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January 8

Algebra Question

Hey,It`s Me The Physics Magazine guy only this time I`ve got an algebra question for you,from an algebra magazine.Well,several acutually.

The lateral area of a right prism,is the sum of the lateral area of two bases.

The lateral area of a right cylinder is the product of the circumference of the base and the height of the cylinder.The surface area is the sum of the lateral area and the areas of the two bases.

The lateral area of a regular pyramid is half the product of the perimeter of the base and the slant height.The surface area is the sum of the lateral area and the area of the base.

The lateral area of a right cone is half of the product of the circumference of the base and the slant height.The surface area is the sum of the lateral area and the area of the base.

Also,the problem from the magazine says you need a calculator to find the lateral aea and surface area of each figure,the figures are a cone,a pyramid an upside down cone,And another pyramid that is 10 meters to 16 meters long.

The next page says the volume of a space figure is the space that the figure occupies.Volume is measured in cubic units.

The volume of a prism is the product of the area of a base and the height of the prism. The volume of a cylinder is the product of the area of a base and the height of the cylinder.

Find the volume of each figure.You may leave answers in terms of pie.[I APOLGIZE,I DON`T HAVE MATHEMATICAL SYMBOLS ON MY COMPUTER.]

Number 14 is a box,number 15 is a cylinder,number 16 is a long box,number 17 is another cylinder lined across.

I have to compare the finding of the volume and the surface area of the prism.What are,the similarites and differences.

The volume of a pyramid is one third the product of the areaof the base and the height of the pyramid.

The volume of a cone is one third the product of the area of the base and the height of the cone.

I also,have to find the volume of a cone two pyramids 7ft to 8 ft wide,2m to 3m and one last one 8cm to 12 cm.

I also have to find the surface area and volume of a sphere with the given radius or diameter.Round answers to the nearest tenth.

A composite space figure combines two or more space figures.The volume of a composite figure is the sum of the volumes of the combined figures.

Find the volume of each figure.When an answer is not a whole number,round to the nearest tenth.

Also,in the magazine I have darts that are thrown at random at each of the boards shown.If a dart hits the board,find the probability that It will land in the shaded area.I have a pyramid,in which the left side is have shaded,a square in which two halves are shaded,and a circle in which there is a shaded triangle. —Preceding unsigned comment added by 68.161.74.60 (talk) 01:43, 8 January 2008 (UTC)

1. Your question will be easier to read if you divide it to paragraphs (using an extra blank line).
2. If you can edit Wikipedia with your computer, you can write mathematical symbols with it. Writing &pi; will give π; Writing <math>\pi</math> will give $\pi$; and below the edit box, there should be a collection of symbols, one of them is the greek letter π (and anyway, it's Pi, not pie).
3. There is no such thing as the "lateral area of a base", hence the lateral area of a right prism is not "the sum of the lateral areas of the two bases". Rather, it is the product of the perimeter of a base and the height.

-- Meni Rosenfeld (talk) 08:40, 8 January 2008 (UTC)

Exciting algebra magazine PMajer (talk) 17:02, 8 January 2008 (UTC)

• Tried to format the question with additional line breaks. --CiaPan (talk) 12:32, 10 January 2008 (UTC)

Numerals and numbers

what is the difference betweenthem?Invisiblebug590 (talk) 09:07, 8 January 2008 (UTC)

Both "numeral" and "number" can mean different things, so this is not easy to answer. Essentially, a "number" is an abstract entity used to describe some quantity, while "numeral" is a written symbol used to refer to numbers. Usually "numeral" just means "digit", such as "1", "3" or "7". More generally, the string of symbols "38" or words "thirty-eight" might also be called numerals. The number is the abstract concept referred to by those symbols. -- Meni Rosenfeld (talk) 09:21, 8 January 2008 (UTC)

To further sharpen this: "38", "thiry-eight", "XXXVIII" and "٣٨" are different numerals. But they all describe the same number. -- Meni Rosenfeld (talk) 13:59, 8 January 2008 (UTC)

See also numeral and number. PrimeHunter (talk) 15:29, 8 January 2008 (UTC)

A problem from Diophantus: Missing a step in understanding the solution

[This is part of Problem 1.2 in Invitation to the Mathematics of Fermat-Wiles] Find two (rational) numbers such that the square of each of them, augmented by the sum of these two numbers, forms a square.

