Wikipedia:Reference desk/Archives/Mathematics/2012 August 31

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August 31

volume

What is the volume (ounces or mls) in a fifth of whiskey? — Preceding unsigned comment added by 71.182.193.144 (talk) 00:33, 31 August 2012 (UTC)

It depends on the mixture, the temperature, the altitude of the sample and your position on the Earth. Moreover, it depends on the orbit of the Earth around the Sun, and on the Sun's motion through our galaxy. All of these things affect the volume. — Fly by Night (talk) 00:57, 31 August 2012 (UTC)

How density varies would only matter if a fifth was a unit mass. It is not, it's a unit of volume. See below. Therefore, if the number of fifths goes up for a given mass, so does the number of ounces and milliliters, in proportion, such that the number of ounces or milliliters in a fifth remains constant. (Note that I didn't engage in a personal attack just because you gave a poor answer.) StuRat (talk) 02:20, 31 August 2012 (UTC)

Not a personal attack, but a plain statement of fact - the first response was inappropriate, being patronising, belittling and worst of all, plain wrong. If you can't give an accurate, polite and helpful answer then don't give any.←86.139.64.77 (talk) 20:46, 1 September 2012 (UTC)

See Fifth (unit). hydnjo (talk) 01:00, 31 August 2012 (UTC)

Resolved

Solving a summation

I don't have the math background to know where to start with this. I have a summation of the usual sort with n on the top and i=0 on the bottom... and then the actual summation formula itself is a constant with i as an exponent. The problem is, I don't want to know what the answer is at a given n, I want to know what n is when I reach a given constant. I want to solve for the n. How do I go about figuring this out? Broba (talk) 07:42, 31 August 2012 (UTC)

Here's a simpler example of what I'm trying to do to illustrate the basics of what I can't figure out: Summation with i=0, the formula inside the summation is 3 * i. I want to know at what value of n the summation equals 18. What's the method you use to arrive at the value of n? Is it possible to do the same if the summation equals say 17? Broba (talk) 07:47, 31 August 2012 (UTC)

Quite simple to do with a computer program, although your example is easy enough to do by hand:

i  3i   ∑
---------
0   0   0
1   3   3
2   6   9
3   9  18

So we don't get 17 for the summation, assuming n is an integer. I don't think non-integer solutions for n (like n = 8/3) make sense in a summation, since, if you add in i = 8/3, how can you justify not adding in i = 7/3, as well, along with the infinite number of other non-integers ? StuRat (talk) 07:58, 31 August 2012 (UTC)

Well, another way to do something like this would be to close form solution of the summation as a function of n (if possible) and then setting it equal to whatever number you want (17 in this case) and solving for n. Like StuRat said, there is no guarantee if n would be a natural integer or not. Are you looking only for natural integers? To illustrate with your example, let $f(n)=\sum _{i=0}^{n}3i={\frac {3n(n+1)}{2}}$. Then you want to solve for n where ${\frac {3n(n+1)}{2}}=17$ and in this case we have a simple quadratic and the answer is $n={\frac {-3+{\sqrt {417}}}{6}}=2.90343...$. There is another solution too which is negative which we ignore. As you can see this is close to n=3 which would give us the sum equals 18. Is this what you are kind of looking for? Someone here might be able to help you furthermore if you give us the sum and how/where this problem arises.68.121.32.26 (talk) 10:32, 31 August 2012 (UTC)

But are non-integer solutions actually allowed in a summation ? To me it doesn't make any sense. As I noted before, if you sum in one non-integer value (the final one), then you would have to sum in all non-integer values, wouldn't you ? This would typically mean the solution is either positive infinity or negative infinity, regardless of n, which isn't very useful. StuRat (talk) 20:22, 31 August 2012 (UTC)

For an exponential function, use $\sum _{i=0}^{n-1}a^{i}={\frac {1-a^{n}}{1-a}}$ Ssscienccce (talk) 15:27, 31 August 2012 (UTC)

Thank you both 68.121.32.26 and Ssscienccce. That is the approach I'm trying to take, although I think my problem has additional complexities. What is the name of this sort of math... I had a lot of trouble knowing what to search for for this.

