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September 8

set

can the infinite set r = {r₁, r₂, ... r_n, ...} where $r_{k+1}={\frac {k+3}{k+2+r_{k}}}$ and $r_{1}={\frac {3}{2}}$ be described by some function $f$ where $f(k)=r_{k}$? —Preceding unsigned comment added by 92.8.193.46 (talk) 01:43, 8 September 2009 (UTC)

You are looking for a closed formula for the function, as the function itself can simply be defined as the mapping f:k-->r_k.Julzes (talk) 01:48, 8 September 2009 (UTC)

Defining s_k = 1/(r_k-1) you get a much nicer recursive relationship, s_k+1 = -(k+3)s_k - 1, with s₁ = 2. You can construct an explicit (but awful) formula for r_k from that:

$r_{k}=1+(-1)^{k-1}\left[(k+2)!\left({\frac {1}{3}}+\sum _{n=2}^{k}{\frac {(-1)^{n}}{(n+2)!}}\right)\right]^{-1}$

Rckrone (talk) 04:04, 8 September 2009 (UTC)

Trivially so: you have already provided an inductive definition of the function you desire. — Carl (CBM · talk) 04:12, 8 September 2009 (UTC)

can all sets of that form where $r_{k}=r_{k}(r_{k-1},r_{k-2},...r_{k-n+1},r_{k-n})$ be written as a function of $k$ in the same manner provided the first $n$ terms are given? —Preceding unsigned comment added by 92.8.193.46 (talk) 11:10, 8 September 2009 (UTC)

Clarify what you mean by 'of that form' in the more complex situation.Julzes (talk) 17:58, 8 September 2009 (UTC)

You may invent ad hoc notations signifying any solution you need. But the sequence (1, 2, 4, 16, 65536, ...) satisfying r _k+1 = 2 ^{r _k}, r ₀ = 1 is not written in terms of elementary functions. Bo Jacoby (talk) 12:38, 8 September 2009 (UTC),

Does it help you to know that

$r_{k}=1+{\frac {(-1)^{k-1}e}{\Gamma (k+3,-1)}}\ ?$

Where Γ is the incomplete Γ-function and e is Euler's constant? (N.B. the incomplete Γ-function is related to the Γ-function which generalises the factorial function to take a complex variable z with Re(z) > 0.) ~~ Dr Dec (Talk) ~~ 15:06, 8 September 2009 (UTC)

Can the choice of numbers and the sequence be changed and still have a closed form like that, Declan? It looks like a special case.Julzes (talk) 02:21, 9 September 2009 (UTC)

I'm not sure, I just though evaluating Rckrone's expression for $r_{k}$ might help some how. I don't really understand the problem in its totality. ~~ Dr Dec (Talk) ~~ 15:38, 10 September 2009 (UTC)

The sorgenfrey plane 2

I've done some more thinking to my question above (the sorgenfrey plane). Let A be all the rational points on the diagonal and B be a {r+sqrt2 | r is rational}. Obviously A and B are both dense and disjoint. Now if you recall how the proof of "regular+countable basis -> normal" goes, we can apply the following theorem: if A and B can be covered by two countable collections {Un} and {Vn} respectively, and that the closure of each Un is disjoint from B, and the closure of each Vn is disjoint from A, then A and B can be covered by disjoint open sets. The above A (rationals) and B (rationals+sqrt2) satisfy this property, hence they can be covered by disjoint open sets. I believe my intuition was right: it is possible to separate two disjoint dense sets, provided both are countable. I don't know what's up with the irrationals; there's more to them than meets the eye. The irrationals are a Baire space in the order topology, but rationals aren't, even though both are densely ordered. My solution is: there's something fishy with the irrationals that our intuition fails to capture. Anyone agree?

To Algebraist: I tried using your example but cant see how it'll work for p/q+1/(q^q). The page Liouville number just says |x-p/q|<1/(q^n), for any n. But I guess it's good enough, it must be that there's always a real number that can be approximated extremely closely by a rational p/q with a low denominator q, no matter how we rig the half open square.

To pma: you must have had one those those moments when two totally unrelated things give exactly the same result right? This and other "miracles" as I call them. I don't believe in miracles, so when I see something like this I tend to give it a deeper thought and try to bring more intuition into it. So please, if you accept logic without question that's fine by me, but things like Godel's incompleteness theorem and Skolem's paradox really puts my faith down. Breath of the Dying (talk) 02:18, 8 September 2009 (UTC)

I agree, in principle; but let's go back: the main thing was very simple: no matter how one fixes an Euclidean nbd I_r for each rational r, the set of all points that belong to infinitely many I_r,

$\cap _{F\in {\mathcal {F}}}\cup _{r\in \mathbb {Q} \setminus F}I_{r}$

is an intersection of countably many dense open subsets of $\mathbb {R}$, hence an uncountable set by the Baire's category argument (${\mathcal {F}}$ being the family of all finite subsets of $\mathbb {Q}$). This somehow is counter-intuitive, since it seems it tell us that a very large part of the irrational numbers are stuffed somewhere very very closely to the rationals. For instance, the trascendental number that Algebr. provided, is approximable by rationals as fast as a lightning. On the contrary, one would expect that being so "close" to the rationals would leave very small room. I think in this case one's intuition is just too much attached to finite or smooth analogies. I'm confident that your doubts will soon disappear. --pma (talk) 06:49, 8 September 2009 (UTC)

The intuition that "being so close to the rationals would leave very small room" is actually correct to the extent that the set of irrational numbers with approximation exponent greater than 2 has Lebesgue measure zero. — Emil J. 11:59, 8 September 2009 (UTC)

Good remark --so maybe what is counterintuitive is just any second category sets with null Lebesgue measure --pma (talk) 14:11, 8 September 2009 (UTC)

mathematics

what is the solution of (integration(differentiation of dx)) —Preceding unsigned comment added by 122.168.71.208 (talk) 10:30, 8 September 2009 (UTC)

Your question as posed does not entirely make sense, but you might find something useful in our article on the fundamental theorem of calculus. Gandalf61 (talk) 10:59, 8 September 2009 (UTC)

Your question is not specific enough to be meaningful, but I'd guess that you're taking integration and differentiation to be inverse operations, as in (square(square root of x)). So what do you think the solution to your question is?86.155.186.70 (talk) 11:06, 8 September 2009 (UTC)

(ec) If you're asking what the indefinite integral of the derivative of f(x) is, it's f(x). If you're asking what $\int f(x)dx$ means, see integral. Where we're calculating the integral over a range (see definite integral), the product of f(x) and dx (see differential) points towards the notion of considering the area under the curve of y=f(x) as the infinite summation of rectangles with height f(x) and width dx (see Riemann sum). In that way, the long-s could be intutitively related to the capital sigma notation. That's the brief overview of the generic notation. If what I've said doesn't help, you'll probably need to come back and be more specific. —Anonymous Dissident^Talk 11:10, 8 September 2009 (UTC)]

Isn't $\int f'(x)\ dx=f(x)+c$ for some constant c? So the antiderivative of the derivative is unique up to an addative constant. ~~ Dr Dec (Talk) ~~ 15:13, 8 September 2009 (UTC)

I might be wrong here, but is +c really necessary when we're considering the antiderivative of the derivative of a function we already know? +c notes an ambiguity – but if we already know the original function, there is no ambiguity. Speaking from a purely theoretical and objective reference frame, the indefinite integral of the derivative of f(x) is f(x), without the +c – because c is already known and is part of the function, if it exists. Contrast this to the position of a person who is trying to integrate the derivative of a function they do not know. Naturally, I'm only assuming that we do know f(x) here. —Anonymous Dissident^Talk 02:23, 9 September 2009 (UTC)

Derivatives don't have to be integrable, in the Lebesgue theory - so it's not always the case that $\int f'(x)\ dx$ even makes sense. But in suitably nice circumstances, this does work. Tinfoilcat (talk) 16:42, 8 September 2009 (UTC)

A derivative of a differentiable function does not have to be Lebesgue integrable, but it is always Henstock–Kurzweil integrable, and its indefinite integral is indeed the original function up to an additive constant. — Emil J. 17:12, 8 September 2009 (UTC)

Although if a function f is only differentiable a.e. then $\int f'(x)\ dx$ may still be still meaningful in the Lebesgue sense, but not necessarily equal to the original function unless f is absolutely continuous. Rckrone (talk) 17:22, 8 September 2009 (UTC)

Piecewise functions

I am getting confused on where to shade...

$h(x):={\begin{cases}3,&{\mbox{if }}-1\leq x\leq 1\\4,&{\mbox{if }}1<x\leq 4\\x,&{\mbox{if }}4\leq x.\end{cases}}$

and how do you graph y=x? Accdude92 (talk) (sign) 13:30, 8 September 2009 (UTC)

I suggest you start by graphing points, and see if you detect a trend. Ray^Talk 14:10, 8 September 2009 (UTC)

(ec) When you have a function outputting a constant, it's as easy as graphing a line like y=10. When the function outputs different constants for different ranges, which line you're drawing differs based on x value. For instance, for "3 if -1≤x≤1", you're graphing y=3 from x = -1 to x = 1. Another way of thinking about it is that you're graphing y=3 for every x value that meets the criterion "x is greater than or equal to -1 and less than or equal to 1". y=x is a line that cuts the origin diagonally – every x value is matched to every y value: (1,1), (2,2), (3,3) and so on. —Anonymous Dissident^Talk 14:19, 8 September 2009 (UTC)

Dihedral group of a polygon

Resolved

Someone asked me to explain the dihedral group of a polygon to them. This is what I said: Dihedral group is the group of symmetries of a polygon by reflection and rotation. If n is odd then besides the n rotations there are n reflections around the n lines originating from the n vertices and bisecting each interior angle. If n is even then besides the n rotations there are n reflections around the perpendicular bisectors of the n edges. So in either case the dihedral group has order 2n.

