because expected value distributes over sum, the expected number of empty boxes is equal to the sum of the expected values of these 0-or-1-valued random variables. That expected value equals the probability that box #i is empty. Because of the symmetry in the problem, this probability is the same for each box. Calling that probability p(n,x,t), where n and x are as above and t is the total number of items removed, the expected number of empty boxes is the sum of n copies of this, which amounts to n·p(n,x,t). So we can fix our attention on one box, and try to determine the probability it is empty after t items have been removed. This can be done combinatorially, since each subset of size t of the original set of nx items is equally likely to have been removed. There are C(nx,t) such subsets, where C(n,k) = n!/(k!(n−k)!) denotes the number of possible combinations when choosing k elements from a set of size n. If we divide the number of combinations that results in a given box being emptied by this number of total combinations, we have the probability we want to determine.

So how many combinations are there, among those C(nx,t) combinations, that contain a marked subset of x items – the items of a given box? Let T be the total original set of nx items, M the marked subset, and let R be a combination of t chosen items containing all of M. Then S = R−M is a subset of t−x items of T−M (where "−M" denotes "set subtraction", also often denoted "\M" and called relative complement). Conversely, given any subset S of t−x items of T−M, the set R = S+M (where "+" denotes set union; in this case the two "summands" are disjoint) is a subset of t items of T containing all of M. So there is a one-to-one correspondence between such sets R and such sets S, and to count the former we can count the latter. But that count is, of course, simply C(nx−x,t−x), and so p(n,x,t) = C(nx−x,t−x)/C(nx,t).

The expected number of empty boxes is then:

n·C(nx−x,t−x)/C(nx,t) = A(n,x)·t!/(t−x)!, where A(n,x) = n·(nx−x)!/(nx)!.

This will not look like a sigmoid when plotted against t.  --Lambiam 23:04, 11 January 2008 (UTC)

Thanks for all that - I think I can probably work it from there (although it's a little more advanced maths than I can do easily). Cheers. Grutness...wha? 23:59, 11 January 2008 (UTC)

Recognizing another form of the derivative

There are two expressions of which I have to find the limit as x approaches 0:

(√(3+x) - √3)/x
and
((1/(2+x)) - (1/2))/x.

I know that direct substitution fails because it produces an indeterminate form in each case, so I basically know that each expression is a disguised form of the derivative. I have a vague recollection that in cases like these, I'm supposed to take the derivative of the minuend of the numerator. I tried, but I got 1/(2√3) and -1/√(x+2) respectively rather than the answers given in the textbook, (√3)/6 and -1/4. How do I find the expression of which to take the derivative to find these limits? Thanks, anon —Preceding unsigned comment added by 141.155.22.61 (talk) 03:22, 11 January 2008 (UTC)

Not the minuend of the numerator, no. See L'Hôpital's rule for what to do in indeterminate cases. Gscshoyru (talk) 03:25, 11 January 2008 (UTC)

Yeah, you could apply L'Hospital. Or then again you could actually bother to understand what's going on. You were on the right track with the "disguised form of the derivative" comment. What's the definition of the derivative? Can you work backwards to find a function, in each case, that makes the definition of derivative look like the limit you're trying to find? --Trovatore (talk) 03:34, 11 January 2008 (UTC)

Trovatore is speaking of the fact that every derivative is a limit, or in other words:

$f'\left(x_{0}\right)=\lim _{x\to 0}\,{\frac {f\left(x_{0}+x\right)-f\left(x_{0}\right)}{x}}$

So, if you know the f's for these "disguised derivates" and know basic differentiation rules, you are done.

If you want another way, the second limit is easy to calculate and doesn't need L'Hôpital. The first one needs no more than algebra, either, if you apply the transformation substitution y=x+3. Pallida  Mors 05:03, 11 January 2008 (UTC)

Taking f'(0) to find the limit of (f(x)−f(0))/x as x tends to 0 should work for both cases. You can also simplify the second form algebraically, which leads you directly to the result. For an algebraic approach to the first case, first multiply the numerator and denominator each by √(3+x) + √3, expand the numerator, and simplify.  --Lambiam 10:32, 11 January 2008 (UTC)

Does the OP (or anyone else) realise that he actually has the right answer to the first problem but in a different form to the book answer? SpinningSpark 20:07, 11 January 2008 (UTC)

Logical simplification

$\neg Q\land T\lor Q$

Does this statement simplify to:

$\neg Q$

or

$\neg Q\lor Q\equiv T$

Depending on whether I match the True with the AND or the OR? This is the last key stage in showing a more detailed statement is a tautology. As you can see one resolves the issue, and one does not. —Preceding unsigned comment added by 91.84.143.82 (talk) 13:21, 11 January 2008 (UTC)

I don't think there is a universally understood convention as to the order of operations here, so the expression is basically meaningless. Indeed, depending on where the parentheses should be, it could mean any one of the options you gave. If you got this expression after some calculation, you need to look back at the calculation and be careful with the parentheses - then you will know which is correct. -- Meni Rosenfeld (talk) 14:12, 11 January 2008 (UTC)

$\neg$ binds stronger than $\land$, $\land$ binds stronger than $\lor$. This is pretty much universally accepted in formal logic, especially in the context of satisfiability, horn clauses, resolution and similar stuff. Compare it to the similar integer arithmetic $-Q*1+Q$, which binds the same. So, your proof is already complete, congrats. —Preceding unsigned comment added by 84.187.80.85 (talk) 20:40, 11 January 2008 (UTC)

Umm ... as the original questioner seems to be unaware of this order of precedence in logic operations, how do we know they have not misapplied the order of operations in a previous step in their work, and mistakenly reached $\neg Q\land T\lor Q$ instead of $\neg Q\land (T\lor Q)$ ? I agree with Meni's suggestion that they go back over their calculation and insert parentheses as a check. Air-dropping an order of precendence rule only into the last step seems dangerous to me. Gandalf61 (talk) 10:10, 12 January 2008 (UTC)

Stem and leaf

In a stem and leaf chart is the max value of the "leaf" equal to the inter quartile range or 150% of the inter quartile range? I've read both but am not sure which is correct. 136.206.1.17 (talk) 17:16, 11 January 2008 (UTC)

The lengths of the leaves are determined by convention so as to be useful; there really is no absolute right or wrong. I am not sure who uses which convention but one thing you may wish to consider is that for a larger sample, it makes sense to increase the proportion of the IQR used. The expectations of the extrememost order statistics from any distribution (at least one where such expectations exist) get farther away from the median of the distribution. This prevents getting points outside the leaves with arbitrarily certain probability just by increasing the sample size, even for the most well behaved distributions (e.g., normal). Baccyak4H (Yak!) 17:51, 11 January 2008 (UTC)

Another nonlinear second-order ODE

${\frac {d^{2}\theta }{dt^{2}}}={\frac {\cos \theta }{\sin ^{3}\theta }}$

Is there a closed-form solution? —Keenan Pepper 21:45, 11 January 2008 (UTC)

Here's what I just got — not the final answer, but quite a lot on the way toward it.

