Wikipedia:Reference desk/Archives/Mathematics/2009 May 6

= May 6 =

Limits of differentiable functions
This problem is taken from a practice AP Calculus BC exam:

79. If f is a differentiable function such that f(3) = 8 and f'(3) = 5, which of the following statements could be false?

$$ \begin{align} (\text{A}) & \lim_{x \to 3} f(x) = 8 \\ (\text{B}) & \lim_{x \to 3^+} f(x) = \lim_{x \to 3^-} f(x) \\ (\text{C}) & \lim_{x \to 3^+} \frac{f(x)-8}{x-3} = 5 \\ (\text{D}) & \lim_{h \to 0} \frac{f(3+h)-8}{h} = 5 \\ (\text{E}) & \lim_{x \to 3} f'(x) = 5 \\ \end{align} $$

The answer is supposed to be E, but I don't see how E can be false. Can you find a function for which this is false? Or prove that the question has no right answer? --Yanwen (talk) 00:49, 6 May 2009 (UTC)


 * A function can be differentiable without its derivative being continuous (E is a statement that the derivative is continuous at x=3). Differentiability class has an example of such a function. --Tango (talk) 01:01, 6 May 2009 (UTC)

$$f(x) = \begin{cases}x^2\sin{(1/x)} & \mbox{if }x \neq 0, \\ 0 &\mbox{if }x = 0\end{cases}$$ is not differentiable at x=0, so it doesn't fit the description of f(x). I guess a way to rephrase the problem is "Does there exist a differentiable function whose derivative is discontinuous?" --Yanwen (talk) 01:12, 6 May 2009 (UTC)
 * But a function such as
 * It is differentiable at 0, the derivative is 0. If it was just x in front of the sine it wouldn't be differentiable but the x2 makes it so it is. You need to go all the way back to the definition of a derivative as a limit in order to work it out, but it really is differentiable. --Tango (talk) 01:15, 6 May 2009 (UTC)
 * How? All of terms of all derivatives have a factor which is either a sine or cosine with the argument 1/x. How can this possibly be evaluated? Elocute (talk) 08:54, 6 May 2009 (UTC)
 * By definition, the derivative of that function at 0 is the limit as x tends to 0 of xsin(1/x). Since sin(1/x) is always between -1 and 1, we have that $$|x \sin(1/x)| \le |x|$$, and so the limit is 0. There's no difficulty with evaluating sin(1/x), since we only need to evaluate it for x≠0. Algebraist 10:48, 6 May 2009 (UTC)
 * It is differentiable, as Tango said before. For any $$x\ne 0$$ we have $$f^\prime(x) = \frac \mathrm{d}{\mathrm{d}x} x^2\sin{(1/x)} = 2x\,\sin\tfrac 1x + x^2\,\cos\tfrac 1x\cdot\tfrac {-1}{x^2} = 2x\,\sin\tfrac 1x - \cos\tfrac 1x.$$ The derivative at zero must be calculated from definition: $$f^\prime(0) = \lim_{h\to 0}\frac{f(h)-f(0)}{h\ -\ 0} = \lim_{h\to 0}\frac{f(h)}h = \lim_{h\to 0}h\sin\tfrac 1h.$$ Compare this to what Algebraist said above: $$(\forall_a |\sin a| \le 1) \implies (\forall_x |x\,\sin\tfrac 1x| \le |x|)$$$$ \implies (\forall_h |\tfrac{f(h)}h| \le |h|) \implies \lim_{h\to 0}\frac{f(h)}h = 0.$$ CiaPan (talk) 14:04, 6 May 2009 (UTC)

"Elocute", when "Tango" wrote "You need to go all the way back to the definition", you really should have heeded those words. You then asked "how?". But the words "You need to go all the way back to the definition", which he had already written, are the answer to "how?". Michael Hardy (talk) 19:59, 6 May 2009 (UTC)