The complications I have is that I tried to use the exponential function that Ssscienccce gave but I have some coefficients that I believe are screwing it up. Non integers are fine (expected probably). This is specifically the form of the problem I'm trying to solve (and yes I could estimate it with a program, I'm not interested in that... I'm looking for the method to arrive at the answer more than any specific answer):

$\sum _{i=0}^{n}100e^{0.05n}=5000$

I'm trying to solve for $n$ here of course. As an aside, I have a hunch there's some analogy to this kind of problem in physics but I haven't grasped exactly which yet. Broba (talk) 17:47, 31 August 2012 (UTC)

Oh I got it now. I just found another version of Ssscienccce's explanation with coefficients. Thanks for you help. Broba (talk) 21:56, 31 August 2012 (UTC)

Resolved

Didn't you mean $\sum _{i=0}^{n}100e^{0.05i}=5000$ ? Bo Jacoby (talk) 06:56, 1 September 2012 (UTC).

Late to the party but whatever... If you can still use some advice, I try to walk you step by step to the solution. Some things, like the 100 make your equation look far more formidable than it really is. I'll compute the "a" for you but only outline the later steps, it's quite straightforward once you see what to compute, and last but not least, I'm doing the stuff mentally atm.

Let's look at your sum of 100exp(0.05i) = 5000. You can cancel out a factor of 100 to get "Sum of exp(0.05i) = 50", and now assume that a = exp(0.05) = 1.051..., so this is a fancy way to write "Sum of a^i = 50".

Ssscienccce's formula gives you a closed form for that kind of summation. Multiply by (1-a) to get something like "1 - foo^bar = 50 (1-a)", add foo^bar, and subtract the constants on the right-hand side to get something like "constant = foo^bar,"

something you can attack with a log operation to get "bar = _foolog(constant)". Sorry for the sloppy terminology.

- ¡Ouch! (^{hurt me} / _{more pain}) 07:23, 3 September 2012 (UTC)

compare all to all

I have a set S and I want to compare every item in S to every other item in S. The comparison is a simple function like f(S1,S2). I claim that it is impossible to do this without S*(S-1) comparisons. I've been told it can be done with S*log(S) comparisons, but not how it can be done that way. Can anyone explain to me how that is possible? — Preceding unsigned comment added by 128.23.113.249 (talk) 14:29, 31 August 2012 (UTC)

Use any of the worst-case $n\log n$ comparison sorts from the table in sorting algorithm#Comparison of algorithms, such as merge sort. Once you sort the set, you can compare any two items by comparing their index in the sorted set without invoking f. This all assumes that f is a bona fide total order.—Emil J. 14:41, 31 August 2012 (UTC)

Please correct me if I'm wrong, but that implies that S is sortable. I'm trying to come up with an example that shows how they are not sortable based on f. What if you have a set of people S. Then f(S1,S2) is a rating of how much S1 likes S2. We can make it simple and assume f(S1,S2)=f(S2,S1) if that is important. Then, you have a weighted graph where each node is an element in S and the edges have a weight, the value of f for the two nodes. If I use f to sort S with, say, merge sort, I won't compare every element in S to every other element and, therefore, will be missing edges in my graph. Does that make sense or am I just rambling nonsense? — Preceding unsigned comment added by 128.23.113.249 (talk) 14:59, 31 August 2012 (UTC)

Your f is not a comparison function (i.e., an indicator function of a total (pre)order). If you allow a completely arbitrary function as f, then there is of course no way of finding all its values for all pairs of elements other than checking all the |S|² pairs.—Emil J. 15:37, 31 August 2012 (UTC)

Thank you. I believe our class argument is actually an argument about how f is defined, not about the set S. It all depends on if f defines total order or just some arbitrary relationship between two items. — Preceding unsigned comment added by 128.23.113.249 (talk) 15:42, 31 August 2012 (UTC)

