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May 13

Unique Factorization Domains

Suppose I have a ring extension of Z, say Z[w].

• How can I tell whether or not Z[w] is a unique factorization domain?
• What if w happens to be a root of 1? Are there tables that tell me the answer?

mike40033 (talk) 01:05, 13 May 2008 (UTC)

I'm learning about this too at the moment, you might be interested in the articles about the Stark-Heegner theorem and the Cyclotomic fields. -- Xedi (talk) 01:21, 13 May 2008 (UTC)

For your last question, Z[w] is a UFD iff it has class number 1, so the table you want is A061653 at the OEIS. According to that, there's a more extensive table in ' L. C. Washington, Introduction to Cyclotomic Fields'. Algebraist 13:48, 14 May 2008 (UTC)

Having checked that book myself, I now have the complete answer to your second question: adjoining a primitive nth root of unity to the integers gives a unique factorization domain iff n is one of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 40, 42, 44, 45, 48, 50, 54, 60, 66, 70, 84, 90. Algebraist 14:08, 14 May 2008 (UTC)

So $\mathbb {Z} [\zeta _{23}]$ is the smallest of these which is not a UFD. How do we find an element with two distinct factorizations? kfgauss (talk) 19:57, 14 May 2008 (UTC)

I think number theory books are likely to have pretty examples, but if you just want any old example: Factor the ideal (2) into prime ideals (2,x) and (2,y). Then x*y is in (2), so x*y = 2*z with z in the ring. 2 is irreducible, but neither x nor y is a multiple of 2, since (2,x) is not (2). By computer x, y, and z can be chosen irreducible (x=1+w^2+w^4+w^5+w^6+w^10+w^11, y=1+w+w^5+w^6+w^7+w^9+w^11, z=w^5+w^6+w^7+w^9+w^10+3*w^11+w^12+w^13+w^15+w^16+w^17), but it's been too long for me to remember if it can always be chosen that way (for arbitrary prime ideals in number rings). According to the computer, {1,2,x,y,z,x*y} are the only divisors up to units of x*y. JackSchmidt (talk) 23:09, 14 May 2008 (UTC)

I don't understand... are you taking w as a primitive 23rd root of unity? Just looking at them, it doesn't seem that x*y=2z... shouldn't x*y start with a 1, for a start? --Tango (talk) 23:29, 14 May 2008 (UTC)

I haven't done the complete calculation (that's what JackSchmidt's computer is for), for the initial 1 cancels with a $\sum _{i=1}^{22}w^{i}$. There's a lot of cancellation going on. Algebraist 00:41, 15 May 2008 (UTC)

Yes, w is a primitive 23rd root of unity. I think number theory texts will have much prettier examples. In principal, no calculations are needed to understand my example (you don't need to know the values of x,y,z), as long as you believe 2 is irreducible in Z[w]. If this argument is not familiar, then reduced cases are probably easier: Factor (2) in Z[sqrt(-5)] as (2,x)*(2,y), then x*y=2*z, 2 is irreducible, and 2 cannot divide x or y. In fact, in this case you can take x=y=1+sqrt(-5), and z=-2+sqrt(-5). JackSchmidt (talk) 03:22, 15 May 2008 (UTC)

Ok, I felt bad and dusted off a number theory book myself. Edwards's "genetic" (historical) introduction to algebraic number theory, "Fermat's Last Theorem" is quite good, and has detailed information on Q[w], both historical and computational. At any rate, it works an example similar to mine, but also notes they worked much too hard, since 47*139 is divisible by 1-z+z^21, but N(47)=47^22, N(139)=139^22, and N(1-z+z^21)=47*139 divides neither. He has already shown that no element of Z[w] has norm 47, so 1-z+z^21 is irreducible, and yet divides neither 47 nor 139, so unique factorization fails. In this example though, 47 is not irreducible. This is page 105, section 4.4, the work of Kummer. JackSchmidt (talk) 03:51, 15 May 2008 (UTC)

Thanks for the great examples! Did you somehow know (2) would fail to be prime, or is it simply a likely guess (which you then confirm with your computer algebra program)? kfgauss (talk) 16:47, 15 May 2008 (UTC)

If you asking "how did Jack think about it?", then yes: (2) almost always factors, and I just asked the computer if this was some bizarre exception. Had it not factored (if someone had perversely asked for 29'th roots of unity instead), then I would have just tried (3), and then when that didn't work, (5). If you are asking, "how should an intelligent person think about this?", then consider a maximal ideal M containing 2R in your ring R=Z[w]. Then R/M is a field in which 1+1=0, since 1+1=2 is in M. Now R is generated by 1 and w, and w is a root of (x^23-1)/(x-1), so R/M is a field of characteristic 2 generated by an element that is a root of x^23-1. Now the field F of 2^11 elements is such a field since 23 divides 2^11-1, and the nonzero elements of F are a cyclic group of order 2^11-1. Since F is a Galois extension of Z/2Z, it is the splitting field of x^23-1 over Z/2Z, and so R/M is contained as a subfield of F (indeed F=R/M). In particular, R/M has at most 2^11 elements, but R/2R has 2^22 elements, so M properly contains 2R. Since R is a Dedekind domain, M divides 2R, and 2R factors. JackSchmidt (talk) 20:26, 15 May 2008 (UTC)

Neat. I notice 29 divides 5^14-1. Thanks for all the details. kfgauss (talk) 07:21, 16 May 2008 (UTC)

Gerschgorin Theorem and Taussky's Result

Okay, so in my numerical analysis class, studying for the finals, I came across the Gerschgorin criterion. One of the exercises is the proof of the criterion which is split up into three parts. I have proved the first two parts but the third one (which I learned later is called the Taussky's theorem), I can't get. Here is everything I have done:

Let $C=(c_{i,j})$ be an arbitrary $d\times d$ singular complex matrix.
Then there exists a nonzero $x$ such that $Cx=0$.
Choose $l\in \{1,2,...,d\}$ such that $|x_{l}|=\max |x_{j}|$.
This maximal element is of course, nonzero.
By considering the $l$th row of $Cx$, I have already proven that
$|c_{l,l}|\leq \sum _{j=1,j\neq l}^{d}|c_{l,j}|$.

Next, taking a dxd matrix B and letting $\lambda$ be an arbitrary eigenvalue of B.
Substituting, $C=B-\lambda I$, I have also shown that $\lambda$ is in a Gerschgorin disc $S_{l}$ where
$S_{l}=\{z\in \mathbb {C} :|z-b_{i,i}|\leq \sum _{j=1,j\neq i}^{d}|b_{i,j}|\}$
And since, $\lambda$ was an arbitrary eigenvalue of B, the entire spectrum of B is contained the union of all the Gerschgorin discs.

Now here is the problem which I can't get. Let us suppose that the matrix B is irreducible and the previous inequality holds as an equality, i.e.
$|c_{l,l}|=\sum _{j=1,j\neq l}^{d}|c_{l,j}|$
for some $l$ between 1 and d. I need to show that this equality implies that
$|c_{k,k}|=\sum _{j=1,j\neq l}^{d}|c_{k,j}|$ for all k between 1 and d.
This would imply that if an eigenvalue is on the boundary of one Gerschgorin disc, then that eigenvalue will be on the boundary of all Gerschgorin discs. Any ideas on how to prove it? I have looked in several books but everyone just states it and calls it a famous result but no one proves it.A Real Kaiser (talk) 05:51, 13 May 2008 (UTC)

Proceed as in your proof of the first step and get to the point $|c_{l,l}||x_{l}|\leq \sum _{j=1,j\neq l}^{d}|c_{l,j}||x_{j}|$. Now if we have the equality you mention, what does that mean about those $x_{j}$'s for which $c_{l,j}\neq 0$? Now use that irreducibility is equivalent to strong connectivity of the directed graph with d vertices with an edge from vertex i to vertex j if and only if $c_{i,j}\neq 0$. kfgauss (talk) 07:16, 13 May 2008 (UTC)

Proving Integrability

On a completely different note, I am trying to prove some properties for this function.
Let $f:[0,1]\rightarrow [0,1]$ be defined as $f(x)={\begin{cases}[1/x]^{-1},&{\mbox{if }}x\in (0,1]\\0,&{\mbox{if }}x=0\end{cases}}$
where [x] is the truncation function (i.e. [x]=the integer part of x). Notice that at each unit fraction, the function has a jump discontinuity and in between, the function is a constant. We have already had some discussion about this function but now I am trying to prove that this function is Riemann integrable. Here are some facts that I proved:

Fact I

$f(x)$ is a non-decreasing function
Proof:
Let $x>y$

$\Rightarrow 1/x<1/y$

$\Rightarrow [1/x]\leq [1/y]$

$\Rightarrow [1/x]^{-1}\geq [1/y]^{-1}$

$\Rightarrow f(x)\geq f(y)$.

Fact II

$f(x)$ is continuous at $x=0$ from the right.

Proof:Let $\epsilon >0$ be given and then let $\delta =\epsilon /2$.

$x-0<\delta$

$\Rightarrow x<\delta$

$\Rightarrow x<\epsilon /2$

$\Rightarrow 1/x>2/\epsilon$

$\Rightarrow [1/x]\geq 2/\epsilon$ (for $\epsilon$ sufficiently small)

$\Rightarrow [1/x]^{-1}\leq \epsilon /2<\epsilon$

$\Rightarrow f(x)<\epsilon$

$\Rightarrow f(x)-0<\epsilon$

$\therefore \lim _{x\rightarrow 0^{+}}f(x)=f(0^{+})=f(0)=0$

Fact III

$f(x)$ is Riemann integrable. This will be shown by proving that for every $\epsilon >0$, we can find a partition p of [0,1] such that the upper sum U(f,p) minus the lower sum L(f,p) using p, is smaller than $\epsilon$.