If you use the definition of a derivative in conjunction with l'Hopital's rule, you can get an interesting result

$$f'(a) = \lim_{x \to a} \frac{f(x)-f(a)}{x-a}$$ Then use l'Hopital's rule to get $$\lim_{x \to a} \frac{f'(x)}{1}$$ so $$f'(a) = \lim_{x \to a} f'(x)$$ But this must be wrong somehow, because of the previous example. What is wrong here?--Yanwen (talk) 20:06, 6 May 2009 (UTC)
 * L'Hopital's rule only applies if the function is differentiable, so you have assumed exactly what you are trying to prove. I don't believe I have ever used L'Hopital's rule, it is rarely useful - forget about it! Go all the way back to first principles:
 * f(x) is differentiable at a if the following limit exists:
 * $$\lim_{\Delta x \to 0} \frac{f(a+\Delta x)-f(a)}{\Delta x}$$
 * Now write that out with the appropriate formula for f(x) (the one for away from 0 - when considering a limit as you approach a point you only need to consider what the function does away from that point), substitute in x=0 and try and simplify it. You'll find it simplifies to $$\lim_{\Delta x \to 0} \Delta x \sin\left(\frac{1}{\Delta x}\right)$$. You now have a product of something going to zero and something bounded, so it goes to zero, ie. it exists. --Tango (talk) 20:20, 6 May 2009 (UTC)
 * Thank you to whoever corrected my mistakes in red pixels! --Tango (talk) 21:12, 6 May 2009 (UTC)
 * I apologize. It was me — but that was quite stupid, and rather offensive, so I remove the red colour. --CiaPan (talk) 09:05, 7 May 2009 (UTC)
 * I thought it was quite funny... --Tango (talk) 13:43, 7 May 2009 (UTC)


 * L'Hopital's rule not only requires the function(s) to be differentiable, it requires the limits involved in the rule to all exist, which will not be the case in general when the derivatives involved are discontinuous. This is what Yanwen was noticing in his derivation.


 * The family of functions $$ f_a(x) = x^a\sin(1/x)$$ is a very nice source of counterexamples. By choosing a appropriately one can get functions that are C^1 and have a discontinuous second derivative at 0, are C^1 but have no second derivative at 0, are C^5 but have a discontinuous 6th derivative at 0, etc. &mdash; Carl (CBM · talk) 12:04, 7 May 2009 (UTC)

Cycloid
If a straight line interescts the same arc of cycloid at two different points, is the smaller arc that is between the two points also a cycloid? if so is it defined by a roulette along the aformentioned straight line? Elocute (talk) 10:32, 6 May 2009 (UTC)
 * No, because the tangent line to a cycloid at the point where it touches the "ground" is vertical. Michael Hardy (talk) 19:57, 6 May 2009 (UTC)

how to do monte carlo for statistics?
let's say there are two functions, a and b where one (at random) is the placibo and one is the drug. The placibo is supposed to randomly return 0's or 1's, and as for the drug...it's supposed to favor 1's by a certain "strength"...

I want to figure out, through monte carlo, how many outputs I should get from A and B to see whether one is a working drug with a given strength?

I'll give you two examples.


 * 1) supposed strength is 100%  -- all 1's, no 0's from the drug function, half 1's half 0's from the placibo function.  How many outputs from A and B should I look at to determine with 98% confidence whether either A or B meets this standard?
 * 2) supposed strenght is 75% -- three quarter 1's, one quarter 0's from the drug function, half 1's half 0's from the placibo function.  How many outputs from A and B should I look at to determine with 98% confidence whether either A or B meets or exceeds this proclaimed strength?

For both of my examples I am less interested in the answer than how to monte carlo the answer!! Thank you. If I get no answers I'll ask the computing desk. 79.122.21.123 (talk) 16:28, 6 May 2009 (UTC)
 * To get "98% confidence" is to mean you have 2% or less chance of having a false positive. In this situation a false positive is the same as a false negative, since you declare one to be the drug and the other is automatically the placebo.  To do this in simulation you would no doubt take the higher of the two values as your actual drug (even if the percents come out to something wierd for your second example like 80% and 95%, there is still a better chance that your placebo got the 80% and your drug got the 95% than the other way around).  A single simualation would be, say, N=100 results from each function. You then run this simulation many times and see if you guess the drug correctly 98% of the time.  If not, you need a higher N.