What I am not sure is whether in case n is odd the reflections are taken around the angle bisectors or not. Can someone clarify this. I searched on the net but couldn't find anything. Thanks--Shahab (talk) 15:35, 8 September 2009 (UTC)

The dihedral group of a regular n sided polygon is generated by the reflections in the perpendicular bisectors of its sides and the bisectors of its interior angles. If n is odd each axis of symmetry connects the mid-point of one side to the opposite vertex. If n is even there are n/2 axes of symmetry connecting the mid-points of opposite sides and n/2 axes of symmetry connecting opposite vertices. In either case there are n axes of symmetry altogether and 2n elements in the symmetry group. Reflecting in one axis of symmetry followed by reflecting in another axis of symmetry produces a rotation through twice the angle between the axes. Gandalf61 (talk) 15:48, 8 September 2009 (UTC)

Thanks for the quick response. This information should also be in the Dihedral groups article.--Shahab (talk) 16:09, 8 September 2009 (UTC)

Metric completion of a field

This question should be elementary, but it's giving me the blues. Suppose we have a field, $\left(K,+,\cdot \right)$ endowed with a absolute value $|\cdot |:K\rightarrow \mathbb {R} _{+}$ that satisfies the three properties:

• AV1: $\left|x\right|=0\iff x=0$
• AV2: $\left|x\cdot y\right|=\left|x\right|\cdot \left|y\right|$
• AV3: $\left|x+y\right|\leq \left|x\right|+\left|y\right|$

Now, suppose $K$ is not metrically complete, i.e., there exist sequences $\left(x_{n}\right)_{n=1}^{\infty }$ that are Cauchy in $K$ with no limit in $K$. We define ${\overline {K}}$ to be the set of Cauchy sequences in $K$ modulo Cauchy sequences that converge to zero. Then, it is natural to define an absolute value, $\left|\cdot \right|^{\star }$ by $\left|\left(x_{n}\right)_{n=1}^{\infty }\right|^{\star }=\lim _{n\rightarrow \infty }\left|x_{n}\right|$. This is all very nice.

My question: how can I prove that $\left|\cdot \right|^{\star }$ is unique, i.e., that it is the only absolute value that we can define on ${\overline {K}}$ that is well-defined, satisfies the axioms AV1-AV3, and restricts to $\left|\cdot \right|$ on $K$ (i.e., $\left|\iota \left(x\right)\right|^{\star }=\left|x\right|$, where $\iota \left(x\right)=\left(x\right)_{n=1}^{\infty }$, the constant sequence)?

I think there's something subtle going on here that's eluding me. I've read Complete metric space, but it didn't quite ease my mind. The first paragraph under the header "Completion" asserts this claim, but doesn't say how it's proven. Can someone give me a hint as to what I'm missing? Thanks in advance. -GTBacchus^(talk) 15:41, 8 September 2009 (UTC)

An absolute value defines a topology, and it is continuous with respect to this topology. Let $|\cdot |'$ be another absolute value on ${\overline {K}}$ which satisfies the properties above. If $x=(x_{n})_{n=0}^{\infty }$ is a Cauchy sequence, then $x=\lim _{n\to \infty }x_{n}$ in ${\overline {K}}$, thus $|x|'=\lim _{n\to \infty }|x_{n}|'=\lim _{n\to \infty }|x_{n}|=|x|^{*}$. Now, the catch with this argument is that I am tacitly assuming that the topology induced by $|\cdot |'$ coincides with the natural topology on ${\overline {K}}$, and I don't quite see why this should be true in general. — Emil J. 16:59, 8 September 2009 (UTC)

Also, I've taken a look at Complete metric space#Completion as you suggest, and I do not see anything related to absolute values there. The generic extension method described there only implies that there exists a unique (uniformly) continuous function $|\cdot |^{*}$ which extends $|\cdot |$. It does not assert that $|\cdot |^{*}$ is an absolute value (though this should be easy enough to prove), and more importantly, it does not assert anything about uniqueness wrt the class of not-necessarily-continuous absolute values. Are you sure that the claim you are trying to prove is even true? — Emil J. 17:07, 8 September 2009 (UTC)

No, I'm not sure. The claim I really want to prove involves a non-Archimedean absolute value (featuring the strong triangle inequality), but I think it might be true in general. Actually, the assertion is simply that the absolute value extends "naturally" to the completion. Uniqueness would be bonus. This comes from a set of lecture notes, and may very well be an ill-posed question. If it's not true, I suppose there's a counter-example?

I've already proven that $|\cdot |^{\star }$ is an absolute value satisfying the same AV1-AV3 axioms. -GTBacchus^(talk) 18:51, 8 September 2009 (UTC)

(ec) What you need is the universal property of the completion $\iota :K\to {\overline {K}}$. For any metric space $K$ and any uniformly continuous map $f:K\to M$ to a complete metric space $M$ there exists a unique uniformly continuous ${\overline {f}}:{\overline {K}}\to M$ such that $f=\iota \circ {\overline {f}}$ (you may say that ${\overline {f}}$ is the unique u.continuous extension of $f$ over ${\overline {K}}$, if you like to think $K$ as a subset of ${\overline {K}}$).

This situation may be described in term of an adjunction, if you like the language of category theory. In any case the universal property allows to extend the absolute value and the algebraic operations from $K$ to ${\overline {K}}$ together with their identities. There are just some small detalis to work out; in case just ask. --pma (talk) 17:07, 8 September 2009 (UTC) PS: as to the unicity matter: precisely, ${\overline {f}}$ is the unique continuous extension, and it turns out to be uniformly continuous with the same modulus of continuity of $f$. --pma (talk) 17:13, 8 September 2009 (UTC)

Ok, so it's the unique continuous extension, but there may be other extensions that aren't continuous? This makes sense to me, and reminds me of a function that appears in Counterexamples in Analysis. There's an additive but non-homogeneous function on $\mathbb {R}$ that sends each number to the sum of the coefficients of its representation in terms of the basis of $\mathbb {R}$ as a vector space over $\mathbb {Q}$. That function, as I recall, is additive, but not remotely continuous. Its graph is dense in the plane, again IIRC. -GTBacchus^(talk) 18:51, 8 September 2009 (UTC)

Sure. So, the whole matter is extremely simple; should something be still unclear to you, we can help you to clarify it. --pma (talk) 19:56, 8 September 2009 (UTC)

A good algebraic number theory book should handle this. I recall that Janusz's Algebraic Number Fields had a very readable and careful discussion, and Lang's Algebraic Number Theory probably covers this adequately as well. Unfortunately I have neither of these books with me and I don't remember the answer! I'll check Milne (available here) if I remember to do so later.

Regarding proving the claim, I got stuck at the same place as Emil, in proving that $(x_{n})_{n=0}^{\infty }=\lim _{n\to \infty }\iota (x_{n})$ in ${\overline {K}}$. I think that if you drop requirement 2 of an absolute value, that $|xy|=|x||y|$, then you get a counterexample in $\mathbb {R} /\mathbb {Q}$ following the plan you mentioned. I think the best bet for finding a counterexample without dropping requirement 2 is in the completion $\mathbb {C} _{p}$ of ${\hat {\mathbb {Q} }}_{p}$. Eric. 216.27.191.178 (talk) 00:56, 9 September 2009 (UTC)

Here's a fairly trivial counterexample. Pick K such that Aut(K/K) is nontrivial (for example, we may take K = Q(i) with the usual complex absolute value, so that K = C is algebraically closed). Fix a nonidentical field automorphism σ of K which is identical on K, and define |x|' = |σ(x)|^*. On the one hand, it is easy to see that |·|' is an absolute value extending |·|. On the other hand, since σ is nonidentical, and K is dense in K, σ is discontinuous, and a fortiori it is not an isometry. Thus, there are x, y such that $|x-y|^{*}\neq |\sigma (x)-\sigma (y)|^{*}=|\sigma (x-y)|^{*}=|x-y|'$, i.e., |·|' ≠ |·|^*. — Emil J. 10:45, 9 September 2009 (UTC)

Interesting. I'm looking at Algebraic Number Theory, by Jürgen Neukirch, page 124, from the section on "Completions". Ahem:

Finally, one proves the uniqueness of the completion $\left({\overline {K}},|\cdot |\right)$: if $\left({\overline {K}}',|\cdot |'\right)$ is another complete valued field that contains $\left(K,|\cdot |\right)$ as a dense subfield, then mapping $|\cdot |-\lim _{n\rightarrow \infty }a_{n}\mapsto |\cdot |'-\lim _{n\rightarrow \infty }a_{n}$ gives a $K$-isomorphism $\sigma :{\overline {K}}\rightarrow {\overline {K}}'$ such that $|a|=|\sigma (a)|'$

How does this jibe with your counterexample, which looks perfectly reasonable to me? (By the way, what's the LaTeX for the angle-shaped "hat"? I'm using \overline in place of it now.) -GTBacchus^(talk) 15:49, 9 September 2009 (UTC)

What he claims is uniqueness up to an isomorphism (preserving the field structure, absolute value, and elements of K). It's perfectly consistent with the example, in fact, his σ is my σ. A hat is \hat in TeX: ${\hat {K}}$. — Emil J. 16:09, 9 September 2009 (UTC)

Ah, so the isomorphism preserves algebraic relations, but non necessarily the topology, because the absolute value in continuous in one case but not in the other? Is that right? -GTBacchus^(talk) 19:11, 9 September 2009 (UTC)

Let me summarize and state the whole thing as I see it: a list of simple facts:

1. Let K be a AV field (field with absolute value, i.e. AV1 AV2 AV3 hold). Then, $d_{K}(x,y):=|x-y|$ makes $K$ a metric space, and a topological field (the field operation are continuous).
2. There exists an AV field ${\hat {K}}$ such that $K$ is a sub AV field of ${\hat {K}}$ (i.e., it is a subfield of ${\hat {K}}$ and its AV is the restriction of the one of ${\hat {K}}$); moreover ${\hat {K}}$ is complete as a metric space and $K$ is dense in ${\hat {K}}$. Any such ${\hat {K}}$ is called an AV field completion of $K$.
3. The AV field ${\hat {K}}$ may be constructed starting from a metric completion $M,{\hat {d}}$ of the metric space $K,d_{K}$; then the AV and the field operations of $K$ extend uniquely to $M$ in such a way to make it an AV field ${\hat {K}}$ whose distance $d_{\hat {K}}$ coincides with the distance ${\hat {d}}$.
4. If ${\overline {K}}$ is another AV field completion of K, there exists a unique homeomorphism $\sigma :{\hat {K}}\to {\overline {K}}$ such that $\sigma _{|K}=id_{K}$; moreover, $\sigma$ is an isomorphism of AV fields.
5. Remark. (EmilJ example) If K is an AV field and H is a super-field of K, then H may admit several AV that make it a completion of K (essentially due to the fact that if K is not complete there is a lot of trascendence in the completion extension).