Proof: For a given $\epsilon >0$, let $\epsilon _{1}=\epsilon /4$, and let N=the largest integer less than or equal to 1/$\epsilon _{1}$. This gives us that 1/N will be the next unit fraction. Hence x=1/N will be the next point (on the right) after $x=\epsilon _{1}$ where $f(x)$ will have a jump discontinuity. Let $\epsilon _{2}=\epsilon _{1}/2N$ and let the partition be

$P=\{0,\epsilon _{1},{\frac {1}{N}}-\epsilon _{2},{\frac {1}{N}}+\epsilon _{2},{\frac {1}{N-1}}-\epsilon _{2},{\frac {1}{N-1}}+\epsilon _{2},...,{\frac {1}{3}}-\epsilon _{2},{\frac {1}{3}}+\epsilon _{2},{\frac {1}{2}}-\epsilon _{2},{\frac {1}{2}}+\epsilon _{2},1\}$

This is a perfectly valid partition, because the division by 2N ensures that none of the points go out of order. The intervals will not overlap. The point of the first fact was to make the Riemann evaluation easier. Since the function is nondecreasing, in order evaluate the area of a Riemann rectangle, for the upper sum, we simply take the value of the function on the right endpoint, and for the lower sum, we simply take the value of the function at the left endpoint. Both facts also show (for my teacher) that the function does NOT become unbounded as x goes to zero. In fact, if we extend the function to the interval [-1,1], the limit from the left as x goes to zero also exists so I can show that the function is actually continuous at x=0. But, that is just something extra.

Now when the difference between the upper sum and the lower sum is taken, the rectangles which fall on the flat region, with the base being of the form $x={\frac {1}{n}}+\epsilon _{2}$ to $x={\frac {1}{n-1}}-\epsilon _{2}$ cancel out because the upper sum is the same as the lower sum. The same thing happens with the last rectangle. The only rectangles that remain are the columns of width $2\epsilon _{2}$ which have the jump discontinuity in the middle.

So,
$U(f,p)-L(f,p)={\frac {1}{N}}\epsilon _{1}+\left({\frac {1}{N-1}}-{\frac {1}{N}}\right)2\epsilon _{2}+\left({\frac {1}{N-2}}-{\frac {1}{N-1}}\right)2\epsilon _{2}+\cdots +\left({\frac {1}{3}}-{\frac {1}{4}}\right)2\epsilon _{2}+\left({\frac {1}{2}}-{\frac {1}{3}}\right)2\epsilon _{2}+\left(1-{\frac {1}{2}}\right)2\epsilon _{2}$

$U(f,p)-L(f,p)={\frac {1}{N}}\epsilon _{1}+\left({\frac {1}{(N-1)(N)}}\right)2\epsilon _{2}+\left({\frac {1}{(N-2)(N-1)}}\right)2\epsilon _{2}+\cdots +\left({\frac {1}{12}}\right)2\epsilon _{2}+\left({\frac {1}{6}}\right)2\epsilon _{2}+\left({\frac {1}{2}}\right)2\epsilon _{2}$

Since height of each rectangle is less than or equal to one, and the number of terms with $2\epsilon _{2}$ is precisely N-1, we have that

$U(f,p)-L(f,p)\leq \epsilon _{1}+\underbrace {2\epsilon _{2}+2\epsilon _{2}+\cdots +2\epsilon _{2}+2\epsilon _{2}+2\epsilon _{2}} _{N-1}$

$U(f,p)-L(f,p)\leq \epsilon _{1}+(N-1)(2\epsilon _{2})\leq \epsilon _{1}+N(2\epsilon _{2})=\epsilon _{1}+N\left({\frac {2\epsilon _{1}}{2N}}\right)=\epsilon _{1}+\epsilon _{1}=\epsilon /2<\epsilon$.

Therefore, the function $f(x)$ is Riemann integrable on the interval [0,1]. The reason, I put all of this up is that I just want to run it by a couple of people here to make sure that I have not made a stupid mistake and to see if this is "rigorous" enough. I will be presenting this to him (and maybe we can finally settle this argument once and for all about this function being Riemann integrable) and he is very particular about rigor. So, do you guys think that am I missing out some particular detail which I should include or is this good enough for a rigorous Mathematical argument? Any comments and suggestion will be welcome about the proof of all three facts. The third one is the most important one, of course. In addition, in the proof of the second fact, in line three, it is obvious that the line is true for a sufficiently small epsilon but how can I write that better? I don't really like it the way I have it but I need it there. Thanks everyone!A Real Kaiser (talk) 09:33, 13 May 2008 (UTC)

In case you don't know, any monotone bounded function on a bounded real interval is Riemann integrable. This is easy to prove (at least assuming Riemann's integrability criterion, which you seem to be doing): pick an equal width partition of mesh $\delta$, and the difference between the upper and lower Darboux sums is $(f(b)-f(a))\delta$. Your proof of III may be a special case of this argument; I don't know, as I haven't read it. Algebraist 10:59, 13 May 2008 (UTC)

Your proof of II is flawed: it is not the case that x < epsilon/2 implies floor(1/x) is at least 2/epsilon for sufficiently small epsilon. You can conclude, however, that floor(1/x) is at least 2/epsilon - 1, which is at least 3/2epsilon (say) for epsilon small enough. Algebraist 11:05, 13 May 2008 (UTC)

Revised Proof of Fact II

Algebraist, you are absolutely right. I just said the opposite of what is true. So here is the revised proof.

Proof:Let $\epsilon >0$ be given and then let $\delta =\epsilon /2$.

$x-0<\delta$

$\Rightarrow x<\delta$

$\Rightarrow x<\epsilon /2$

$\Rightarrow 1/x>2/\epsilon$

$\Rightarrow {\frac {1}{x}}-1>{\frac {2}{\epsilon }}-1$

$\Rightarrow [1/x]>{\frac {1}{x}}-1>{\frac {2}{\epsilon }}-1={\frac {2-\epsilon }{\epsilon }}$

$\Rightarrow [1/x]^{-1}<{\frac {\epsilon }{2-\epsilon }}<\epsilon$ because $\forall \epsilon <1$ we have that $2-\epsilon >1$

$\Rightarrow f(x)<\epsilon$

$\Rightarrow f(x)-0<\epsilon$

$\therefore \lim _{x\rightarrow 0^{+}}f(x)=f(0^{+})=f(0)=0$

Fact IV

Here is another thing I thought of. I will try to prove that $2x\geq f(x)\geq x$ for all $x\in [0,1]$.

Proof:
First, let {x}=x-[x] be the fractional part of a real number x.

Now we have that for all $x\in \mathbb {R} ,\exists ![x]$ and $\{x\}$ such that $x=[x]+\{x\}$ and $0\leq \{x\}<1$.

Now let us consider all $y\in \mathbb {R}$ such that $y\geq 1$.

We now have that $y=[y]+\{y\}<[y]+[y]=2[y]$.

$\Rightarrow y<2[y]\Rightarrow y/2<[y]$

$\Rightarrow {\frac {1}{2x}}<[1/x]$ for all $x\in [0,1]$

and using the fact that $[y]\leq y$ for all $y\Rightarrow {\frac {1}{2x}}<[1/x]\leq {\frac {1}{x}}$

$\Rightarrow 2x>[1/x]^{-1}\geq x$

$\Rightarrow 2x>f(x)\geq x$

and then using the squeeze theorem, I can show that the limit of f(x) from the right as x goes to zero is indeed zero and this function is most definitely not unbounded in [0,1]. I think that after these four facts combined, there is no question that f(x) is bounded and Riemann integrable in [0,1].

Fact V

In fact, we can also find the exact value of the integral as follows:

Since, the function as a jump discontinuity at every unit fraction, with 1/2 being the first one from the left as x goes to zero, and in between two consecutive unit fractions, the function is constant

(i.e. $f(x)={\frac {1}{n}}{\text{ for }}x\in \left({\frac {1}{n+1}},{\frac {1}{n}}\right]{\text{ for all }}n\in \mathbb {N}$), we can just find the area underneath the curve by drawing rectangles

with length ${\frac {1}{n}}-{\frac {1}{n+1}}$ with height ${\frac {1}{n}}$ and then add up the areas of all such rectangles.

$\int _{0}^{1}f(x)dx=\sum _{n=1}^{\infty }\left(\left({\frac {1}{n}}-{\frac {1}{n+1}}\right){\frac {1}{n}}\right)=\sum _{n=1}^{\infty }\left({\frac {1}{n^{2}}}-{\frac {1}{n(n+1)}}\right)$

$=\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}-\sum _{n=1}^{\infty }{\frac {1}{n(n+1)}}={\frac {\pi ^{2}}{6}}-\sum _{n=1}^{\infty }\left({\frac {1}{n}}-{\frac {1}{n+1}}\right)={\frac {\pi ^{2}}{6}}-1.$

The first sum is just from the Riemann Zeta function (the values are known) and the second sum is a telescoping series. We can split up the series because it is absolutely convergent. So, what do you guys think? Should I change something or is this enough? Any more stupid mistakes in what I wrote or my reasoning? Come on guys, don't be shy. I just need your input and some constructive criticism.A Real Kaiser (talk) 23:23, 13 May 2008 (UTC)

What purpose are you producing this for? I ask because this will affect the level of detail required in your arguments. For example, if asked to show this function was Riemann integrable, I would say 'clearly increasing and bounded, hence integrable' or possibly 'clearly measurable and almost everywhere continuous, hence integrable', and leave it at that, but I'd expect a bit more if this was set as an exam question or something. Algebraist 11:08, 14 May 2008 (UTC)

(per post on my talk) If this is for convincing a very confused person, I can't guess what might work (the function's bounded above by 1! How can it have infinite integral?). If I was going for full first-year-analysis style all-the-details rigour, though, I would be a bit more careful with V: it's not quite trivial that the Riemann integral has the right limit-related properties to give you that infinite sum. Algebraist 20:47, 14 May 2008 (UTC)

Ummm... Based on the discussion at Algebraist's Talk, it would be enough to prove the integral isn't infinite... Maybe... Using a property of multiplicative inversion that you've been exposed to repeatedly,

$x\in (0,1]\Rightarrow 1/x\in [1,\infty )$

Since we have the basic properties of the floor function: $[u]\leq u\land (\forall v\in [1,2),[v]=1)$ (showing that $[1,\infty )$ is mapped somewhere to its right and then that at least {1} is in the image of that map, so the interval in the next step is best possible), then we get

$\Rightarrow [1/x]\in [1,\infty )$

Using the multiplicative inversion property again:

$\Rightarrow [1/x]^{-1}\in (0,1]$

Now,

$((x=0\Rightarrow f(x)\in \{0\})\land (x\in (0,1]\Rightarrow f(x)\in (0,1]))$

$\Rightarrow (x\in \{0\}\cup (0,1]\Rightarrow f(x)\in \{0\}\cup (0,1])$

and we may simplify this result to

$0\leq f(x)\leq 1,{\mbox{ if }}x\in [0,1]$

Finally we get, for $a,b\in [0,1]$:

$\int _{a}^{b}0\,dx\leq \int _{a}^{b}f(x)\,dx\leq \int _{a}^{b}1\,dx$

(If more detail is needed in that last step, note that any partition of the interval (a,b) produces a Riemann sum for f which is termwise ordered exactly as written.)