 * If you don't want to run a bunch of simulations for each N, you can track which function was the higher one at each N for each simulation, thus you can see how accurate all of the N's are. Anythingapplied (talk) 18:25, 6 May 2009 (UTC)


 * Though your formulation sounds natural, I think you have not given enough information to be able to determine with any confidence at all the answers, even given unboundedly many trials. The problem is that you have only specified what happens when {a,b} contains a working drug and a placebo.  You have not specified what happens in the other cases or what the probability is amongst these cases.  In somewhat natural terms, you haven't specified whether {a,b} contains maliciously crafted b such that trials of b are not independent.
 * Do you mean to say that {a,b} contains two bernoulli processes, and you want to decide if the parameter for either a or b meets or exceeds a given probability? You want to do this by sampling both a and b repeatedly, and use this to give an estimator of the mean of each?
 * If so, then you might want to ask whether you really want two variables or one. Would it be sufficient to test a independently, and test b independently?
 * If so, then you are looking at a reasonably standard statistics question, which is basically a shifted version of checking if a coin is fair. If you are looking for placebos, then this is exactly the question you are asking, but looking for other means is also similar. JackSchmidt (talk) 18:30, 6 May 2009 (UTC)
 * By the way, my reading of your problem is assuming you really meant "whether" not "which". Anythingapplied's answer is extremely suitable if you mean "which", not "whether". JackSchmidt (talk) 18:34, 6 May 2009 (UTC)
 * (Yes, I did really mean whether)

but... I tried reading your answers, and also the telling if a coin is fair article. It is too hard for me, too many fomulas, and you haven't told me how to monte carlo. anythingapplied did tell me how to monte carlo, but he thought I meant which and not whether... I'll reiterate the question to make it explicit: The question is not about which but whether either a or b functions, which each spit out 0's and 1's, meets the criteria for example that it will spit out 75% ones, and how many tests I should do to figure this out with 98% confidence. Specifically I want to know how to Monte Carlo the number, not use a formula to figure it out! If I don't get an answer here I'll ask at the computing desk.... Here are more examples:
 * 1) supposed strength of 100%. How many results should I see from a and b -- given that one of them, though I don't know which one, is a control spitting out ones and zeros randomly (evenly), and that the other one may or may not meet the standard of spitting out 100% ones -- if I want to know with 98% confidence WHETHER the standard (100% ones) is being met by one of them...?
 * 2) supposed strength of 75%. How many results should I see from a and b -- given that one of them, though I don't know which one, is a control spitting out ones and zeros randomly (evenly), and that the other one may or may not meet the standard of spitting out 75% ones -- if I want to know with 98% confidence WHETHER the standard (75% ones) is being met by one of them...?
 * 3) supposed strength of 60%. How many results should I see from a and b -- given that one of them, though I don't know which one, is a control spitting out ones and zeros randomly (evenly), and that the other one may or may not meet the standard of spitting out 60% ones -- if I want to know with 98% confidence WHETHER the standard (60% ones) is being met by one of them...?

for all of the above two things apply:
 * 1) If the standard isn't being met by ONE of them, it is being met by NEITHER of them, since I know that one will spit out zeros and ones randomly and evenly. So you're really only just testing one of the two, the other is a ruse, it is irrelevant, you don't really need it...
 * 2) I am less interested in the answer than how to monte carlo the answer. I want to come up with the answer doing millions of iterations of monte carlo, not using a formula...  That's what this whole question is about.

Thank you! 79.122.21.123 (talk) 19:52, 6 May 2009 (UTC)


 * I think the question JackSchmidt is asking is what *do* the drugs do if they aren't one placebo and one effective drug? Would they be two placebos? Or a placebo and a partially effective drug? Or what? You don't seem to have answered that, and I don't think we can help you until you do. --Tango (talk) 20:12, 6 May 2009 (UTC)


 * I don't understand the question. If it isn't one placebo and one drug-candidate that meets the strength threshold, then it's one placebo and one drug-candidate that has failed to meet the strength threhold.  If half of headaches disappear within 1 hour by themselves, and you're testing something (against placebo) for whether it makes 90% of headaches disappear within 1 hour, it doesn't matter what you call the latter if it fails to meet that critereon.  I can try to think more about what to call it if it fails, but is it really that important?
 * Actually I think this is a good concrete example.
 * How would I monte carlo to arrive at the number of trials I should do of a and b, where one makes a headache disappear half the time within 1 hour, and the other may or may not make at least 90% of headaches disappear within an hour, if I want to learn with 98% confidence whether this strength threshold is being met by the candidate drug!!!

