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July 13[edit]

Dominion probability[edit]

What's the probability of getting a 5-2 copper split in the two opening hands of Dominion, in either order? (The deck starts with 7 coppers and 3 estates, and you draw 5 cards per hand, discarding all 5 after your turn.) --134.10.113.106 (talk) 00:25, 13 July 2011 (UTC)[reply]

There are 10 choose 5 = 252 possible first hands, and of those, 7 choose 5 = 21 are all copper. So 8.33% probability of your first hand being all copper. The same likelihood of your second hand being all copper, and since they're disjoint possibilities, they add to get 16.66% probability.--Antendren (talk) 01:17, 13 July 2011 (UTC)[reply]

Non-associativity of addition with infinite series[edit]

I've seen examples illustrating the non-associativity of addition with infinite series, but they all illustrate the property for divergent series only. Are there any examples of conditionally-convergent illustrating this property also?--Leon (talk) 07:17, 13 July 2011 (UTC)[reply]

No. Associating terms in a series is effectively selecting a sub-sequence of the sequence of partial sums. For any convergent series (even just conditionally convergent), the sequence of partial sums converges, and thus any sub-sequence will converge to the same value.--Antendren (talk) 07:28, 13 July 2011 (UTC)[reply]

It may be worth emphasizing that conditionally convergent series of reals can be rearranged (by permuting the terms) to achieve any real value. This is the Riemann rearrangement theorem. 173.75.158.107 (talk) 11:45, 13 July 2011 (UTC)[reply]

OK, and commutativity: the addition of terms of any conditionally convergent series cannot be commutative, owing to the Riemann rearrangement theorem; is this true of divergent series also? To cut to the chase, can we say the following about the arrangement of terms in series?

Unconditionally convergent: always commutative and associative.
Conditionally convergent: never commutative, always associative.
Divergent: never commutative, not always (but sometimes) associative.

(if my reasoning is both incorrect and unclear, I think that since a conditionally convergent series, which is always associative, can be rearranged into a divergent one by a permutation, that some divergent series may inherit that property) --Leon (talk) 14:17, 13 July 2011 (UTC)[reply]

Depends what you mean by a divergent series being associative or commutative. For example, consider the series 1 + 1 + 1 + 1 + 1 + 1 + ... No matter how you associate terms or rearrange terms, the series will still diverge to positive infinity. You might call this series associative and commutative (of course, some would call this series convergent).--Antendren (talk) 21:08, 13 July 2011 (UTC)[reply]

Given that by associating terms of this series to get 1 + 2 + 3 + 4 + … or 1 + 2 + 4 + 8 + …, is it really correct to say that it is associative?--Leon (talk) 13:22, 14 July 2011 (UTC)[reply]

Well, that depends on if you say those series also sum to positive infinity, or if you are using a summability method to assign a finite value to them. If the former, it's not a counterexample to associativity. If the latter, we're getting into an area I don't know much about, so I can't speak on it.--Antendren (talk) 02:51, 15 July 2011 (UTC)[reply]

Cute way to evaluate a series[edit]

The series $\sum _{n=0}^{\infty }{\frac {n+1}{2^{n}}}$ can be evaluated by arranging the dyadic fractions into a triangular array:

{\begin{matrix}1&{\frac {1}{2}}&{\frac {1}{4}}&{\frac {1}{8}}&{\frac {1}{16}}&\cdots \\&{\frac {1}{2}}&{\frac {1}{4}}&{\frac {1}{8}}&{\frac {1}{16}}&\cdots \\&&{\frac {1}{4}}&{\frac {1}{8}}&{\frac {1}{16}}&\cdots \\&&&{\frac {1}{8}}&{\frac {1}{16}}&\cdots \\&&&&\cdots &\cdots \end{matrix}}

The sum down the $n$ th column is $(n+1)/2^{n}$ , so the value of the series is the sum of the entries in the array. The sum of the $n$ th row is $1/2^{n-1}$ , so the sum of the series is $\sum _{n=0}^{\infty }{\frac {1}{2^{n-1}}}=4$ .