With a little computation, we can compute and limit the bounds.

$0=\int _{a}^{b}0\,dx\leq \int _{a}^{b}f(x)\,dx\leq \int _{a}^{b}1\,dx=b-a\leq 1-0=1$

and I don't know how to make its finiteness more transparent. -- Fuzzyeric (talk) 03:41, 18 May 2008 (UTC)

Sum

"Let ABCDEFGH be an eight digit number where A,B,C,D,E,F,G,H are the digits 1,2,3,4,5,6,7,8 in some order. For example, A=4,B=1,C=3,D=2,E=5,F=6,G=7,H=8 gives 41325678. FInd other eight digit numbers in which the three digit number ABC is divisible by 7, the three-digit number BCD is divisible by 6, CDE is divisible by 5, DEF is divisible by 4, EFG is divisible by 3 and FGH is divisible by 2."

I know that E is 5 but don't konw the others. Can someone pls give me some hints as the maths exam is round the corner(to be exact, tomorrow)? tks. —Preceding unsigned comment added by Invisiblebug590 (talk • contribs) 10:26, 13 May 2008 (UTC)

How many other numbers does it want? It looks like there are quite a lot. I suggest starting at the end and working back. FGH being divisible by 2 tells you something about H, but doesn't restrict F or G, for example. You can then work your way up narrowing down your options bit by bit. --Tango (talk) 10:42, 13 May 2008 (UTC)

The "easy" way to solve this is a brute force computer program. You only have 8! = 40320 permutations to evaluate. My program has determined that there are five solutions to your question. Note that BCD, DEF and FGH are all divisible by 2, so that uses up all but one of your even mumbers for D, F & H. BCD and EFG are both divisible by 3, so B+C+D+E+F+G is also divisible by 3. Since the sum 1..8 = 36 then A+H will also be divisible by 3 which reduces the possibilities a lot. -- SGBailey (talk) 11:10, 13 May 2008 (UTC)

If it's any help, the way I found a solution to this was to observe that there are only two choices for F, one of which leads to a unique choice for G. Fixing F and G at these values allows everything else to be worked out without too much difficulty. Algebraist 11:13, 13 May 2008 (UTC)

E=5, DEF is divisible by 4 therefore EF=52 or 56. EFG is divisible by 3 therefore of the div 3s (522, 525, 528, 561, 564, 567) we can try 528, 561, 564, 567. -- SGBailey (talk) 22:22, 13 May 2008 (UTC)

The above was a reply to a comment that got itself deleted, Hence the apparent non-sequitur. -- SGBailey (talk) 22:25, 13 May 2008 (UTC)

The Answer is 73485612. FIgured it out in 15 min. This is how you do it (not very mathematic but it works):

the last 4 digits (EFGH) are even, so H is 2,4,6, or 8. You know in order to use all 8 numbers the last 5 will be every-other even odd even odd even (keep that in mind), so G will be either 1,3,5,or 7. Even next. So depending on what H is, F isnt that. E isthe same but with odd. You start by figuring out the last 4 first because to know the DEF, EF needs to be divisible my 4, and EFG needs to be divisible by 3. so if you make a little chart using this info. you have over 120 possiblities now. Now figure out the EFG set that E is 5, F is even and G makes the sum of the three a multiple of 3. You get 561H, or ABCD561H ( H is 2,4, or 8). Now you have 73248 left to choose from. This part is easier. D will be even, ABC will add to a multiple of 7 and BCD adds to 3 or a maultiple of 3. Looking at this you realize that the only 3 numbers that will give you a multiple of 7 is 734 (14). Now you just eliminated the one that ended with 4. 2 options left. D is 2 or 8. you see what will give you x3. so 7+4+8, 7+3+2, 3+4+8, 7+4+2, 7+3+2, 3+4+2. 348 works. Now you have your answer. A3485612 or 73485612. Its a lot easier to do than to hear.--Xtothe3rd (talk) 01:45, 14 May 2008 (UTC)

As mentioned above, there appear to be 5 answers to the problem, not just the two Xtothe3rd mentions. -- SGBailey (talk) 08:04, 14 May 2008 (UTC)

Unless I've missed the point, 73485612 is invalid as the 3-digit number 734 isn't divisible by 7.…81.159.11.144 (talk) 11:50, 14 May 2008 (UTC)

Yes. Xtothe3rd appears to be under the impression that ABC is a multiple of seven iff A+B+C is a multiple of seven. In fact ABC is a multiple of seven iff 2A+3B+C is a multiple of seven. Algebraist 13:41, 14 May 2008 (UTC)

ABCDEFGH = 23185674, 27385614, 37145286, 41325678, 41385672 I think. -- SGBailey (talk) 10:13, 16 May 2008 (UTC)

Logic circuit design

Please provide symbol/math tags in TeX/HTML for logic gates like NOT, AND, OR, XOR,... Currently, users are uploading image files to depict circuit diagrams. A XML version would be immediately useful beyond Wiki.Anwar (talk) 14:35, 13 May 2008 (UTC)

Feature requests should go to [1], this page is for asking maths questions. --Tango (talk) 18:21, 13 May 2008 (UTC)

Homeless Statistics in China

Hi, I am an academic student currently researching on China's demographic. I am having a difficult time locating the most current information concerning "the homeless percentage in China overall and by region or province and cities vs. rural." I would really appreciate it if anyone can provide me with the most up to date data or point me to the right direction on where to search.

Thank you for your time and help. Your prompt reply will be greatly appreciated.

Sincerely, username: darjeelinguru

Hi. This is really a question for the Humanities Reference Desk. This seems like difficult information to come by. You might start by looking at the page Demographics of the People's Republic of China, which has some related information, but not what you ask for specifically. Then you might look at the references and links at the bottom of that page. Best of luck.

Also, to sign your name automatically, type four tildes in a row (i.e. ~ four times in a row). kfgauss (talk) 18:29, 13 May 2008 (UTC)

We certainly don't make it any easier to choose a reference desk section for such questions by including the ambiguous term "statistics" as a sub-topic for the mathematics section. -- Meni Rosenfeld (talk) 18:40, 13 May 2008 (UTC)

We certainly don't but I think that (a little bit of) common sense goes a long way. Finance and Economics questions should go on the Hum desk but if I had a question on the Black-Scholes model, I'd definitely come here. But I have been here for a while so it would be easier for me to figure out what goes where. Zain Ebrahim (talk) 07:25, 14 May 2008 (UTC)

I would expect China to suppress any statistics on homelessness in their country, and to say "we don't have any homeless people here". If you point out a few of them, they would likely be gone the next day. StuRat (talk) 00:05, 17 May 2008 (UTC)

Mathcad functions

Using the Mathcad statistical functions how would I convert percentile to a z-score and vice versa? 71.100.14.205 (talk) 16:32, 13 May 2008 (UTC)

I don't know what functions Mathcad has built-in, but see error function and (more specifically) probit. --Tardis (talk) 16:03, 14 May 2008 (UTC)

Pi

When was pi first used in England how was it measured? —Preceding unsigned comment added by 82.71.27.169 (talk) 18:26, 13 May 2008 (UTC)

I couldn't be sure, and I am sorry if I am stating the obvious, but have you looked at Pi?

And what precisely do you mean by England?…81.159.11.144 (talk) 11:55, 14 May 2008 (UTC)

I wasn't aware "England" was an ambiguous term... --Tango (talk) 14:24, 14 May 2008 (UTC)

It is when projected back over at least two thousand years of history. Algebraist 15:24, 14 May 2008 (UTC)

Well, he could be referring to England itself, to Britain or the UK generally, or in the English national football team, to which of course the answer is never since their combined IQ isn't likely to get into the double digits. -mattbuck (Talk) 15:26, 14 May 2008 (UTC)

Ah, well that's not a matter of precision, it's a matter of accuracy. "England", "Britain" and "UK" are all different things, there is no ambiguity, just correct and incorrect usage. As for a 2000 year history - I'm not sure Western Europe did much significant maths back then, ideas about Pi probably only reached England a few hundred years ago and the borders have been pretty much fixed for over 1000 years. --Tango (talk) 15:53, 14 May 2008 (UTC)

My point (and it might well be false) was that there were Romans in (what is now) England two millennia ago, Romans tended to drag around Greeks to do any thinking that needed doing, and the Greeks were into geometry. Algebraist 16:06, 14 May 2008 (UTC)

True, there may well have been people using Pi back then. I was thinking of people studying it, which I don't believe anyone in England at that time did (at least not successfully enough to be remembered), since just using it would count as measuring it. --Tango (talk) 20:01, 14 May 2008 (UTC)

It is the case, regrettably, that "England" is often used in a casual, slipshod way by people who mean a wider geographical unit. This seems to cause particular offence to Scots, who are members of a nation, while those in the principality of Wales just add the slight to their general moroseness (it's possible that I am over-generalising here). Not being privy to the OP's reason for wanting to know, it's only a suspicion that he/she actually meant something wider. And of course the Romans reached Scotland and Wales anyway.…81.159.11.144 (talk) 23:19, 14 May 2008 (UTC)

It depends what you mean by "used" and "measured". We know that Archimedes wrote about geometric methods of estimating π and presumably his works were being copied out by at least some scribes somewhere in England in the first millenium - but did they understand what they were copying ? And the masons designing and building Anglo-Saxon churches must have had a working knowledge of the geometry of circles - but does that count as "using" π ? The earliest example I can find of an English mathematician who studied π is John Wallis who discovered the Wallis product in 1655. Gandalf61 (talk) 10:18, 15 May 2008 (UTC)