If you want, you could consider 90% a number Marketing has dictated, where less than that and it's worthless, they already have 88% and 85% and 80% drugs, they can't market those as 90% but if you get them one that's 90% effective (with 98% confidence), they can market that. So you want to know how many trials you should do against placebo (which will get you 50% gone headaches). —Preceding unsigned comment added by 79.122.21.123 (talk) 20:33, 6 May 2009 (UTC)

Is this phrasing (analogy) quite clear? And again, I'm less interested in the number than the method I should use to monte carlo the number for myself, since I'm finding the maths too hard. Thanks! 79.122.21.123 (talk) —Preceding undated comment added 20:31, 6 May 2009 (UTC).


 * If this is meant to be based on a realistic study then you shouldn't know the probabilities for the placebo in advance - that's why you have a control group taking a placebo in the first place. Have you read Statistical hypothesis testing? That's what you are trying to do - test a hypothesis. --Tango (talk) 21:11, 6 May 2009 (UTC)


 * sigh. No.  It's not meant to be based on a realistic study.  It's meant to be based on two functions, a and b, and a threshold of x%.  For that threshold, how do I determine how many trials will let me ascertain with 98% confidence level whether either of a and b spit out random 0's and 1's that are weighted to at least that threshold?  Here is an example with some pseudocode:

a{ return with 50% chance 0 with 50% chance 1 }

b return with 20% chance 0 with 80% chance 1 }

There should be difference numbers for the 80% threshold test (at 98% confidence). If I do the respective number of tests for the 55% threshold, the 60%, the 65%, the 70%, the 75%, and the 80%, in all of these b should pass...  The reason it should pass is that the REALITY, the TRUTH is that it gives 80% ones.

There should be a way of finding this out with 98% confidence.

There should also be a way of having it FAIL the test for 85%, or for 90%, or for 95% ones, since in fact as you can see from the pseudocode above it does not actually favor ones so strongly.

I should be able to come up with numbers for these different thresholds. How do I do this using the monte carlo method? If there are still no replies tomorrow I will give my ideas, but monte carlo is really easy and someone should be able to explain using five lines of pseudocode. 79.122.21.123 (talk) 22:14, 6 May 2009 (UTC)


 * (ec)I think the tricky part is he is actually trying to *design* a test of a statistically hypothesis, but without working out the details carefully. Rather he wants to use monte-carlo simulation to measure the performance of his tests.  Since it is just a Bernoulli process (i.i.d. {0,1} variables) and he is trying to estimate the mean, in principal one could figure this out quite precisely.  However, instead he wants to say (I think) "Ok, I will try the following test: we sample 100 times, and see if the sample mean is above 90%; if so we declare that the true mean is 90% or above.  Now let's run this test on some simulated samples.  First I'll try a 50% effective drug (the placebo; apparently normalized to be 50% effective).  Is it the case that only 2% or less of the time I declare the placebo to be a 90% effective drug? Yay! Test is ok at discarding placebos.  Ok, now I'll run the test on simulated samples from a 90% effective drug.  Do I declare it to be an effective drug at least 98% of the time? Oh no! Test is crappy at detecting effective drugs.  Guess I'll need to adjust."
 * Well, that's my naive version. Probably he wants something a little less ad hoc, but not *much* less ad hoc.  Sadly I would have to work out the theoretical solution first before giving advice on how to come up with it via simulation, so someone with more experience or more confidence will have to answer.  JackSchmidt (talk) 22:18, 6 May 2009 (UTC)


 * Yes, thank you, you have my intentions about right. However, your specific guess as to an algorithm is diffferent from my conception in two ways:
 * You imply that I would test each test length (your example: 100) a single time. However I know that would not be enough tries to give a good estimation (because of short-term variability): I would not expect it to converge on the true number I would arrive at with a formula.  However, if I did it 1million or 100million times, it would almost certainly be very close to the number I would get from a formula.
 * You suggest that I want to first sample a and b, and then run simulations. In fact, I want to run simulations first, so that I can get a table of different test lengths for different strengths, to choose the test length corresponding test strength I feel is most appropriate.
 * One other thing I don't like about your formulation is "is it the case that only 2% or less of the time I declare the placebo to be a 90% effective drug". This is much easier to achieve, since the placebo is normalized to 50%, than keeping from declaring a 70% effective drug to be 90% effective.  To give examples: maybe 2,000 samples would be enough to rule out accidentally choosing the 50% placibo as the "90% drug", but you would need 7,000 samples to be 98% confident that a 70% drug would not accidentally pass for 90%....  So in fact the test is more stringent: it is not only telling which of a, b is the 90% effective drug, it is telling whether either are.  (The answer to this question is in reality, by the truth of the matter, "no" for a 70% drug.  So "neither a or b is a 90% drug" should be the result for running the appropriate test length -- at least, this should be the test result at least 98% of the time!!)  —Preceding unsigned comment added by 94.27.231.41 (talk) 09:13, 7 May 2009 (UTC)

