Hi. On most graphing calculators there is a limit to the amount of numbers it can display before resorting to scientific notation. For example, on my TI-84 Plus, it can calculate 2³³ just fine, but then at 2³⁴, it displays the answer in scientific notation. My question is, is there a method to determine the maximum number the calculator can calculate before resorting to scientific notation? I presume it has to be a number under E10, but I'd like to know if there is a definite mathematical way to do this. Thanks. Vic93 (t/c) 00:00, 21 February 2009 (UTC)[reply]

Try reading the manual. Algebraist 00:01, 21 February 2009 (UTC)[reply]

It is entirely up to the people who designed the calculator. Usually things like that are constrained by how the number is stored in memory, but there is more than one way to keep track of that kind of number. Black Carrot (talk) 03:28, 21 February 2009 (UTC)[reply]

For an idea of how it might work, try skimming our article on floating-point numbers. Black Carrot (talk) 03:34, 21 February 2009 (UTC)[reply]

To experiment, I usually try entering something like (ten nines) 9999999999 + 1 and seeing what happens. The detailed answer is in the manual (bottom of PDF page 24, printed page 21).

Be aware there's a difference between how the calculator stores numbers and how it displays them. (See the manual PDF page 659, printed page 656.) Internally, the calculator is always calculating in scientific notation, using up to 14 digits in the significand and up to 2 digits in the exponent. --Bavi H (talk) 08:30, 22 February 2009 (UTC)[reply]

The maximum number displayed is 4. Above that it says a suffusion of yellow. --Trovatore (talk) 08:38, 22 February 2009 (UTC)[reply]

If it's true that God can't make two plus two equal five, and that everything that statement implies is true, could God still have made the value of pi less than 3.14159 or more than 3.1416, as easily as He might have made the gravitational constant less than 6.674 m³kg^-1s² or more than 6.675 m³kg^-1s² (ignoring the anthropic principle, i.e. assuming it weren't necessary for the modified universe to support sentient life)? NeonMerlin 08:58, 21 February 2009 (UTC)[reply]

This depends somewhat on what is meant by π. If π denotes the mathematical constant, then to change its value is just as hard as changing the value of 2+2, and so impossible according to mainstream christian theology. If, on the other hand, π is taken to mean something physical, say the ratio of circumference to diameter of physically existing circles, then an omnipotent power could change its value, or make it vary from place to place, and indeed (since space is not perfectly flat) this is believed to be in fact the case. Algebraist 09:54, 21 February 2009 (UTC)[reply]

Algebraist, is π a constant beyond its definition of the ration of the circumference to diameter of a circle? If it is not defined by that, then by what is it defined? If, say, an omnipotent power changed this ratio, would not then the value of π change with it? --Aseld ^talk 10:04, 21 February 2009 (UTC)[reply]

What Algebraist means is that for circles in Euclidean geometry, defined as the set of points in a certain plane having a certain distance (Euclidean norm) to a certain point in that plane, it follows from the axioms of Euclidean geometry that 3.14159 < π < 3.1416 (by Archimedes' approach but with more than 96 sides), just as it follows from the Peano axioms that 2 + 2 = 5. As these axioms don't have anything to do with the universe per se, an omnipotent God couldn't change them any more than he could change the rules of tic-tac-toe. But reality, the universe, which clearly is not exactly modelled on Euclidean geometry, is quite another matter. —JAO • T • C 10:22, 21 February 2009 (UTC)[reply]

I think you're using the wrong Peano axioms there. Algebraist 10:25, 21 February 2009 (UTC)[reply]

Now, how on Earth should I know what 2 + 2 is? —JAO • T • C 10:31, 21 February 2009 (UTC)[reply]

Not wrong, just different! --Tango (talk) 14:00, 21 February 2009 (UTC)[reply]

There's no geometry where circumference/diameter is some constant other than π. The circumference of a circle is 2π sin r in elliptical geometry and 2π sinh r in hyperbolic geometry, where π is the same constant that shows up in Euclidean geometry. There's no constant π' that would enable you to write those formulas as 2π'r. A factor of π shows up in the field equation of general relativity. π is genuinely universal, it's not at all limited to flat geometry. (Are there weirder geometries that could genuinely be said to have a "different π"? I don't know.) -- BenRG (talk) 15:08, 21 February 2009 (UTC)[reply]

The natural geometry on a cone would be interesting from that perspective. Away from the point at the top, it's flat, so circles that don't include that point and are small enough not to wrap around the cone would have the regular value of pi. Circles centred on that point would have a different ratio of circumference to diameter, but it would be constant. Circles centred elsewhere that either include the top point or wrap around the cone would do something weird... I would have to think about it. --Tango (talk) 15:14, 21 February 2009 (UTC)[reply]

You may be interested in a discussion that took place on the Science desk a few weeks ago: http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Archives/Science/2009_January_25#Cosmic_gods --Tango (talk) 14:00, 21 February 2009 (UTC)[reply]

The article Philosophy of mathematics might also be of interest, in particular the section on mathematical realism. 207.241.239.70 (talk) 06:07, 27 February 2009 (UTC)[reply]

Which of the following does the OP mean by "God can't make two plus two equal five"?

• There exists a mighty being who needs to be identified as "God". Any combination of inabilities and abilities can be proposed for such a being and none is verifiable, so

the only answer to the question is "perhaps". OR

• There is no God, hence no godly abilities exist. The mathematical definitions of addition and pi are understood and agreed upon by all reputable sources. Consensus wins and the answer is No. Cuddlyable3 (talk) 14:17, 27 February 2009 (UTC)[reply]

Hi there - I'm a little confused about a vector calculus question, or rather the hint they're giving me for the question, was wondering if I could grab a little help!

If ${\displaystyle \nabla \cdot {\textbf {J}}=0}$ in volume V and ${\displaystyle {\textbf {J}}\cdot {\textbf {n}}=0}$ on the surface S enclosing V, show that

${\displaystyle \int _{V}{\textbf {J}}dV=0}$. Hint: use ${\displaystyle {\frac {\partial {(x_{i}J_{j})}}{\partial {x_{j}}}}}$.

I'm really not sure where the hint is pointing me! Differentiating gives ${\displaystyle (\nabla \cdot J)x+(J\cdot \nabla )x=0+(J\cdot \nabla )x=(J\cdot \nabla )x}$ but what then?

Thanks a lot for any help, Spamalert101 (talk) 11:12, 21 February 2009 (UTC)spamalert[reply]

I reckon you were pretty close but I think we can take your identity a little further,

${\displaystyle {\frac {\partial {(x_{i}J_{j})}}{\partial {x_{j}}}}=(J\cdot \nabla )x_{i}=\sum _{j}J_{j}{\frac {\partial {x_{i}}}{\partial {x_{j}}}}=J_{i}}$

This lets us express their integral in terms of the derivative they suggested.

${\displaystyle \int _{V}J_{i}dV=\int _{V}\nabla \cdot (x_{i}{\textbf {J}})dV}$

Do you see how to finish the proof now? Vespertine1215 (talk) 16:43, 21 February 2009 (UTC)[reply]

I just need to know what this question wants me to do

Find total no. of distinct ways the product " XY³Z³ " can be written without using exponent.

I haven't understood what the sum means so i'd be glad if anyone can explain what this question means. Vineeth h (talk) 13:27, 21 February 2009 (UTC)[reply]

Does it mean how many distinct ways are there to order an X, three Y's and three Z's? (eg. XYYYZZZ, XYYZZZY, XYZZZYY, etc.) --Tango (talk) 14:04, 21 February 2009 (UTC)[reply]

Well Tango according to that the answer should come 140 but it isn't 140!!! So i guess thats not it!