Numb3rs

An episode of Numb3rs involved a (false) proof for the Riemann hypothesis that was used as a ransom with the intent of breaking encryption. Did they get it mixed up with the P = NP problem, or are they that closely related? — DanielLC 23:33, 13 May 2008 (UTC)

It is suggested that a proof for the Riemann hypothesis may be a key factor in understanding and breaking the pattern of prime numbers, what would in turn speed up our best factorization algorithms of very large numbers. These are frequently used in cryptography for public keys. I guess that's their whole point. — Kieff | Talk 23:42, 13 May 2008 (UTC)

Is the idea that it's still exponential time, but faster than normal, or did they simultaneously (almost) prove that P = NP without even mentioning it? Also, it would have to be something in the way it was proved, or they'd just assume the Riemann hypothesis to be true and then just see if it works. In might not be formal, but if you can break encryptions, who cares. — DanielLC 23:59, 13 May 2008 (UTC)

I think they said something non-specific about the method of the proof giving an easy way to factor numbers. See [2] by Chris Caldwell and the links [3] and [4] in one of the replies. PrimeHunter (talk) 00:07, 14 May 2008 (UTC)

There's no evidence that factoring is NP-complete, and most complexity theorists think it isn't, so factoring being in P doesn't imply P=NP. I don't think it would even affect most theorists' faith in P≠NP very much; factoring is considered much easier than known NP-complete problems (though a polytime algorithm would still be very surprising). -- BenRG (talk) 01:06, 14 May 2008 (UTC)

P=NP doesn't have that much to do with practical ease of breaking encryption. In particular, being in P doesn't mean a problem is easy. Even a modest-looking polynomial like $n^{20}$ is more difficult in practice than an exponential like $2^{n}$. And it is believed that, if a proof for P=NP is found, it will involve huge polynomials, with degrees which could easily surpass millions. So even if the Riemann hypothesis is related to easier breaking of factorization-based systems, it doesn't mean it is in any way related to P=NP. -- Meni Rosenfeld (talk) 08:31, 14 May 2008 (UTC)

May 14

what is the diffrent with prism and pyramid?

--24.78.51.208 (talk) 00:17, 14 May 2008 (UTC)

Maybe you can find what you want in prism (geometry) and pyramid (geometry). PrimeHunter (talk) 01:55, 14 May 2008 (UTC)

A prism has two triangular faces and three rectangular faces; a pyramid has four triangular faces and one square face. Stifle (talk) 09:55, 14 May 2008 (UTC)

Not necessarily. You just described a triangular prism and a square pyramid. A prism is a solid that has a uniform cross-section, whereas a pyramid has a base and edges from the base which connect to some point. I hope my terminology is correct. --WikiSlasher (talk) 12:14, 14 May 2008 (UTC)

So basically, a pyramid has a vertex that's connected to all other vertices by an edge, whereas a prism does not have such a vertex. (There are other polyhedrons that have a vertex connected to all other vertices though.) – b_jonas 07:55, 16 May 2008 (UTC)

regression toward the mean

The eugenics article says: [Galton] reasoned that, since many human societies sought to protect the underprivileged and weak, those societies were at odds with the natural selection responsible for extinction of the weakest; and only by changing these social policies could society be saved from a "reversion towards mediocrity," a phrase he first coined in statistics and which later changed to the now common "regression towards the mean."[18] Isn't this a completely different concept from "regression towards the mean"? My understanding of regression is that it is symmetric with respect to both time and direction. So although extreme data points will regress to the mean, the total number of extreme data points does not increase or decrease on average. --MagneticFlux (talk) 03:00, 14 May 2008 (UTC)

Yes, the strict sense of what was meant in the quote is not the same as the current "regression towards the mean".

Ironically, one could plausibly argue that if Galton did fully understand the current meaning, he may have been less keen on his eugenical ideas (then again, speaking as no expert, maybe he already did understand, and/or was less keen). RttM implies that the impact of any eugenical activity would be less than one would naively think, and thus can actually be seen to be a mild discouragement to it. Then again, if one watches sports commentators ramble on about the current superlative players du jour, one can sympathize with a eugenicist's misunderstanding. Baccyak4H (Yak!) 14:16, 14 May 2008 (UTC)

Spearman-Rho, and Kruskal-Wallis tests

I was wondering, why is there a 6 in the spearman-rho formula, and why is there a 12 in the formula for a Kruskal-Wallis test? --Agester (talk) 23:27, 14 May 2008 (UTC)

Regarding Kruskal-Wallis, our article says that: "Notice that the denominator of the expression for $K$ is exactly $(N-1)N(N+1)/12$". Using the notation from the article, we can work out the value for the denominator:

$\sum _{i=1}^{g}\sum _{j=1}^{n_{i}}(r_{ij}-{\bar {r}})^{2}=\sum _{k=1}^{N}(k-{\bar {r}})^{2}$

$=\sum _{k=1}^{N}(k^{2}-2k{\bar {r}}+{\bar {r}}^{2})$

$=\sum _{k=1}^{N}k^{2}-2{\bar {r}}\sum _{k=1}^{N}k+N{\bar {r}}^{2}$

$=\sum _{k=1}^{N}k^{2}-2{\bar {r}}(N{\bar {r}})+N{\bar {r}}^{2}$

$=\sum _{k=1}^{N}k^{2}-N{\bar {r}}^{2}$

$={\frac {N(N+1)(2N+1)}{6}}-N{\frac {(N+1)^{2}}{4}}$

$={\frac {(N-1)N(N+1)}{12}}$

Notice that the transformation I did from $\sum _{k=1}^{N}(k-{\bar {r}})^{2}$ to $\sum _{k=1}^{N}k^{2}-N{\bar {r}}^{2}$ is exactly the same as in the article computational formula for the variance, and works because ${\bar {r}}=(N+1)/2$ is the average of the integers 1 through $N$. As for the very first equality, where I ditched the $r_{ij}$, this is because in the summation over $i$ and $j$, $r_{ij}$ varies exactly over the integers 1 through $N$, so the equality is simply changing the order of the summation (which is valid).

I don't know about Spearman-Rho, as our article doesn't give an explicit explanation for the formula. Eric. 86.153.201.165 (talk) 03:28, 15 May 2008 (UTC)

Ok, I've looked at the article on Spearman-rho formula in a bit more detail. It says "In principle, ρ is simply a special case of the Pearson product-moment coefficient in which the data are converted to rankings before calculating the coefficient." So if we look at Pearson product-moment correlation coefficient, we find the following formula:

$\rho ={\frac {1}{N}}\sum \left({\frac {X_{i}-\mu _{X}}{\sigma _{X}}}\right)\left({\frac {Y_{i}-\mu _{Y}}{\sigma _{Y}}}\right)$

This is the information we need. The formula on the Spearman-rho page:

$\rho =1-{\frac {6\sum d_{i}^{2}}{N(N^{2}-1)}}$

is the same as the previous formula, with some simplifications made. In our Pearson product-moment correlation coefficient, $\mu$ is the mean and $\sigma$ is the standard deviation of our population $X_{i}$ or $Y_{i}$. The additional information we have, is that the $X_{i}$ are exactly the numbers 1 through $N$, in some order; similarly for the $Y_{i}$. Because the mean $\mu _{X}$ and standard deviation $\sigma _{X}$ do not depend upon the order, we can compute these values. We have:

$\mu ={\frac {1}{N}}\sum _{i=1}^{N}X_{i}={\frac {1}{N}}\sum _{i=1}^{N}i={\frac {N+1}{2}}$

$\sigma ^{2}={\frac {1}{N}}\sum _{i=1}^{N}(X_{i}-\mu _{X})^{2}={\frac {1}{N}}\cdot {\frac {(N-1)N(N+1)}{12}}={\frac {(N^{2}-1)}{12}}$

Notice that our $\mu$ here is the same thing we had called ${\bar {r}}$ in our discussion above of Kruskal-Wallis. And our $\sigma ^{2}$ (i.e., variance) is the same as the denominator of $K$ which we calculated above for the Kruskal-Wallis formula, except divided by $N$. Now that we know what $\mu$ and $\sigma$ are, we can proceed to simplify down the Pearson product-moment correlation coefficient:

$\rho ={\frac {1}{N}}\sum \left({\frac {X_{i}-\mu _{X}}{\sigma _{X}}}\right)\left({\frac {Y_{i}-\mu _{Y}}{\sigma _{Y}}}\right)={\frac {1}{N}}\sum \left({\frac {X_{i}-\mu }{\sigma }}\right)\left({\frac {Y_{i}-\mu }{\sigma }}\right)$

$={\frac {1}{N\sigma ^{2}}}\sum (X_{i}-\mu )(Y_{i}-\mu )$

$={\frac {12}{N(N^{2}-1)}}\sum (X_{i}-\mu )(Y_{i}-\mu )$

It should be possible to continue in this way to get to the end formula (I haven't checked; also, keep in mind that $d_{i}=X_{i}-Y_{i}$):

$=1-{\frac {6\sum d_{i}^{2}}{N(N^{2}-1)}}$

but in any case we can already where the 6 (and the $N^{2}-1$, as well) is coming from. Eric. 86.153.201.165 (talk) 05:41, 15 May 2008 (UTC)

I've tried to keep up with your math. It's not making sense to me (Spearman-Rho). I tried various calculations with the information you provided but i just couldn't get back to the original formula (posted on the Spearman-Rho article).
However, for the Kruskal-Wallis test formula, my text book represents it a bit more different than on the article / the one you presented. It gives me (sorry I'm still not too good with wikipedia's math function thingy): H_obt = [ 12 / N(N+1) ][ Σ (R)²/n_i ] - 3(N+1)

I don't have time at the moment to answer your questions, but I'll see if I can get back to them in a few days and also fill in the step I skipped at the end (or maybe someone else will). Can I ask, did the part I wrote out above make sense, or is there something in particular you didn't follow? Don't worry about having trouble with using LaTeX: here's how you would typeset your formula:

