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July 8

Power of Sums

I remember one of my teachers (a long time ago) using a recursive generator to come up with a formula for power sums. For example, if he wanted to know what the polynomial form of $S(n)=\sum _{k=1}^{n}k^{4}$ is (assuming that we already know the constant, first, second, and the third power sum formula), he would plug those in and obtain the fourth power sum formula. Does anyone know where I can find it or what it is called? Is there a page here on wikipedia for it? In addition, is there anyway, any cool techniques for finding out such expression of sums of powers that are not recursive and do not depend on the knowledge of any previous powers. For example, if I wanted the formula for the sum of 20th power of natural integers, how would I find it without first having to find all such expressions for previous powers? Thanks--76.79.202.34 (talk) 01:45, 8 July 2008 (UTC)[reply]

Yes.

If S(n,a)= product of 1/a! x n(n+1)(n+2)(..)(n+a)

Then S(n,a+1) - S(n-1,a+1) = S(n,a)

AND also Sum from n=1 to n=n of S(n,a) = S(n,a+1)

((or very similar I'm a bit rusty))

Note S(n,a+1) is a polynomial of order a+2 ie one order greater than S(n,a)..

See Figurate number which is closely related to Binomial coefficient.

The actual method is probably described somewhere... but I can't find it.. if you can't see how to do it from the above I can explain in more detail.87.102.86.73 (talk) 02:26, 8 July 2008 (UTC)[reply]

I don't know about the second part of your question.87.102.86.73 (talk) 02:28, 8 July 2008 (UTC)[reply]

Look at Faulhaber's formula. And then at Bernoulli number. Michael Hardy (talk) 02:29, 8 July 2008 (UTC)[reply]

product of primes..

This probably comes up a lot.: If N=P1 x P2 (where Pnumber is a prime) is it possible that N=P3 x P4 (where P3 is not equal to P1, P2)?

I tried but couldn't disprove it, nor could I easily think of an example.. Can this be resolved?87.102.86.73 (talk) 06:38, 8 July 2008 (UTC)[reply]

No - see Fundamental theorem of arithmetic --tcsetattr (talk / contribs) 06:41, 8 July 2008 (UTC)[reply]

Good. I was beginning to think I had a disproof ie

if P1P2=P3P4

Then P1P2/P3=P4

writing P1=P3+k (k is an integer) gives:

(P3+k)P2/P3=P4

P2+kP2/P3=P4

kP2/P3=P4-P2

ie kP2/P3 = INTEGER

so there for k=nP3 ?? (n is integer)

which would mean that P1=P3+nP3=(1+n)P3

so therefor P1 is not a prime..

Therefor a contradiction , and...

I question marked the part I am only 99% sure of, the proof isn't the same as those at the above links? is the proof ok.?87.102.86.73 (talk) 07:17, 8 July 2008 (UTC)[reply]

This is more-or-less the same as the usual proof. Unfortunately, all the hard work is done in the bit you're not sure of, which is not obvious and has to be proved. It's usually called Euclid's lemma. Algebraist 10:05, 8 July 2008 (UTC)[reply]

I'll have a look at that then.. I did attempt to prove that bit (not shown) using remainders, but it felt like I was a dog chasing my tail... 87.102.86.73 (talk) 13:12, 8 July 2008 (UTC)[reply]

Thanks.87.102.86.73 (talk) 13:59, 8 July 2008 (UTC)[reply]

What the..

In cartesian three-dimensional space, there is a point A located at (1, 2, 2), a point B located at (-3, 1, -1), and a point C located at (0, 2, 1). One of my sibling's questions asks what's the shortest distance from B to AC? I would have thought just the length of BC, no? Square root of 14? I figure that because C is the "lowest" point on AC. It is definitely not AB, as AB is longer than BC (basic geometrical axiom). I used vectors to show that there is no point E on AC such that BE is perpendicular to AC, so I'm inclined to think it's just the length of BC, √14?

I know this is homework, but we've been working on this for a while. —Preceding unsigned comment added by 124.191.112.93 (talk) 08:06, 8 July 2008 (UTC)[reply]

dot product..

I'd guess that

The shortest distance from B to AC will be when the line from B to AC is perpendicular (at right angles) to AC...