Firstly, the equation can be written as:

${\dot {\theta }}{\frac {d{\dot {\theta }}}{d\theta }}={\frac {d\left({\dot {\theta }}^{2}\right)}{2d\theta }}={\frac {\cos \theta }{\sin ^{3}\theta }},$

from which:

${\dot {\theta }}^{2}=2\int {\frac {d\sin \theta }{\sin ^{3}\theta }}=-{\frac {6}{\sin ^{4}\theta }}+C_{1}$

and:

$t=\pm \int {\frac {d\theta }{\sqrt {C_{1}-{\frac {6}{\sin ^{4}\theta }}}}}$.

Substitute $\alpha =\tan \theta$, $\sin ^{2}\theta ={\frac {\alpha ^{2}}{1+\alpha ^{2}}}$ and $d\theta ={\frac {d\alpha }{1+\alpha ^{2}}}$ and multiply both the numerator and the denominator by $\alpha ^{2}$:

$t=\pm \int {\frac {\alpha ^{2}d\alpha }{\left(1+\alpha ^{2}\right){\sqrt {C_{1}\alpha ^{4}-6\left(1+\alpha ^{2}\right)^{2}}}}}.$

From the Wolfram Integrator we get an aswer, but a messy one. With some little help from Maxima, I got:

$t=\pm {\frac {F\left(\beta \left|{\frac {a_{-}}{a_{+}}}\right.\right)-\Pi \left({\frac {1}{a_{-}}};\beta \left|{\frac {a_{-}}{a_{+}}}\right.\right)}{\sqrt {6a_{+}}}}+C_{2},$

where $a_{\pm }={\sqrt {\tfrac {C_{1}}{6}}}\pm 1$ and $\beta =i\,\operatorname {arsh} \left(\alpha {\sqrt {a_{-}}}\right)=\arcsin \left(\alpha {\sqrt {-a_{-}}}\right)$. $F$ is the incomplete elliptic integral of the first kind and $\Pi$ of the third kind. Now solve for $\beta$, $\alpha ={\tfrac {\sin \beta }{\sqrt {-a_{-}}}}$ and finally $\theta =\arctan \alpha$… ${\ddot {\smile }}$

For some specific $C_{1}$-s we can go a bit further.

For $C_{1}=0$ the Integrator gives (OK, that's an easy integral to do by hand also):

$\mp t={\frac {\alpha }{2{\sqrt {6}}\left(\alpha ^{2}+1\right)}}-{\frac {i}{2{\sqrt {6}}}}\arctan \alpha +C_{2},$

an equation I don't know by which special functions to solve.

$C_{1}=6$ is a more interesting case: by some standard substitutions (completing the square etc) the radical can be removed and the rational function integrated, as the Integrator naturally knows:

$\mp t={\frac {-\arctan {\frac {-i\alpha }{\sqrt {2\alpha ^{2}+1}}}-{\frac {1}{\sqrt {2}}}\arctan {\frac {i\alpha {\sqrt {2}}}{\sqrt {2\alpha ^{2}+1}}}}{\sqrt {6}}}+C_{2}={\frac {1}{\sqrt {6}}}\left[\arctan(i\xi )-{\tfrac {1}{\sqrt {2}}}\arctan \left(i\xi {\sqrt {2}}\right)\right]+C_{2},$

where $\xi ={\tfrac {\alpha }{\sqrt {2\alpha ^{2}+1}}}$. As $\arctan(ix)={\tfrac {i}{2}}\ln {\tfrac {1+x}{1-x}}$:

$e^{\pm 2{\sqrt {6}}i\left(t+C_{2}\right)}={\frac {1+\xi }{1-\xi }}\left({\frac {1+\xi {\sqrt {2}}}{1-\xi {\sqrt {2}}}}\right)^{-{\tfrac {1}{\sqrt {2}}}}\implies e^{\pm 2{\sqrt {3}}i\left(t+C_{2}\right)}=\left({\frac {1+\xi }{1-\xi }}\right)^{\tfrac {1}{\sqrt {2}}}\left({\frac {1-\xi {\sqrt {2}}}{1+\xi {\sqrt {2}}}}\right)^{\tfrac {1}{2}}.$

This is a cubic equation in $\xi$, which is, in principle, solvable. I used to have an error here. Now, I believe, it's correct, but I'm too tired to solve it right now. After that we could find $\alpha =\pm {\sqrt {\tfrac {\xi }{1-2\xi }}}$ and, again, $\theta =\arctan \alpha$.

As you can see, I haven't paid any attention to the domains my substitutions are valid in, because my main aim was finding some solutions, not a complete analysis the whole solution set. Good night! — Pt _(T) 04:08, 12 January 2008 (UTC) (minor fixes (some $C_{2}$-s had been omitted) — Pt _(T) 04:18, 12 January 2008 (UTC), a bigger fix 04:54, 12 January 2008 (UTC))

If there only would be an additional minus sign in the equation - then a simple solution would be arccos(t). Icek (talk) 04:31, 12 January 2008 (UTC)

Thus, one solution is $\theta =\arccos(it)$ and, as the equation is time-invariant, $\theta =\arccos(it+C)$ is actually a family of solutions. By the way, I've just discovered a serious error in my second equation! ${\ddot {\smile }}$ I'm working on it. — Pt _(T) 16:42, 12 January 2008 (UTC)

Okay, could someone explain that very first step in much more detail? I've never seen it before, and it seems like a powerful technique. I see that the formal manipulation of symbols is correct (${\frac {d{\dot {\theta }}}{dt}}={\frac {d\theta }{dt}}{\frac {d{\dot {\theta }}}{d\theta }}$ is just the chain rule), but I can't figure out what ${\frac {d{\dot {\theta }}}{d\theta }}$ means, exactly. $\theta$ is a function of $t$, so ${\dot {\theta }}$ is also a function of $t$, and I don't see how you can turn it into a function of $\theta$ in order to take the derivative.