$H_{\text{obt}}=\left[{\frac {12}{N(N+1)}}\sum {\frac {R^{2}}{n_{i}}}\right]-3(N+1)$

Eric. 144.32.89.104 (talk) 14:35, 19 May 2008 (UTC)

I did follow the exact steps you provided. And i came up with the same results as you did until the last part (you misplaced an N in the denominator but i caught on with what you were going with):
$={\frac {12}{N(N^{2}-1)}}\sum (X_{i}-{\frac {N+1}{2}})(Y_{i}-{\frac {N+1}{2}})$
I tried changing the μ you provided, but it came out to a long quadratic equation that didn't match anything close to d_i² where d_i = X_i - Y_i. --Agester (talk) 21:25, 20 May 2008 (UTC)

Ah, right, I did miss an $N$ (fixed now). Recall from the above that:

$N\mu ^{2}=\sum _{i}X_{i}^{2}-{\frac {N(N^{2}-1)}{12}}$

Ok, continuing, we have:

$\rho ={\frac {12}{N(N^{2}-1)}}\sum _{i}(X_{i}-\mu )(Y_{i}-\mu )$

$={\frac {12}{N(N^{2}-1)}}\left[\sum _{i}X_{i}Y_{i}+N\mu ^{2}-\mu \sum _{i}X_{i}-\mu \sum _{i}Y_{i}\right]$

$={\frac {6}{N(N^{2}-1)}}\left[2\sum _{i}X_{i}Y_{i}-2N\mu ^{2}\right]$

$={\frac {6}{N(N^{2}-1)}}\left[2\sum _{i}X_{i}Y_{i}+\left({\frac {N(N^{2}-1)}{6}}-\sum _{i}X_{i}^{2}-\sum _{i}Y_{i}^{2}\right)\right]$

$=1-{\frac {6}{N(N^{2}-1)}}\left[\sum _{i}X_{i}^{2}+\sum _{i}Y_{i}^{2}-2\sum _{i}X_{i}Y_{i}\right]$

$=1-{\frac {6}{N(N^{2}-1)}}\sum _{i}d_{i}^{2},$

which is what we wanted. Now, the other thing we want to show is that $K=H_{\text{obt}}$, where

$K={\frac {12}{N(N+1)}}\sum _{i=1}^{g}n_{i}({\bar {r_{i}}}-{\bar {r}})^{2}$

${\bar {r}}=\mu$

${\bar {r_{i}}}={\frac {1}{n_{i}}}\sum _{j=1}^{n_{i}}r_{ij}$

Let $S=\sum _{i=1}^{g}n_{i}({\bar {r_{i}}}-{\bar {r}})^{2}$. Then

$S=\sum _{i=1}^{g}n_{i}\left[{\bar {r_{i}}}^{2}+\mu ^{2}-2\mu {\bar {r_{i}}}\right]$

$=\mu ^{2}\sum _{i=1}^{g}n_{i}-2\mu \sum _{i=1}^{g}n_{i}{\bar {r_{i}}}+\sum _{i=1}^{g}n_{i}{\bar {r_{i}}}^{2}$

$=N\mu ^{2}-2N\mu ^{2}+\sum _{i=1}^{g}n_{i}{\bar {r_{i}}}^{2}$

$=\left[\sum _{i=1}^{g}n_{i}{\bar {r_{i}}}^{2}\right]-N\left({\frac {N+1}{2}}\right)^{2}$

$=\left[\sum _{i=1}^{g}n_{i}{\bar {r_{i}}}^{2}\right]-{\frac {N(N+1)(N+1)}{4}}.$

If we assume that your $R$ is

$R=\sum _{j=1}^{n_{i}}r_{ij}=n_{i}{\bar {r_{i}}},$

then hopefully it should be clear from here how to find that $K=H_{\text{obt}}$. Eric. 144.32.89.104 (talk) 13:04, 22 May 2008 (UTC)

May 15

Notation

What does $(k,n]$ mean in mathematics. How does it differ from $[k,n]$ or $(k,n)$? Thanks, User:Zrs_12 —Preceding unsigned comment added by 12.213.80.54 (talk) 02:53, 15 May 2008 (UTC)

It's interval notation. Digger3000 (talk) 03:18, 15 May 2008 (UTC)

Basically, ( or ) mean exclusive. [ and ] mean inclusive. So (k,n] is from k to n, but not including k. Paragon12321 (talk) 03:40, 15 May 2008 (UTC)

Having read the above, I investigated whether it would be possible to create a redirect pages called (a,b] etc. We aren't allowd to include [ or ] in page names for wiki syntax reasons. If you type (a,b] into the search box, it returns pages with a and b on them. If you type "(a,b]" it returns "not found - maybe the string isn't indexed". Is there any way we can make it so that folk asking Zrs_12's question can find an answer? -- SGBailey (talk) 07:48, 15 May 2008 (UTC)

Probably not. However, if a user knows this has something to do with intervals he will have no problem finding Interval (mathematics), and if he knows the names of the symbols, the relevant information can be found at Bracket.

That said, square brackets and parentheses being quite ubiquitous symbols, such notations can mean different things in different contexts. I am most familiar with [a,b] being the lcm, and (a,b) being the gcd or even an inner product. I don't know any alternative for the hybrid [a,b), though. -- Meni Rosenfeld (talk) 08:04, 15 May 2008 (UTC)

[a,b] is also used for the Lie Bracket and related concepts. (a,b) is used for any ordered pair, which includes numerous things (groups and topological spaces spring to mind). They're both very convenient notations and are very widely used for all kinds of things (generally unambiguously, though, which is impressive!). I also can't think of any other meaning for (a,b], though. --Tango (talk) 12:45, 15 May 2008 (UTC)

Variance from cdf

How do I calculate an approximate variance given an approximation to the cumulative distribution function at various points? The cdf is numerically calculated and often non-smooth, so I don't want to use finite-differences or such to get the pdf. The cdf is defined at ~100 points, and is close to zero at the lowest point and close to 1 at the highest (within about 100 times the machine epsilon). Thanks. moink (talk) 17:56, 15 May 2008 (UTC)

I'm assuming you mean finitely many points. Say a < b and you know F(a) and F(b) and no value of F at points between a and b. I'm not sure what you mean by using finite differences to get the pdf. The probability that the distribution assigns to the interval (a, b] is F(b) − F(a). The very crudest thing you could do is concetrate that probability at b. That smells a lot like something that could be called "finite differences". If you distributed that probability uniformly over the interval (a, b), that would the same as using linear interpolation to get a approximation of the cdf at points between a and b, and then the approximate density in that interval would be (F(b) − F(a))/(b − a). I think the expression for the variance would then bear some algebraic simplification BEFORE you plug in the actual numbers. Maybe that's too much like "finite differences" for you. A subtler approach would make the density non-uniform on the interval in some way that depends on the various values of F not only at those two points. Maybe I'll return to this question later when I have some time. Michael Hardy (talk) 18:38, 15 May 2008 (UTC)

What you said was enough for me to convert my mean-calculating code to variance-calculating code. I suppose in a way a substep could be likened to finite differences, but numerically integrating a non-smooth function is much less problematic than numerically differentiating it. Here's my Matlab code:

function sig=calcvar(cdf,mu,thresh)

for i=1:size(cdf,1)
  sig(i)=0;
  for k=1:length(thresh)-1
    sig(i)=sig(i)+(0.5*(thresh(k+1)+thresh(k))-mu(i))^2*(cdf(i,k+1)-cdf(i,k));
  end
end

Note that it actually does it over a bunch of cdf's, and that the variable thresh is a series of "thresholds" at which I know the cdfs. Thanks. moink (talk) 18:50, 15 May 2008 (UTC)

You'll get better accuracy (without additional complication) by replacing the $\left({\frac {t_{k}+t_{k+1}}{2}}-\mu \right)^{2}$ term with ${\frac {\int _{t_{k}}^{t_{k+1}}(x-\mu )^{2}f_{k}\,dx}{F_{k+1}-F_{k}}}={\frac {(t_{k+1}-\mu )^{3}-(t_{k}-\mu )^{3}}{3(t_{k+1}-t_{k})}}$. Since the integral is easy, there's no point in approximating it with a one-point evaluation at the midpoint (times the interval length). This is particularly important here because the correct formula is always greater than the approximation — by ${\frac {1}{12}}(t_{k+1}-t_{k})^{2}$ — so error will accumulate (though it will be reduced by a finer mesh of points).

Of course, selecting $f_{k}:={\frac {F_{k+1}-F_{k}}{t_{k+1}-t_{k}}}$ as Michael suggested is exactly finite differences for ${\frac {dF}{dx}}$. Here it doesn't really matter, though, because your final quantity is $\sigma ^{2}=\int _{-\infty }^{\infty }f(x)(x-\mu )^{2}\,dx$ and your initial quantity is $F(x)=\int _{-\infty }^{x}f(t)\,dt$; they're both (first-order) integrals of $f(x)$, so noise doesn't get inherently amplified passing from one to the other. If you expect your data is actually in error, however, it might be wise to pass some sort of smooth curve through it (splines or Bézier curves come to mind, or some sort of least squares regression if you have a model for the CDF) whose variance could then be calculated exactly. --Tardis (talk) 15:52, 16 May 2008 (UTC)

Thank you! This is exactly the kind of suggestion I was looking for. I will make the change to my code. I did notice it seemed to be underestimating; now i understand why. moink (talk) 16:14, 16 May 2008 (UTC)

The difference between the two methods is about 3% on average, on the particular data set I'm looking at. moink (talk) 16:43, 16 May 2008 (UTC)

The biggest family-tree you've ever seen...

I was kind-of a fan of the Fox series New Amsterdam (apparently I was the only one, it got canceled after 8 episodes), which was centered around a character named John Amsterdam. In 1642, this fella was a Dutch soldier who saved the life of a young native American girl (in a battle in the place that would become New York) who blessed/cursed him with immortality (he will become mortal again when he meets his true love). Unlike many other immortal people in fiction, he was able to father children. In one episode he states that he has fathered a total of 63 children. When he said that it struck me that he probably was the ancestor of a large chunk of the population in New York.