If you call this point F and say F = A +nAC then you need to solve AC.BF=0

You might need to extend the line AC beyond the ends at A and C.. Did the question say anything about this? Have you got the dot product?87.102.86.73 (talk) 08:36, 8 July 2008 (UTC)[reply]

Alternatively find the distance squared (BF²) as a function of n, and differentiate to find the minimum..87.102.86.73 (talk) 08:59, 8 July 2008 (UTC)[reply]

Both give the same answer - that the point is way off the end past C... which should help you guess that C is actually nearer to B than A..87.102.86.73 (talk) 09:01, 8 July 2008 (UTC)[reply]

No, you can't extend the line. I realise that if you extend the line by a factor of 3.5, and then join B to that new end of the line, you'll get a right angle. There is no such point E on AC such that BE is perpendicular to AC. 87.102.86.73, you said that "C is actually nearer to B than A." That's what I think, too. I'm inclined to think the shortest distance from B to AC is just the length of BC. Thanks for your help, by the way. —Preceding unsigned comment added by 124.191.112.93 (talk) 14:33, 8 July 2008 (UTC)[reply]

HK ≅? H × K

If $H,K$ are subgroups of some group $G$ , there is a theorem from abstract algebra that says when $HK\cong H\times K$ , but it has a supposition I can’t seem to remember or find: I thought it had something to do with normality, but trying to prove it myself it seems like it’s that $H\cap K=\{e\}$ and one is a subset of the centralizer of the other. I used the map defined by $(h,k)\mapsto hk$ , which is trivially surjective. It is injective as long as $H\cap K=\{e\}$ . It is a homomorphism as long as everything in $H$ commutes with everything in $K$ . Any idea if that can be restated in terms of normality, or if that’s just wrong? GromXXVII (talk) 11:00, 8 July 2008 (UTC)[reply]

(In the presence of the assumption that

H\cap K=\{e\}

) that condition's equivalent to the statement that both H and K are normal in HK. This form of the theorem is listed at direct product#Group direct product. Of course, if just one is normal, we get a semidirect product. Algebraist 11:06, 8 July 2008 (UTC)[reply]

Algebraist's answer is probably sufficient. I'll ramble on about commuting and normalizing and hopefully explain something you'll find useful on the way.

Centralizers are very strange creatures. If H ≤ C_G(K), then K ≤ C_G(H). Sometimes it can be helpful to rewrite centralizer relations in terms of commutator relations. Both of these conditions are equivalent to [H,K] = 1, which is more obviously symmetric. Normalizers can also be stated in terms of commutator relations: H ≤ N_G(K) if and only if [H,K] ≤ K. This shows that if H normalizes K and K normalizes H, then [H,K] ≤ H ∩ K. In particular, if H and K intersect trivially, then the only way they can normalize each other is if they commute. This then is how to switch around between centralizers, normalizers, and commutators.

In terms of thinking directly about normal subgroups, it is important to distinguish between the set HK = { hk : h in H, k in K }, and the subgroup <H,K> generated by HK. HK is a subgroup if and only if HK=KH. If HK=KH for every subgroup K, then H is called a permutable subgroup. One way to ensure that HK=KH is to have Hk=kH for every k in K. In particular, every normal subgroup is permutable. This is basically where things could go wrong if you did not have normality. You defined a homomorphism from H × K to <H,K>, but it need not be surjective unless HK=<H,K>, that is, unless HK is a group.

I think it is important when trying to understand groups to really get a feel for how different HK and <H,K> are. For instance, if G is a finite nonabelian simple group, then it has subgroups H, K of order 2 such that HK has order 4, but <H,K> = G has much larger order (say 60 for the alternating group on five points, or say over 10⁵³ for the monster group, or larger, etc.). JackSchmidt (talk) 14:15, 8 July 2008 (UTC)[reply]

Addition of probability over time (ie. POP)

Hi. First of all, none of this is homework, I would just like to know the calculation methods for these probabilities. Specificly, probability of precipitation (POP). Below I will provide some examples, and am asking how to solve them.