Also, could other versions of this technique be applied to higher-order equations, for example ${\frac {d^{3}x}{dt^{3}}}=f(x)$ ? —Keenan Pepper 16:07, 12 January 2008 (UTC)

Well, ${\dot {\theta }}$ is a function of $t$ and, if $\theta$ has an inverse, then $t$ can be thought of as a function of $\theta$, thus: ${\dot {\theta }}[t(\theta )]$.

I don't know what this method is called, I found it once in a differential calculus textbook. I don't have it around at the moment, so I cannot tell you much more about it. Generalization to higher orders seems not to be straighforward.

Anyway, my first integral went wrong, it should be:

${\dot {\theta }}^{2}=2\int {\frac {d\sin \theta }{\sin ^{3}\theta }}=-{\frac {1}{\sin ^{2}\theta }}+C_{1}\implies t=\pm \int {\frac {d\theta }{\sqrt {C_{1}-{\frac {1}{\sin ^{2}\theta }}}}}.$

This is an easy integral (thanks to the $\pm$ we can be liberal with the signs):

$t=\pm \int {\frac {\sin \theta d\theta }{\sqrt {C_{1}\sin ^{2}\theta -1}}}=\pm \int {\frac {d\cos \theta }{\sqrt {C_{1}\left(1-\cos ^{2}\theta \right)-1}}}=\pm {\frac {1}{\sqrt {C_{1}}}}\int {\frac {d\cos \theta }{\sqrt {\left(1-{\tfrac {1}{C_{1}}}\right)-\cos ^{2}\theta }}}.$

From the tables (the Integrator gives a mess I'm not willing to clean up):

$t=\pm {\frac {1}{\sqrt {C_{1}}}}\arcsin {\frac {\cos \theta }{\sqrt {1-{\tfrac {1}{C_{1}}}}}}+C_{2}.$

Denoting ${\tfrac {1}{\sqrt {C_{1}}}}=\tau$ and $\pm t-C_{2}=\pm \left(t-t_{0}\right)$:

$\theta =\arccos \left[\pm {\sqrt {1-\tau ^{2}}}\sin {\frac {t-t_{0}}{\tau }}\right].$

Taking the limit $\tau \to \infty$ and choosing the $+$ from the $\pm$, we get:

$\theta =\arccos \left[i\left(t-t_{0}\right)\right],$

the solution found previously. — Pt _(T) 17:40, 12 January 2008 (UTC)

I've been lazy again. Firstly, I'd forgotten my lessons on trigonometric equations:

$\theta =\pm \arccos \left[\pm {\sqrt {1-\tau ^{2}}}\sin {\frac {t-t_{0}}{\tau }}\right]+2n\pi \quad (n\in \mathbb {Z} ).$

Secondly, separation of variables often doesn't give us the constant solutions:

$\theta =A=\mathrm {const} \implies {\frac {\cos A}{\sin ^{3}A}}=0\implies A={\frac {\pi }{2}}+n\pi \quad (n\in \mathbb {Z} ).$

Fortunately, those solutions have been already covered by $\tau =1$. But there may still be some singular solutions... Any advice on how to find them or justify my belief that there aren't any? — Pt _(T) 18:12, 12 January 2008 (UTC)

By the way, the method is quite useful in physics. In principle, every homogeneous 2nd order ODE can be reduced to an inhomogeneous 1st order one. For example, applying the substitution ${\ddot {x}}={\tfrac {d\left({\dot {x}}^{2}\right)}{2dx}}$ to the Newton's II law in one dimension for a particle with mass $m$ in a potential energy field $U$, we get the conservation of energy:

$m{\ddot {x}}+{\frac {dU}{dx}}=0\implies {\frac {d\left({\tfrac {m{\dot {x}}^{2}}{2}}+U\right)}{dx}}=0,$

which can (for a known potential energy) be easily solved for $x(t)$ to obtain the particle's motion. (The first integration constant is the total energy.) — Pt _(T) 22:26, 15 January 2008 (UTC) (corrected 01:11, 16 January 2008 (UTC))

January 12

100 sum

How do I write 100 using all the digits from 1 to 9, not necessarily in their natural order, with only one written symbol which denotes an operation?

Also I wish to know how many ways are there to write statements which equals to 100 using all the digits in their natural order such as:

100=123-45-67=89

(pls list down the ways)

this is not homework, i saw this question on a maths magazine.Invisiblebug590 (talk) 10:51, 12 January 2008 (UTC)

$9+8+7+6+5+4+3+2+1+\sum _{k=1}^{55}1$. Welp, —Preceding unsigned comment added by Damien Karras (talk • contribs) 15:41, 12 January 2008 (UTC)

That uses 1 twice, 5 three times, 10 operation symbols, and gives the value 1585. Algebraist 15:53, 12 January 2008 (UTC)

That's what they want you to think!!! 81.153.211.247 (talk) 17:02, 12 January 2008 (UTC)

81.153.211.247 changed $\sum _{k=1}^{55}k$ to $\sum _{k=1}^{55}1$ after the comment by Algebraist. PrimeHunter (talk) 19:12, 12 January 2008 (UTC)

It depends on the rules. I would say exponentiation written as a^b requires no "written symbol which denotes an operation", but I don't know whether 100 can be reached with exponentiation and a single written operation. If a is a sequence of digits then a_b can mean a interpreted as a base b number, for example 100₂ = 4. If this is allowed then there are many easy solutions. PrimeHunter (talk) 16:28, 12 January 2008 (UTC)

Oh nuts, and I thought I was so clever having found 321489base567. You beat me to it SpinningSpark 16:42, 12 January 2008 (UTC)

That, of course, is not a solution to the problem as stated which required the digits to be in there natural order. In fact, their are no solutions of the form abaseb. SpinningSpark 16:50, 12 January 2008 (UTC)

Many easy solutions PrimeHunter? I have only a small single digit (base ten) number for total solutions. SpinningSpark 17:18, 12 January 2008 (UTC)

I was referring to the first question where one written operation is allowed and any order of digits is allowed. 98+2_b gives 6! possible values of b. And there are other possible patterns. PrimeHunter (talk) 17:30, 12 January 2008 (UTC)

Ah, I see. I was counting the change of base as being the allowed operation. SpinningSpark 17:59, 12 January 2008 (UTC)

By the way, is there a prize in the magazine for solving this, and if you win it, will you share with Wikipedia? SpinningSpark 16:53, 12 January 2008 (UTC)

Raffle

Say we have a raffle with 100 tickets and 3 prizes. One and one ticket is sold until all the prizes are won. How many tickets can you expect to sell?