And what is the given answer then? 18? --pma (talk) 16:37, 21 February 2009 (UTC)[reply]

Well, you have 2XYYYZZZ / 2, 3XYYYZZZ / 3 etc. so you could argue there are a countably infinite number of ways of writing this product with using an exponent. Pretty meaningless question without clearer constraints on which expressions are allowed. Gandalf61 (talk) 18:11, 21 February 2009 (UTC)[reply]

Even if you take "product" as meaning you can only multiply, not divide, you still have things like (-X)(-Y)YYZZZ. That keeps it finite, but still quite a bit bigger - 140*(7*6+7*6*5*4+7*6*5*4*3*2)=829080. --Tango (talk) 18:36, 21 February 2009 (UTC)[reply]

If I've got a surface defined as a function of R2, and several level curves of that surface, is there a computationally simple way to approximately recreate the original surface from its level sets? I'm trying to find a way to turn a cell-shading of a drawing into a smooth gradient. Black Carrot (talk) 16:55, 21 February 2009 (UTC)[reply]

Presumably you mean a function "on", rather than "of", R2, and you mean R². If ƒ is any real-valued function on R² and g is any strictly increasing function from R to R, then ƒ and the composition g o ƒ will both have exactly the same level sets. Thus many different functions can share the same level sets. Michael Hardy (talk) 20:51, 21 February 2009 (UTC)[reply]

I think this a classic computer graphics problem and there is extensive literature on such matters. I'm guessing that do don't actually have the level sets in their entirity, just some points lying on them. You will probably want some sort of surface fitting algoritm, say fitting a Bézier surface to the data. Inverse distance weighting seems to be one technique. Kriging is one well known method used in geostatistics with a strong statistical basis. --Salix (talk): 23:50, 21 February 2009 (UTC)[reply]

That's a good start. I just tried out inverse distance weighting. It hasn't quite worked yet, but I think I can fix it. The primary difficulty is that it's designed to interpolate a small, sparse set, whereas I'd like to interpolate hundreds of pixels along a sequence of curves. Some of these curves are different lengths (imagine two concentric circles) so the number of pixels along them are different, and that throws it off a bit. BTW, is it just me or does that formula look a lot like gravity when p=2? Black Carrot (talk) 04:23, 22 February 2009 (UTC)[reply]

Try googling for "surface reconstruction contour" there are many methods for this, I don't really know enough on the subject to advise on the best. Although there does seem some interesting connections with Voronoi diagrams. --Salix (talk): 08:19, 22 February 2009 (UTC)[reply]

On the Wikipedia entry of Even and odd functions, even function can be written as ${\displaystyle f(x)={\frac {1}{2}}[f(x)+f(-x)]\,}$ and odd function can be writen as ${\displaystyle f(x)={\frac {1}{2}}[f(x)-f(-x)]\,}$. How are these formulas dervived in the first place from the definiton of odd and even functions? 142.244.175.223 (talk) 00:14, 22 February 2009 (UTC)[reply]

Just apply the definitions - replace f(-x) by the appropriate thing and simplify, you'll find you end up with just f(x) afterwards. --Tango (talk) 01:11, 22 February 2009 (UTC)[reply]

Please: Don't use the same notation—the letter ƒ—for two different functions. What the article says is that

${\displaystyle f_{\text{even}}(x)=1/2[f(x)+f(-x)]\,}$

where ƒ is one function and ƒ_even is another (and similarly for odd functions). The function called ƒ is in general neither even nor odd. Michael Hardy (talk) 05:28, 22 February 2009 (UTC)[reply]

Yes, but I think in the OP they where not different functions... The question was: how is that f even implies f=f_even and f odd implies f=f_odd, and this is explained in Tango's answer. However I think that the OP also means: how are the decomposition formulas derived. An answer is: you want to write a function f(x) as a sum of an even function e(x) plus an odd function o(x):

f(x)=e(x)+o(x).

It turns out that there is exactly one choice for the pair e(x), o(x), because from the definiton of odd and even functions you must also have, for all x

f(-x)=e(-x)+o(-x)=e(x)-o(x)

and from the system of the two you get e(x) and o(x) as in the decomposition formulas. --84.221.198.10 (talk) 09:24, 22 February 2009 (UTC)[reply]

So let me get this straight: for even functions we have ${\displaystyle f(x)=f(-x)\,}$ and we can decompose this into ${\displaystyle f(x)=1/2f(-x)+1/2f(-x)\,}$ and replacing f(x) with f(-x) we can get since f(x) is an even function ${\displaystyle f(x)=1/2f(x)+1/2f(-x)\,}$, is this logic even right? 72.53.7.177 (talk) 09:25, 22 February 2009 (UTC)[reply]

Unimpeachable. But as Michael Hardy remarks the point is to write in the form "even + odd" functions that are in general neither even nor odd. Example: f(x)=e^x decomposes into the sum of cosh(x) plus sinh(x).--84.221.198.10 (talk) 09:45, 22 February 2009 (UTC)[reply]

Indeed. Continuing from where Michael started, any function f(x) can be used to construct an even function

${\displaystyle f_{\text{even}}(x)={\frac {f(x)+f(-x)}{2}}\,}$

and an odd function

${\displaystyle f_{\text{odd}}(x)={\frac {f(x)-f(-x)}{2}}\,}$

such that

${\displaystyle f(x)=f_{\text{even}}(x)+f_{\text{odd}}(x)}$

If f(x) is already an even function then f(x)=f(−x), so we have

${\displaystyle f_{\text{even}}(x)={\frac {f(x)+f(-x)}{2}}={\frac {f(x)+f(x)}{2}}=f(x)}$

${\displaystyle f_{\text{odd}}(x)={\frac {f(x)-f(-x)}{2}}={\frac {f(x)-f(x)}{2}}=0}$

Gandalf61 (talk) 11:17, 22 February 2009 (UTC)[reply]

an odd fuction plus an odd function is odd... That's odd, isn't it? Well, so it's not odd that it's odd. That's odd, isn't it?(...) pma (talk) 13:36, 22 February 2009 (UTC)[reply]

Can Graham's number be expressed in terms of an output of the Ackermann function with manageable inputs? NeonMerlin 07:08, 22 February 2009 (UTC)[reply]

It seems unlikely, since one can be defined in terms of powers of 3, and the other one in terms of powers of 2. Ctourneur (talk) 20:02, 22 February 2009 (UTC)[reply]

A more reasonable question would be to try to find k such that A(k) ≤ G ≤ A(k+1). There is (as far as I know) no reason to expect one of these to be an equality. As for what the k is, I have no idea. Staecker (talk) 22:19, 22 February 2009 (UTC)[reply]

The value of k there is very, very large: it is certainly much, much larger than 3 -> 3 -> 27 (see Conway chained arrow notation). It is in fact bigger than 3 -> 3 -> (3 -> 3 -> (3 -> 3 -> ( ... (3 -> 3 -> 27) ... ))), where there are 60 sets of parentheses. Eric. 131.215.158.184 (talk) 06:01, 23 February 2009 (UTC)[reply]

Is there a closed form with a fixed number of terms for the sequence defined by t₀ = 1 and t_n = t_{n - 1} + 2^{t_{n - 1}}? In case anyone's wondering, it originates here. NeonMerlin 08:11, 22 February 2009 (UTC)[reply]

I doubt there's anything useful. There's nothing in the Sloane's entry. Algebraist 09:40, 22 February 2009 (UTC)[reply]

Please can anybody explain very popularly in simple way the technique of fast fourier transform?