This trick works for any series of the form $\sum (n+1)x^{n}$ . Although we can certainly evaluate this by less elementary methods, this is the most elegant way I've seen to compute the series. My question is, presumably this trick is well-known. Does it have a home in some article in Wikipedia? (Maybe in the article about the series itself, but where would that be?) Sławomir Biały (talk) 12:00, 13 July 2011 (UTC)[reply]

More generally, assuming everything converges

\sum _{n=0}^{\infty }\left(\sum _{i=0}^{\infty }f_{i}\right)g_{n}=\sum _{i=0}^{\infty }f_{i}\left(\sum _{n=i}^{\infty }g_{n}\right).

This is basically changing the "order of summation", a discrete version of order of integration (calculus). I found this at wolfram.com which seems to cover it, look under the heading "Double infinite summation", but I can't find it here. Maybe this will help refine the search a bit.--RDBury (talk) 13:49, 13 July 2011 (UTC)[reply]

Yes it's easy to justify the interchange in order of summation. (For instance, we could clobber it with Tonelli's theorem.) The question is whether writing this single series as a double series is a well-known/notable trick. Sławomir Biały (talk) 14:36, 13 July 2011 (UTC)[reply]

Just thinking that a more general term would be easier to find. Another generalization is

\sum _{n=0}^{\infty }a_{n}x^{n}={\frac {1}{1-x}}\left(a_{0}+\sum _{n=1}^{\infty }(a_{n}-a_{n-1})x^{n}\right)

which is like step 1 of a binomial transform. The full binomial transform would give you a finite sum in your example. I'm sure people like Euler and Gauss used this kind of thing all the time, they may have though it too trivial to write out though. It's exactly the kind of thing you need to work with hypergeometric series for example.

Hitting it with Tonelli's theorem seems like hunting for rabbits with nuclear tipped cruise missiles.Naraht (talk) 19:10, 13 July 2011 (UTC)[reply]

That was Sławomir's point. — Fly by Night (talk) 19:26, 13 July 2011 (UTC)[reply]

If you generalize the technique you get

\sum _{k=0}^{\infty }f_{k}g_{k}=

{\begin{matrix}f_{0}g_{0}&+f_{0}g_{1}&+f_{0}g_{2}&+f_{0}g_{3}&+f_{0}g_{4}&\cdots \\&+(f_{1}-f_{0})g_{1}&+(f_{1}-f_{0})g_{2}&+(f_{1}-f_{0})g_{3}&+(f_{1}-f_{0})g_{4}&\cdots \\&&+(f_{2}-f_{1})g_{2}&+(f_{2}-f_{1})g_{3}&+(f_{2}-f_{1})g_{4}&\cdots \\&&&+(f_{3}-f_{2})g_{3}&+(f_{3}-f_{2})g_{4}&\cdots \\&&&&\cdots &\cdots \end{matrix}}

=f_{0}\sum _{k=0}^{\infty }g_{k}+(f_{1}-f_{0})\sum _{k=1}^{\infty }g_{k}+(f_{2}-f_{1})\sum _{k=2}^{\infty }g_{k}+\dots

which is the first equation in Summation by parts#Newton series, putting n=infinity. So the answer I'm going with for the judges is "summation by parts". Not sure where Tonelli's theorem comes in with that idea though.--RDBury (talk) 21:42, 13 July 2011 (UTC)[reply]

Nice. Sławomir Biały (talk) 22:09, 13 July 2011 (UTC)[reply]

Need name for a distribution[edit]

Hi. The function $f(x)=\exp(x)/(1+\exp(x))^{2}$ defines a probability density function. Does anyone know a name for it? Robinh (talk) 21:36, 13 July 2011 (UTC)[reply]

Try logistic distribution.--RDBury (talk) 21:57, 13 July 2011 (UTC)[reply]

That's it! Easy if you know. Best wishes, Robinh (talk) 23:30, 13 July 2011 (UTC)[reply]