I can find the answer in R with Monte Carlo simulation:

> a<-replicate(1e6,max(sample(100,3)))
> mean(a)
[1] 75.76514

And the pdf:

> table(a)/length(a)
a
       3        4        5        6        7        8        9       10 
0.000009 0.000017 0.000031 0.000061 0.000095 0.000124 0.000173 0.000240 
      11       12       13       14       15       16       17       18 
0.000271 0.000341 0.000415 0.000490 0.000592 0.000621 0.000702 0.000852 
      19       20       21       22       23       24       25       26 
0.000935 0.001054 0.001223 0.001284 0.001379 0.001580 0.001800 0.001801 
      27       28       29       30       31       32       33       34 
0.002051 0.002104 0.002342 0.002460 0.002743 0.002983 0.003021 0.003263 
      35       36       37       38       39       40       41       42 
0.003510 0.003664 0.003782 0.004103 0.004300 0.004583 0.004832 0.005045 
      43       44       45       46       47       48       49       50 
0.005392 0.005691 0.005824 0.006133 0.006435 0.006764 0.007052 0.007013 
      51       52       53       54       55       56       57       58 
0.007601 0.007802 0.008294 0.008482 0.008876 0.009179 0.009520 0.009832 
      59       60       61       62       63       64       65       66 
0.010120 0.010691 0.010846 0.011362 0.011675 0.012062 0.012457 0.012694 
      67       68       69       70       71       72       73       74 
0.013018 0.013750 0.014184 0.014458 0.014838 0.015364 0.015773 0.016187 
      75       76       77       78       79       80       81       82 
0.016798 0.017136 0.017520 0.018039 0.018802 0.019151 0.019553 0.020083 
      83       84       85       86       87       88       89       90 
0.020735 0.021073 0.021243 0.022291 0.022502 0.023258 0.023667 0.024278 
      91       92       93       94       95       96       97       98 
0.024394 0.025330 0.025982 0.026277 0.027101 0.027910 0.028286 0.028522 
      99      100 
0.029578 0.030246

It would be fun to have the exact values, and a formula for m tickets and n prizes.

Any suggestions?

--Δεζηθ (talk) 14:45, 12 January 2008 (UTC)

Working from the back end, so to speak, the probability that the 100th ticket is NOT a prize is;

$P_{1}={\frac {97}{100}}$

The probability that there is not a prize in the last two tickets;

$P_{2}={\frac {97}{100}}{\frac {96}{99}}$

and generally for r tickets from the end;

$P_{r}={\frac {97}{100}}{\frac {96}{99}}{\frac {95}{98}}...={\frac {(m-n)!}{(m-n-r)!}}{\frac {(m-r)!}{(m)!}}$

The expected number of tickets is the value of r making P_r closest to 0.5
SpinningSpark 17:49, 12 January 2008 (UTC)

A better strategy is to not announce the prizes till you have sold all the tickets. SpinningSpark 17:52, 12 January 2008 (UTC)

What's the probability of ending the raffle with the rth ticket's sell? In other words, what are the chances of giving away the third (or mth) prize with the rth ticket?

This is clearly the product of two probabilities: The probability of having given already m-1 prizes with the sell of r-1 tickets, times the probability of selling the last lucky ticket out of the n-(r-1) remanining numbers. Hence,

$Q(r)=P(n,m,r-1,m-1)\times {\frac {1}{n-(r-1)}}$

...where $P(n,m,r-1,n-1)$ is the probability (under the hypergeometric model) of taking m-1 out of m marked items, from a total reference of n items, with r-1 extractions.

A simple expression for this product is given by

$Q(r)={\frac {m(n-m)!(r-1)!}{n!(r-m)!}}.$

Letting n=100 and m=3, values for the predicted chances for the OP's original inquiry can be computed. I must say that these values astonishingly match those estimated by the original poster; I praise the OP's simulation's performance.

The expected number of tickets sold can be computed with

${\tilde {r}}=\left[\sum _{r=m}^{n}r\times {\frac {(m(n-m)!(r-1)!)}{n!(r-m)!}}\right].$

It comes as no surprise that, for n=100 and m=3, this amounts to 75.75. Pallida  Mors 20:45, 13 January 2008 (UTC)

By the way, my silicon friend tells me that the expected number of tickets can be simplified into

${\tilde {r}}={\frac {m(n+1)}{m+1}}.$ Pallida  Mors 22:55, 13 January 2008 (UTC)

Dividing a slice of cake across the radius to end up with two equal volumes.

Your collective kind indulgence sought in solving this question that's been bugging me- and everyone I've bored with it.

Imagine a piece of cake, a segment, of radius R and angle at the apex (a). There is icing along the sector of circumference C. The fact that I like this icing, but my friend doesn't, causes us to agree to cut the cake on a line perpendicular to R at half a. We wish to each have equal volumes of cake sponge- The cake is of even height. The icing cannot be scraped off, nor can the cake be cut at half the height (above the plate). These sneaky suggestions have already been put forward! If we cut the cake half way between the apex and the icing, then clearly I get too much cake in my (almost) trapezius and my friend gets too little in their triangle. How far up or down the R, at half a, should we cut the cake to ensure equal amounts of cake? The icing is neglible (though tasty). —Preceding unsigned comment added by Geoffgraves (talk • contribs) 21:06, 12 January 2008 (UTC)

Well, I'd do it by integration. say we cut the cake down the middle at a/2, then by symmetry the area of cake (and hence volume) at the midpoint on one side is the midpoint of the whole thing. Integrate over this. I'm thinking it and will get back to you. Mmmm.... cake.... - mattbuck 21:16, 12 January 2008 (UTC)

OK, working with half the slice, we find that,

$I=-R\int \limits _{0}^{\arccos(1-\cos {\frac {a}{2}})}\sin ^{2}ydy+{\dfrac {R^{2}}{4}}\sin a.$

Use this to find the midpoint. - mattbuck 21:48, 12 January 2008 (UTC)

I don't think you need calculus for this. If a is small then one of the two pieces will be an isosceles triangle whose area you can work out with high school trigonometry. The area of the full slice is also easy to work out. Set the former equal to half the latter and solve for the position of the cut. I get $h=R{\sqrt {\frac {a}{4\tan(a/2)}}}$, where h is the distance from the apex to the plane of the cut, and a is in radians. -- BenRG (talk) 07:46, 13 January 2008 (UTC)

I also get that formula, and get $108.604^{\circ }$ as the maximum a for which a straight cut can separate the slice into icingless and iced portions of equal area. --Tardis (talk) 17:38, 14 January 2008 (UTC)

January 13

Bayes question

I'm trying to classify images appearing in HTML into photographs / non-photographs.