Diophantus provided this solution:

We assume that the smallest number is x and the largest x+1, so that x²+(2x+1)=₳. But we also need (x+1)²+(2x+1)=₳. We set.

$x^{2}+4x+2=(x-2)^{2}$

and find (1/4, 5/4).

Where I'm stumped is where does $x^{2}+4x+2=(x-2)^{2}$ come from? The left hand side is the expansion of (x+1)²+(2x+1), but why is the right hand side $(x-2)^{2}$? 68.183.18.54 (talk) 17:53, 8 January 2008 (UTC)

I don't think it comes from anywhere. We just want $x^{2}+4x+2$ to be a square, so we arbitrarily choose it to be $(x-2)^{2}$. We could just as well take $(x-3)^{2}$ and get (0.7, 1.7). The choice of the numbers being x and $x+1$ is likewise arbitrary. -- Meni Rosenfeld (talk) 19:01, 8 January 2008 (UTC)

I know it's not really relevant, but after digging into the Unicode charts (since my browser won't display it), I'd just like ask why you're using the symbol for a former currency of Argentina in your equation? —Ilmari Karonen (talk) 21:31, 8 January 2008 (UTC)

Because it looks like a square. 68.183.18.54 (talk) 21:57, 8 January 2008 (UTC)

Not quite, your browser just displays a square when it doesn't recognize the character. If you don't have the proper fonts installed this will happen with many characters.

You can display a square in LaTeX using $\Box$. -- Meni Rosenfeld (talk) 22:18, 8 January 2008 (UTC)

Or in Unicode text using □ (U+25A1, WHITE SQUARE). —Ilmari Karonen (talk) 22:46, 8 January 2008 (UTC)

Ah, I couldn't remember \Box, which is the bigger part of why I picked that character. I hadn't realized that I was grabbing an inappropriate symbol out of the symbols list... oops. 68.183.18.54 (talk) 00:15, 9 January 2008 (UTC)

Thanks for this comment -- I was thinking that my browser was messed up, because it was displaying currency symbols where I knew they shouldn't be... =) Tesseran (talk) 03:18, 10 January 2008 (UTC)

January 9

fuzzy logic 2

fuzzy logic in washing machines09:38, 9 January 2008 (UTC)

See Fuzzy Logic. -- Meni Rosenfeld (talk) 09:42, 9 January 2008 (UTC)

And don't forget to clean the filter afterwards.  --Lambiam 21:36, 9 January 2008 (UTC)

While you guys are making fun, the question does have basis in reality, even if the questioner has mistaken the Ref Desk for a search engine. In fact, it's not even necessary to go off Wikipedia, the fuzzy logic article states "Fuzzy logic can be used to control household appliances such as washing machines (which sense load size and detergent concentration and adjust their wash cycles accordingly)". --LarryMac | Talk 21:44, 9 January 2008 (UTC)

Links to fuzzy logic didn't seem to help with the OP's previous question on the issue. -- Meni Rosenfeld (talk) 08:42, 10 January 2008 (UTC)

How to write and chart indifference curves?

I'm currently working on the article on AD-AS model. I'd like to create a chart drawing the model. It's very easy to draw, but I want it to be drawn to scale. What programs are good for drawing indifference curves used in Economics? I only have an introductory education in economics combined with my own independent research. How indifference curves are used isn't covered until the intermediate level.