Resolved

quotation[edit]

how can this possibly be right:
"All the taxes paid over a lifetime by the average American are spent by the government in less than a second." That would mean after 1 second, there would have to be a one-lifetime deficit, and a year has 31 million seconds. So, even if a lifetime is a paltry "20 years", that would be 20 * 30 = 600 million times that average annual earnings in defiicit. If the annual earnings is even $10,000 then that would mean there would be a 6 trillion dollar deficit. In fact the deficit is only 40billion (a huge difference of over 100-fold, especially considering that a lifetime is surely more than 20 years and the average salary surely more than 10k per annum). so how can that quotation possibly be right? 188.222.102.201 (talk) 23:12, 13 July 2011 (UTC)[reply]

"Spending" here probably means budget spending, not deficit spending. The annual budget is about three and a half trillion, or ten billion per day, four hundred million per hour, seven million per minute, one hundred thousand per second. A person paying $2500 tax per year would total one hundred thousand in a 40 year career. So it seems about right to me.--RDBury (talk) 23:37, 13 July 2011 (UTC)[reply]

I think you're confusing average with total, because your calculation doesn't take into account the size of the population. As you say, assuming a 20 year lifetime, the quote says the government spends 600 million lifetimes in 20 years. But there are 300 million people in the US, so we generate 300 million lifetimes in 20 years. That's only a 2:1 ratio of spending to earning. Since lifetimes are closer to 80 years, the quote actually gives us a ratio of roughly 8:1.--Antendren (talk) 23:43, 13 July 2011 (UTC)[reply]

Working lifetimes are not so long. —Tamfang (talk) 22:51, 15 July 2011 (UTC)[reply]

Rewriting an absolutely convergent series[edit]

Related to the above question, an absolutely convergent series is said to also be "unconditionally convergent", ie, satisfy the property that any permutation of the terms of the series will still yield the same sum. But this isn't the only way to rewrite the sum. What if you take a series and convert it to an arbitrary well-ordered sum, with the obvious requirement that there be a bijection between the terms of the well-ordered sum and the original series? Is that guaranteed to converge to the same value as the original series? --COVIZAPIBETEFOKY (talk) 23:32, 13 July 2011 (UTC)[reply]

Yes. You can start by proving it for series of all non-negative terms, by induction on the order-type of the sum: if the well-ordered sum converges to something less than the original series, grab finitely many terms from the original series which sum to more than the well-ordered sum, deriving a contradiction. If the well-ordered sum converges to something greater than the original series, necessarily the order-type is a successor ordinal

\alpha +1

. But then since finite sums are commutative, you can move the last summand to the front of the series, and get a well-ordered sum of order-type

\alpha

which converges to more than the sum of the original series, contradicting the inductive hypothesis.

Next, you can basically repeat the proof that absolute convergence implies unconditional convergence: grab finitely many terms that bring the sum of the original series within epsilon, bound the remainder of the well-ordered sum by the sum of the absolute values. I suppose this requires first showing that finite rearrangement does not change well-ordered sums, and that a well-ordered sum is bounded by the sum of its absolute values, but those are simple enough.--Antendren (talk) 00:23, 14 July 2011 (UTC)[reply]

Thanks, that mostly makes sense. I'm not clear on why the order type of a sum which converges to something greater than the original series must be a successor. --COVIZAPIBETEFOKY (talk) 01:11, 14 July 2011 (UTC)[reply]

Sorry, it's based on the inductive hypothesis again. Since the sum at a limit ordinal is just the supremum of the sums at previous ordinals, if it happened at a limit ordinal, there must be some previous ordinal where the partial sum is greater than the sum of the original series. But the partial sum at this ordinal corresponds to the rearrangement of a subseries of the original series, and the sum of a subseries (of non-negative terms) is at most the sum of the original series, so this contradicts the inductive hypothesis.--Antendren (talk) 01:29, 14 July 2011 (UTC)[reply]

Another way of putting this is to compare the well-ordered sum with the Lebesgue integral (treating your sequence as a function from a discrete measure space where every point has mass 1). You can prove by induction on α that the well-ordered sum up to α equals the integral of the function restricted to ordinals less than α, using additivity of the integral at successors and dominated convergence at limits. Algebraist 08:53, 14 July 2011 (UTC)[reply]

Thanks a lot! --COVIZAPIBETEFOKY (talk) 13:03, 14 July 2011 (UTC)[reply]

Resolved