Two possible indicators (among many others) I can use are whether the image tag has an alt attribute and whether the image has a title attribute set.

In my data set I've found that about 90% of photographs have an alt attribute set; and about 60% of non-photographs have an alt attribute set. Also, about 20% of photographs have a title attribute set; and about 5% of non-photographs have a title attribute set.

How can I calculate the probability of an image being either photograph or non-photograph for every possible (true/false) configuration of the two input variables?

I think that this is just Bayesian conversion of Pr(H|D) to Pr(D|H), but I've not found a simple explanation of how to do that with more than one input variable. (And yes, let's assume that the title/alt attribute presences are independent).

--Clairvoyant walrus2 (talk) 00:16, 13 January 2008 (UTC)

You will need a prior to do anything Bayesian. In other words you need to know the probability of your images being photos/non-photos in the absence of alt/title evidence. And rather than assuming independence, why not just gather data on the prevalence of each of the four combined possibilities? Algebraist 00:27, 13 January 2008 (UTC)

I can get priors easily enough... why not get all 4 possibilities? There aren't really just two input variables, I just described it that way for simplicity, there are actually a couple dozen. --Clairvoyant walrus2 (talk) 02:37, 13 January 2008 (UTC)

How'd you end up with the name "Clairvoyant walrus2"? Was "Clairvoyant walrus" already taken? Anyway, the usual Bayesian rule says that $P(A|B)={\frac {P(B|A)P(A)}{P(B|A)P(A)+P(B|{\bar {A}})P({\bar {A}})}}$. Substitute $B=C\wedge D$ and use independence of $C$ and $D$ (which means that $P(C\wedge D|K)=P(C|K)P(D|K)$) to get $P(A|C,D)={\frac {P(C|A)P(D|A)P(A)}{P(C|A)P(D|A)P(A)+P(C|{\bar {A}})P(D|{\bar {A}})P({\bar {A}})}}$. Now you can replace $A$ with "x is a photo" and $C$ with "x (does / does not) have an alt attribute" and $D$ with "x (does / does not) have a title attribute". -- BenRG (talk) 08:07, 13 January 2008 (UTC)

Wikipedia:Reference desk/Archives/Mathematics/2007 December 14#Advice for integrating disparate similarity measures? indicates that Clairvoyant walrus is the same as Clairvoyant walrus2. The latter account was created the day after the last edit by the former. Maybe due to a forgotten password? PrimeHunter (talk) 08:38, 13 January 2008 (UTC)

Ok, I see now. Thanks. The answer had been staring me in the face. I implemented it in my application and it works well. --Clairvoyant walrus2 (talk) 04:17, 14 January 2008 (UTC)

Standard deviation and confidiance interfulls

When doing questions for an exam and comparing them with a friend we realise that we had different formulas for standard deviation and thus were getting different answers.

mine was

$\sigma ={\sqrt {\frac {\sum _{i=1}^{N}(x_{i}-{\overline {x}})^{2}}{N}}}.$

where as his was

$\sigma ={\sqrt {\frac {\sum _{i=1}^{N}(x_{i}-{\overline {x}})^{2}}{N-1}}}.$

both of us had found different books showing each, which one is correct. The question was to find the confidence interval. —Preceding unsigned comment added by 136.206.1.17 (talk) 11:10, 13 January 2008 (UTC)

The former is the correct way to calculate the standard deviation of a given population. The latter is the unbiased ("best" in some way) way to estimate the standard deviation of a population given only a sample of it. You may need one or the other depending on the circumstances - I don't think "confidence intervals" gives enough context. -- Meni Rosenfeld (talk) 13:12, 13 January 2008 (UTC)

Are you sure that the latter is unbiased, Meni? If I understand our article correctly, its square is an unbiased estimator of the variance. However, the formula given needs a slight correction to get an unbiased estimate of the standard deviation. --NorwegianBlue ^talk 18:34, 13 January 2008 (UTC)

You are of course correct. -- Meni Rosenfeld (talk) 18:52, 13 January 2008 (UTC)

Cyclic pentagon

As part of the solution of a newspaper puzzle, I have 10 sets of 5 distinct internal angles of a cyclic pentagon. In each case, the angles could be put into 12 distinguishable sequences of occurrence round the pentagon (e.g. ABCDE, BCDEA, AEDCB are non-distinguishable in the sense of giving the same figure, turning over if necessary). My question is twofold:

1) Can any sequence of 5 positive numbers summing to 540° be drawn as a cyclic pentagon with internal angles in that order, and if so, how? 2) Can it be established without drawing whether or not the centre of the circumscribing circle is inside the pentagon?

For example, one of my sets is (171°,161°,131°,44°,33°). Having to assess 120 pentagons seems wildly excessive for the purposes of the puzzle. 86.152.78.37 (talk) 16:34, 13 January 2008 (UTC)

Denoting by O the center of the circle, $\alpha =\angle COD$ and so on, you have the following equations:

$\left({\begin{array}{ccccc}1&1&0&0&1\\1&1&1&0&0\\0&1&1&1&0\\0&0&1&1&1\\1&0&0&1&1\end{array}}\right)\left({\begin{array}{c}\alpha \\\beta \\\gamma \\\delta \\\epsilon \end{array}}\right)=\left({\begin{array}{c}2A\\2B\\2C\\2D\\2E\end{array}}\right)$

For which the solution is:

$\left({\begin{array}{c}\alpha \\\beta \\\gamma \\\delta \\\epsilon \end{array}}\right)={\frac {2}{3}}\left({\begin{array}{rrrrr}-1&2&-1&-1&2\\2&-1&2&-1&-1\\-1&2&-1&2&-1\\-1&-1&2&-1&2\\2&-1&-1&2&-1\end{array}}\right)\left({\begin{array}{c}A\\B\\C\\D\\E\end{array}}\right)$

I think a sufficient and necessary condition to being a legal pentagon is that all of those internal angles are positive. Therefore, a solution does not always exist, but you can easily find when it does. Also, once you have calculated the internal angles it is trivial to construct the pentagon. I also think that the centre of the circumscribing circle is insideoutside the pentagon iff one of the internal angles is greater than 180°, which is also easy to check. -- Meni Rosenfeld (talk) 18:24, 13 January 2008 (UTC)

Assuming my geometry is correct, our silicon masters are more than willing to provide solutions to the problem. For the set you give there is no legal permutation. -- Meni Rosenfeld (talk) 18:32, 13 January 2008 (UTC)

Thanks. I understood as soon as I realised that your "internal angles" are at the centre, whereas mine are at the vertices. I think too that you mean that the centre of the circle is outside the pentagon iff one of α etc exceeds 180°.