For examples of the kinds of curves I'm talking about, search for "AD-AS model" on Google images or see Supply and demand, Indifference curve, and Keynesian cross. I've never taken a course on Calculus and it's been a long time since I took Pre-Calc. I think I can figure out indifference curves if someone could post the basic mathematical functions representing Supply and demand. If you could post that function on Supply and demand, it would also be very useful. Zenwhat (talk) 10:06, 9 January 2008 (UTC)

Supply and demand depend on enormously many factors in complicated ways. There is no simple "one-size-fits-all" function which can accurately describe this phenomenon in generality. Of course, you can try using some approximation - for demand, I think an exponential function, $D=\alpha e^{-\beta P}\;\!$, can give a reasonable fit. For supply, I have no idea.

I don't know about drawing programs, but some here do so you will eventually get an answer. -- Meni Rosenfeld (talk) 10:35, 9 January 2008 (UTC)

It seems possible to use one of MATLAB compatible programs to get the graph (for example, GNU Octave), or some sort of spreadsheet... --Martynas Patasius (talk) 11:14, 9 January 2008 (UTC)

I had attempted before to use exponential functions to draw supply and demand using a basic spreadsheet. The supply curve can be drawn through a simple exponential function (y=x^2). I was going absolutely nuts, though, trying to figure out how to draw the demand curve. I'm going to post this same question in humanities, to see if there are any economists there that can answer. Zenwhat (talk) 14:45, 9 January 2008 (UTC)

$y=x^{2}\;\!$ isn't exponential, it's quadratic. It's trivial to find a function that will look like the one on those graphs, the hard part is to find one which will actually reflect reality in some way. Did you understand what I suggested for the demand? -- Meni Rosenfeld (talk) 14:54, 9 January 2008 (UTC)

Of course, if all you want is something that just looks nice, use something like $y=a(x-b)^{2}+c$, where $b\leq 0$ for supply and b is large for demand. -- Meni Rosenfeld (talk) 14:59, 9 January 2008 (UTC)

I feel a bit confused about your question. Indifference curve is a microeconomic term. Though you could find easily the answer to how to draw them, I'm not sure if that's what you want, since you are apparently working on Macroeconomic stuff. Meni has given you some answers for drawing something that could look like aggregate demands or supplies. That's basically all you need to know, since no one really cares to calculate such curves, they are basically an abstraction to understand macroeconomic equilibrium.

Just for the record, drawing indifference curves is easy: take the Cobb-Douglas case:

$u(x,y)=x^{\alpha }y^{1-\alpha }.$

Then, for utility level U, the indifference curve is given by

$y=(Ux^{-\alpha })^{\frac {1}{1-\alpha }}.$

For drawing such curves, I use a CAS [Mathematica], but I remember struggling a bit with spreadsheets before, and it works. Pallida  Mors 17:04, 9 January 2008 (UTC)

Oh, a point I forgot: you may want to get an inverse expression: Although we economists rotate ordinate/abscissa axes, poor programs aren't economists. And let me give you an additional hint on formulations: Take

$P=AX^{B}+C.$

Just play with those coefficients until you find a nice curve. For demand, A and B must have different signs. Together with A, B determines convexity. Pallida  Mors 17:20, 9 January 2008 (UTC)

Pallida, indifference curve is a microeconomic theory, but they're used extensively in Macroeconomics. See New Keynesian economics and Mainstream economics (Keynesianism built on microeconomic models). Hence, I slapped a factual inaccuracy tag on Indifference curve. Also, I think I misspoke when I called the Keynesian cross an indifference curve. That's all correct, right? Isn't Thanks for your help! Zenwhat (talk) 22:44, 9 January 2008 (UTC)

Ehm... I haven't really see indifference curves in "new" macroeconomic texts. I have seen of course other microeconomic topics, like utility maximization, representative-agent models, etc. It's alright if you're interested in indifference curves. Bear in mind though that they don't represent utility functions by themselves, they just represent a contour, or level set, of the utility function. Thus, Keynesian cross is not related to such curves. Pallida  Mors 16:34, 10 January 2008 (UTC)