Note that there is an analogous situation in the context of completion of normed linear spaces, including the remark (that becames much simpler: there is a lot of linear independence passing to the completion.) Do you agree? There are possibly typos here and there.--pma (talk) 20:13, 9 September 2009 (UTC)

Is "homeomorphism" a typo? If not I'm confused about that since Emil J gave a convincincing argument that a map like that would be discontinuous, but maybe there's something I'm not getting about the topology involved here. Rckrone (talk) 23:19, 9 September 2009 (UTC)

From Merriam-Webster: "a function that is a one-to-one mapping between sets such that both the function and its inverse are continuous and that in topology exists for geometric figures which can be transformed one into the other by an elastic deformation".Alan (talk) 23:27, 9 September 2009 (UTC)

I think the question was not, "what does homeomorphism mean", but rather "is homeomorphism in this context a typo for homomorphism, in this case, of valued fields?" A homomorphism (no 'e') preserves algebraic structure, but is not necessarily continuous. Or, if it is necessarily continuous in the case of valued fields, that's a theorem. I'm also wondering about the continuity issue. -GTBacchus^(talk) 05:00, 10 September 2009 (UTC)

Not a typo: I wrote "a unique homeomorphism $\sigma$" to emphasize that $\sigma$ is unique even in the class of all homeomorphism that extend the identity of K forgetting the algebraic structure; I could have even written "a unique continuous $\sigma$": this unique continuous element $\sigma$ is then automatically an isometry of metric spaces and an isomorphism of AV field (in particular, a homomorphism). On the contrary, if we try to characterize $\sigma$ algebrically, we meet difficulties, exactly because (in general) there are a lot of non-continuous automorphisms of ${\hat {K}}$ that extend the identity of K, as EmilJ's remark showed. The completion of an absolute valued field is primarily a completion of a metric space: the algebraic extension follows the metric extension. There are other mixed topological-algebraic structures, where a topological property is characterized in a purely algebraic way (first example that cames to my mind: a complex-valued homomorphism of a Banach algebra is necessarily continuous); but this is not the case. Still I do not understand why you seem to be unsatisfied with a unicity property stated in terms of the topological structure. Here the natural thing is to think in terms of completion of metric spaces: in this more general and simpler category, completion is unique up to a unique isometry (or if you like, up to a unique homeomorphism that is automatically an isometry). Once this is clear, take it to the context of valued fields: the same $\sigma$ will be automatically a valued field isomorphism also. --pma (talk) 06:23, 10 September 2009 (UTC)

I'm sorry if I'm being dense about this (no pun), and I thank you for your patience. In your numbered points above, here's my understanding:

1. Start w/ an AV-field K, on which the AV and field operations are continuous.
2. There exists some metric completion of K that is an AV-field (with AV and field ops continuous) that contains K as a dense sub-AV-field.
3. You can get such a completion by completing K as a metric space to obtain K-hat, and then uniquely extending the AV and operations - all of which presumably remain continuous.
4. If there's some K-bar (different from K-hat) that does what K-hat does (completes K metrically, and extends the AV and algebra so they're still continuous), then K-bar and K-hat are equivalent via a unique homeomorphism that respects all the structure.
5. Instead of starting with a metric completion, we could take a super-field, which extends the algebra first, and add topological structure to it afterwards. Then we can get away with a discontinuous AV. Even though it restricts to the old AV on K, and gets along w/ the field operations, it's not a metric completion of K, so point #4 doesn't apply. That's why #3 is a more natural way to do it.

Is that all correct? -GTBacchus^(talk) 07:29, 10 September 2009 (UTC)

I understand where I was confused (thanks pma) and I think it's the same place, so I will give it a shot. Referring to Emil J's example, the map σ as a regular-old field isomorphism from K to itself is discontinuous with the topology induced by |.|*. However as AV fields, K with metric |.|* is distinct from K with metric |.|'. Call them K₁ and K₂. σ as a map between AV fields from K₁ to K₂ is continuous since open sets according to the topology induced by |.|* get mapped to open sets according to the topology induced by |.|' (instead of using |.|* at both ends like before), which can be seen by the fact that |x|* = |σ(x)|'. In other words σ preserves the AV as the transition is made from the one metric to the other. Rckrone (talk) 08:13, 10 September 2009 (UTC)

One way to think about it is that there are different ways to embed the metric completion of K into the superfield K, two of them being K₁ and K₂, and σ is the AV field isomorphism that relates them. They both contain the same sets of elements, so merely as fields they are identical, but with the added structure of a metric they are distinct. Rckrone (talk) 08:38, 10 September 2009 (UTC)

agreed. Also, they have the same set of elements, and they are isomorphic, but not via the identity map.--pma (talk) 09:24, 10 September 2009 (UTC)

(ec) Correct. As to point 1, note that the continuity of the structure mappings (sum, product, multiplicative inverse, absolute value) follows as a consequence of the AV field axioms and the choice of the topology on K (the metric topology of the distance d_K(x,y):=|x-y|), so it doesn't need to be assumed. In particular, the distance d_K makes

• $+:K\times K\to K$ a Lipschitz function;
• $\cdot :K\times K\to K$ a locally Lipschitz function (it's Lip on every disk); also, $x\to ax$ is Lipschitz (of constant |a|) from K to K;
• $(\cdot )^{-1}:K\setminus \{0\}\to K$ (multiplicative inverse) a locally Lipschitz function: precisely, Lipschitz on the complement of every nbd of 0 (just by AV2, because |1/x-1/y|=|x-y|/|xy|);
• $|\cdot |:K\to \mathbb {R}$ (the abs value) a Lipschitz function, just by AV3.

As to point 3, if M is any completion of K as metric space (hence a plain metric space with no algebraic structure) the above structure maps (+, $\cdot$, inverse, $|\cdot |$), thanks to the universal property of metric completion, have unique continuous extensions respectively to maps:

• ${\hat {+}}:M\times M\to M$ ;
• ${\hat {\cdot }}:M\times M\to M$ ;
• ${\hat {(\cdot )^{-1}}}:M\setminus \{0\}\to M$ ;
• ${\hat {|\cdot |}}:M\to \mathbb {R}$ ,

that a priori are only continuous maps: but then all the AV field axioms holds for them by continuity, since they hold on the dense subset K. In particular the distance of M is, of course, just ${\hat {|x-y|}}$. Technical detail: notice that, while the continuous extension of the sum and of the AV is an immediate application of the universal property of metric completion, it is a bit less immediate for the product and the inverse, because they are only locally uniformly continuous; the latter is possibly not even defined on a complete domain, M\{0}. But that difficulty is very easily bypassed: first extend uniquely by continuity the product map (respectively, the multiplicative inverse map) in the closed sets of MxM (resp., of M\{0}) where it is Lipschitz, then glue together the extensions using the unicity property, and observe that the result of the gluing is still locally Lip, hence continuous.

As to point 4, we can make a stronger statement putting a comma pregnant with meaning in the middle (making it a non-restrictive clause): K-bar and K-hat are equivalent via a unique homeomorphism , which respects all the structure. That is, the identity map on K extends to a unique homeomorphism between them, and, this unique homeomorphism turns out to be an AV field isomorphism. Is that OK? (note: I used everywhere the term AV field just because I'm not sure about the exact term, maybe you know). I think the other people above also agreed with this picture: maybe they may correct or add something. --pma (talk) 09:18, 10 September 2009 (UTC)

That's awesome. Thank you so much. Thank you all, who have commented here. I'm, like, smarter now. :) I'll check back for any further notes or corrections here, but I really feel comfortable with what we've just walked through, including the technical points about the product and multiplicative inverse maps not being uniformly continuous on the whole field.

Neukirch, in the book I consulted, calls the structure we're talking about a "valued field"; I don't know how standard that terminology is. -GTBacchus^(talk) 11:07, 10 September 2009 (UTC)

Poker and math

What problem, if any, do computers have in playing poker? Could a computer playing online poker win against human players? --Mr.K. (talk) 16:12, 8 September 2009 (UTC)

The combinatorial work necessary for optimal estimates of the strengths of others hands is hard, probably intractable. Coming up with good heuristics is also difficult. Dealing with the non-mechanistic issues (bluffing, modeling the other players) is also highly non-trivial. But yes, they can, and they do, regularly. My success at Sam Fox Strip Poker was limited (as was, to be honest, my motivation, due to the 320x200 greyscale images). --Stephan Schulz (talk) 16:34, 8 September 2009 (UTC)

Limit poker is much easier to do than no-limit, I believe. Have a look at Computer poker players Tinfoilcat (talk) 16:38, 8 September 2009 (UTC)

Yeah that's right. It's all about the pot odds (and more importantly the implied pot odds). If you need to put in x% of the pot to call and you expect to make a winning hand more than x% of the time then you've got good poot odds and you should call. (Obviously you might not make you hand and lose, but if you made the same call 1,000 times then you'd expect to make a profit). Like Stephan says: it's the bluffing that is hard to get around. You can make a really big bet so that the pot odds are so bad that the player wouldn't normally call, e.g. if you bet the twice the pot then the pot odds are will be so bad that you'd need to be able to make a winning hand 2 times in 3 (or better) to be able to call. But that's the bluff: I could have rubbush cards, but make a big bet so that the other player doesn't have the pot odds to call. ~~ Dr Dec (Talk) ~~ 17:13, 8 September 2009 (UTC)

Yes, limit is easier for a computer. There are far fewer options to consider (you can fold, call or raise by a fix amount, rather than fold, call or raise by any amount of your choice [as long as it is over the minimum raise]). --Tango (talk) 20:49, 8 September 2009 (UTC)

Hermite polynomials and Hermite Interpolation

Does anyone know if there is a connection between Hermite polynomials and Hermite interpolation? ...Other than the association with Charles Hermite.

When interpolating a polynomial it can help to use a basis other than simple powers of x so I thought the Hermite polynomial might be useful for Hermite interpolation in the same way as the Newton basis polynomials are used in the Divided differences algorithm.