Since I did not really get any algorithms and said I would post my tentaive algorithm today if I got no responses, here it is (pseudocode):

would the numbers produced by this algorithm mean what they say they do?? Thank you.... —Preceding unsigned comment added by 94.27.231.41 (talk) 09:36, 7 May 2009 (UTC)

also, to anyone who might be wondering why I use both an imposter and a placibo. Well, in fact, what I did not include in the test is that the drug and placibo would be randomized, to make it blind. The role of the imposter is really like the role in medicine of the "best drug on the market".... it is a kind of threshold the placibo and drug are being compared against. I want my test to make the imposter fail 98% of the time (together with the placibo) but the drug to pass 98% of the time -- that it is, if it really does meet the threshold. The margin of error is 2%. So that means I don't consider it a bad result if a drug that in fact, in reality, gives 79% 1's, passes for an 80% effective drug. However a 77% drug should not. (98% of the time). Does my algorithm do what I think/say it does? Thank you! 94.27.231.41 (talk) 09:40, 7 May 2009 (UTC)


 * You might be interested in Sequential analysis, in the real world when testing if say a batch of light bulbs is okay one can cut the tests needed quite a bit compared to setting the number of tests in advance. The same idea has started to be applied in medical testing so trials are terminated earlier if they are obviously good or bad. Dmcq (talk) 22:17, 7 May 2009 (UTC)

Non-discrete valuation ring
Hey, I'm trying to come up with a non-DVR. All examples I have found online and such are pretty complicated, perhaps above my knowledge at the time. But, I think I just came up with a very simple one and I'm just asking for some verification.

Take any field K and define the trivial valuation on it, so |0| = 0 and |x| = 1 for x neq 0. This is a nonarchimedean valuation I believe. And the set $$\{|x| : x \in K, x \neq 0\} = \{1\}$$ is not isomorphic to $$\mathbb{Z}$$ as groups, since it has only 1 element. So, it is a non-discrete valuation ring??? Thanks for any help. StatisticsMan (talk) 18:11, 6 May 2009 (UTC)


 * Hrm. This is probably cheating.  Fields are valuation rings in a technical sense, but they are even more discrete than DVRs.  Your value group is 0 ≤ Z, and typically you still call a valuation discrete if its value group is discrete, not just if it is Z (though Z is the only interesting case).
 * If you are comfortable with group rings, then they give very natural examples of valuation rings with given value group. JackSchmidt (talk) 18:19, 6 May 2009 (UTC)


 * That makes sense. The definition in my notes is not real clear so I am not sure if it is saying a subgroup of the group of integers or specifically the group of integers.  I will ask my professor, but it does make sense since the point is that the set is discrete.  I am mostly comfortable with group rings and have worked with them in at least two different classes.  I will think about that.  Thanks StatisticsMan (talk) 19:04, 6 May 2009 (UTC)


 * Hopefully I'll remember to write up the value group -> valuation ring example. It is fairly easy, and in several textbooks, but I remember thinking it was utterly amazing and unexpected when I first learned it.  It happens to not be on wikipedia as far as I can tell, but I'm pretty sure it is in Bourbaki (the ref for the article) and in Fuchs and Fuchs-Salce (where I learned it).  I think the case of G = Z × Z is even "easy" to describe (probably it is a subring of k(x,y)), but the valuation probably sounds pretty cooked up without the general framework. JackSchmidt (talk) 21:07, 6 May 2009 (UTC)


 * Thanks for the help. I talked to my professor and she said my example is a correct example of a non-DVR.  And, the reason is a field is a local ring since its unique maximal ideal is {0} but the definition of a DVR says the maximal ideal must be nonzero. StatisticsMan (talk) 17:27, 7 May 2009 (UTC)