By the way, it looks like the required images have already been created for Lithuanian Wikipedia (lt:AS-AD modelis, Image:IS LM model.svg). Probably you would only have to translate them into English (or ask the author to do so). --Martynas Patasius (talk) 01:25, 10 January 2008 (UTC)

Those aren't the same, I don't think. Anyone, feel free to correct me if I'm wrong here (I'll check my textbook), but the basic model of AD/AS looks just like the basic model of Supply and demand:

In the Keynesian model (probably the one he used in his General Theory), the demand curve is logarithmic rather than linear, because it's almost flat on the leftward-half of the graph, then shoots back up on the rightward half, so it looks more like this. I really ought to get a textbook on Intermediate economics. Zenwhat (talk) 01:49, 10 January 2008 (UTC)

Well, I have always see those curves as convex functions, but I can't recall any reason for it besides a graphical tradition. Anyway, Monetarists and Keynesians would argue a lifetime about the form of the aggregate supply, so I wouldn't worry too much about a specific formulation. For most people, they represent, as I said above, an abstraction. That said, there exist specialists dedicated to calibrate and estimate IS-LM and AD-AS models. Pallida  Mors 16:34, 10 January 2008 (UTC)

Non-recurring and non-terminating numbers(irrational numbers)

How do we prove or see that a number such as pi is non-recurring and non-terminating? What is the formula for it? On another note, when were the Roman, Mayan and Greek numerals first used/invented/discovered? How were they discovered? DISCLAIMER: this is not homework nor any competition question. It is only for my own reference and reading but I could not find the answer. I would appreciate if anyone can help me. —Preceding unsigned comment added by Invisiblebug590 (talk • contribs) 12:48, 9 January 2008 (UTC)

See A simple proof that π is irrational, Roman numerals, Maya numerals and Greek numerals. -- Meni Rosenfeld (talk) 13:02, 9 January 2008 (UTC)

I saw a proof of e's irrationality a while back - it wasn't particularly nice. mattbuck (talk) 13:12, 9 January 2008 (UTC)

Simple proof that e is irrational - Suppose e is rational. Then e = a/b for some integers a and b. So b!e = a(b-1)! is an integer. But

$b!e-\sum _{k=0}^{b}{\frac {b!}{k!}}=\sum _{k=b+1}^{\infty }{\frac {b!}{k!}}<\sum _{k=1}^{\infty }{\frac {1}{(b+1)^{k}}}={\frac {1}{b}}$

$\Rightarrow \sum _{k=0}^{b}{\frac {b!}{k!}}<b!e<\sum _{k=0}^{b}{\frac {b!}{k!}}+{\frac {1}{b}}$

Therefore b!e cannot be an integer, so e cannot be rational. I think that proving that e is transcendental is much more difficult. Gandalf61 (talk) 15:18, 9 January 2008 (UTC)

I don't think much is known about how some of the old numeral systems were invented (I call it invented and not discovered). We have articles about Roman numerals, Maya numerals, Greek numerals. PrimeHunter (talk) 16:54, 9 January 2008 (UTC)

We have articles on Proof that e is irrational and Proof that π is irrational. There is a chapter on irrational numbers in Hardy & Wright's Introduction to the Theory of Numbers, which includes proofs that general n^th roots; e and its rational powers; and π are irrational. It also says that e^π is known to be irrational. Interestingly, it is still not known whether π^e is irrational. Gandalf61 (talk) 14:38, 10 January 2008 (UTC)

Second-order nonlinear ODE

Does the differential equation

${\frac {d^{2}x}{dt^{2}}}=-\sin x$

have an analytic solution? It arises in the equation of motion of a rigid pendulum, such as a pendulum ride at an amusement park.