Neither article in Wikipedia states such a relationship but there is a link to an external article on Hermite interpolation from the article on Hermite polynomials... but that could just have been places in error.

Thanks,

Yaris678 (talk) 19:38, 8 September 2009 (UTC)

Well, I do not see any connection at all between Hermite polynomials and interpolation problems. --pma (talk) 20:10, 8 September 2009 (UTC)

Derivative of a function between matrices

Hi there refdesk!

I've been completing a couple of old exercises in analysis and I've come upon the following:

Define f: M_n → M_n by f(A) = A³. Show that f is differentiable everywhere, then find its derivative.

Now I've realised I haven't a clue how to differentiate a function between matrices (M_n the set of real n x n matrices), nor even how the derivative is defined. Could anyone point me in the direction of an explanation or give me a very brief rundown of how to solve a question like this, which sounds like it's probably very simple once you know how!

Thanks a lot,

Typeships17 (talk) 20:38, 8 September 2009 (UTC)

Matrix calculus might help. --Tango (talk) 20:44, 8 September 2009 (UTC)

Doesn't the space of n × n matrices have a manifold structure? The space of n × n matrices would be an n²-dimensional manifold. So write a general matrix as M := (m_i,j), and then compute M³ = (m_i,j)³. Then you just need to check that each of the entries of M³ is differentiable as a function of (m_1,1,…,m_i,j,…,m_n,n). This is clearly true since each entry with just be a homogeneous polynomial of degree 3. ~~ Dr Dec (Talk) ~~ 20:50, 8 September 2009 (UTC)

Yes, it can be thought of as a family of n² functions each defined on $\mathbb {R} ^{n^{2}}$ and differentiated accordingly. --Tango (talk) 21:33, 8 September 2009 (UTC)

Thanks a lot guys, I'll try that link out now. In addition, I was wondering if anyone could explain to me why the derivative $Df\vert _{x}(h)$ of a function between Rⁿ and R^m is orthogonal to x in not-too-technical terms: I can get my head around the concept where m=1 and you can look at level sets, but I'm not sure what geometrical argument you can apply when mapping to more than 1 dimension. Thanks very much, Typeships17 (talk) 23:39, 8 September 2009 (UTC)

sorry but it is absolutely not clear what you mean. What's x? What's $Df\vert _{x}$? --pma (talk) 09:48, 9 September 2009 (UTC)

I think I know what s/he might be getting at. If f : R^m → Rⁿ is a submersion with x₀ a regular point of f and y₀ a regular value of f then the tangent space to {x | f(x) = y₀} at x₀ is given by the kernel of the differential of f evaluated at x₀ (i.e. the Jacobian matrix evaluated at x₀). In the case when n = 1 then the differential if just the gradient vector of f and its kernel is just the orthogonal space of the gradient vector. ~~ Dr Dec (Talk) ~~ 22:37, 9 September 2009 (UTC)

I've tried to answer what I think the question was talking about. See here. Let me know if I've made any errors. ~~ Dr Dec (Talk) ~~ 12:10, 10 September 2009 (UTC)

Consider the big laugh expansion: $(A+H)^{3}=A^{3}+(HAA+AHA+AAH)+o(\|H\|)$: your $f$ is differentiable at any $A$ and the Fréchet differential is $Df(A)H=HAA+AHA+AAH$. This holds for $A$ in any Banach algebra of course.--pma (talk) 07:04, 9 September 2009 (UTC)

thanks for the correction... ehm, I made an error counting to 3 ;-) --pma (talk) 13:51, 9 September 2009 (UTC)

Is this a hyperbola or...

Draw line segments connecting (a,0) to (0,1), (a-1,0) to (0,2) etc. up to (1,0) to (0,a). Is the curve that is approximated a hyperbola (well, half of one), or is it something else? Donald Hosek (talk) 21:57, 8 September 2009 (UTC)

You mean an envelope like this? It certainly looks like a hyperbola to me, although I can't guarantee it. --Tango (talk) 22:08, 8 September 2009 (UTC)

As the very article you linked states, it's a parabola, not a hyperbola. Algebraist 22:22, 8 September 2009 (UTC)

Don't stop when y gets down to 0; keep going. Likewise keep going in the opposite direction. If you do that, you'll see that the curve does not have asymptotes. So it cannot be a hyperbola. Michael Hardy (talk) 02:09, 9 September 2009 (UTC)

Indeed, to get an equilateral hyperbola one has to take all line segments whose X and Y intercepts have constant $product$, not $sum$. BTW, using a ruler, this fact gives a more precise way of drawing a hyperbola, compared with drawing by points. --pma (talk) 15:13, 9 September 2009 (UTC)

(ec) I think you're talking about the envelope of the family of lines given by

$y=\left({\frac {1+t}{t-a}}\right)x+(1+t)\ ,$

where, for each value of t with 0 ≤ t ≤ a – 1, we get the equation of a line. Well, let

$F((x,y),t):=y-\left({\frac {1+t}{t-a}}\right)x-(1+t)\ .$

Then for a fixed t₀ the corresponding line is given by

$\{(x,y)\in \mathbb {R} ^{2}:F((x,y),t_{0})=0\}.$

In this case, the envelope is given by solving F = F_t = 0 in terms of x and y. Doing this gives the envelope, or discriminant:

${\mathcal {D}}_{F}(t)=\left({\frac {(a-t)^{2}}{a+1}},{\frac {(1+t)^{2}}{a+1}}\right).$

This curve has the property that all of the lines are tangent to it. The question is now: is this curve part of a hyperbola? ~~ Dr Dec (Talk) ~~ 22:25, 8 September 2009 (UTC)

One way to visualize that it's a parabola rather than a hyperbola without finding the explicit function is to consider what happens when you continue drawing lines past where you stopped: (0, 0) to (0, a+1), (-1, 0) to (0, a+2),... (-n, 0) to (0, a+1+n),... The slope of these lines go through the vertical and approach 1. You can also continue in the other direction to see a similar trend. Rckrone (talk) 22:49, 8 September 2009 (UTC)

Yes, that's a(n arc of) parabola. Let's call s>0 and t>0 respectively the X and the Y intercept of a segment in the family. If c is the constant value of s+t in the family of segments, following Declan's approach you'll get

$x^{1/2}+y^{1/2}=c^{1/2}$;

and if you rationalize and change coordinates after a π/4 rotation (i.e. x=(u+v)/2, y=(u-v)/2) you get a normal form of a parabola.

BTW, you can generalize this picture a bit. Given any $\alpha >0$ and $c>0$, the envelope of the family of all segments within the first quadrant with x and y intercepts satisfying

$s^{\alpha }+t^{\alpha }=c^{\alpha }$

is a curve of equation

$x^{\frac {\alpha }{\alpha +1}}+y^{\frac {\alpha }{\alpha +1}}=c^{\frac {\alpha }{\alpha +1}}$

(yes, $\alpha =2$ gives 2/3).

You do not need any differential calculus; it's a plain consequence of the Hölder inequality in $\mathbb {R} ^{2}$ (and for your particular case $\alpha =1$, it's just a Cauchy-Schwarz).--pma (talk) 08:28, 9 September 2009 (UTC)

hey!now that's all in the first link, too! ;-) --pma (talk) 13:54, 9 September 2009 (UTC)

September 9

Effect of type I vs Type II error cost difference

What work has been done on how an appropriate p-value for null hypothesis testing changes when type I errors are more or less costly than type II? (What piques my interest about this question is its implications for understanding the evolution of apophenia.) NeonMerlin[1] 05:21, 9 September 2009 (UTC)

To be precise, p-values are not affected by such considerations: those values are merely calculated from statistics taken from the sample; what type I/II error considerations can do, is affect confidence levels: Bear in mind that p-value can be defined as the least significance level under which your null hypothesis is rejected (here significance level is 1-α, where α stands for confidence level).
In regard with type I/II error issues, set any other considerations aside (such as sample size, which has an inverse relation with both error probabilities), the thumb rule is that the more relatively costly type I error is, the less significance level you are willing to set, or in other words, the more probable is that the statistic from a sample will fall into the non-rejection interval (or the more probable is that the p-value is greater than the (prefixed) significance level. Pallida  Mors 18:21, 10 September 2009 (UTC)

Why angle side angle?

http://imgur.com/MpXTh.jpg

It's given that angle 1 is equal to angle 2, angle 3 is equal to angle 4, point D is the midpoint of line BE, and that line BC is equal to segment DE. BC is equal to BD via substitution, thus triangle ABD is congruent to triangle EBC via ASA, but it's bugging the hell out of me. Shouldn't it be AAS, seeing as how BE is outside of B and C? How would BC help me anyway? Ugh. I hope I phrased that in an understandable way. --Glaesisvellir (talk) 22:05, 9 September 2009 (UTC)

Once you have two angles, you can easily calculate the third using the fact that the sum of all the angles in a triangle is 180 degrees. That means it really doesn't make any difference whether you have ASA or AAS. --Tango (talk) 22:34, 9 September 2009 (UTC)

Thank you. The only problem is, what if the measurements of the other two angles are not listed, and measuring them is not an option? --Glaesisvellir (talk) 00:50, 10 September 2009 (UTC)

If you're trying to prove two triangles are congruent, such as the ones in the image you linked, you can argue that since two of the angles of each triangle are equal, then the third angles of each triangle must be equal. Referring to the image, that argument is $m\angle A=180^{\circ }-m\angle 1-m\angle 3=180^{\circ }-m\angle 2-m\angle 4=m\angle E$. Rckrone (talk) 03:53, 10 September 2009 (UTC)

September 10

Weaker variant of Schur for sum of two = irreducibles: two eigenvalues?

Hello,

I was considering endomorphisms of \mathbb{C} G- modules, and something funny happened with the eigenvalues. I was wondering if this were true:

If V is the \mathbb{C} G-module V with V=A+B, with A and B two isomorphic irreducible modules, then every endomorphism f has at most two complex eigenvalues. Moreover, every vector v\in V is either an eigenvector of f, or is in an invariant subspace with respect to f of dimension two.

It looks like it true, but I've never seen it before so that is why I am in doubt.. I was also wondering if there were more general things like this that are known and very important.