I don't recall coming across anything like this before, but I'd have thought that the condition for a pentagon (and hexagon, ...) to be cyclic would be of interest, as an extension to all triangles being cyclic and quadrilaterals only if opposite angles sum to 180°. 86.152.78.37 (talk) 23:49, 13 January 2008 (UTC)

The difference originates from the fact that for an odd number of sides, the system of equations is regular, thus there is always a "solution". However, if the numerical solution includes negatives, it means that the polygon intersects itself. For the triangle, it is so simple it can never intersect itself, so there is always a legal solution. For a pentagon, you still can always find a cyclic one, but it might intersect itself. The condition that it doesn't intersect itself manifests as a bunch of inequalities. For an even number of sides, such as a square, the system is singular, and thus the vertex angles must satisfy some condition (in this case, $A+C=B+D$) if you want any sort of solution. -- Meni Rosenfeld (talk) 10:09, 14 January 2008 (UTC)

Proving Limits

Imagine I evaluate following expression$\lim _{x\rightarrow 3}{\frac {x^{3}+3x^{2}-9x-27}{x^{3}-9x}}$ 

and get 2 as its limit. I factored both the numerator and the denominator as much as I could and that's the result I got. What I would like to know is prove it indeed is 2. I suppose I could check with a graphic calculator, but I would like to be able to do it using Heine's method. It needn't be the above expression, that was just an example. Any limit will do it, just in case someone thinks I am here just to have my homework solved for me. That's not the case. Thank you very much in advance. -- Ishikawa Minoru (talk) 17:46, 13 January 2008 (UTC)

I have not heard of Heine's method, but L'Hôpital's rule can be useful here.

The way to prove a limit depends on what you are allowed to use. If you want to go all the way back to the δ-ε definition, it should go as follows: Let ε>0. Calculate $\left|{\frac {x^{3}+3x^{2}-9x-27}{x^{3}-9x}}-2\right|=\left|{\frac {3-x}{x}}\right|={\frac {|x-3|}{|x|}}$ (valid for $x\neq 0,3$). Now you need to find some $\delta >0$ such that if $|x-3|<\delta$ then ${\frac {|x-3|}{|x|}}<\epsilon$. Can you do it? -- Meni Rosenfeld (talk) 17:58, 13 January 2008 (UTC)

It would be a mistake to use L'Hopital's rule in this problem in certain contexts. For one thing, you may be finding limits in order to prove facts about derivatives, and then using derivatives in L'Hopital's rule. The fact that you get 0 when you plug in x = 3 tells you that (x − 3) is a factor, so you get

$\lim _{x\to 3}{\frac {(x-3)({\text{something}})}{(x-3)({\text{something}})}}$

and just use long division to find the two "somethings". You use the word "prove", which suggests either you don't want only to evaluate the limit, or you don't realize what the correct way of using that word is. Maybe I'll say more later if you clarify further. Michael Hardy (talk) 00:12, 14 January 2008 (UTC)

I find it unlikely that this particular problem is related to the evaluation of a derivative. Again, it all depends on what is allowed to be used, we certainly want to avoid any circularities. That said, I am personally more comfortable with taking derivatives for L'hopital's than with polynomial long division for factoring. -- Meni Rosenfeld (talk) 09:14, 14 January 2008 (UTC)

Assuming you're allowed basic properties of limits, factorising is useful:

${\frac {x^{3}+3x^{2}-9x-27}{x^{3}-9x}}={\frac {(x-3)(x^{2}+6x+9)}{(x-3)(x^{2}+3x)}}={\frac {x^{2}+6x+9}{x^{2}+3x}}$

Algebraist 18:29, 13 January 2008 (UTC)

...and that equals $={\frac {x^{2}+6x+9}{x^{2}+3x}}={\frac {(x+3)^{2}}{x(x+3)}}={\frac {x+3}{x}}=1+{\frac {3}{x}}$
CiaPan (talk) 06:32, 16 January 2008 (UTC)

Jiang Chun-Xuan

Can someone at this desk take a look at the above article. It was created by a newbie and needs cleanup but I don't know enough to do it myself. Theresa Knott | The otter sank 18:03, 13 January 2008 (UTC)

MATH QUESTION ON SPECIAL RIGHT TRIANGLES AND HOW TO CALCULATE THEIR AREA'S

Math question on special right triangles and how to calculate their area's
Now then, isn't that easier on the eyes?  ;-)

Okay, for Geometry we have this problem to do and I completly forgot how to do it. It has something to do with a special right triangle and calculating it's area. I don't wanna know just the answer but how to get to it. The question is a diagram of a 45-45-90 triangle and the 'taller' side is 73. The hypotenuse and base are unlabeled. The right angle is where the 'taller' side and bottom base meet. —Preceding unsigned comment added by 80.148.25.183 (talk) 19:34, 13 January 2008 (UTC)

If it's a 45-45-90 triangle, then there's a very specific relationship between the length of one leg and another. What is it? Gscshoyru (talk) 19:37, 13 January 2008 (UTC)

Look here if you need more help. hydnjo talk 01:14, 14 January 2008 (UTC)

Have you looked at Special_right_triangles#45-45-90_triangle ? --Ybbor^Talk 01:43, 14 January 2008 (UTC)

Yes, that's much better! hydnjo talk 01:56, 14 January 2008 (UTC)

January 14

WWII dollars

It says that the financial cost of WWII in 1944 dollars is one trillion. How much would that be today? —Preceding unsigned comment added by Jwking (talk • contribs) 18:31, 14 January 2008 (UTC)

I humbly suggest Science or Humanities as more apropriate desks for this question. Anyway, according to this, such amount will correspond to 11.49 trillions in 2006 dollars. Pallida  Mors 19:06, 14 January 2008 (UTC)

It really depends on how you are calculating the historical exchange rate. This site discusses a variety of ways to calculate it; depending on what index you use, it can be anywhere between $9.6 trillion and $60 trillion. The numbers are only meaningful if you are considering things correctly; I'd lean towards the relative share of GDP (the $60 trillion figure) because that makes more sense in terms of big government spending than does the CPI (the $11.4 trillion figure above), which compares things like how much a loaf of bread costs at any given time. --24.147.69.31 (talk) 00:08, 16 January 2008 (UTC)

Parametric equations

I have just been introduced to parametric equations and I'm having a bit of a hard time finding the derivative of one.