His last kid (well, latest) is called Omar and was born in 1943. If we assume that he had kids with some regularity between 1642 and 1943, that would mean he fathered a kid every four years ((1943-1642)/63=4). My question is this: approximately how many living descendants would he have? Making some reasonable assumptions, we could say that everyone of his kids had two kids at the age of 25, who themselves had two kids, and so on. I tried calculating this, but I couldn't quite nail down the sum properly (haven't done real maths in a looooong time), could you guys help me? I realize this is a fairly trivial question, but I thought it would be fun to find out. I also realize that the answer will be highly approximate, but I'm really just looking for an order of magnitude here (1,000? 10,000? 100,000? 1,000,000?). Cheers! --Oskar 19:46, 15 May 2008 (UTC)

I get $\sum _{k=1}^{63}2^{{\frac {61+4k}{25}}-1}\approx 27950$. So that's about 30000 or so, using your formula and ignoring possible intermarriages between his descendants. —Ilmari Karonen (talk) 20:03, 15 May 2008 (UTC)

Which is, of course, just the size of the last generation, and doesn't include surviving members of the previous generations. Assuming the three latest generations are still alive on average means multiplying the figure by 1 + 1/2 + 1/4 = 1.75, giving a total of about 48912. Four surviving generations would make that 52406. That rounds to about 50000 either way, given the crudeness of the approximation we're dealing with, and I wouldn't be too confident about that 5. Or about the number of zeros either. But for a quick back-of-the-envelope calculation, which is what you were asking for, it should be within a few orders of magnitude. —Ilmari Karonen (talk) 20:08, 15 May 2008 (UTC)

One of the biggest problems with this kind of calculation is that you're assuming his descendants all marry people that aren't descendants. If two descendants get married, their two kids end up being counted twice, and if a large chunk of the population are descended from him then the chances of that happening become quite large, so you get an overestimate. Although 2 children per family is probably an underestimate, since that would involve 0 population growth. Perhaps we're lucky and it cancels out! These problems become more apparent if you go back further - consider the people that claim to be descended from Jesus. If Jesus had two children and they each had two children when they're about 25, 2000 years later you end up with about 10²⁴ descendants alive today - 15 orders of magnitude larger than the total world population! Over a 300 year period, the effects are significantly less, but it still shows how enormously wrong such a simple model can end up being. --Tango (talk) 20:59, 15 May 2008 (UTC)

As a computational problem: One child every 4 years from 1642 to 1943 and each descendent having exactly 2 children with an unrelated spouse at their 25th birthday, I get 336,261 descendents through 2008. All the above caveats of course apply. Dragons flight (talk) 04:51, 16 May 2008 (UTC)

On the other hand, even though Tango is right in that calculating the number of descendants that way is an overestimate, it is true that many people who lived ages ago have lots of descendants because everyone has two parents. You don't have to be immortal for this. I think, if you take a human from a few thousand years ago, then most probably he or she either has no descendants or a significant proportion of the contemporary population is his or her descendant. On the other hand, if you count only male-line descendants, this is far from true, because everyone has only about two male-line ancestor in any generation. – b_jonas 07:46, 16 May 2008 (UTC)

Surely everyone has precisely one male-line ancestor in each generation? --Tango (talk) 13:18, 16 May 2008 (UTC)

I believe b_jonas means something like 'people of whom you are a male-line descendant', in which case you get two per generation, barring incest. Algebraist 13:45, 16 May 2008 (UTC)

You'd only be a male-line descendant of two people per generation if you're male: otherwise zero (well, I suppose it depends on your definition). Anyhow, b_jonas is touching on the identical ancestors point, a point in history where every person then alive is either the ancestor of everyone now alive, or of noone now alive; our article says this likely happened 5000 to 15000 years ago. (Related are most recent common ancestor, mitochondrial Eve, and Y-chromosomal Adam.) Eric. 86.153.201.165 (talk) 00:10, 17 May 2008 (UTC)

The difference in mine and Dragons flight's results appears to stem from the original poster's assertion that (1943-1642)/63 = 4, which of course isn't true (it's actually 43/9 = 4.777…). Besides, the correct divisor isn't 63 (the number of children) but 62 (the number of gaps between children), giving an average interval of about 4.8548 years. Combining this figure with the other assumptions given, I get a total number of 202,689 descendants, of which the latest generation of course makes up exactly half, or 101,344. With three surviving generations, the number of surviving descendants would then be about 177,353. —Ilmari Karonen (talk) 01:11, 19 May 2008 (UTC)

Generated and induced σ-algebras

Let X be a nonempty set and ${\mathcal {S}}$ a collection of subsets of X. We let ${\mathcal {S}}^{\sigma }$ denote the σ-algebra generated by ${\mathcal {S}}$, and take ${\mathcal {C}}\cap Y$ to mean $\{C\cap Y\ |\ C\in {\mathcal {C}}\}$ for any subset Y of X and collection of subsets ${\mathcal {C}}$. Is it true that

$\left({\mathcal {S}}\cap Y\right)^{\sigma }={\mathcal {S}}^{\sigma }\cap Y$?

The right side is of course the induced σ-algebra on Y, and that it includes the left is clear. But does the other direction follow? If not, what sort of additional hypothesis would suffice?  — merge 20:51, 15 May 2008 (UTC)

Yes, the reverse inclusion holds. Proof is pretty easy: consider the collection of subsets A of X such that $A\cap Y$ is in the LHS; show that this collection contains every element of ${\mathcal {S}}$ and is a σ-algebra on X. Algebraist 22:54, 15 May 2008 (UTC)

Oh. Let ${\mathcal {A}}$ be that collection. Then $X\in {\mathcal {A}}$, and if $A\in {\mathcal {A}}$ then $A^{c}\cap Y=Y-A=Y-(A\cap Y)$, so $A^{c}\in {\mathcal {A}}$. Moreover, if $\{A_{n}\}\,$ is a sequence of sets in ${\mathcal {A}}$ then $(\cup _{n}A_{n})\cap Y=\cup _{n}(A_{n}\cap Y)$, so that $\cup _{n}A_{n}\in {\mathcal {A}}$. Thus ${\mathcal {A}}$ is a σ-algebra in X, and it clearly contains ${\mathcal {S}}$ since ${\mathcal {S}}\cap Y\subset \left({\mathcal {S}}\cap Y\right)^{\sigma }$. It follows that ${\mathcal {S}}^{\sigma }\subset {\mathcal {A}}^{\sigma }$, or in other words that

${\mathcal {S}}^{\sigma }\cap Y\subset \left({\mathcal {S}}\cap Y\right)^{\sigma }.$

Did I get that right?  — merge 01:00, 16 May 2008 (UTC)

Yeah, that's what I'd have written if I'd felt up to hacking out all that tex. Algebraist 09:45, 16 May 2008 (UTC)

Hah. Thank you! The TeX is the easy part; it's coming up with the tricks that's hard. I'm not very clever, but can usually struggle through when pointed in the right direction.  — merge 10:02, 16 May 2008 (UTC)

If I might be allowed some musing-space, I think the problem with doing this sort of question is that there's no such thing as a 'typical element' of a generated sigma-algebra. If you want to prove a statement about open sets, or arbitrary elements of the group/vector space/whatever generated by some generating set, you can write 'let U be open' or 'let v be an arbitrary linear combination of spanning vectors' or whatever, but it's never useful to write 'let B be a Borel set', because there's nothing very sensible you can say about what an arbitrary Borel set looks like (the best you can do involves recursion to the first uncountable ordinal). Thus the only way to show that all elements of ${\mathcal {S}}^{\sigma }$ have some property is to show that the set of things with that property contains all of ${\mathcal {S}}$ and is a σ-algebra. And maybe use Dynkin's lemma, though thankfully we were spared that here. Algebraist 11:14, 16 May 2008 (UTC)

You've hit on the real problem, so your comments are very much appreciated. The same thoughts have been going through my mind in working with σ-algebras: they are structurally similar to topological spaces, but seemingly much less tractable for the reason you stated. The situation is aggravated by the lack of exposition on this matter, which can easily leave the newcomer adrift. So I have a couple of further questions:

• Is there a source you'd recommend that provides good training in this area (not just for σ-algebras, but for expanding one's bag of tricks with this kind of reasoning)--perhaps a book in the set theory area?
• More deeply, do you think there is a way to remedy this situation somehow in the theory itself? Perhaps this is "impossible", in the same sense that it's impossible to exhibit a well-ordering of the reals. But on the other hand, perhaps this is a weakness of the theory, and one could modify or supplement it with constructions, methods or tools that would make reasoning easier.

 — merge 12:15, 16 May 2008 (UTC)

I can't recommend any books; sorry. On your second point, you might be interested in the Borel hierarchy I alluded to earlier, which provides a sort of classification of the Borel sets (or the sets of any generated σ-algebra). It's handy for some proofs, though the only one I can currently recall is that the Borel σ-algebra on R has the cardinality of the continuum (and is thus much smaller than the Lebesgue σ-algebra). Algebraist 13:05, 16 May 2008 (UTC)

May 18

How can Pi exist.

I was thinking a while back about this. If Pi is 3.14 and so on but NEVER ends then it doesnt make sense. Look at it like this(yes im resorting to childish terms) remember how in school you learned fractions, with a pie. 1 was a whole pi, 2 half and so on. Take Pi over 1. that would still be pi but it how would that look? 3 whole pies, not the complcated part, then u have the .14159................... the pie would never stop going around, filling up the tray. It would be very very small increments but it would still be moving. Therefore it would eventually fill up,seeings that Pi is infinitum. This has been meesing with my mind for a while now. Everyone i explain this to understands where im coming from, but cant give an answer

76.111.150.73 (talk) 13:20, 18 May 2008 (UTC)Rob

Well, I'm afraid your perfect record has been broken, because now there's someone you've explained this to who doesn't understand where you're coming from, namely me.