Many thanks, Evilbu

Homomorphisms behave nicely with respect to direct sum decompositions. In your case, where V is the direct sum of two copies of A, End(V) is isomorphic to the 2x2 matrix ring with entries in End(A). Since you work over C, this is really the 2x2 matrices over C. Now at least it's obvious that there can be at most two eigenvalues. Tinfoilcat (talk) 16:46, 10 September 2009 (UTC)

maths?

what is maths —Preceding unsigned comment added by 59.96.142.2 (talk) 17:05, 10 September 2009 (UTC)

math is using numbers to get new numbers.Accdude92 (talk) (sign) 17:07, 10 September 2009 (UTC)

Mathematics is the science and study of quantity, structure, space, and change. Mathematics is both the queen and the handmaiden of all science. --LarryMac | Talk 17:42, 10 September 2009 (UTC)

I'd say she's also science's drinking buddy, confidant, and occasional tennis doubles partner. -GTBacchus^(talk) 17:22, 11 September 2009 (UTC)

Mathematics is an esoteric discipline whose higher reaches are guarded jealously by those in the know so that it is hard for new entrants to join them. Thus in some ways mathematics, like other learned professions, provides a good living for its practitioners.86.152.79.120 (talk) 22:33, 10 September 2009 (UTC)

That is the opposite of the truth. Lots of popularizations get written by mathematicians, but far fewer people are willing to read them than mathematicians might wish. Michael Hardy (talk) 21:00, 11 September 2009 (UTC)

Is it possible for "maths" not to use numbers and still be properly considered as maths? 202.147.44.84 (talk) 22:30, 10 September 2009 (UTC)

Yes. Algebraist 22:33, 10 September 2009 (UTC)

Yes. Topology and geometry don't involve a great deal of numbers (they pop up from time to time, but mostly just to index things). --Tango (talk) 22:43, 10 September 2009 (UTC)

Someone has to properly ingrain the idea in humans' minds that mathematics is not solely about numbers, equations or formulas. Although number theory is a branch of mathematics, it is not concerned with "the arithmetic of big numbers" - such operations may be done with a computer (some cannot but that is irrelevant to mathematicians). Furthermore, although mathematics does deal with equations in a vague sense (see algebraic geometry, field theory and algebraic number theory), a mathematician is not really concerned with "solving equations" but rather how the solution sets behave. In algebraic geometry, one is concerned with the "geometric" properties of solution sets; in field theory, one is sometimes concerned with algebraic extensions of fields (an algebraic extension of a field (F) is simply another field containing F whose every element is the solution of a non-zero polynomial with coefficients in F). For instance, a non-mathematician may ask which numbers can be a solution to an equation. A mathematician will then formalize this: Define a number to be algebraic over the rational numbers if it is the solution to some non-zero polynomial with rational coefficients. The mathematician will now investigate this in greater depth by analysing some non-trival (first) examples of algebraic numbers. For instance, $x^{2}-2=0$, has one solution equating to ${\sqrt {2}}$; this is "non-trivial" since it "cannot be written as a fraction" (again, this idea must be formalized). After this, a mathematician will attempt to modify this example and investigate what one may obtain - what about, $x^{2}+1=0$, he/she will ask. I urge the OP to attempt this - it is remarkable how this question leads to a totally new branch of mathematics. Lastly, "formulas" are not of interest in mathematics - if at all they serve a purpose it is how one obtains them and not what they state. In differential geometry, these so-called "formulas" exist but the way in which they are obtained leads to new insights in differential topology, for instance.

I have told you what mathematics is not but I have not told you what it is. I would love to do so but I think that the best basic examples of this may be found in Group (mathematics) and Vector space - wonderfully written articles which may be understood by even someone who cannot count (by the way, one need not know how to count to do mathematics - I would not know how to count if my reputation as a mathematician (as viewed by "society") were not at stake). Ideas relating to numbers are basic to mathematics and existed more than 2000 years ago. If I had to give a "definition" for mathematics (which numerous people have attempted to do), I would define it as: "The logical deduction of truth from a set of axioms." The more experience I have with the subject, the further this definition changes in my mind - I am sure that I will never be able to find a definition which suits me perfectly. --PST 05:06, 11 September 2009 (UTC)

Isn't your description a bit extreme in the opposite direction? I'd say that some formulas really are of interest, et cetera. Also, would a person who cannot count know the meaning of "two vectors" and "17th century"?

My personal definition for mathematics used to be along the lines of "The logical deduction of truth from a set of axioms", but lately I'm of the opinion that "mathematics" is a name for two distinct things - one is a method of obtaining truths (using axioms, proofs and the like), and one is a body of knowledge in subject matters which are traditionally explored with this method (quantity, structure, ...). Theoretical computer science is explored with the mathematical method, but is not usually considered to be included in the mathematical body of knowledge. -- Meni Rosenfeld (talk) 06:05, 11 September 2009 (UTC)

By my comment I implied that "counting" is simply an assignment of symbols and sounds to commonsense. In particular, it is not essential that mathematics be based on an agreement upon universal terminology; each and every person may define his own truth and do mathematics. It is just a question of mathematical interest - can a deep theory emerge from this, for instance. Thus, if a student finished high school without knowing how to count, I would not form any opinion of his/her mathematical ability but I would ask him/her for a substitute that he/she has developed.

With regards to formulas, I do not know whether they are really of interest except for the method that obtained them. For instance, one can give a formula for the area of a circle but that does not mean much except for the applications that it may produce. The general ways in which one may obtain that formula (which I found interesting the first time I saw them) are of actual interest. Although a formula can sometimes reflect the method which produced it, the method will have deep insights which the formula cannot encapsulate. --PST 02:19, 12 September 2009 (UTC)

Factor in a load of things, then factor out what it's all about. That's maths by my WP:OR, though it doesn't say what it's all for. Dmcq (talk) 10:55, 11 September 2009 (UTC)

I'd say mathematics is the study of abstract structures. And two abstract structures are the same precisely if they're isomorphic. Hence mathematics is the study of truths that are preserved by isomorphism. Michael Hardy (talk) 21:01, 11 September 2009 (UTC)

I think that Michael's definition above concurs with what Cantor once famously quoted - "Mathematics is not the study of objects but the relations between them." I agree that mathemaics is the study of truth preserved by isomorphism, as well. Nevertheless one runs into the point that Meni Rosenfield stated above - "Theoretical computer science is explored with the mathematical method, but is not usually considered to be included in the mathematical body of knowledge." If one thinks about it, all fields require the logical deduction of truth but the nature of the "axioms" change. For instance, scientists will logically deduce truth from experimental evidence to build a theory. Although my next comment will be debatable, I think that it is the case - all bodies of knowledge are in some sense mathematical in nature. This is perhaps explained by the fact that there are many mathematicians in history who have explored other fields (to name a few, Emanuel Lasker and John Nash, but there are far more). In effect, every field is mathematics once the "truth" has been decided. Deciding the truth is what is unique to the field. --PST 02:19, 12 September 2009 (UTC)

Then where does Philosophy come into play then? Because isn't that like the general case of all the sciences? After all, all science originated from philosophy... so would it make more sense to say that math is philosophical in nature, or that philosophy is mathematical in nature?--12.48.220.130 (talk) 20:56, 16 September 2009 (UTC)

You can choose the one you like more among these: Definitions of mathematics --78.13.142.34 (talk) 20:52, 13 September 2009 (UTC)

Paper towels

If you know the radius of the roll and the radius of the inner cylinder, what is the formula for the length? Thickness is harder to measure, so I guess what I really want is a formula for how much is left when the radius is half what it started out to be.Vchimpanzee · talk · contributions · 22:23, 10 September 2009 (UTC)

The length of the paper towel on the roll will depend upon the thickness of an individual sheet of paper; there is no way around that. If you find it difficult to measure the thickness of one sheet, you could try measuring the thickness of ten or so sheets (say, wrapped around the roll to make it easier) and divide by ten.

Once you know the thickness of an individual sheet, there are a few ways to estimate the length. One way might be to measure the volume of the paper towels (by measuring the volume of a cylinder with the larger radius and subtracting the volume of the inner cylinder), and use that and the thickness to get the length. If you are careful, you will notice that this method does not depend upon the height of the roll (if you leave the height h as a variable when computing the volume, you will notice the h cancels out at the end when computing the length of the paper towel).

As for your second question, "how much is left when the radius is half...", I don't understand what you are asking. Do you mean, what proportion of the paper towel has not been used when the outer radius has been halved? Eric. 216.27.191.178 (talk) 23:02, 10 September 2009 (UTC)

Usually they tell you on the package how many sheets are in a full roll, and the size of each sheet. That lets you figure out the length of a full roll (call this L). The length will also be (approximately) proportional to the area of the annulus that you see when you look at the roll from the end. Let R be the radius of the full roll and r be the radius of the inner cylinder. Then the area of the annulus is $A_{1}=\pi (R^{2}-r^{2})$. The thickness of the annulus is of course R-r. When enough towels are used up that the outer radius of the roll is half what it started at (i.e. it is R/2), the area of the annulus is of course $A_{2}=\pi \left(({R/2})^{2}-r^{2}\right)$. The length of that partial roll will be ${A_{2} \over A_{1}}L$. If what you are really asking is the length once the thickness is half what it started as, then just find the new outer radius $R_{2}=r+{{R-r} \over 2}$ and similarly calculate the new area and new length. —Preceding unsigned comment added by 70.90.174.101 (talk) 23:21, 10 September 2009 (UTC)

A similar question was posed here. Clearly, a common mathematical thought induced by the closet --"When I'll get out, I'll post on the RD/M" ;-) --pma (talk) 06:32, 11 September 2009 (UTC)

September 11

Matrices and equations

I know that one may use matrices to solve linear equations of the form

${\begin{cases}a_{1}x_{1}+a_{2}x_{2}=c_{1}\\a_{3}x_{1}+a_{4}x_{2}=c_{2}\\\end{cases}}$

Where the c and a are constants. There are a number of methods, but the one I know best is Cramer's rule, which makes the solving of x₁ and x₂ a relatively simple matter:

$x_{1}={\frac {\begin{vmatrix}c_{1}&a_{2}\\c_{2}&a_{4}\end{vmatrix}}{\begin{vmatrix}a_{1}&a_{2}\\a_{3}&a_{4}\end{vmatrix}}}={{c_{1}a_{4}-a_{2}c_{2}} \over a_{1}a_{4}-a_{2}a_{3}}$

$x_{2}={\frac {\begin{vmatrix}a_{1}&c_{1}\\a_{3}&c_{2}\end{vmatrix}}{\begin{vmatrix}a_{1}&a_{2}\\a_{3}&a_{4}\end{vmatrix}}}={{a_{1}c_{2}-c_{1}a_{3}} \over a_{1}a_{4}-a_{2}a_{3}}$

This is all well and good, but Cramer's rule hits a stumbling block as soon as we have an x₁ or x₂ with a degree ≠1. My question is this: can matrices be used to solve systems of equations that are not linear, such as those containing polynomials and more complex expressions? For example, could a matrix be used to solve this?