You may want to check out Closest pair of points problem Ssscienccce (talk) 16:19, 31 August 2012 (UTC)

solutions to an assignment

Hello I am an independent learner looking through material here: http://www.hutter1.net/ethz/uaiethz.htm and in particular trying the assignment here: http://www.hutter1.net/ethz/assign1.pdf but sadly there are no solutions to this assignment on that website. Some questions are easy math questions but some are hard. Could someone kind create solutions or refer to elsewhere? — Preceding unsigned comment added by Bulkc (talk • contribs) 18:05, 31 August 2012 (UTC)

That's a bit much to ask of us, I'm afraid. We also won't do your homework for you. However, if you do it, and post your answers here, we might be willing to check it for you and point out any mistakes. StuRat (talk) 20:28, 31 August 2012 (UTC)

Ok. The first two questions and the probability questions are easy any way. I think I can do too the question MDL-ML, part (i) of MDL-Ber and some of KC-KC myself. What about the rest? For example can you show me how to do KC-KC part (iv) which does not follow from the rest I think. Or inequality K(xy) < K(x,y) — Preceding unsigned comment added by Bulkc (talk • contribs) 20:51, 31 August 2012 (UTC)

For (iv), let $T$ be the machine that, on input $a$, first runs $U(a)$. Then, if the output is a pair $(x,f)$, runs $f$ on $x$ and outputs $f(x)$. Now, if $U(a)=(x,f)$, then $T(a)=f(x)$. This shows that $K_{T}(f(x))\leq K(x,f)$, which is less than $K(x)+K(f)$ by part (iii). Since $K\leq ^{+}K_{T}$, the result follows.

For $K(xy)\leq ^{+}K(x,y)$, let $T$ be the machine that, on input $a$, first runs $U(a)$. Then, if the output is a pair $(x,y)$, $T$ outputs $xy$. Then follow the same reasoning as above.--121.73.35.181 (talk) 00:08, 1 September 2012 (UTC)

Thank you! I can do now these questions, and some other ones, due to your advice about Turing machines. But, I am still having some trouble. What about AP-CM (ii) and (iii)? Or MDL-Ber (iii), (v) and (vi)?--Bulkc (talk) 06:14, 1 September 2012 (UTC)

Actually, I just did MDL-Ber (v) using (iv)! — Preceding unsigned comment added by Bulkc (talk • contribs) 06:32, 1 September 2012 (UTC)

Woops, had one of my inequalities backwards. Fixed.--121.73.35.181 (talk) 10:28, 1 September 2012 (UTC)

• After much work and reading through the material, I think I can solve all questions except for AP-CM. I think this is hardest. — Preceding unsigned comment added by Bulkc (talk • contribs) 02:58, 2 September 2012 (UTC)

Here's (i): Fix $\epsilon >0$. Let $A_{t}$ be the set of infinite sequences $x$ with $M(0|x_{<t})-\mu (0|x_{<t})>\epsilon$. Note that $\sum _{t}\mu (A_{t})\epsilon ^{2}$ is less than the sum in theorem 4.5, so in particular is finite. So $\sum _{t}\mu (A_{t})$ is finite, and thus $\limsup _{t}A_{t}$ has $\mu$-measure 0. This contains all sequences where the limit is infinitely often above $\epsilon$. Union over all rational $\epsilon$ to get the set of all sequences where the limit is non-zero, and it still has $\mu$-measure 0.--121.73.35.181 (talk) 22:40, 2 September 2012 (UTC)

Thank you very much for your help. Let me just ask for solutions to the ones that I'm still not wholly sure about:

KC-Cmp (i)

AP-RC, especially inequality $K(x|l(x)){\stackrel {+}{<}}KM(x)$

AP-CM (ii), (iii), and maybe (iv) but I think I can figure to do that one.

I think I would like solutions to all of KC-KC too because I find Kolmogorov complexity confusing still. --Bulkc (talk) 22:56, 4 September 2012 (UTC)