Are you asking how it is that the decimal representation 3.141592654... can go on forever and never quite equal exactly π, or are you making a little joke about π and pie? —Steve Summit (talk) 13:44, 18 May 2008 (UTC)

You know that the decimal expansion of 1/3 is 0.333..., and that of 1/7 is 0.14285714..., right? Do these give you the same problems in comprehension? Oliphaunt (talk) 13:48, 18 May 2008 (UTC)

I think I see what you're saying. Pi is the sum of an infinite number of numbers (3+1/10+4/100+1/1000+...), and you're wondering how come that doesn't add up to infinity? It's a good question. Put simply, the numbers are getting smaller and smaller fast enough that you never get past a certain amount (pi). Consider this sum: 1+1/2+1/4+1/8+... That, it turns out, adds up to 2. You can see how it works like this: Consider a box 2 metres long. You put something 1 metre long in the box, that leaves 1 metre left. You then put something 1/2 a metre long in the box, that leaves 1/2 a metre left. Then you add 1/4, and that leaves 1/4 left, and so on. Because the amounts you're adding are getting smaller and smaller, there is always room to add the next one. That amount left gets smaller and smaller, and after an infinite number of steps it's 0, so the total length of all the things is the length of the box: 2. Pi is a little trickier, because it doesn't follow a nice pattern (it's irrational), but the concept is the same. Does that make sense? --Tango (talk) 13:58, 18 May 2008 (UTC)

Ohh! One could make a very similar example for pi using the Method of exhaustion. Unfortunately the article doesn’t have the standard illustrations, but using the particular summation as the area added in each n-gon you get the same illustration. GromXXVII (talk) 16:44, 18 May 2008 (UTC)

Take a look at the articles on 0.999... and Zeno's paradoxes too, because I think they are related to your question. • Anakin ^(talk) 15:13, 18 May 2008 (UTC)

No I wasnt making a joke on the words, i understand that even after many many decimal places its still not even to be considered, but still it gos on FOREVER so there fore why wouldnt all the decimals eventually add up to a whole? —Preceding unsigned comment added by 76.111.150.73 (talk) 23:48, 18 May 2008 (UTC)

76.111.150.73 (talk) 23:49, 18 May 2008 (UTC)Rob

Why do you expect the decimals to add up to 1 as opposed to, say, 1/2? If an infinite number of decimals can add up to a finite number (which they can, as I demonstrated above), then can add up to any finite number (less than or equal to 1). In this case, they add up to the fractional part of pi. --Tango (talk) 23:56, 18 May 2008 (UTC)

I dont expect it to add to anything, If it were infinite then why would it not continiously go and go and go. like after years and years wouldnt it be 4 or even as you said 3.5 or anything like it??

Im not trying to argue im trying to understand 76.111.150.73 (talk) 00:07, 19 May 2008 (UTC)Rob

Sorry, I misunderstood you there - I thought you were expecting it to stop at a whole, you mean eventually reach a whole and keep going? Did you understand my explanation above about putting things in a box? --Tango (talk) 00:11, 19 May 2008 (UTC)

Actually the box thing is just about exactly what i mean. Only problem is a box has a closing point. Pi and numbers do not so wouldnt it overflow? Is this even a rational question or am i just making myself look stupid. 76.111.150.73 (talk) 00:14, 19 May 2008 (UTC)Rob

No, it's not a dumb question. I, myself, have also pondered this same situation. Just think of it like this: There are some infinte series ($\lim _{n\to \infty }\sum _{k=0}^{n}{\frac {1}{2^{k}}}=2$ as Tango said) that converge to a real number from an infinite series of terms. On the other hand, you have series that diverge. Such as $\lim _{n\to \infty }\sum _{k=1}^{n}k=\infty$. Pi is kind of like a convergent series in that there is an infinite quantity of numbers that does not amount to infinity. Does that help at all? Zrs 12 (talk) 01:27, 19 May 2008 (UTC)

A slight misunderstanding. Tango referred to the series $\sum _{k=0}^{\infty }{\frac {1}{2^{k}}}=2.$ The series $\sum _{k=1}^{\infty }{\frac {1}{k^{2}}}$ is the Basel problem. It also converges, but not towards 2, but rather towards ${\frac {\pi ^{2}}{6}}\approx 1.64493.$ Bo Jacoby (talk) 10:16, 19 May 2008 (UTC).

Oops, sorry about that. Typo. I fixed it now. -- Zrs 12 (talk) 19:37, 19 May 2008 (UTC)

Sorry I misunderstood your question at first. You asked, "after years and years wouldn't it be 4 or even as you said 3.5?" Well, no.

My explanation wasn't very good... The box has a closing point simply because I already knew the answer so picked a box the right size (I was just demonstrating that my answer was right, rather than calculating the answer). There was no end to the items I was adding to the box, but they still fit in the box because they got smaller. If I had a box pi metres long and added items 3 metres long, then a tenth of a metre long, then 4/100 m long, the 1/1000 m long, and so on, there would still always be room for the next item and it would never overflow (I used the example of powers of two because it's easier to see, but it works the same way). --Tango (talk) 14:45, 19 May 2008 (UTC)

We start with 3. *If* we added a bunch of 9's we could get to 4, but we don't: the first thing we add is a 1, or 1/10. So now we've got 3.1.

If we added a bunch more 9's we could get to 3.2, but we don't: the next thing we add is a 4, or 4/100. So now we've got 3.14.

If we added a bunch more 9's we could get to 3.15, but we don't: the next thing we add is a 1, or 1/1000. So now we've got 3.141.

If we added a bunch more 9's we could get to 3.142, but we don't: the next thing we add is a 5, or 5/10000. So now we've got 3.1415.

If we added a bunch more 9's we could get to 3.1416. We do add one 9, but after that we add a 2, so we're stuck with 3.141592. And so on.

Not only will we never get to 3.5 or 4, it looks like we'll never get to 3.2, or even 3.1416.

Remember that "I'm thinking of a number between 1 and 100" game? You guess 50, and I say it's higher. You guess 75, and I say it's lower. So now you know it's between 50 and 75. Each guess you make narrows it down -- and your guesses will never get outside of the interval 50..75.

Suppose after a few more guesses you get it narrowed down to the interval 61..63. You triumphantly guess "62", but I say "too high". I cheated: the number I'm thinking of isn't an integer, and evidently it's some fraction between 61 and 62.

So you guess "61.5" and I say "too high", and you guess "61.3" and I say "too low", and you guess "61.4" and I say "too high", and you guess "61.35" and I say "too low", and you guess "61.37" and I say "too high", and you guess "61.36" and I say "too low", and you guess "61.365" and I say "too low", and so on. We can go on like that forever, but unless I cheat again, the number will never get outside the range 61..62, and in fact it will never get very far from 61.365. —Steve Summit (talk) 04:36, 19 May 2008 (UTC)

Your question is a valid one, but it has a good answer. Go look at Zeno's paradox (the "dichotomy paradox" section), which is essentially equivalent to the question you've raised. kfgauss (talk) 05:38, 19 May 2008 (UTC)

It might be worthwhile looking at the ratio test, for example, for some sense of general rules when adding little pieces does or does not result in infinity. --Fangz (talk) 10:22, 19 May 2008 (UTC)

Ok I guess i understand now, the increase is only fractions of the last fraction. I think i get it. Thanks. BringBackZero (talk) 17:28, 19 May 2008 (UTC)Rob

By the way if this question seems irrational to some of you, its coming from a 15 year old, so dont flip out BringBackZero (talk) 17:34, 19 May 2008 (UTC)Rob

Not at all - it's a difficult thing to get your head round. I didn't learn the underlying maths properly until the first year of University. --Tango (talk) 19:02, 19 May 2008 (UTC)

Ha! If it seems "irrational." LOL. Irrational numbers are strange, but e^{i π} = -1, which may be even stranger (though I'm not sure we should jump the gun on what it may or may not prove). --Prestidigitator (talk) 22:04, 19 May 2008 (UTC)

Parity of powers

In parity (mathematics) it states that a fractional number, by definition is not odd or even. But I was considering the fact that square numbers have the same parity as their square root. I.e., square an odd number, you get an odd number, and square an even number, you get an even number. The same seems to apply to other powers. E.g.,

powers of 2: 2, 4, 8, 16, 32, 64... are all even, and

powers of 3: 3, 9, 27, 81, 243, 729... are all odd.

The only case this definitely doesn't work for is $x^{0}$, because you still get 1 when x is even. But I was wondering, why shouldn't this work for fractional or negative powers. E.g., √2 is irrational, but if the pattern holds it should be even. Similarly for $2^{-1}=0.5$. I haven't been able to find any conflicts where a power of an even number is equal to a power of an odd number, so is there any other reason why this might not work? • Anakin ^(talk) 13:59, 18 May 2008 (UTC)

If $x=\log _{2}3$, then $2^{x}=3$, and you have a power of 2 which is clearly odd. I don't see how "a is even if there is an odd integer b and rational numbers c and d where $c\neq 0$ such that $x=2^{c}b^{d}$" is a plausible extension of "a is even if there is an integer k such that $a=2k$". -- Meni Rosenfeld (talk) 14:24, 18 May 2008 (UTC)

(ec). The rule you desecribed above works if the powers are integers and greater than zero. For example, it is possible to solve for x if 2^0.5 = 3^x but 2 is even and 3 is odd. Also, even and odd have quite specific definitions, you should have a look at them. Zain Ebrahim (talk) 14:27, 18 May 2008 (UTC)

Thank you for these answers. The fact that $x=\log _{2}3$ gives a power of 2 which is odd does clearly invalidate my rule. That kind of means that x can't be even or odd (and it is irrational), which makes it less of a surprise that √2 can't be even or odd. The fact that an even number must be divisible by 2 is also a blow, because √2 (1.41421...) clearly isn't. This gives me a few things to think about though, thanks. • Anakin ^(talk) 15:07, 18 May 2008 (UTC)

There are similar concepts that sometimes come up. For instance in $\mathbb {Z} \left[{\frac {1+{\sqrt {-1}}}{2}}\right]=\left\{a+b{\frac {1+{\sqrt {-1}}}{2}}|a,b\in \mathbb {Z} \right\}$ there is another form of parity: in whether or not the coefficients are integral, or have a denominator of 2. GromXXVII (talk) 16:52, 18 May 2008 (UTC)

Coin toss

If you have 7 coins and you toss them together, what is the probability of getting 3 heads? What is the probability of getting all heads?