${\begin{cases}3x^{2}-2y+5z=356\\4x+y-10z^{\frac {1}{2}}=145\\6x+11y+15z=512\\\end{cases}}$

Thanks for the help. 58.168.51.217 (talk) 07:41, 11 September 2009 (UTC)

To a great extent, no. Solving polynomial systems involves much, much more mathematics, both in the theory and algorithms. Special systems, of course, can be reduced to linearity, e.g. (changing your example a little) you can solve this

${\begin{cases}3x^{2}-2y+5z^{\frac {1}{2}}=356\\4x^{2}+y-10z^{\frac {1}{2}}=145\\6x^{2}+11y+15z^{\frac {1}{2}}=512\\\end{cases}}$

first as a linear system in x² ,y , z^1/2. --84.221.209.213 (talk) 09:14, 11 September 2009 (UTC)

Can you point me toward that mathematics? And I think I'm seeing what you've done here. What we find for z (for example) through Cramer's is really the square root of the z we're looking for. Is that correct? 58.168.51.217 (talk) 10:54, 11 September 2009 (UTC)

Correct. But are you more interested in explicit computations, algorithms, approximation of solutions, or more in the qualitative aspect: what is the shape and the structure of the set of solutions, in particular, existence results and so on. The two aspect are linked, of course. Another point is, what kind of functions a have you in your mind: polynomials, analitic functions, differentiable functions, continuous. Each has a whole theoried behind: commutative algebra algebraic geometry differential geometry nonlinear functional analysis degree theory fixed point theory...

Maybe the most simple interesting result, which is satisfactory both theoretically and for the practical point of view, is the contraction principle. --131.114.73.84 (talk) 12:09, 11 September 2009 (UTC)

If you don't put in additional restrictions then even if all the coefficients are integers and there is only one equation there is no general solution. This is Hilbert's tenth problem. Dmcq (talk) 12:25, 11 September 2009 (UTC)

...yes: talking about integers solutions. That's another main issue: does the OP want integer, real or complex solutions? That changes completely theory and methods.--131.114.73.84 (talk) 13:23, 11 September 2009 (UTC)

There is also the issue that even single-variable polynomial equations may have no easy solution, e.g. x⁵−x−1 = 0 has no solution in radicals. — Carl (CBM · talk) 13:31, 11 September 2009 (UTC)

As an applied mathematician, I would point you to Newton's Method. If you just want numerical values for x, y, z that solve the nonlinear system, that is you best bet. If you want to know if you can express the solution algebraically (i.e. something like x = sqrt(5)+log(4)), then you'll have to ask the pure mathematicians...

BTW: Cramer's rule is a pretty ineffcient way to solve a linear system of equations. Gaussian Elimination is much better. Cramers rule requires you to find n+1 nxn determinants - the most efficient way to calculate a determinant is ... Gaussian Elimination! So better to use that straightaway.195.128.250.134 (talk) 23:25, 11 September 2009 (UTC)

Irrational perfect set

How can I find a nonempty perfect set in $\mathbb {R}$ which has no rational numbers in it. The book in which I found this problem has not discussed measure theory as yet, and so I do not want to use it in any way. Thanks--Shahab (talk) 07:45, 11 September 2009 (UTC)

for instance play with the binary expansions: fix an irrational number c, and consider all numbers 0<x<1 whose binary expansion has x_2k=c_k for all k, the other digits being free.--pma (talk) 08:03, 11 September 2009 (UTC)

Re OP: what do you mean by "perfect"? If you mean perfect set, the set of all irrationals is an example. If you mean perfect closed set, you can do it by a Cantor-set like construction, where you add an extra technique that uses an enumeration of the rationals. I don't want to say more in case this is a homework problem; once you see how the hint works you will be fine. — Carl (CBM · talk) 13:34, 11 September 2009 (UTC)

Perfect sets are always closed. The set of all irrationals is not perfect. — Emil J. 13:50, 11 September 2009 (UTC)

You're right, of course; I am so used to saying "perfect closed set" that I have to think about whether the terms are redundant. In any case, we have three solution techniques now. — Carl (CBM · talk) 13:53, 11 September 2009 (UTC)

More abstractly, you could begin with the inclusion of the Cantor space 2^ω into the Baire space ω^ω and compose this with a homeomorphism between Baire space and the irrational numbers. — Carl (CBM · talk) 13:40, 11 September 2009 (UTC)

No Carl this isn't homework. (Unfortunately I don't go to school). Anyway I thought about pma's hint and came up with the following reasoning. I want to confirm whether this is correct. Let c be an irrational number in [0,1] and $E:=\{x\in [0,1]:x_{2k}=c_{k}\}$. Clearly as the 2kth digit of any $x\in E$ doesn't follow a pattern so x is irrational. Moreover any neighborhood of such an x is bound to contain a $y(\neq x)\in E$ as other then the even placed digits all others may be freely chosen. So all points of E are its limit points. Also if y is any limit point outside E then let 2t be the first position from the left in the decimal expansion of y such that y_2t doesn't coincide with c_t, and let e be the rational number in (0,1) with 1 at the 2t position and 0's elsewhere. Then (y-e,y+e) doesn't contain any element of E. Hence E is closed. --Shahab (talk) 19:05, 11 September 2009 (UTC)

Correct! (Alternatively, you can prove that E is homeomorphic to the Cantor space 2^ω quoted by Carl, certainly a compact space with no isolated points.) --pma (talk) 20:41, 11 September 2009 (UTC)

Life expectancy of the oldest

Today's passing away of Gertrude Baines made me think of the kind of question my old physics prof liked to ask: A general question that requires an understanding of what is relevant, which assumptions to make, and how to apply the correct mathematical methods. Maybe I shouldn't post this here, because what I'm looking for is not a mathematically precise result, but rather an estimate of the order of magnitude. I wanted to figure this out myself, but I need to do other things now; and because it is connected to a current event I rather posted it here for others who might enjoy it as long as it's fresh:

What is the life expectancy of someone who just became the oldest person? Or, asked differently, how long is the expectation value for the time a person holds the title "oldest person"? Of course, you could just go to Oldest people#Oldest living people since 1955 and divide the observed time by the number of people, but how would you go about it if you didn't have that information? If you don't like ignoring what you already know, you might go the other way and ask: Given the observed life expectancy of the oldest, what conclusions can we draw about Senescence § Theories of aging? — Sebastian 20:23, 11 September 2009 (UTC)

The fact that they are the oldest person isn't relevant. You would calculate their life expectancy in the same way as with anyone else, using a life table. Life tables aren't particularly accurate at the upper end due to the very small sample size. If someone is the older person ever, then there aren't any data points to base the expectation on. You could try and extrapolate based on the last few years you do have data for, but that wouldn't be very reliable at all. --Tango (talk) 20:35, 11 September 2009 (UTC)

What the conditional expected lifetime is, given that the person is the oldest person, but not given the age, is a perfectly valid question in the right context. But I think the "right context" for me might involve knowing more than we're given here. Michael Hardy (talk) 20:57, 11 September 2009 (UTC)

I think the events are independent. How does someone dying 1000 miles away affect the chances of another person dying? Even if they were both in some kind of cohort initially. An insurance company might revise it's estimate of the probability based on the new evidence, but I think the actual probability remains the same.--RDBury (talk) 21:23, 11 September 2009 (UTC)

Well, yes, you could work out the probability distribution of the age of the current oldest person and use that, combined with the life tables, to get the life expectancy of a randomly selected oldest person, but I'm not sure why you would want to. --Tango (talk) 21:41, 11 September 2009 (UTC)

Jeanne Calment's answer at age 120 was "A very short one!".John Z (talk) 22:41, 11 September 2009 (UTC)

I guess she already said that at age 90, when she negotiated the reverse mortgage.  ;-) — Sebastian 07:45, 12 September 2009 (UTC)

The general statistical methods for such problems are described in extreme value theory. Robinh (talk) 06:23, 12 September 2009 (UTC)

Yup, that's basically what I've been looking for. I'm a bit surprised that it takes its own theory and that it was discovered so late - I expected it to be rather simple exercise. — Sebastian 07:45, 12 September 2009 (UTC)

Using discrete crime data points to estimate relative rates by geography

I have a set of datapoints that represent individual murders-- each datapoint has a longitude and latitude. How can I produce an estimate for any geographic point, such that the map could be painted with an overlay representing the estimated murder rate for a given area.

Obviously, there will be limits to how accurate the estimates can be, just trying to find a good ballpark sort of measure.

I had thought about doing something along the lines of:

Estimated rate at a point ~= Sum over all data points(1/distance to murder)

But this seems sort of arbitrary, in that, it's something I just made up. Perhaps murder rate should use an inverse square rule instead. Or maybe there is some more advanced procedure for estimating a location-based-rate using a random sampling of data points.

Advice greatly welcomed.

(also know, murder rate isn't the actual issue being explored, it's just easier to describe) --Alecmconroy (talk) 20:47, 11 September 2009 (UTC)

Since when did this forum get so morbid? Anyway, you probably want to find some sort of data analysis/data mining software. This kind of problem occurs in business all the time, though more people who might want to by a product than people who get murdered, and there is commercial software out there to help with it. I don't know if I'm allowed to mention a specific product but try Googling "mapping software" as a first step. It's an interesting problem but it's already being worked by people who are actually getting paid to do it.--RDBury (talk) 21:37, 11 September 2009 (UTC)

Yes, that is an interesting Rorschach isn't it-- when asked to come up with a meaning for a binary discrete samples, I just think murder. Murder has a certain binariness to it in a way that "crime" or "disease" or "profit" just lack. lol I guess I am morbid.