Also, some of my friends insist on rewriting the equation this way:

$-{\frac {1}{\sin x}}d^{2}x=dt^{2}$

and then trying to integrate twice or something. I tried to tell them that this is a meaningless manipulation of symbols, but they still think this approach might be fruitful. How can I convince them that this will never work? —Keenan Pepper 16:41, 9 January 2008 (UTC)

Of course the equation has an analytic solution, and you can find the coefficients of its Taylor expansion using any of your favorite methods. To the best of my knowledge, it is not elementary, though.

I don't think this string of symbols is any less meaningful than, say, $y\ dy=dx\;\!$. The only difference is that the latter allows a solution via a simple transformation, and the former does not. The only way to convince them that this is futile is to let them try and fail (and point out any specific errors in their attempts). -- Meni Rosenfeld (talk) 17:14, 9 January 2008 (UTC)

Also, check out Pendulum (mathematics) and Jacobi's elliptic functions. The solution, for $x(0)=0,x'(0)=1$ is given in Mathematica notation as 2JacobiAmplitude[t/2,4], but I don't know how that translates to traditional notation. -- Meni Rosenfeld (talk) 17:29, 9 January 2008 (UTC)

Mathematica's TraditionalForm and TeXForm give me essentially $2\,\operatorname {am} \!\!\left(\left.{\frac {t}{2}}\right|4\right)$. --Tardis (talk) 17:34, 9 January 2008 (UTC)

I meant analytic in the sense of [1]. It's enough to know that it's a Jacobi elliptic function. What my friends were doing couldn't possibly produce that, so it's obviously wrong. =) Thanks! —Keenan Pepper 20:56, 9 January 2008 (UTC)

I figured this was the intention, but it's just been so long since I've last encountered "analytic" used this way. -- Meni Rosenfeld (talk) 21:01, 9 January 2008 (UTC)

[edit conflict] See pendulum (mathematics): no. As far as integrating, one can rewrite the equation as ${\frac {dv}{\sin x}}=-\,dt$, then express x (which is really an angle) in terms of v via the conservation of energy argument outlined in the main pendulum article and integrate. However, the expression is $\sin x={\sqrt {1-\left({\frac {rv^{2}}{2g}}+\cos \theta _{0}\right)^{2}}}$ where r is the length of the pendulum, g is of course gravity, and $\theta _{0}$ is its maximum displacement, so the integration over v (which is really an angular velocity) might be a bit interesting. As for explaining to your friends, you might point out that ${\frac {d^{2}x}{dt^{2}}}={\frac {d(dx/dt)}{dt}}$, and so you can't actually get the two $dt$s out: the original notation is a confusion of $\left({\frac {d}{dt}}\right)^{2}$, where multiplication is used to represent operation in the usual fashion. --Tardis (talk) 17:34, 9 January 2008 (UTC)

I disagree with the assertion that the notation ${\frac {d^{2}x}{dt^{2}}}$ is a confusion of $\left({\frac {d}{dt}}\right)^{2}x$. The Leibniz notation can be given meaning for higher orders just like for first order. We are comfortable with dt being an infinitesimal change in t, and $dx=x(t+dt)-x(t)$ being the corresponding change in x. $dt^{2}$ is just that - the product of dt with itself. Then $d^{2}x=d(dx)$ (d itself is an operator, thus in its context multiplication means application, or composition) is the infinitesimal change in dx, $d^{2}x=dx(t+dt)-dx(t)=(x(t+2dt)-x(t+dt))-(x(t+dt)-x(t))=x(t+2dt)-2x(t+dt)+x(t)\;\!$. You can then plainly see that ${\frac {d^{2}x}{dt^{2}}}$ indeed gives the second derivative. -- Meni Rosenfeld (talk) 18:13, 9 January 2008 (UTC)

Hm, you're right. I was too focused on finding a "square root" of the whole object; it didn't occur to me that one of the ²s might be a true squaring and the other an operator power. Thanks! --Tardis (talk) 21:12, 9 January 2008 (UTC)