Try Fast fourier transform? By popular, do you mean simple and feel that article is too complex? Basically the transform analyses the data to determine the amplitude and phase of signals of various frequencies, which information has many uses and for which it is possible to inverse transform back to the original data. -- SGBailey (talk) 07:20, 23 February 2009 (UTC)[reply]

Hello. How do I get long division symbols and synthetic division symbols on Office Word 2007? I could not find them in Equations Editor. Thanks in advance. --Mayfare (talk) 22:11, 22 February 2009 (UTC)[reply]

Regarding long divisions, it is apparently possible with the use of the old friend Math Editor (available through the "Insert Object" feature) (See this online discussion). I couldn't find the appropriate symbol in the new "integrated" editor. Could that be possible? :S
As for the synthetic division, if this is what you mean, then I guess you can implement it building a table and inserting the proper symbols in each cell of the table. Pallida  Mors 14:27, 23 February 2009 (UTC)[reply]

They might not exist. In particular, MS Word 2007 appears to use the encoding described in UTN #28 ([[1]]) which mentions on page 14 that an encoding for the long division enclosure is not chosen yet. Note that this is as of nearly 3 years ago though.GromXXVII (talk) 20:52, 23 February 2009 (UTC)[reply]

Hi there - I'm trying to finish off a question and seem to be missing something, I don't think it's particularly hard, I think I'm just being stupid and missing something obvious (hopefully);

having shown that for u(x) the unique solution to Laplace's equation in the volume V with u=f(x)on the surface S enclosing V, and v any function with continuous 1st partial deriv.s in V which =0 on S, we have ${\displaystyle \int _{V}\nabla v\cdot \nabla udV=0}$, I want to show for any w with continuous 1st partial deriv.s in V with w=f on S,

${\displaystyle \int _{V}|\nabla w|^{2}dV\geq \int _{V}|\nabla u|^{2}dV}$. I've tried subbing in v=w-u into the original equation to get ${\displaystyle \int _{V}\nabla w\cdot \nabla udV=\int _{V}|\nabla u|^{2}dV}$, which is sortof halfway there, but I'm not sure where to go next. If anyone could point out where I was being stupid... thanks! Otherlobby17 (talk) 12:53, 23 February 2009 (UTC)Otherlobby17[reply]

I can't see a way to massage your last equation into the desired result but with the only justification that it leads to the answer, I'd start from

${\displaystyle w({\vec {x}})=u({\vec {x}})+v({\vec {x}})}$

Then by taking gradients,

${\displaystyle \nabla w=\nabla u+\nabla v}$

Squaring,

${\displaystyle |\nabla w|^{2}=|\nabla u|^{2}+|\nabla v|^{2}+2\nabla u\cdot \nabla v}$

And finally integrating over the volume, we get

${\displaystyle \int _{V}|\nabla w|^{2}\mathrm {d} V=\int _{V}|\nabla u|^{2}\mathrm {d} V+\int _{V}|\nabla v|^{2}\mathrm {d} V+2\int _{V}\nabla u\cdot \nabla v\mathrm {d} V}$

Now, by the equation you have already proved the last term vanishes and as ${\displaystyle |\nabla v|^{2}}$ is strictly positive your inequality follows. Vespertine1215 (talk) 17:29, 23 February 2009 (UTC)[reply]

Brilliant, thanks! Otherlobby17 (talk) 17:41, 23 February 2009 (UTC)Otherlobby17[reply]

Talking about inner product spaces, is it clear the relation between "minimizing the distance from a given subspace" and "orthogonality"? It's just this. --pma (talk) 21:09, 23 February 2009 (UTC)[reply]

I suspected the two may have been linked ;) A further part of the question asks me to relate the 'capacity' of an object ${\displaystyle -\int _{S}{\frac {\partial \phi }{\partial n}}dA}$, where the potential φ(x) satisfies Laplace’s equation in the volume outside the object, ${\displaystyle \phi =1}$ on S and ${\displaystyle \phi \to 0}$ at ${\displaystyle \infty }$, before showing the capacity of a cube is bounded by ${\displaystyle 2\pi a<C<2{\sqrt {3}}\pi a}$ for cube of sides a (using the in/circumcircle). Are these 2 related in a similar way too then - capacity and minimizing this distance? Thanks for your time! Otherlobby17 (talk) 7:31, 24 February 2009 (UTC)Otherlobby17

Well, as you probably know, you also have, for your potential ${\displaystyle \phi }$ (by Green identities)

${\displaystyle -\int _{S}{\frac {\partial \phi }{\partial n}}dA=\int _{\mathbb {R} ^{3}\setminus V}|\nabla \phi |^{2}\mathrm {d} x}$,

and in fact you can define the capacity of the body V as the infimum of the energy ${\displaystyle \scriptstyle \int _{\mathbb {R} ^{3}\setminus V}|\nabla u|^{2}\mathrm {d} x}$ over all smooth function ${\displaystyle u}$ taking the value 1 on the bundary of ${\displaystyle V}$ and 0 at infinity. So an upper bound for the capacity is just the energy of any of these functions, or more generally, the infimum of the energy taken in a suitable smaller class of functions (is it clear? Infimum over a smaller class gives a larger value). Now, at least in the case of a convex body ${\displaystyle Q}$, a reasonable class is: functions of the distance from ${\displaystyle Q}$, that is, functions of the form ${\displaystyle \scriptstyle u(x)=f(\mathrm {dist} (x,Q))}$, with a smooth ${\displaystyle \scriptstyle f:[0,\infty ]\to \mathbb {R} }$. For a cube ${\displaystyle \scriptstyle V=[0,a]^{3}}$ this makes particularly simple the expression of the energy of ${\displaystyle u}$. The domain of integration ${\displaystyle \scriptstyle R^{3}\setminus V}$ can be partitioned into three subdomains, corresponding respectively to all points projecting on faces, edges, vertices: you should find:

${\displaystyle \int _{\mathbb {R} ^{3}\setminus V}|\nabla u|^{2}\mathrm {d} x=6a^{2}\int _{0}^{\infty }f'(t)^{2}dt+6\pi a\int _{0}^{\infty }tf'(t)^{2}dt+4\pi \int _{0}^{\infty }t^{2}f'(t)^{2}dt=}$

${\displaystyle \int _{0}^{\infty }(6a^{2}+6\pi at+4\pi t^{2})f'(t)^{2}dt}$.

Now you should check this computation, and then minimize over all ${\displaystyle f}$ with ${\displaystyle f(0)=1,f(x)=o(1)}$ for ${\displaystyle \scriptstyle x\to \infty }$ (try it, it is worth; in case ask for details). As to the lower bound, there is a non trivial fact: an Euclidean ball minimizes the capacity among all bodies with the same volume. Thus a lower bound for the capacity of a cube is the capacity of a ball with same volume.--pma (talk) 17:03, 24 February 2009 (UTC)[reply]

Note: concerning the minimum problem

${\displaystyle \textstyle \min \int _{0}^{\infty }(6a^{2}+6\pi at+4\pi t^{2})f'(t)^{2}dt}$

over all smooth functions with ${\displaystyle \scriptstyle f(0)=1}$ and ${\displaystyle \scriptstyle f(\infty )=0}$, the Euler-Lagrange equations for the minimizer ${\displaystyle f}$ give:

${\displaystyle \textstyle (6a^{2}+6\pi at+4\pi t^{2})f'(t)=c}$ ,

where the constant ${\displaystyle c}$ is to be determined by the constraint ${\displaystyle \scriptstyle \int _{0}^{\infty }f'(t)dt=f(\infty )-f(0)=-1}$.

You don't need to compute ${\displaystyle \textstyle f}$, because the energy only depends on ${\displaystyle \textstyle f'}$. You don't either need to know that the minimum problem really admits a minimizer, and that the minimizer satisfies the Euler-Lagrange equation (although these are both necessary issues in order to conclude that ${\displaystyle \textstyle f}$ is a minimizer!). You can follow a perfectly rigorous heuristic argument: just plug the ${\displaystyle \textstyle f}$ found by the Euler-Lagrange equation into

${\displaystyle \textstyle \int _{0}^{\infty }(6a^{2}+6\pi at+4\pi t^{2})f'(t)^{2}dt}$,

for in any case this is an upper bound for C. You'll get:

${\displaystyle \textstyle a\left(\int _{0}^{\infty }{\frac {dt}{6+6\pi t+4\pi t^{2}}}\right)^{-1}}$,

that is ${\displaystyle \textstyle (8.929..)a}$ ; actually, a little better than ${\displaystyle \textstyle 2{\sqrt {3}}\pi a=(10.88..)a}$ (still I wonder how your bounds have been obtained) --pma (talk) 22:10, 24 February 2009 (UTC)[reply]

As a last remark, notice that with minor changes you get a bound on the capacity C(Q) of any convex polyhedron Q --pma (talk) 11:39, 25 February 2009 (UTC)[reply]

I suspect that this problem does not rely on something as complicated as the argument that the capacity is minimized for a sphere; it is a matter of recognizing that the cube of side length a contains a sphere of radius a/2, and is contained inside a sphere of radius ${\displaystyle {\sqrt {3}}a/2}$. Recalling that the capacity of a set is always smaller than the capacity of a containing set, it remains only to find the capacity of a sphere of radius r.