Suppose I am using $x=f(t)$ and $y=g(t)$.

I am using the formula $y-f(a)=f'(a)(x-a)\,\!$ to find the tangent to a curve. For $f(a)$, am I right in saying that I want to find $g(a)$? 172.142.94.249 (talk) 20:20, 14 January 2008 (UTC)

The formula $y-f(a)=f'(a)(x-a)\,\!$ is for "normal" functions, not for parametric curves. The formula you need is $f'(a)(y-g(a))=g'(a)(x-f(a))\;\!$. -- Meni Rosenfeld (talk) 20:34, 14 January 2008 (UTC)

Thank you; everything is working out fine now. I wonder though, since I haven't been taught this and I can't find it in my textbook, is there a way of finding a tangent to a parametric curve at a particular point without using the formula you gave me? 172.142.94.249 (talk) 20:44, 14 January 2008 (UTC)

Depends on what you are allowed to use. The tangent is the line which gives a first order approximation to the curve. Suppose you have $t\approx 0$. Then your first order approximations for the coordinates of the point corresponding to $a+t$ are $x=f(a+t)\approx f(a)+f'(a)t$ and $y=g(a+t)\approx g(a)+g'(a)t$, so the point on the tangent corresponding to t satisfies $x=f(a)+f'(a)t\;\!$ and $y=g(a)+g'(a)t\;\!$. If you eliminate t from these equations, you will have the equation of the tangent. -- Meni Rosenfeld (talk) 20:53, 14 January 2008 (UTC)

January 15

Game of Life

In Conway's game of life, what is the shortest-known period glider gun (psudo glider guns, made from filling gaps in a glider stream do not count)? Thanks, *Max* (talk) 02:28, 15 January 2008 (UTC).

According to this page, it's the period 22 gun I and Jason Summers found in 2000. —David Eppstein (talk) 02:39, 15 January 2008 (UTC)

I may have discovered a new period 20 diagonal rake. How do I verify whether it is known? *Max* (talk) 02:41, 15 January 2008 (UTC).

c/12 diagonal rakes based on Corderships have been known for some time.

The first c/4 diagonal puffer was found by Hartmut Holzwart in February 2004 and is period 28; it emits tubs and beehives but some other variations have since been found. In July 2005 David Bell used it as part of a breeder, by crashing the output of a c/2 rake into it to form a sequence of switch engines:

x = 155, y = 102
77boo$77bobo$77bo$80bo$80boo$80boo$79boo$75booboo$75bo4bo$75bob4o$79b
oo$79boo$79boo$81bob3o$77bobbo$76b5o6bo$76bobboo5bobo$74bobo10bo$73b3o
4bo$73bobboo$77b3o$77boo10bo$79bo8bobo3bo$77bobo8bobobbobo$70boo4boo
11bo4bo$69boo5boobo$71bo4booboobobboo$76b3obobobbobo$67boo3boobboo7bo
10bo$66boo3boo15bo6bobo3bo$68bobb3o14boo5bobobbobo$70b4o14boo6bo4bo$
69bobo15boo$68boo13booboo$68bobo12bo4bo$67bo15bob4o14bo$68boo17boo13bo
bo3bo$87boo13bobobbobo$87boo14bo4bo$89bo$85bobbo4bo$84b5o4bo$84bobboo
4bo16bo$82bobo12bo11bobo3bo$81b3o4bo7bobo10bobobbobo$81bobboo11bo12bo
4bo$85b3o$55bobo27boo$54bobbo29bo$53boo30bobo11bo17bo$52bo25boo4boo12b
obo3bo11bobo3bo$51b4o22boo5boobo10bobobbobo10bobobbobo$50bo4bo6bo16bo
4booboobo8bo4bo12bo4bo$50bobbo5boobo21b3obobo$50bobbo5bo15boo3boobboo$
51bo7bo14boo3boo$52b4obo6boo10bobb3o24bo17bo$53bo3bo4bo4bo10b4o23bobo
3bo11bobo3bo$54bo6bo15bobo25bobobbobo10bobobbobo$54b5obbo5bo8boo28bo4b
o12bo4bo$61b6o9bobo$54b5o16bo$54bo21boo$53bo3bobbo24boo26bo17bo$11bobo
12boo24b4obo25bo4bo23bobo3bo11bobo3bo$10bobbo10bo4bo21bo7boboo19bo29bo
bobbobo10bobobbobo$9boo7bo4bo26bobbo5bobbo6bo12bo5bo24bo4bo12bo4bo$8bo
6boobo4bo5bo20bobbo5bobbo4bobo12b6o9boo$7b6obbo7b6o21bo4bo4boo6boo5bo
20b4o$4boo7bo3b4o30b4o18bobo19booboo$3bo3b3obo4bo35bo21boo5bo14boo22bo
17bo$bbo3bo3boobobboobobbo30boo24bobo5bo31bobo3bo11bobo3bo$bbo5boo3bo
5bo6boo26bobbo22boo5bo31bobobbobo10bobobbobo$bb3o3b4obo7bobobb3o26bobo
27bobo32bo4bo12bo4bo$11bo9bo6bo57boo$bb3o6bobo7bo6bo5bo73boo$bo5bo5bo
8bo4bo4bobo3boobboobboobboobboobboobboobboobboobb3oboo27bobbo3b3o$o3b
oobo3boo11boo7boo3boobboobboobboobboobboobboobboobboobb4obbo27boo6bo
10bo17bo$o3bo6b3o62boboo35bo10bobo3bo11bobo3bo$o3boobo3boo11boo7boo3b
oobboobboobboobboobboobboobboobboobboo50bobobbobo10bobobbobo$bo5bo5bo
8bo4bo4bobo3boobboobboobboobboobboobboobboobboobboo34boo15bo4bo12bo4bo
$bb3o6bobo7bo6bo5bo75bobbo$11bo9bo6bo80boo$bb3o3b4obo7bobobb3o78bo$bbo
5boo3bo5bo6boo50boo27bo26bo17bo$bbo3bo3boobobboobobbo54booboo25bo4bo
20bobo3bo11bobo$3bo3b3obo4bo61b4o6boo19b3o21bobobbobo10bobo$4boo7bo3b
4o58boo6boob3o41bo4bo12bo$7b6obbo7b6o59b5o$8bo6boobo4bo5bo59b3o$9boo7b
o4bo95boo$10bobbo10bo4bo88boob4o16bo$11bobo12boo91b6o15bobo3bo$120b4o
16bobobbobo$141bo4bo4$148bo$147bobo3bo$147bobobbobo$148bo4bo!