Any way, thanks to the commenters, I think I'm on the right trail. RDBury (or others)-- do we know offhand of any such software that would be easily available to do this? I actually already have the data in google maps, so if it can do it, that would be extra awesome-- but I can't find any mention it in the documentation right now. --Alecmconroy (talk) 09:20, 12 September 2009 (UTC)

Sounds like you want some kind of 2D histogram. Or, more sophisticatedly, density estimation. Yaris678 (talk) 21:47, 11 September 2009 (UTC)

Differential equation

Just wondering if anyone could give a hint (or can solve)

d2[fn(x)]/dx2 = -k/[fn(x)]2

Starts at f(0)=real number, f'(x)=0. Any ideas? (not advanced student)83.100.250.79 (talk) 22:45, 11 September 2009 (UTC)

It's a one dimensional motion with an inverse square law force. Multiply by d[fn(x)]/dx and integrate. You'll get the conservation of energy, that gives a first order autonomous equation, that you can integrate again. --78.13.143.41 (talk) 23:18, 11 September 2009 (UTC)

Got stuck at "Multiply by d[fn(x)]/dx and integrate." - left hand side integration by parts is giving me more dⁿ[f(x)]/dxⁿ ...83.100.250.79 (talk) 23:38, 11 September 2009 (UTC)

Use substitution. --COVIZAPIBETEFOKY (talk) 02:52, 12 September 2009 (UTC)

Substitute what? There's not a lot there.83.100.250.79 (talk) 11:00, 12 September 2009 (UTC)

I'm not going to do all your homework for you. Both integrals can be evaluated using the technique of substitution. Figure out what u should be in each integral, such that the expression inside the integral is of the form f(u)du. --COVIZAPIBETEFOKY (talk) 15:23, 12 September 2009 (UTC)

It's not homework, it's curiosity. And I'm looking for a link to an answer, or more helpful answer, I'm familiar with integration by substitution, but I'm stuck. Is this a well known differential. Can someone link to a method to solve it. Thanks.83.100.250.79 (talk) 15:34, 12 September 2009 (UTC)

If you do the multiplication suggested then on the LHS you will have a function (d(fn(x))/dx) and its derivative - that kind of integral is generally done by substituting u=(the function) (in this case, u=d(fn(x))/dx). On the RHS you have an expression involving a function (fn(x)) and the derivative of that function, so you should substitute u=fn(x). I haven't actually tried it, but that ought to work. --Tango (talk) 16:24, 12 September 2009 (UTC)

I still don't understand the multiplication method... I found a substitution: I found this [2] which gives the answer - I suppose that's what user:78... meant when they said "its one dimensional motion with inverse square law" - I found it by using that as a search term.

It'll probably make sense in a few moments once I've worked it through...83.100.250.79 (talk) 16:51, 12 September 2009 (UTC)

You just multiply both sides by d[fn(x)]/dx before you integrate, it makes for easier integrals (which you can do using the substitutions I described). --Tango (talk) 02:09, 13 September 2009 (UTC)

September 12

Simplify integral

Can this be simplified? ${\frac {d}{dv}}\int _{-\infty }^{\infty }\phi _{1}(v_{1})\left(\int _{-v_{1}}^{v-v_{1}}\phi _{2}(v_{2})dv_{2}\right)dv_{1}$

Probably some bad choice of variables here, I'll rewrite it first.

${\frac {d}{dv}}\int _{-\infty }^{\infty }f(y)\left(\int _{-y}^{v-y}g(x)dx\right)dy$

Thanks in advance--Yanwen (talk) 03:54, 12 September 2009 (UTC)

I get f*g where * means convolution. I'm kinda rusty with this type of problem so take for what it's worth.--RDBury (talk) 05:05, 12 September 2009 (UTC)

This appears to be an exercise (I'm guessing homework) in using the Leibniz integral rule and the first fundamental theorem of calculus. 67.122.211.205 (talk) 19:09, 12 September 2009 (UTC)

Ahha. Leibniz integral rule was what I was missing.

${\frac {d}{dv}}\int _{-\infty }^{\infty }f(y)\left(\int _{-y}^{v-y}g(x)dx\right)dy$

$\int _{-\infty }^{\infty }{\frac {\partial }{\partial v}}f(y)\left(G(v-y)-G(-y)\right)dy$

$\int _{-\infty }^{\infty }f(y)g(v-y)dy$--Yanwen (talk) 01:39, 13 September 2009 (UTC)

what is the value of k that must be added to 7,16,43,79 so that they are in proportion?

what is the value of k that must be added to 7,16,43,79 so that they are in proportion? —Preceding unsigned comment added by 122.168.68.145 (talk) 09:32, 12 September 2009 (UTC)

5.--RDBury (talk) 14:16, 12 September 2009 (UTC)

only if 16=19...Tinfoilcat (talk) 16:19, 12 September 2009 (UTC)

In proportion to what? --Tango (talk) 16:20, 12 September 2009 (UTC)

My interpretation was that you're supposed to add a constant k to each term and end up with a bit of a geometric series (i.e. al, al^2, al^3, al^4). This problem does not have a solution Tinfoilcat (talk) 16:28, 12 September 2009 (UTC)

Perhaps they don't have to be consecutive terms from a geometric series? Or could it just be that one must divide the next, which divides the next, etc.? --Tango (talk) 16:41, 12 September 2009 (UTC)

The OP probably means that the first and second number should be in the same ratio as the third and fourth. In this case, RDBury's answer is correct. -- Meni Rosenfeld (talk) 19:49, 12 September 2009 (UTC)

As Meni stated, RDBury's answer is correct BUT, doesn't this strike anyone as a HW problem? hydnjo (talk) 21:08, 12 September 2009 (UTC)

Fourier Transforms

Resolved

Could anyone let me know whether I'm doing this wrong?

For $f(x)=\left\{{\begin{array}{lr}cos(x):|x|<{\frac {\pi }{2}}\\0\,\,\,\,\,\,\,\,\,\,\,\,\,:|x|>{\frac {\pi }{2}}\end{array}}\right.$, do we have ${\tilde {f}}(k)=\int _{\frac {-\pi }{2}}^{\frac {\pi }{2}}f(x)e^{-ikx}\,dx$ (that's the definition I'm using, so I know that bit's correct) $=Re(\int _{\frac {-\pi }{2}}^{\frac {\pi }{2}}e^{ix}e^{-ikx}\,dx)=Re[{\frac {e^{(1-k)ix}}{(1-k)i}}]_{\frac {-\pi }{2}}^{\frac {\pi }{2}}=(*)\,Re[{\frac {-i\cdot 2i}{1-k}}\,sin(1-k)({\frac {\pi }{2}})]=2{\frac {sin((1-k)({\frac {\pi }{2}}))}{1-k}}$. Is that correct?

Because I thought ${\tilde {f'}}(k)=ik{\tilde {f}}(x)$, but then $f'(x)=-sin(x)$ so ${\tilde {f'}}(k)={\tilde {-sin(x)}}={\frac {-1}{i}}Im[(\int _{\frac {-\pi }{2}}^{\frac {\pi }{2}}e^{ix}e^{-ikx}\,dx]$ which unless I'm being stupid becomes 0 at $(*)$, doesn't it? But then that doesn't equal $ik\cdot 2{\frac {sin((1-k)({\frac {\pi }{2}}))}{1-k}}$, does it? Perhaps I'm being dense.

Following that, I'm trying to show 'by considering the F.T. of the above function and its derivative', that $\int _{0}^{\infty }{\frac {{\frac {\pi ^{2}}{4}}-cos^{2}(t)}{({\frac {\pi ^{2}}{4}}-t^{2})^{2}}}=\int _{0}^{\infty }{\frac {t^{2}-cos^{2}(t)}{({\frac {\pi ^{2}}{4}}-t^{2})^{2}}}={\frac {\pi }{4}}$, but I haven't a clue how to do it! I know it's related to Parseval's relation, but other than possibly moving the 2 integrals together and canceling a $({\frac {\pi ^{2}}{4}}-t^{2})$, then showing it's equal to 0 (which still doesn't solve the problem fully), i'm not sure how to move along, or whether that's even a good idea. Could anyone suggest anything perhaps?

Thanks a lot,

Delaypoems101 (talk) 23:50, 12 September 2009 (UTC)

Let me point out one error you made in computing the Fourier Transform of $f(x)$. It is true that $\cos(x)=Re\{e^{ix}\}$; however

$\int \cos(x)e^{-ikx}dx=\int Re\{e^{ix}\}\;e^{-ikx}dx\neq Re\{\int e^{ix}\;e^{-ikx}dx\}$.

Instead use the relation $2\cos(x)=e^{ix}+e^{-ix}$ to compute the Fourier transform integral (Hint: ${\tilde {f}}(k)$ is sum of two shifted sinc functions). Abecedare (talk) 00:11, 13 September 2009 (UTC)

Yeah, we can caluculate the Fourier transform directly using integration by parts twice.