The capacity of a sphere can be found by considering the function ${\displaystyle V(x)={\frac {r}{|x|}}}$ for ${\displaystyle |x|\geq r}$, which is the capacitary potential for such a sphere centered at the origin. A bit of calculation should give you that this capacity is ${\displaystyle 4\pi r}$. Ray (talk) 14:25, 27 February 2009 (UTC)[reply]

Right, most likely this is how the problem was intended... but proving the monotonicity of the capacity (as defined in the OP) requires also a non trivial variational argument, so I liked more a direct and more precise computation, at least for the upper bound --the computation of the upper bound that I provided is all elementary, in particular does not uses the minimality of the sphere.. --pma (talk) 17:54, 27 February 2009 (UTC)[reply]

What I like is that the same computation with u(x)=f(dist(x,Q)) works for any bounded convex Q in ${\displaystyle \scriptstyle \mathbb {R} ^{3}}$. Minimizing over all f with f(0)=1 and ${\displaystyle \scriptstyle f(\infty )=0}$ we get as before an upper bound for the capacity of ${\displaystyle Q}$:

${\displaystyle \textstyle 2K\left(\int _{0}^{\infty }{\frac {dt}{\pi t^{2}+t+A/K^{2}}}\right)^{-1}}$,

where A and K are respectively the area and the total mean curvature of ${\displaystyle \scriptstyle \partial Q}$. For the cube, ${\displaystyle A=6a^{2}}$ and ${\displaystyle K=3\pi a}$. For a polyhedron, ${\displaystyle \scriptstyle K={\frac {1}{2}}\sum _{e}l_{e}\alpha _{e}}$, sum extended over all edges e, where ${\displaystyle l_{e}}$ is the length of the edge ${\displaystyle e}$ and ${\displaystyle \alpha _{e}}$ is the angle between the normals to the adjacent faces to e. This upper bound still has homogeneity 1, since the total mean curvature has the dimension of a lenght. And ${\displaystyle \scriptstyle {\frac {A}{K^{2}}}}$ is a shape coefficient (adimensional). --pma (talk) 23:42, 27 February 2009 (UTC)[reply]

If x is a rational number, let me define ${\displaystyle \|x\|=\max\{|p|,|q|\}}$, where p and q are coprime integers such that x = p/q. Does this thing have a standard name? — Emil J. 15:18, 23 February 2009 (UTC)[reply]

Not that I know of. Does it have any uses? Things only get named if they are used. --Tango (talk) 15:45, 23 February 2009 (UTC)[reply]

It is a convenient measure of the complexity of x. Such an x requires at most 2 lg(∥x∥) bits, and the cost of arithmetic with such x is conveniently bounded in terms of ∥x∥. I don't recall a name being given to it. JackSchmidt (talk) 16:33, 23 February 2009 (UTC)[reply]

What are you using it for? —Ben Kovitz (talk) 15:55, 23 February 2009 (UTC)[reply]

Well, I need it in a proof I am working on now. In fact, I use a generalization of the concept to rational vectors: ${\displaystyle \|(x_{1},\dots ,x_{k})\|=\max\{q,|qx_{1}|,\dots ,|qx_{k}|\}}$, where q is the smallest positive integer such that ${\displaystyle qx_{1},\dots ,qx_{k}\in \mathbb {Z} }$. Basically, the idea is that on the one hand, ${\displaystyle \log \|{\vec {x}}\|}$ is polynomially related to the number of bits required to write the vector ${\displaystyle {\vec {x}}}$ down (in particular, there are only finitely many vectors with ${\displaystyle \|{\vec {x}}\|\leq n}$ for a given n and k), and on the other hand, it's easier to work with in the given context than the actual number of bits. The vector definition is admittedly a bit weird, however I seem to recall that I've seen the basic definition just for rational numbers used somewhere, but I can't remember how it was called, that's why I am asking. So far I've called it a "norm" of ${\displaystyle {\vec {x}}}$ (whence the ${\displaystyle \|\|}$ notation), which is short and sweet, but misleading, as it does not really satisfy the usual axioms of a norm. — Emil J. 16:21, 23 February 2009 (UTC)[reply]

Look at Height of a polynomial. In your case the polynomial would be qx − p. Michael Hardy (talk) 22:19, 23 February 2009 (UTC)[reply]

Height has a nice ring to it. Black Carrot (talk) 23:22, 23 February 2009 (UTC)[reply]

It's usually called the multiplicative height of x, denoted H(x). There's a related quantity called the absolute logarithmic height h(x) which is just the logarithm of H(x). Chenxlee (talk) 23:31, 23 February 2009 (UTC)[reply]

Thanks! This is what I was looking for. — Emil J. 11:13, 24 February 2009 (UTC)[reply]

Hey guys, I've been working on a proof, and was wondering if you guys could check it to ensure the reasoning is valid:

File:Inversefunction.GIF

Start with a continuous increasing one-to-one inverse function f^-1, as shown in my crudely drawn image above.

As can be seen, the rectangle with sides b and f^-1(b) can be partitioned into three areas.

Meaning,

${\displaystyle bf^{-1}(b)=A_{I}+A_{II}+A_{III}\,}$

Area one is a rectangle of sides a and f^-1(a), while Area two is cleary the definite integral from a to b of f^-1. By virtue of an inverse function being a reflection of the original function about the line y=x, Area three is the definite integral from f^-1(a) to f^-1(b) of f(y).

Substituting:

${\displaystyle bf^{-1}(b)=af^{-1}(a)+\int _{a}^{b}f^{-1}(x)\,dx+\int _{f^{-1}(a)}^{f^{-1}(b)}f(y)\,dy\,}$

Rearranging and evaluation yields

${\displaystyle \int _{a}^{b}f^{-1}(x)\,dx=bf^{-1}(b)-af^{-1}(a)-F(f^{-1}(b))+F(f^{-1}(a))\,}$

where F(t) is an antiderivative of f(t).

This can be rewritten as

${\displaystyle \int _{a}^{b}f^{-1}(x)\,dx=g(b)-g(a){\text{ , }}g(t)=tf^{-1}(t)-F(f^{-1}(t))\,}$

Now here's the step I'm not quite sure about:

By what I believe would be the converse of the Fundamental Theorem of Calculus part two,

${\displaystyle \int f^{-1}(x)\,dx=g(x)=xf^{-1}(x)-F(f^{-1}(x))+C\,}$

Is this step allowed? Did I make any other mistakes? Any input is appreciated. Tuvwxyz (T) (C) 23:27, 23 February 2009 (UTC)[reply]

I think you can get what you got by integration by parts and some substitutions:

Take ${\displaystyle u=f^{-1}(x),dv=dx}$; then ${\displaystyle v=x}$; we will not compute ${\displaystyle du}$ for simplicity; but note that ${\displaystyle x=f(u)}$:

${\displaystyle \int f^{-1}(x)\,dx=xf^{-1}(x)-\int xdu=xf^{-1}(x)-\int f(u)du=xf^{-1}(x)-F(u)+C=xf^{-1}(x)-F(f^{-1}(x))+C}$ --Spoon! (talk) 01:37, 24 February 2009 (UTC)[reply]