On the first day, the POP is 60%. On the second day, it's 40%. What is the total POP for the two days (the chance % that it will rain within the 2 day period)?
For each 6-hour period of the day, the POP is 40%. What is the POP for the whole day?
The POP for a two-hour period is 65%. For the second hour of the period, it's 45%. What is the POP for the first hour?
For a week, the daily POP for each day is 20%. What is the POP for the whole week?
For 3 days, the daily POP is 50% ± 20%. What is the POP uncertainty for the entire 3-day period, and what would be the total POP were there no uncertainty?
For a 3-hour period, the POP is 70%. For the first two hours, the total POP is 40%. What is the POP for the last hour?
For a 6-hour period, the total POP is 70%. Considering that both the first 3 hours and the last 3 hours have the same POP, what is that POP for each 3-hour period?

Thanks. ~A H 1^(TCU) 16:35, 8 July 2008 (UTC)[reply]

You do not have sufficient information to answer any of these questions. Algebraist 17:08, 8 July 2008 (UTC)[reply]

I agree, and would like to add a comment that often the POP stated in weather forcasts does not make any sense mathematically. For example, they may list the POP for the day as 20% and then in the hour by hour forcast list the POP for 5 consecutive hours as 20% each. If we take these numbers as precise, then it would mean that if it rains that day it MUST rain in each of these five hours. One can argue that the numbers are rounded to the nearest full 10% and then there is no logical consequence of this type. But still, these numbers just don't make sense (given what weather is like) and show a lack of understanding of probability by the people who put up these forcasts. In effect, we should not take these POP numbers very seriously. Oded (talk) 17:25, 8 July 2008 (UTC)[reply]

Precipitation is bad example because the probability of precipitation on Tuesday is generally not independent of the probability of precipitation on Wednesday. If they were indedpendent, then the probability of precipitation over Tuesday or Wedesday,

P(T\lor W)=P(T)+P(W)-P(T)*P(W)

, however the assumption of independence is a bad one for this kind of real world process. Dragons flight (talk) 17:17, 8 July 2008 (UTC)[reply]

I guess it would not hurt to give some illustrative examples. Suppose that the probability to have rain on at least one of the days Tue and Wed is 1/2, and suppose that we know that the probability that it rains on Tue is equal to the probability that it rains on Wed. What could that probability be? Well, in the extreme cases, it could be 1/2 (in which case it rains on Tue if and only if it rains on Wed) or it could be 1/4 (in which case if it rains on Tue than it does not rain on Wed). These are the two extremes, and it could theoretically be any number in between.

Now consider a more general example. Suppse that the probability that it rains on a specific week is

p

, and the probability that it rains on the

j

^th day of the week is

p_{j}

. What can these numbers

p_{j}

be? Well, the only constraints are

0\leq p_{j}\leq p\leq 1

and

\sum _{j=1}^{7}p_{j}\geq p

. Oded (talk) 19:11, 8 July 2008 (UTC)[reply]

I've been led (by references that aren't to hand) to believe that POP is meant to be a probability that it will be precipitating at any given event in the spacetime region in question, rather than that any given point will see any precipitation during the interval. (Put differently, it means that there will be precipitation over the given proportion of the area in question on average: if it rains everywhere in Dallas for 12 hours and then rains in just 10% of the city for 12, the POP for (all of) Dallas that day is 55%.) If the rain is very scattered and intermittent, it may very well rain "everywhere" even if the POP is low; alternatively, even a 99% POP may mean no rain for a particular area that is no more than 1% of the region.

With this definition of POP, it is easy to combine them via a weighted average of their disjoint spacetime volumes, and to separate them by solving those averages for the constituents. But you can never acquire a "chance of any rain" from a POP unless it's 0 or 1. Of course, all of this assumes that the forecasters are correct, and that they are merely averaging over the spacetime region in question rather than also making errors and averaging over uncertainties. --Tardis (talk) 03:13, 9 July 2008 (UTC)[reply]

I doubt that this is what POP means. Observe that the POP for the day is always (at least that is what I have seen) at least as large as the highest POP for any hour in the day. With the definition that you are suggesting, it would instead be some average and therefore tend to be in the middle. (But the actual meaning of POP would fit better in the Science reference desk.) Oded (talk) 03:32, 9 July 2008 (UTC)[reply]