${\hat {f}}(\xi )=\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\!\!\cos(x)e^{-ix\xi }\ dx={\frac {2\cos \left({\frac {\pi }{2}}\xi \right)}{1-\xi ^{2}}}$

As for the first integral, well, you have some problems. For example, the integrand has an even-ordered pole when $t=\pm {\frac {\pi }{2}}$. This singularity is non-removable since the numerator of the integrand is always positive for real t. In fact

$\int _{0}^{y}{\frac {{\frac {\pi ^{2}}{4}}-\cos ^{2}(t)}{({\frac {\pi ^{2}}{4}}-t^{2})^{2}}}\ dt\to \infty \ {\mbox{as}}\ y\to \infty \ .$

As for the second integral, well, again:

$\int _{0}^{y}{\frac {t^{2}-\cos ^{2}(t)}{({\frac {\pi ^{2}}{4}}-t^{2})^{2}}}\ dt\to \infty \ {\mbox{as}}\ y\to \infty \ .$ ~~ Dr Dec (Talk) ~~ 01:19, 13 September 2009 (UTC)

Oh god, how stupid of me, there's no minus in the numerator - sorry about wasting your time! Do things work better with $\int _{0}^{y}{\frac {{\frac {\pi ^{2}}{4}}\cos ^{2}(t)}{({\frac {\pi ^{2}}{4}}-t^{2})^{2}}}\ dt$ and $\int _{0}^{y}{\frac {t^{2}\cos ^{2}(t)}{({\frac {\pi ^{2}}{4}}-t^{2})^{2}}}\ dt$?Delaypoems101 (talk) 02:08, 13 September 2009 (UTC)

No matter - got it sorted: thanks all! Delaypoems101 (talk) 03:41, 13 September 2009 (UTC)

September 13

Path problem

Someone told me about this puzzle:

			o	o
		o	o	o	A
o	o	o	o	o	o	o
o	o	o	o	o	o	o
o	o	o	o	o	o	o
	B	o	o	o
		o	o

The object is to go from A to B, passing through all "o"'s but one. Not sure how to do it. Is it even possible? It seems like I can only miss out an even number of "o"'s. Professor M. Fiendish, Esq. 03:41, 13 September 2009 (UTC)

What are the movement rules? Algebraist 03:50, 13 September 2009 (UTC)

Only horizontally and vertically but not diagonally. Professor M. Fiendish, Esq. 04:06, 13 September 2009 (UTC)

Can an 'o' be traversed more than once? —Anonymous Dissident^Talk 04:39, 13 September 2009 (UTC)

No. Professor M. Fiendish, Esq. 04:52, 13 September 2009 (UTC)

Seems to be impossible. If we color the board with a checker pattern, any path must alternate between white and black squares. A and B are the same color so any path connecting them has to go through an odd number of 'o's. Rckrone (talk) 06:32, 13 September 2009 (UTC)

That's assuming you're not allowed to travel on the blank squares. Is that right? Otherwise it would be pretty easy. Rckrone (talk) 06:37, 13 September 2009 (UTC)

Oops, I checked the puzzle again and there were four rows of complete "o"'s.

			o	o
		o	o	o	A
o	o	o	o	o	o	o
o	o	o	o	o	o	o
o	o	o	o	o	o	o
o	o	o	o	o	o	o
	B	o	o	o
		o	o

Is it possible now? Professor M. Fiendish, Esq. 07:42, 13 September 2009 (UTC)

No. A and B are now on different colors, so any path has to go through an even number of 'o's, but now there are also an even number of 'o's on the board. Rckrone (talk) 08:04, 13 September 2009 (UTC)

Just as a point of interest, this is basically a graph theory problem. The board is a bipartite graph, and so for a pair of vertices A and B, the lengths of any two paths between A and B have to have the same parity. Rckrone (talk) 08:14, 13 September 2009 (UTC)

The parity argument fails if you're allowed to go through the same 'o' more than once. I assume that's something you forgot to include because it's pretty trivial otherwise. This kind of parity argument is very common in puzzles and recreational mathematics. If you wanted to pass through every o, just once, the parity argument fails and it turns out you can do it.--RDBury (talk) 08:17, 13 September 2009 (UTC)

You want A and B on different colours. And the problem is quite easily solved going though the o's once and only once. I've put the path below as where a path of light would go reflected by some mirrors :)

			/	\
		/	/	\	A
/	.	/	/	\	/	\
\	.	.	/	\	/	.
/	.	.	.	.	\	.
\	.	.	.	\	\	/
	B	\	/	/
		\	/

Dmcq (talk) 20:22, 13 September 2009 (UTC)

Just read the question again. It says to leave out one 'o'. Funny condition. It is impossible then. The version with only three lines of o's between is possible then though. Dmcq (talk) 20:26, 13 September 2009 (UTC)

I think Rckrone has already proven that the 3-line version is impossible. -- Meni Rosenfeld (talk) 20:53, 13 September 2009 (UTC)

Table puzzle

Sorry for the influx of puzzles, but I read this in a puzzle book somewhere. Any ideas?

There is a 6-by-6 table. Fill each cell in the table with 1, 2, 3, 4, 5, 6, 7, 8 or 9 to the following rules (R2 meaning row 2 and C6 meaning column 6):

R1: The sum of numbers placed in this row is 21.

R2: The sum of numbers placed in this row is 39.

R3: This row is filled in by six consecutive numbers, in order.

R4: In this row, its largest and smallest numbers are placed next to each other.

R5: There are exactly two 9's in this row.

R6: Sum of all the numbers placed in this row is 21.

C1: There are exactly two 9's in this column. Its smallest number is 2.

C2: This column is filled in by six consecutive numbers, in order.

C3: There are exactly two 3's in this column. The sum of numbers placed in this column is 21.

C4: There are exactly two 8's in this column. There is no number 4 in this column.

C5: There are exactly two 6's in this column. The sum of numbers placed in this column is 40.

C6: There is no number 7 in this column.

Every cell must be occupied! Professor M. Fiendish, Esq. 11:17, 13 September 2009 (UTC)

Q. Does R1 imply 123456 or can i use 1 twice etc?77.86.47.174 (talk) 18:52, 13 September 2009 (UTC)

Note that R5 says there are two 9's in this row. So, there seems to be no rule as far as only one of each digit per row. That's why I stopped thinking about it. It would probably take a very long time to figure out. 39 would also be 4, 5, 6, 7, 8, 9 if no repeats were allowed. Note also that C5 says the sum is 40, which is impossible with no repeats. StatisticsMan (talk) 19:29, 13 September 2009 (UTC)

IMO this isn't a very mathematical problem. The rules are complex and of different types so it seems impossible to try anything but a trial and error approach. You can narrow it down to 2 possibilities for row 3 and column 2 pretty easily, but after that it seems like the possibilities increase exponentially so you'd need a computer to explore them all. I'm not really stumped so much as I think I have better things to do with my time. Do you happen to know if the solution is supposed to be unique?--RDBury (talk) 02:04, 14 September 2009 (UTC)

Not unique. Probably not even close. Just by trying random ideas, I hit on this:

 231861
 941799
 456789
 363x9y
 977869
 383124

Which satisfies all the rules for many different x,y combinations. one of them has to be the "smallest" in the row, meaning a 1 or a 2, since it's next to the 9, but the other one can be almost anything as long as x!=4, x!=8, and y!=7. And if you make x=8, you get a whole new family of solutions by modifying R1/C4 and R1/C6, still plenty of ways to make them add up to 9 without disturbing anything. This ruleset is busted. 69.245.227.37 (talk) 03:48, 14 September 2009 (UTC)

Sample space

Alright, so I'm in a graduate statistics course and this is a very basic question, first question in book. But, I got counted off for it and I have talked to the professor and the TA and they don't seem to understand at all what I'm saying. So, I'll just give you the question but not my answer or the professors and see what you think. Also, I give you the definition of sample space in the book.

1.1 For each of the following experiments, describe the sample space.

(d) Record the weights of 10-day-old rats.

The set, S, of all possible outcomes of a particular experiment is called the sample space for the experiment.

Thanks for your ideas. StatisticsMan (talk) 16:26, 13 September 2009 (UTC)

I'm not a statistician, but based on the information you've given I would say the sample space is the positive real numbers. Or it could be $\mathbb {R} _{+}^{n}$, where n is the number of rats you are measuring. I can't tell if "a particular experiment" refers to each of the rats individually or the rats as a group. --Tango (talk) 16:44, 13 September 2009 (UTC)

Well, my answer was the nonnegative integers, whereas the professor says the answer is all nonnegative reals. However, allow myself to explain my answer because I did explain what I was doing. In the previous part of the problem, part (c), which is a similar problem, I said basically that the sample space could be all nonnegative reals. But, if we are measuring things, perhaps we pick some unit and we measure to the nearest unit. So, I could also see the answer being nonnegative integers. So, then on this part I just said it COULD be nonnegative integers again. The point I want to make here is it specifically says "Record the weight". It is not "The actual weight". If it were actual weight, of course the answer is all nonnegative or positive reals... of course perhaps we could put some upper bound like 100 pounds. Any number in there is possible. But, it says record the weight. So, someone comes to me and says record the weights. I grab a scale which measures in grams to the nearest gram. The only possible weights I am going to get are integers. It's not possible for 1.2847858 to come up. So, the sample space in that case is integers, is it not???? That is my experiment, that is the sample space that goes with it. All reals is actually wrong in that case, is it not? Or, all reals is right but P(not an integer) = 0. Does this make sense? StatisticsMan (talk) 01:47, 14 September 2009 (UTC)

I see what you are saying, but unless your level of precision happens to be to the nearest whatever unit you are using (or more), it won't be integers. If we are measuring in kilograms and precision is to the near gram, say, we will probably get non-integer answers. Yes, strictly speaking it is still a discrete set, but we don't know what discrete set it is, so we have to go with the smallest set that contains all possible answers, and that is the positive reals. (Definitely not the non-negative reals - you can't have a rat with a weight of 0.) --Tango (talk) 02:03, 14 September 2009 (UTC)

September 14

chess variables and states

Is it more proper to think of the game of chess as having 32 variables with 65 states (64 in-play states and 1 out-of-play state) or as having 64 variables with 33 states (32 occupied and 1 unoccupied state)? -- Taxa (talk) 02:19, 14 September 2009 (UTC)

There isn't one objectively proper way to think about it. It's just a question of what the most useful scheme is for what you happen to be doing with the information. Is there a particular problem you're trying to solve? Rckrone (talk) 02:31, 14 September 2009 (UTC)

I'm just trying to figure the best way to think of it. I mean if you do the same for tic-tac-toe then it would seem the positions (squares) always remain while the players vary so positions as variables seems better. -- Taxa (talk) 03:04, 14 September 2009 (UTC)

Okay, if you are not going to buy that then how about the idea that there are programs that can reduce many-valued logic equations to minimum form. If you had a computer big and fast enough to assign it the task of reducing all known winning games to minimum form then you might come up with a way to win at chess all the time. -- Taxa (talk) 03:33, 14 September 2009 (UTC)

Most serious chess programs these days use bitboards. 67.122.211.205 (talk) 04:30, 14 September 2009 (UTC)