I know |x| means absolute values, but I also seen ||x|| notation being used, what's the difference between the two? 23:27, 23 February 2009 (UTC)

||x|| means the norm of x. For a scalar, this is usually just the absolute value, but for vectors it can be different things. The standard norms (I forget the names) are the square root of the sum of the squares of the absolute values of the components, the sum of the absolute values of the components, and the maximum of the absolute values of the components. -mattbuck (Talk) 23:51, 23 February 2009 (UTC)[reply]

Those norms are called L² (aka Euclidean, aka standard), L¹ (aka taxicab, aka Manhattan), and L^∞ (aka uniform, aka sup) respectively. See L^p space for the family these are part of. Algebraist 10:53, 24 February 2009 (UTC)[reply]

How would one show rigorously that the limit of the maximum of a finite set of continuous functions is the maximum of the limit, i.e. ${\displaystyle \lim _{x\to a}max\{f_{1}(x),f_{2}(x),...,f_{n}(x)\}=max\{f_{1}(a),f_{2}(a),...,f_{n}(a)\}}$? I've tried fiddling around with the inequalities of the limits a little bit but can't see a nice way to get this result out - thankyou! Mathmos6 (talk) 01:09, 24 February 2009 (UTC)Mathmos6[reply]

Hint 1 - Try the epsilon-delta definition of continuity. The epsilon you'll need for the limit will turn out to be the same (this doesn't always happen, but max is a fairly special function). Hint 2 (more complicated) - This is a special case of a more general result that the limit of a continuous function is the value of the function with the limit plugged in. You'll have to prove this (or quote it from your textbook) and show that max is continuous. 98.216.49.28 (talk) 04:29, 24 February 2009 (UTC)[reply]

I am slightly confused by something. In the work I am doing, I have encountered the notion of ${\displaystyle PSU(3,2^{n})}$. The PSU obviously refers to the projective special unitary group... but I am only familiar with this over the complex numbers. How does this work over a finite field? Is there just one, or are there many, equivalents of conjugation here? SetaLyas (talk) 16:04, 24 February 2009 (UTC) 15:51, 24 February 2009 (UTC)[reply]

There's some kind of description in Unitary group#Finite fields. — Emil J. 16:01, 24 February 2009 (UTC)[reply]

Yeah I know, I've read it. It seems kinda badly written, and just confused me even more when I first saw it. It doesn't really explain what the special unitary group is... that's why I asked the specific questions that I did. thanks though! SetaLyas (talk) 16:10, 24 February 2009 (UTC)[reply]

If I understand it correctly, the role of conjugation is played by the automorphism ${\displaystyle \alpha (x)=x^{q}}$ (which is the unique nontrivial automorphism of ${\displaystyle \mathbf {F} _{q^{2}}}$ fixing ${\displaystyle \mathbf {F} _{q}}$; notice that ${\displaystyle \mathbf {F} _{q^{2}}/\mathbf {F} _{q}}$ is a Galois extension of degree 2). So, yes, there is just one of them. — Emil J. 16:40, 24 February 2009 (UTC)[reply]

thanks Emil J! so is there no description of SU over a finite field that doesn't involve these field extensions and Galois theory? SetaLyas (talk) 16:45, 24 February 2009 (UTC)[reply]

Well, I don't think you really need Galois theory, as everything can be seen directly. The automorphism ${\displaystyle \alpha }$ of ${\displaystyle \mathbf {F} _{q^{2}}}$ is given by an explicit formula, we have ${\displaystyle \alpha (\alpha (x))=x}$ and ${\displaystyle \mathbf {F} _{q}=\{x\in \mathbf {F} _{q^{2}}\mid \alpha (x)=x\}}$ by simple algebra, and that should be all you need to know. — Emil J. 17:01, 24 February 2009 (UTC)[reply]

I think I've found where my confusion arises. So ${\displaystyle U(3,2^{n})}$ actually denotes the group of 3x3 matrices over the field ${\displaystyle F_{q^{2}}}$ (not over the field ${\displaystyle F_{q}}$) such that ${\displaystyle \alpha (A)=A^{-1}}$? SetaLyas (talk) 17:11, 24 February 2009 (UTC)[reply]

Yes. And as mentioned in the article, here either ${\displaystyle 2^{n}=q}$ or ${\displaystyle 2^{n}=q^{2}}$, the notation varies in the literature. — Emil J. 17:42, 24 February 2009 (UTC)[reply]

To be clear, this is true when α(A) is the matrix whose (i,j)th is the qth power of the (j,i)th entry of A; that is, it is the conjugate transpose. The field with q elements plays the role of the real numbers, and the field with q² elements plays the role of the complex numbers, and the qth power map plays the role of complex conjugation. JackSchmidt (talk) 20:39, 24 February 2009 (UTC)[reply]

I'm trying to come up with an equation for this using basic algebra. Can anyone guide me in this? First step I figured out I have to simplify the expression n+(n-1)+(n-2)+...+(n-n). By just looking at the previous expression, it turns out to equal n*n+n-[sum of all numbers between 0 and n including n]...which goes back full circle because I have to prove the term in the square brackets. I'm lost and I feel like I'm making this too complicated. 128.163.165.39 (talk) 17:44, 24 February 2009 (UTC)[reply]

You'll find it simpler if you ignore the algebra for a bit, and start at the beginning using numbers. What you're looking at are the Triangular Numbers, and the first few terms are 1, 3, 6, 10, 15, 21, 28, 36. Try using the difference between successive terms to calculate the equation that governs it. I'll give you a hint - it is a quadratic. -mattbuck (Talk) 18:11, 24 February 2009 (UTC)[reply]

Try Triangular numbers. --Tango (talk) 18:31, 24 February 2009 (UTC)[reply]

Add the first and last numbers together, then add the 2nd and penultimate numbers, then the 3rd and 3rd to last and so on. Do you spot a pattern (it's a really easy pattern!)? Use that pattern to work out a formula. --Tango (talk) 18:31, 24 February 2009 (UTC)[reply]

Although what you have shown seems to have gotten you back to where you started, you can use it to find the formula. If we let ${\displaystyle S_{n}=0+1+2+...+n}$, then you have shown that ${\displaystyle S_{n}=n^{2}+n-S_{n}}$. Hence ${\displaystyle 2S_{n}=n^{2}+n}$ and so ${\displaystyle S_{n}={\frac {n^{2}+n}{2}}}$. --Matthew Auger (talk) 20:16, 24 February 2009 (UTC)[reply]

Read Knuth's Concrete Mathematics, it's a really good book and explains this well. (It also has some advanced material in the later chapters, feel free to skip that part, don't let it scare you away.) – b_jonas 09:43, 25 February 2009 (UTC)[reply]

Arithmetic progression —Preceding unsigned comment added by 129.67.37.225 (talk) 15:40, 25 February 2009 (UTC)[reply]

Another valuable technique to learn when looking for patterns in sequences and series is proof by mathematical induction. Zunaid 12:02, 27 February 2009 (UTC)[reply]

I am trying to figure out the length across an ellipse at any given angle theta (Θ). The ellipse is given as 2a across the major axis, and 2b across the minor axis. I am basically trying to figure out how to make a plot of the lengths as I go from the major axis and vertical axis being at the same direction, to the major and vertical axis that are 90 degrees apart.

This question is related to non destructive evaluation. I need to find the lengths of an ellipsoidal crack when the crack is in different orientations. The lengths will then help me find the intensities of the x-rays that are going through the part and the crack at different orientations.

Thanks —Preceding unsigned comment added by 12.216.236.227 (talk) 19:47, 24 February 2009 (UTC)[reply]

I think you want the equation for this ellipse in polar coordinates, that is

${\displaystyle \rho ={\sqrt {a^{2}\cos ^{2}(\theta )+b^{2}\sin ^{2}(\theta )}}}$

Then if you mean the length of the chord at angle ${\displaystyle \textstyle \theta }$, it's of course ${\displaystyle \textstyle 2\rho }$.--pma (talk) 22:56, 24 February 2009 (UTC)[reply]

I have an investment question regarding rental property.