What is the method of solving this? Can you do this: 7C3/2^7 for the first question? —Preceding unsigned comment added by 68.99.185.240 (talk) 17:32, 18 May 2008 (UTC)

Hello. Yes, you are correct that that would work. For the second, just replace 7C3 with 7C7 = 1. -mattbuck (Talk) 17:52, 18 May 2008 (UTC)

Note that, assuming ideal coins so Pr(heads) = 1/2 for one toss, ${\tbinom {7}{3}}/2^{7}$ is the probability of getting exactly three heads, not at least three heads. --Prestidigitator (talk) 17:12, 19 May 2008 (UTC)

probability

If you draw a 5-card hand, what is the probability of getting exactly three 7's?

I was wondering if you could simply find the number of combinations for which you have three 7's, and divide it by the total number of 5-card combinations? So... (4C3*48C2)/(52C5)??? Does this mean that we are considering each 5-card hand to have equal probability of coming up? Is the problem more complex than that? —Preceding unsigned comment added by 68.99.185.240 (talk) 19:23, 18 May 2008 (UTC)

Yes, that should work: indeed the probability of each hand is the same. GromXXVII (talk) 20:30, 18 May 2008 (UTC)

Test

My friend has a vocabulary test. His study sheet has 75 words that might be on the test. 15 of the words will be randomly selected to actually be on the test. And among the 15, he is required to only define 6. I know that in order to have a 100% chance of getting all six, you only need to know 66 words. However, it seems there is a 99.99999% chance of getting all 6 with only 65 and so on. How do you calculate the probability as a function of target (number of words you want to get right) and work (number of words you need to study)? --MagneticFlux (talk) 20:08, 18 May 2008 (UTC)

Looks like a hypergeometric distribution to me. 75 words with the number you memorize “defective”, and a sample of 15. GromXXVII (talk) 20:34, 18 May 2008 (UTC)

May 19

chemical mathematics

what is chemical math? and specifically is graph energy a subcategory? also, what is the definition of both 'chemical math' and 'graph energy' and where are their applications? —Preceding unsigned comment added by 79.127.9.254 (talk) 04:00, 19 May 2008 (UTC)

Chemical math - do you mean stoichiometry? Neıl 龱 15:00, 19 May 2008 (UTC)

For a basic chemistry class I don't think you need to know any math beyond algebra. StuRat (talk) 19:31, 19 May 2008 (UTC)

Days of the Week for BC

I found a website that allows me to calculate the day of the week for any given AD date. (http://katzentier.de/_misc/perpetual_calendar.htm) However this doesn't include BC, and I don't know how to calculate it...should I just use the calendar worky-thingy as if it were BC or is there something/somewhere else I need to do/can go?

Lady BlahDeBlah (talk) 18:42, 19 May 2008 (UTC)

Try entering BC dates as negative numbers. X BC is the same as -X+1 AD (the +1 is because there was no year 0, it goes from 1 BC to 1 AD), so 20 BC gets entered as -19. --Tango (talk) 19:00, 19 May 2008 (UTC)

The thing on the website won't let me enter the '-'... Lady BlahDeBlah (talk) 19:16, 19 May 2008 (UTC)

Also note that the farther back in time we go, the fewer civilizations will be using the familiar 7-day week. I doubt if American pre-Columbian civilizations would have a 7-day week, unless by sheer coincidence. I wonder if we have an article on this ? StuRat (talk) 19:27, 19 May 2008 (UTC)

Yes, of course, but this is really more for my benefit, as in 'Just wondering'. I think the site only has 4 spaces for the year, so I can do something like 86 BC, but not like 2270 BC. Lady BlahDeBlah (talk) 19:50, 19 May 2008 (UTC) EDIT: Nope, I tried it with the minus and it doesn't like it. Blargh!!! Lady BlahDeBlah (talk) 19:53, 19 May 2008 (UTC)

The site explains the method - you'll just have to do it by hand (or google for another calculator). --Tango (talk) 20:17, 19 May 2008 (UTC)

Claus Tønderings' Calendar FAQ might provide some good information about this kind of thing. Note especially that while the seven-week cycle was strictly followed much before 1 AD, the Julian calendar was introduced only in 45 BC, so you can't reliably calculate when a date before that was. – b_jonas 21:00, 19 May 2008 (UTC)

This page will compute the Julian day for any date back to 4713 BC (using the Julian calendar for all dates prior to the October 1582 switchover). To get the day of the week from the Julian day, round to the nearest integer (rounding up if it's halfway between) and take the remainder modulo 7. 0 is Monday, 1 is Tuesday, and so on. This isn't quite as convenient as one might hope for, but it'll do in a pinch. -- BenRG (talk) 21:13, 19 May 2008 (UTC)

I believe you can probably just add a multiple of 400 years to get an equivalent AD date. --Prestidigitator (talk) 21:15, 19 May 2008 (UTC)

That works for Gregorian dates, but I think it's much more common to use the Julian calendar for very old dates, in which case you need to add a multiple of 28 instead, and use a Julian day-of-week calculator. -- BenRG (talk) 21:52, 19 May 2008 (UTC)

Note that the choice of Julian or Gregorian calendar depends not just on date but on location too. Only Catholic countries adopted the Gregorian calendar in 1582 - many Protestant countries were deeply suspicious of its Catholic origins. Britain only adopted it in 1752, and Greece continued to use the Julian calendar until 1923. Gandalf61 (talk) 08:46, 20 May 2008 (UTC)

May 20

Neg. Binom. GLM question

In doing a negative binomial glm in R, I found two oddities. Doing the analysis of deviance I get the message "Dispersion parameter for the negative binomial(0.553) family taken to be 1." I interpret this to mean no dispersion parameter, as in the case of a quasilikelihood, is estimated. Is this correct?

The second question is more detailed. In comparing two (nested, neg. binom.) models, the anova comparison, anova(model1,model2), shows a different log-likelihood than taking the deviance(model1)-deviance(model2). Not very different, mind you, about 0.1, which is more that I could attribute to rounding error. Any ideas? --TeaDrinker (talk) 01:57, 20 May 2008 (UTC)

Suggestions.

1. Are you using glm or something from the MASS package? (Or something else???) Whichever you are, look at the help page for the relevant function, especially where it talks about "family". But I would suspect your hunch is true regardless, that the dispersion is assumed fixed.
2. Two thoughts. You will want to make sure the fitting method for both models is the same, i.e., don't use REML for one and ML for the other; in fact, you should probably use ML for both and not REML (which is the default for some functions, BTW), as in many cases the comparison of REML-fitted models can be meaningless. Second, check the "type" argument of the relevant anova method. Make sure it will do the equivalent test to the comparison of the deviances, as some "types" do different tests.

Check also the R-Help mailing list if you need to. The readers there are very helpful and knowledgeable—just make sure you read the manuals that came with your R distribution, the relevant function help pages, and search the archives of that list before posting a question there. If you don't, they can be quite, er, uncivil.  ;-) Baccyak4H (Yak!) 03:20, 20 May 2008 (UTC)

Partial differential equation

Heh...does anyone wanna help me out and solve the partial differential equation given in the last problem here? Since it IS a PDE, it is quite a lot of grunt work for not much benefit. But if you were to get (1/2)sin(3x)e^(3t)+(1/2)sin(3x)e^(-3t), I would be happy and grateful. My answer is a solution to the PDE, but I'd like to know if it's the general solution, as was asked for. --M1ss1ontomars2k4 (talk) 02:42, 20 May 2008 (UTC)

I think you got it. Obviously $u_{xx}=-9u$, so $u_{tt}=9u$ which has the general solution $u(x,t)=a(x)e^{3t}+b(x)e^{-3t}$ (or equivalently $u(x,t)=c(x)\cosh(3t)+d(x)\sinh(3t)$). The initial conditions narrow that down to $u(x,t)=\sin(3x)\cosh(3t)$ just like you have it. --Prestidigitator (talk) 05:59, 20 May 2008 (UTC)

Standard deviations and probabilities

Hi. I'm working on Donald Bradman, trying to get the article to FA. As long ago as Dec 2006, this edit was made, with no edit summary or discussion at the article talk page. It's never been challenged or referenced.

As the requirements of FAC mean that every claim needs to be referenced, and the user seems to have left the Project (with no email address enabled) I wondered whether the probability figures he's used derive from the (cited) standard deviation figures. As my maths is rubbish, I have no idea.

So, three questions:

1. Do the probability figures derive from the SD ones?
2. Are they accurate?
3. If we're so far at a "yes" and "yes" scenario, how can I possibly cite the probabilities?

Cheers --Dweller (talk) 10:41, 20 May 2008 (UTC)

Yes, the probabilities derive from the number of SDs. They are accurate (except that the first should be 185,000, I think) subject to the following caveats: firstly, it is not clear on what basis they have been rounded to their current values. Without access to the book in question, I don't know the precise SD figures, but assuming the current figures are accurate to 1 decimal place only, the probability figures are being given far too precisely: for example, the Don's probability should be '1 in somewhere between 147,000 and 233,000'. Secondly, they depend on the assumption that the underlying distribution is approximately normal. I do not know if this is a sensible assumption.

As for sourcing, I doubt you'll be able to find these specific figures anywhere if they're not present in Davis' work (which I don't have access to). Algebraist 11:30, 20 May 2008 (UTC)

Thanks. I don't think Davis includes the probabilities, but getting hold of a copy of the book has been a bit of a problem! Your comment of "'1 in somewhere between 147,000 and 233,000'"... does that derive from a log chart or something that I could reference? --Dweller (talk) 11:36, 20 May 2008 (UTC)

I couldn't find precise enough normal distribution tables online (though they doubtless exist somewhere), so I used this tool, linked from normal distribution. Algebraist 11:48, 20 May 2008 (UTC)

Fields

For the field of integers mod p, I cannot find an inverse element for all cases. $x*x^{-1}=1$ where x = 0 does not seem to work. What have I missed? 81.187.252.174 (talk) 11:10, 20 May 2008 (UTC)

0 is the only element with no multiplicative inverse. You've missed nothing. –King Bee ^{(τ • γ)} 11:29, 20 May 2008 (UTC)

Hurf, didn't read the definition properly. Blargh burp 81.187.252.174 (talk) 12:14, 20 May 2008 (UTC)