Transform the second order equation into two first order equations by introducing the velocity like this

${\frac {dx}{dt}}=v$

${\frac {dv}{dt}}=-\sin x$

Note that the energy

$E={\frac {1}{2}}v^{2}-\cos x$

is conserved:

$dE=v\cdot dv+\sin x\cdot dx=v\cdot (-\sin x\cdot dt)+\sin x\cdot (v\cdot dt)=0\cdot dt=0$

This gives the following relationship between velocity and displacement

${\frac {dx}{dt}}=v=(2\cdot E+2\cdot \cos x)^{\frac {1}{2}}$

and the following relationship between time and displacement

$t=\int (2\cdot E+2\cdot \cos x)^{-{\frac {1}{2}}}dx$

Bo Jacoby (talk) 22:26, 9 January 2008 (UTC)

Cutting Sandwiches

In one of my many physics-induced daydreams i was thinking about cutting sandwiches. Specifically, can you cut a standard square sandwich twice (that is 2 cuts with a knife) and end up with more than 4 bits? Also noticed that something like a zig-zag sandwich _could_ be cut into more than 4 bits, so is there a rule or proof surrounding the number of bits you can cut sandwiches into? One of the more 'interesting' questions on here i expect.. ;) Benbread (talk) 20:21, 9 January 2008 (UTC)

I'll assume that the knife cuts can be modelled as straight lines. The relevant feature of a square is that it is convex. Thus, after one cut, each side of the cut is the intersection of convex sets (a square and a half-plane) and is thus itself convex; in particular, it means that it is connected, thus you are only left with 2 pieces. Cutting again is equivalent to cutting each piece separately, and since each of those is convex, once again you get only 2 pieces from each. So no, for a convex shape, like a square, you can only get up to $2^{n}$ pieces with n cuts. For non-convex shapes it could get more complicated. -- Meni Rosenfeld (talk) 20:47, 9 January 2008 (UTC)

This had me thinking. For any sufficiently sane set, you can look at its convex hull minus the set itself, and count how many pieces this is made of. This gives us a "non-convexity index", being 0 for convex sets, 1 for simply nonconvex ones, and so on. Has anyone heard of such a thing before? I'm pretty sure the original question can be answered in terms of this non-convexity index of the sandwich. -- Meni Rosenfeld (talk) 20:54, 9 January 2008 (UTC)

I'm sure this is true, provided that you define "sufficiently sane" as meaning: the cuttability index can be expressed in terms of the non-convexity index. Otherwise, consider a convex-shaped sandwich from which a fractal bite has been taken by a hungry Koch snowflake in such a way that the non-convexity index equals one. (If necessary we patch up the sandwich a bit at the intersection of the bite and the pre-bite boundary of the sandwich.) By just a single cut in the right way along some edge of one of the straight-edged finite iterations converging to the fractal, we can get infinitely many bits.  --Lambiam 21:34, 9 January 2008 (UTC)

Yep, that's how I define it! :-) Seriously though, I do not understand your construction. A koch-bitten sandwich obviously has an infinite non-convexity index, and I don't see how you intend to patch it into an index of 1 while maintaining its fractal nature. -- Meni Rosenfeld (talk) 22:45, 9 January 2008 (UTC)

As a simpler illustration of Lambiam's point, consider a set in the shape of the letter E (with the middle arm shorter than the other two): it has a non-convexity index of one but can be cut into four pieces with a single cut. Conversely, consider a set in the shape of a gear wheel with 100 short teeth: it has a non-convexity index of 100, but cannot be cut into the expected 100+2 pieces, because no single line intersects more than a few teeth. --mglg_(talk) 23:01, 9 January 2008 (UTC)