In an online game I have the option to invest in two different properties, with the following considerations:

Property 1 will pay for itself in 6 years, and will require about 3 years of saving to purchase

Property 2 will pay for itself in 4 years, but will require about 11 years of saving to purchase

The catch is that property 1 costs $200,000 and property 2 costs $750,000

Assume that there is no inflation, and that income and costs remain constant for the duration, and that property values will not change.

Which is more efficient for me to purchase?

I am leaning toward the cheeper one, but am not sure since I can only get 1 property at that price. 12.216.168.198 (talk) 23:16, 24 February 2009 (UTC)[reply]

Assuming no inflation and 0% interest rates (pretty realistic at the moment, actually!), and assuming that by "pay for itself in X years" you mean it generates a profit of price/X each year, then the answer is going to depend on the planned length of the investment. Obviously, if you are going to invest for less than 11 years, option 2 makes you nothing, so option 1 would be better. Over a very long period of time (100's of years, say), the amount of time it takes to save up and then pay off the purchase costs are irrelevant and you want the one with the greater return, which is option 2. There will be some point inbetween where the best choice switches from one to the other. I'll do some maths and see if I can work out what it is... --Tango (talk) 23:32, 24 February 2009 (UTC)[reply]

The switch is around 16 or 17 years. If you're investing for less than 16 years, take option 1, more than 17 years, take option 2. (Between 16 and 17 years would require me to do the arithmetic more carefully!) --Tango (talk) 23:42, 24 February 2009 (UTC)[reply]

Thanks for the help. —Preceding unsigned comment added by 12.216.168.198 (talk) 23:54, 24 February 2009 (UTC)[reply]

How many types of functions are there? I have polynomial, exponential, and trigonometric so far but I was wondering if there are any more.

Here, a function is a different "type" if it can not be expressed in terms of functions of other types in a finite number of terms. Inverse functions do not count as different types. So hyperbolic functions are not of a different type because they can be expressed in terms of exponential functions and polynomial functions. ${\displaystyle \sinh(x)={\frac {e^{x}-e^{-x}}{2}}}$ can be expressed as ${\displaystyle f(x)=e^{x}}$ and ${\displaystyle g(x)={\frac {x}{2}}}$ and ${\displaystyle h(x)=-x}$: ${\displaystyle \sinh(x)=g(f(x)-f(h(x)))}$

Trigonometric functions are a different type from polynomial functions even though ${\displaystyle \sin(x)=x-{\frac {x^{3}}{6}}+{\frac {x^{5}}{120}}\cdot \cdot \cdot }$ because it requires an infinite amount of terms.

Are there any other types?--Yanwen (talk) 23:52, 24 February 2009 (UTC)[reply]

The question "how many types of functions are there?" is quite ambiguous, just as the questions "how many languages are there?" or "how many colors are there?". Also, mathematicians are constantly making new functions as the need arises. Putting those issues aside, there are some other things you want to consider: like, what do you mean when you say "function"? All of your examples are functions from reals to reals (or complex numbers to complex numbers), but perphaps you're also interested in functions from integers to integers (like the factorial function, Euler's totient function, divisor function, and many number theoretical functions, as well as functions from computer science like Busy beaver function and the Ackermann function), integers to reals, or more abstract mathematical structures (sets, groups, rings, fields, topological spaces, metric spaces, etc. etc.). Also, what restrictions do you want to impose on these functions? Your example functions are all infinitely differentiable. Are you interested in continuous functions, uniformly continuous functions (though note that polynomials of degree above 1 are not uniformly continuous on the real line), differentiable functions, holomorphic functions, homomorphisms, etc.? My goal is not to overwhelm you (depending on your mathematical background), but to emphasize that there are many, many different kinds of functions used in mathematics.

Maybe if you're looking for more "types" of functions, maybe you will want to think about what type of function each of the following is: square-root, Riemann Zeta function, Gamma function, sawtooth function, Dirichlet function, question mark function, Weierstrass function. I hope you find something interesting/new in there somewhere.

Here's an example of one type of function that you might want to add to your list: Rational functions. Combining rational functions with trigonometric ones gives you, for example, the sinc function.

You may find the Stone-Weierstrass theorem interesting. Eric. 131.215.158.184 (talk) 05:09, 25 February 2009 (UTC)[reply]

There are also difficulties in deciding which functions are of the same type. For example, I would consider exponential and trigonometric functions to be of the same type, since they are readily interdefinable, but that only works over the complex numbers. Algebraist 09:30, 25 February 2009 (UTC)[reply]

But it seems to me that the original question about "how many types of functions are there" is quite precise and not ambiguous (although the term "type of function" is unhappy, because it does not refer to an equivalence relation; rather, he has in mind a functional dependence). Anyay, the OP says that a function is of "different type" wrto a set of functions ${\displaystyle \scriptstyle {\mathcal {F}}}$, if it can not be obtained by ${\displaystyle \scriptstyle {\mathcal {F}}}$ with finitely many operations (he thinks to sum, product, composition, inverse; and we can decide whether to allow derivatives, and antiderivatives too; one could even allow solutions of algebraic or differential equations with coefficients in ${\displaystyle \scriptstyle {\mathcal {F}}}$). Now, to decide if a certain function f is or is not in the enlarged class ${\displaystyle \textstyle {\mathcal {F}}}$ generated by a certain class of functions ${\displaystyle \scriptstyle {\mathcal {F}}}$, by means of a certain set of operations, may be an extremely difficult algebraic problem; I do not know much about the spectacular results that have been obtained in this direction. Nevertheless, usually you can quite easily show that starting with a countable ${\displaystyle \scriptstyle {\mathcal {F}}}$ there is always a "new" analytic entire function f which is not in the enlarged class ${\displaystyle \textstyle {\mathcal {F}}}$. The standard argument, without going into the details, is from their order at infinity: the functions in ${\displaystyle \textstyle {\mathcal {F}}}$, are all o(f), what implies that f is actually "new". In other words you can produce new functions for an uncountable ordinal of times, so to speak. On the other hand, if you allow limits, the scenario completely changes; you enter in the realm of the approximation theory of functions, where density theorems like the one quoted by Eric allow to generate all functions in suitable classes, by means of very simple base systems. At least, I hope you liked the notation for enlarged classes. --pma (talk) 10:15, 25 February 2009 (UTC)[reply]

Hi, I am trying to work out a proof that every principal ideal domain is a Noetherian ring (assuming it is true). I was able to prove that every Euclidean domain was a Noetherian ring, but that was with the assistance of the degree function. I set a~b if a = b*u for some invertible u and it turns out that members of the same equivalence class have the same degree. Since every ideal in a Euclidean domain is principal (which I could prove), the proof that every Euclidean domain is a Noetherian ring follows. Any ideas on whether every PID is Noetherian? Thanks a lot for your help in advance. —Preceding unsigned comment added by 58.161.138.117 (talk) 09:32, 25 February 2009 (UTC)[reply]

Yes. The general proof is (I think) easier than your proof for EDs. Just suppose that you have an ascending chain of ideals, and note that the union of the chain must again be an ideal, and so is principal. More generally, Noetherianity is equivalent to all ideals being finitely generated. Algebraist 10:16, 25 February 2009 (UTC)[reply]

Thanks for your response! But how does what you said show that any ascending chain of ideals terminates? I might just not be getting it... —Preceding unsigned comment added by 58.161.138.117 (talk) 10:29, 25 February 2009 (UTC)[reply]

Algebraist has already given the proof! If it doesn't seem obvious to you, try thinking a little harder. It might just be that you are sleepy!--PST 11:42, 25 February 2009 (UTC)[reply]

Hey pst!! how are you doing, welcome back! ;) --pma (talk) 11:53, 25 February 2009 (UTC)[reply]

Glad to see the usual folk at the reference desk! ;) --PST 01:21, 26 February 2009 (UTC)[reply]

Think about where the generator of the union came from. — Emil J. 13:54, 25 February 2009 (UTC)[reply]

By the way, domains where just the f.g. ideals are principal are called Bézout domains. So non-Noetherian ones are sort of non-Noetherian PIDs.John Z (talk)

I understand that the only non-trivial natural number to be simultaneously square and pyramidal is 4900. From what I understand, the original proof is non-elementary but there have been some elementary proofs since. I've got through several sheets of paper playing around with congruences and I'm getting nowhere so I'm ready to give up. Strictly speaking, I'm into applied mathematics but I was told about this problem and am genuinely curious. I've even had my old number theory textbook out trying to find inspiration!