Finally in October 2005 the first clean c/4 diagonal rake was constructed, by David Bell. It has period 4508 and occupies roughly a 4500 by 4000 bounding box. I think not much more than that is known about c/4 diagonal puffer and rake technology. So if you have a new period, and with a rake no less, I think it would be new and interesting. But you might want to check with Holzwart or Bell, or if you care to share your pattern here I could do so. —David Eppstein (talk) 03:37, 15 January 2008 (UTC)

What format is that code in? Algebraist 07:33, 15 January 2008 (UTC)

It's in rle (run-length-encoded) format [1] [2] [3] [4]. A "b" stands for a blank space, an "o" stands for a live cell, a "$" stands for the end of a row of cells, and a number prior to any of these three characters is a repetition count. You should be able to copy and past it into most life programs such as xlife, golly, etc. —David Eppstein (talk) 18:31, 15 January 2008 (UTC)

Thanks. *Max* (talk) 19:43, 15 January 2008 (UTC)

^That post (originally signed User:131.111.8.103) was from me signed out. Can you please not co-opt my posts, even if they are unedifying? Algebraist 06:36, 16 January 2008 (UTC)

File:Conway's Life Rake.jpg

Actually, it's period 60. It emitts 1 glider every 20 generations, but it destroys 2/3 of them. It's speed is c/2. *Max* (talk) 01:02, 16 January 2008 (UTC).

That's not a c/4 diagonal rake, it's a c/2 orthogonal rake. According to the "c/2 orthogonal spaceships, puffers, rakes" section of the Life status page I linked to above, a rake with that period is already known. —David Eppstein (talk) 02:32, 16 January 2008 (UTC)

Iterated automorphisms

Let Aut(G) be the automorphism group of the finite group G. Consider the directed graph whose vertices correspond to finite groups (up to isomorphism), such that there is an edge from G to H iff $\operatorname {Aut} (G)\cong H$. What is known about the structure of this graph? For example, are there any nontrivial cycles? Is the number of connected components finite or infinite? —Keenan Pepper 22:47, 15 January 2008 (UTC)

No answers, but another question: does this graph have any infinite (nonrepeating) paths? (btw, I assume by nontrivial you mean a cycle which is not a loop. Is that correct?) Algebraist 09:02, 16 January 2008 (UTC)

There are a number of loops, e.g. the dihedral group of order 8, that don't come from complete groups. I don't know any larger cycles though. 79.67.220.203 (talk) 11:45, 16 January 2008 (UTC)

A Second Opinon

Sorry,If I spelled the word opinon wrong,basicilly I`m not the guy whose been posting those Physics Magazine Questions. But,I`m doing that same thing. And I`m someone,whose not doing it for the prize money just for fun.You see I`m doing The Embriodery One,what I need to know is specific examples which I wrote that show that The Embriodery One,would be sucessful.I got my own theroy,basicilly I just want a second opionon. Also,there`s some Math in this but I think I can figure that part out. —Preceding unsigned comment added by 68.161.128.203 (talk) 23:22, 15 January 2008 (UTC)

Are you attempting to say this was not posted by you, but by someone with the same IP address? For the rest, sorry, but I have got no clue what you are talking about. The Embroidery One? One of the Old Ones? Is it a relative of the Flying Spaghetti Monster?  --Lambiam 23:32, 15 January 2008 (UTC)

Possibly it is related to Nylonathatep, the Laddering Horror. Algebraist 07:01, 16 January 2008 (UTC)

Fresh from getting laughed off of the science desk for trying to conscript us to solve a problem so he could make $50, this guy comes here and tries to pass off the idea that he's a different guy with the same IP, spelling, and question as the last guy. Buddy, if you think we're that stupid, how are we supposed to help you solve the problem? --24.147.69.31 (talk) 00:00, 16 January 2008 (UTC)

January 16

Could someone check my work please?

"Find the equation of a tangent line to the graph of f(x) = cos x that can be used to approximate the value of cos(π/6 + 0.1). Then, find an approximation of cos(π/6 + 0.1)."

Tangent-line equation: f'(x) = -sin x

f(x₀) ≈ f'(c)(x₀ - c) + f(c)

where x₀ = the number whose f( ) value we're trying to approximate = π/6 + 0.1, and
c = a convenient number close to x₀ on the tangent line = π/6

cos (π/6 + 0.1) ≈ -(sin(π/6))(π/6)(π/6 + 0.1 - π/6) + cos(π/6)
cos (π/6 + 0.1) ≈ -(π/6)(1/2)(0.1) + (√3)/2
cos (π/6 + 0.1) ≈ -π/120 + (√3)/2
cos (π/6 + 0.1) ≈ -(π + 60)/120

But somehow it seems a little weird to have π, though it's a cornerstone of radian notation, in a y-value ... Am I on the right track? Thanks, anon.

cos (π/6 + 0.1) ≈ -(sin(π/6))(π/6)(π/6 + 0.1 - π/6) + cos(π/6)

Where did this extra factor of π/6 come from? Also, how did you manage to get rid of the √3 in the last step? —Keenan Pepper 02:14, 16 January 2008 (UTC)

Think of it this way: the slope of y = cos(x) at x = π/6 is -sin(π/6), which is -1/2. So the tangent line to y = cos(x) at x = π/6 is the line with slope -1/2 that passes through (π/6, cos(π/6)). The point-slope formula tells us that the equation of this line is

$y-\cos \left({\frac {\pi }{6}}\right)=-{\frac {1}{2}}\left(x-{\frac {\pi }{6}}\right)$

$\Rightarrow y=\cos \left({\frac {\pi }{6}}\right)-{\frac {1}{2}}\left(x-{\frac {\pi }{6}}\right)$

Now find the value of y when x = π/6 + 0.1 and you have an approximation to cos(π/6 + 0.1). Gandalf61 (talk) 07:42, 16 January 2008 (UTC)

About Pi

Where does the expression pi origionate from? —Preceding unsigned comment added by 207.224.29.240 (talk) 03:12, 16 January 2008 (UTC)

What do you mean, "the expression pi"? If you mean $\pi$, in its context as the ratio of a circle's circumference to its diameter, check out Pi - where the reason for its use is explained in the lead section. If you mean $\Pi$, in the context of an operator to multiply several things together, see Multiplication#Capital pi notation, and while it doesn't explain why Pi was chosen, it would be to stand for "product" just like the sigma of Sigma notation stands for "sum". For other uses of pi, see Pi (disambiguation). Confusing Manifestation(Say hi!) 03:30, 16 January 2008 (UTC)