Here's an alternative measure of concavity: Convex sets have concavity zero. If the convex hull of set S, minus S, has concavity n, then S has concavity n + 1. Otherwise S has infinite concavity. Then the letter E has concavity 2, because the convex hull minus the E is a backwards C shape that has concavity 1. The Koch snowflake has infinite concavity. Does that do the trick? —Keenan Pepper 00:33, 10 January 2008 (UTC)

I have the impression that for a shape whose boundary is a non-intersecting differentiable loop the concavity (which you could the wiggle count) is half the number of times the curvature around the loop crosses zero.  --Lambiam 01:56, 10 January 2008 (UTC)

The example of E certainly shows I was being a bit naive with my thinking (regardless of sandwiches, I would want it to have an index of 2, I think). I'll have to ponder some more whether Keenan's suggsestion reflects my intuition. -- Meni Rosenfeld (talk) 08:51, 10 January 2008 (UTC)

Thanks everyone for your responses they're humorous if anything else - I think yo've gone beyond my understanding but what i can pick out is very interesting indeed, thank you all :) -Benbread (talk) 08:56, 10 January 2008 (UTC)

The sandwich is supposed to stay fixed during the whole operation of k cuttings, right? If so, your question is equivalent to ask how many connected components are there, at most, in $R^{2}$minus k straight lines, and the answer is that if the lines are generic (=no two of them are parallel, no three of them pass for the same point), then the number of pieces is maximum and it is

${k \choose 0}+{k \choose 1}+{k \choose 2}={\frac {k^{2}}{2}}+{\frac {k}{2}}+1,$

more generally, the complement of the union of k affine hyperplanes in $R^{d}$ has at most exactly

${k \choose 0}+{k \choose 1}+...+{k \choose d}={\frac {k^{d}}{d!}}+O(k^{d-1})$

connected components, and the maximum number is reached by any generic choice of the hyperplanes. You can easily prove it by induction. For instance, you can cut a watermelon in $R^{3}$, with k=0,1,2,3,4 katana cuttings into respectively 1, 2, 4, 8, 15 pieces. Be quick. Notice that the sequence of powers of 2 stops as soon as you pass k=the dimension. I also remember a movie with Charles Bronson, where a bunch of pieces of watermelon were obtained by means of a gun machine. PMajer (talk) 12:38, 10 January 2008 (UTC)

Silly me, I also assumed the sandwich is fixed but didn't pay attention to $2^{n}$ not being tight. Anyway, I don't think it follows from the original question that it must be fixed, and if not, going all the way to $2^{n}$ is trivial. -- Meni Rosenfeld (talk) 12:44, 10 January 2008 (UTC)

Just to speak, a much harder question seems: how many ways are there to cut his sandwich with k cuts. Say that we consider equivalent cuttings with the same graph. I don't know the answer. With 4 cuts there is only one way, but with 5 it looks like a mess and I'm not even able to count them. Not to speak of higher dimensions...PMajer (talk) 13:11, 10 January 2008 (UTC)

I think you're missing the point here somewhat. Would a six-dimensional sandwich be doubly as delicious as a three-dimensional one? mattbuck (talk) 14:15, 10 January 2008 (UTC)

yes and even more, provided your tongue has five-dimensional boundary

I went ahead and thought through what happenes with 3 or more cuts, and in the case of 3 the maximum number of zones I can get is 7, (basically the three lines create a triangle inside the sandwitch) but in the case of 4 there begins to be inefficiencies, I don't believe it is possible for another line to cut all 7 zones created by the first 3 lines. A math-wiki (talk) 19:06, 10 January 2008 (UTC)

right, and you can easily prove this, because the fourth line meets at most once each of the preceding ones, therefore it is divided by them into 4 segments at most, and for each segment you get a new piece of sandwich. Thus 4 cuts can make at most 7+4=11 pieces. See the general formula above. —Preceding unsigned comment added by PMajer (talk • contribs) 19:48, 10 January 2008 (UTC)

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 , 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)

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)

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 daollars 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)

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)