Is anyone able to provide an outline of a proof? I haven't found anything on the net so far.

Thanks, Readro (talk) 23:13, 25 February 2009 (UTC)[reply]

The wikipedia article quotes the original result by G.N.Watson (1918). As you are saying there is a more recent and elementary proof: I've found:

• Ma, De Gang. An elementary proof of the solutions to the Diophantine equation ${\displaystyle 6y^{2}=x(x+1)(2x+1)}$. (Chinese) Sichuan Daxue Xuebao 1985, no. 4, 107--116
• Anglin, W.S. The square pyramid puzzle. Amer. Math. Monthly 97 (1990), no. 2, 120--124 [2]

It really looks elementary and short. --pma (talk) 12:17, 26 February 2009 (UTC)[reply]

This is not a homework question; it is a U.I.L. question and neither I nor my coaches can figure it out. "A donut shop sells 6 different types of donuts. How many different ways are there to buy a dozen?" We've tried many combinations and permutations, but still can't figure it out. TIA for a response, Ζρς ι'β' ^¡hábleme! 00:36, 26 February 2009 (UTC)[reply]

This is extremely basic probability. I don't know what the UIL is, but if your coaches are supposed to be teaching you mathematics and cannot figure it out, then there is a problem. Think about it in terms of dice. Say you rolled a die twice - how many possible different results are there? If you can't figure it out in your head, try drawing a diagram. Then just scale the results up - it's the exact same thing. --Aseld ^talk 01:25, 26 February 2009 (UTC)[reply]

Well number the dozen from 1-12. The first donut can be chosen in six possible ways (because there are six different types of donuts). Similarly, each donut from 2-12 can be chosen in six possible ways. Now, just multiply six by itself 12 times so the answer is 6¹² possible dozens. (P.S Good to see some Spanish on the English WP!). --PST 01:26, 26 February 2009 (UTC)[reply]

You probably want to identify combinations that only differ by ordering. A jam doughnut, 10 ring doughnuts and a jam doughnut isn't really different from two jam doughnuts and 10 ring doughnuts. --Tango (talk) 01:44, 26 February 2009 (UTC)[reply]

Tango is right. This is where the problem arises for us. For this problem, "ABBA" is the same as "AABB" or "ABAB". Ζρς ι'β' ^¡hábleme! 01:50, 26 February 2009 (UTC)[reply]

I don't care what order the clerk puts my donuts into the box, I just care that I have what I ordered. Here's a good way to think about this problem: imagine you have 12 plain donuts lined up on a table, and you get to put 5 marks between them (i.e. 00|00000|00|0|00|). Everything before the first mark gets turned into donut type 1, everything between mark 1 and 2 gets turned into donut type 2, etc. You can convince yourself that there is a one to one correspondence between ways to put the marks on the table, and ways to order your donuts. How many ways is that?76.126.116.54 (talk) 01:53, 26 February 2009 (UTC)[reply]

I don't understand what you are trying to say. Ζρς ι'β' ^¡hábleme! 01:57, 26 February 2009 (UTC)[reply]

A standard trick for basic combinatorics questions is to come up with a type of diagram that corresponds to each thing you're trying to count. 00|00000|00|0|00| means order 2 of type 1, 5 of type 2, 2 of type 3, 1 of type 4, 2 of type 5 and 0 of type 6. |||||000000000000 means order 12 of type 6. So you're trying to count the number of ways to insert the |'s between the 0's. This can be expressed via the appropriate binomial coefficient. 76.126.116.54 (talk) 02:02, 26 February 2009 (UTC)[reply]

Say you have a dozen doughnuts and five chocolate bars, and you arrange them in a line. You can have any order you like, but there must be 5 chocolate bars. So how many ways of doing that are there? Now, consider that all doughnuts before the first chocolate bar are type one, all between 1 and 2 are type 2, etc. You now have a dozen doughnuts, divided into six types by the placement of your chocolate bars. Do you see where this is going? -mattbuck (Talk) 02:04, 26 February 2009 (UTC)[reply]

Sorry, I'm really horrible at probability, but we tried all kinds of combinations and permutations. None of them agreed with the answer key; however, the answer on the answer key was verfied by a U.I.L. official. Also, a coach from another school explained this to one of our students (the student couldn't remember what the coach had said today), but the student did remember that it was not a normal probability question (combinations, permutations, etc.). Ζρς ι'β' ^¡hábleme!

Using 76.126.116.54's technique, I find 6188. Is that what the answer sheet gave? Eric. 131.215.158.184 (talk) 08:34, 26 February 2009 (UTC)[reply]

You may like to study the article on multiset. (The answer is 12376). Bo Jacoby (talk) 08:48, 26 February 2009 (UTC).[reply]

I don't believe that 12376 is incorrect. See later in thread. -- SGBailey (talk) 14:58, 26 February 2009 (UTC)[reply]

You don't believe that 12376 is correct, right? (Neither do I) --NorwegianBlue ^talk 22:51, 26 February 2009 (UTC)[reply]

This isn't a probability question, it's a combinatorics question. Instead of working with a line of 12 objects split into six categories, we're going to work with a line of 17 split into two - 0s and |s for lack of a better terminology. 0s will represent the doughnuts, and |s will represent a change in doughnut type. Since we have six types of doughnuts, we therefore need 5 changes, so there will be 5 |s. Added to the 12 doughnuts, that makes a line of 17 objects. Instead of working directly on how many doughnuts, work out how many ways there are to arrange the 5 |s and 12 0s. This is equivalent to saying "I have 17 objects, and I want to pick five of them. How many ways of doing this are there?". If you pick one particular set of five, then because the 0s are identical and so can be interchanged (ditto the |s), you've uniquely determined where the 0s are. They'll be split into six groups (including groups of 0 in the case of ||), and this is one possible doughnut pick. Now you need to assure yourself that there aren't any picks which cannot be represented by a unique distribution of 0s and |s, and you'll find there's a 1:1 correspondence between possible doughnut combinations, and possible combinations of 12 0s and 5 |s. -mattbuck (Talk) 13:55, 26 February 2009 (UTC)[reply]

The R (programming language) has functionality for this:

> summary(blockparts(rep(12,6),12))
                                                            
[1,] 12 11 10 9 8 7 6 5 4 3 ... 0  0  0  0  1  0  0  0  0  0 
[2,] 0  1  2  3 4 5 6 7 8 9 ... 1  0  0  0  0  1  0  0  0  0 
[3,] 0  0  0  0 0 0 0 0 0 0 ... 0  1  0  0  0  0  1  0  0  0 
[4,] 0  0  0  0 0 0 0 0 0 0 ... 0  0  1  0  0  0  0  1  0  0 
[5,] 0  0  0  0 0 0 0 0 0 0 ... 1  1  1  2  0  0  0  0  1  0 
[6,] 0  0  0  0 0 0 0 0 0 0 ... 10 10 10 10 11 11 11 11 11 12
> 

Each column shows a possible box of donuts. So the first is 12 donuts of type 1, the second column shows 11 of type 1 and one of type 2, and so on to the last column which is 12 of type 6. HTH, Robinh (talk) 10:56, 26 February 2009 (UTC)[reply]

By the way, the Greek numeral for 12 is ιβʹ, not ι'β'. Sorry for being OT. — Emil J. 12:36, 26 February 2009 (UTC)[reply]

The pick any 5 out of 17 seems like a good approach and ₁₇C₅ gives 17! / ( 5! * 12! ) which is 6188. Do a simpler problem: 3 doughnuts and 3 types of doughnuts, so we need to change type twice, so we have 5 items (3 doughnuts with two type changes) and need to choose 2 of them. We could choose 12,13,14,15,23,24,25,34,35,45 which is 10 things. ₅C₂ gives 5! / ( 2! * 3! ) which is indeed 10. Thus I believe the 6188 answer. -- SGBailey (talk) 14:58, 26 February 2009 (UTC)[reply]

As far as I can see, exactly this question is answered in Combinations; Number of combinations with repetition, which agrees with the results of SGBailey and Eric. According to that article, the answer is choose(n+k-1,k), where n is the number of items on the menu, and k is the number you buy.

choose(6+12-1,12)= 6188

--NorwegianBlue ^talk 15:22, 26 February 2009 (UTC)[reply]

What values for a, b, c, d and m are the "fit best" for the formulas below? Is there a software I can plug it in, and it will calculate for me?

2c+d=2
c+2a=3
b+2m=3
2a+b+2m=5
3c=4
2c+a=5
c+a+b+m=5

--Sonjaaa (talk) 16:11, 26 February 2009 (UTC)[reply]

No exact solution exists. What do you mean by '"fit best"'? Algebraist 16:16, 26 February 2009 (UTC)[reply]

Hi Sonjaaa. I guess "fit best" is in the sense of linear least squares; then you want the Moore-Penrose pseudoinverse. The software is very common; but if it is only that system one can try and do it by hands... not me! --pma (talk) 18:35, 26 February 2009 (UTC)[reply]

Well, (with maple): a=1, b=2, c=3/2, d=-1, m=1/2. This gives as output in your system: [2, 7/2, 3, 5, 9/2, 4, 5], making the minimal distance ${\displaystyle \scriptstyle {\sqrt {6}}/2}$ from the data [2,3,3,5,4,5,5]. --pma (talk) 19:15, 26 February 2009 (UTC)[reply]

I'm very curious about the card game... where does that system come from? --84.221.68.150 (talk) 20:33, 26 February 2009 (UTC)[reply]

I wish I had a card game as homework too...--131.114.73.83 (talk) 13:03, 27 February 2009 (UTC)[reply]

Dominion (card game) --Sonjaaa (talk) 20:29, 27 February 2009 (UTC)[reply]

There are also linear regression calculators on line. Here is one, should you need it again [3]. You have to choose the option LSQ-RREF and introduce the augmented matrix, that is in your case, (ordering the variables a, b, c, d, m):

0 0 2 1 0 2

2 0 1 0 0 3

0 1 0 0 2 3

2 1 0 0 2 5

0 0 3 0 0 4

1 0 2 0 0 5

1 1 1 0 1 5

--pma (talk) 21:06, 27 February 2009 (UTC)[reply]

these are not homework, though probably that's exactly what they look like. The give away is if I was a stats student I should have a clue how to approach these kind of problems. I've been coming up with these stats questions but don't have the math to solve them. Most start with: say you have a 100 sided die. Here's a couple: On average how many unique numbers will NOT have appeared after 100 rolls? On average, how many rolls would it take to see each number at least once. And lastly, on average, after 100 rolls, what number roll will produce the last unique number, not a repeat. Can someone come up with a mathematical solution or is that pretty much only solvable if you modelled it with a random number generator? I reckon the first two are probably solvable. Not sure about number three. Vespine (talk) 09:06, 27 February 2009 (UTC)[reply]

The second problem is the Coupon collector's problem. McKay (talk) 09:41, 27 February 2009 (UTC)[reply]

I'm trying to work out this problem (which is not homework): Suppose ${\displaystyle A:\mathbb {X} \to \mathbb {Y} }$ is a bounded linear operator, where ${\displaystyle \mathbb {Y} }$ is a Banach space, with ${\displaystyle \|I-A\|<1}$ (and ${\displaystyle I}$ the identity operator). The exponential of ${\displaystyle A}$ is defined as ${\displaystyle e^{A}=\exp(A)=\sum _{n=1}^{\infty }A^{n}/n!}$ and the logarithm is defined as ${\displaystyle \log(A)=-\sum _{k=1}^{\infty }{\frac {1}{k}}(I-A)^{k}}$. As both of these series are absolutely convergent, it is clear that they do indeed converge. My problem is to show that ${\displaystyle \exp(\log A)=A}$.

My first idea is to show that the exponential function (as an operator from ${\displaystyle L(\mathbb {X} ,\mathbb {Y} )}$ into itself) is continuous (a relatively simple task), so that the problem reduces to showing ${\displaystyle \exp {\Bigl (}-\sum _{k=1}^{n}{\frac {1}{k}}(I-A)^{k}{\Bigr )}=\exp {\Bigl (}-\sum _{k=1}^{n}{\frac {1}{k}}\sum _{j=0}^{k}{\binom {k}{j}}(-A)^{j}{\Bigr )}\to A}$.

I thought perhaps changing the order of summation could be helpful, though doing this didn't really reveal any form that appeared useful to me. Does anyone have any useful suggestions? Nm420 (talk) 19:04, 27 February 2009 (UTC)[reply]

I haven't followed up this idea at all, but it might be useful to look at some proofs that those power series define functions from R to R. It's possible that the ideas easily generalize. Algebraist 19:12, 27 February 2009 (UTC)[reply]

Exact; let's write A=I-x so it is exp(log(I-x))=I-x ; this is true in the sense of real functions (|x|<1), therefore in the sense of formal power series, therefore in any Banach algebra (again with ||x||<1). No need of computation... (and I guess you mean ${\displaystyle \mathbb {X} =\mathbb {Y} }$) --pma (talk) 19:50, 27 February 2009 (UTC)[reply]

In many fields of mathematics, objects are generalized/based from/on Euclidean space. So if you prove it for Euclidean space, it should be easy to generalize (note that not all objects are based on Euclidean space). --PST 12:41, 28 February 2009 (UTC)[reply]

pma's solution is very slick, but if you actually want to do a calculation, then I suggest setting ${\displaystyle B=I-A}$ and then showing that ${\displaystyle \exp(\log(I-B))=I-B}$. Both sides of this are well-defined formal power series in ${\displaystyle B}$, and you should be able to show that the coefficient of ${\displaystyle B^{n}}$ is the same on both sides by reversing the order of summation, as you expected, though I haven't worked it out, so I'm not sure how complicated it turns out to be. (Note also that your definition of ${\displaystyle \exp(A)}$ is missing ${\displaystyle I+}$ from its beginning.)Ctourneur (talk) 14:12, 28 February 2009 (UTC)[reply]

What day of the week were Nov 11, 1918 July 4, 1776 June 10, 1215 ? 65.167.146.130 (talk) 20:25, 27 February 2009 (UTC)[reply]

Where in the world are you asking for? Different countries adopted things like the Gregorian calendar or Julian calendar at different times, to say nothing of the ambiguities offered with just a year number BC or AD etc... 78.151.212.201 (talk) 22:03, 27 February 2009 (UTC)[reply]

Check out this page and choose a "day of the week" calculator. Using the top of the list link I got Monday, Thursday, and Wednesday -hydnjo talk 02:07, 28 February 2009 (UTC)[reply]

Let S be an infinite set of strings over the alphabet {0,1} and S_k denote the strings in S that have exactly k 1's. Let's say I can prove that for every k, there is a Turing machine T(k) that recognizes exactly the set S_k. The proof might be nonconstructive--that is, I can't necessarily prove that any specific machine M recognizes S_k; I can only prove the existence of such an M. 1) Is it still appropriate to say that each S_k is a decidable set? 2) Is it appropriate at all to say that S itself is a decidable set? 3) is it pretty clear that S is not necessarily a recursive set? Question is motivated by this article which I do not yet understand. Thanks. 76.195.10.34 (talk) 19:18, 28 February 2009 (UTC)[reply]