 * Cool. Definitely a field is a valuation ring and not a DVR, and is *way* simpler than any non-discrete valuation ring (valuation ring with non-discrete value group).  Hopefully at some point I'll have both the inclination and the reliable source at hand to add the general example, but sounds like you read the problem correctly and don't need the other examples. JackSchmidt (talk) 17:48, 7 May 2009 (UTC)


 * Ah, so maybe this is my problem. I needed a valuation ring that is not a DVR and I assumed that made it a non-discrete valuation ring.  Are you saying these are not necessarily the same? StatisticsMan (talk) 20:18, 7 May 2009 (UTC)


 * Weirder than that. I am saying that "not-DVR" and "non-discrete valuation ring" are different, that is, "not" is not associative. "VR but Not (DVR)" is different than "(not-D) VR", but only in a silly technical sense.  Of course the language here is so finicky that I made sure to indicate more precisely what I meant: "VR but not-DVR" is different from "VR with non-discrete value group", though I think the enwiki article may even cloud that issue by claiming that the trivial group is not discrete.  I've got a little free time in the library and am going to clean up some things off my todo list. JackSchmidt (talk) 15:13, 8 May 2009 (UTC)

Group homomorphisms
I am trying to count the number of homomorphisms from one group to another ($$\mathbb{Q}_p^\times \to \mathbb{Z}_p$$) and I think I get it but I was just wondering if I could get a little help in general if I have a group that is a product of others and I am finding the homomorphisms from it to another group.

Example: Find the homomorphisms from $$\mathbb{Z} \times \mathbb{Z} / 5 \mathbb{Z} \to \mathbb{Z}$$.

Can we just look at the parts? As in, look at homomorphisms from $$\mathbb{Z} \to \mathbb{Z}$$ first? 0 must be mapped to 0 and then 1 could be mapped to any number I believe and since 1 generates the group, this fully determines the homomorphism. Then, look at homomorphisms $$\mathbb{Z} / 5 \mathbb{Z} \to \mathbb{Z}$$. Here, 0 must be mapped to 0 again. But, 1 can only be mapped to 0 since otherwise its image would be of order 5, which is impossible, right? Then, putting it together, we have that any homomorphism $$\mathbb{Z} \times \mathbb{Z} / 5 \mathbb{Z} \to \mathbb{Z}$$ will map (0, 0) to 0 and (1, 0) can be mapped to any integer and (0, 1) must be mapped to 0. And, these determine the homomorphism completely. Is this all correct?

Thanks StatisticsMan (talk) 19:00, 6 May 2009 (UTC)


 * Yes. I'll assume you are only talking about abelian groups (or modules, but not non-abelian groups).  Then the set of homomorphisms from A to B form an abelian group (or module) called Hom(A,B).  Hom has the nice property that Hom( A × C, B ) = Hom( A, B ) × Hom( C, B ).  The reason is kind of clear:  f((a,c)) = f((a,0) + (0,c)) = f((a,0)) + f((0,c)), so you only need to know f1:A→B:a→f((a,0)) and f2:C→B:c→f((0,c)), since f((a,c)) = f1(a) + f2(c).  Your calculations of Hom(Z,Z)=Z and Hom(Z/5Z,Z)=0 are correct, as well as your summary. JackSchmidt (talk) 21:04, 6 May 2009 (UTC)

Okay, thanks. StatisticsMan (talk) 21:57, 6 May 2009 (UTC)


 * I am back for more help :) I don't get this.  What I know is $$\mathbb{Q}_p^\times \cong \mathbb{Z} \times \mathbb{Z}_p \times \mathbb{Z} / (p - 1)\mathbb{Z}$$, (p > 2), as groups.  Again, I want homomorphisms from there to $$\mathbb{Z}_p$$.  If I consider elements as ordered triples, (a, b, c), then I know (0, 0, 1) must be mapped to 0 since it is an element of order p - 1 and $$\mathbb{Z}_p$$ has no elements of that order or order that divide that except for 0.  But, that is all I have.  What about the element (1, 0, 0) for example?  I can't think of any restrictions.  I know that the image of a homomorphism is a subgroup, but it seems to me that all multiples of any numbers should be a subgroup, as far as I can tell.  I found in a book that the only closed subgroups of $$\mathbb{Z}_p$$ are sets of the form $$p^n \mathbb{Z}_p$$.  And, with (0, 1, 0), I also have no idea.  It seems again that 1 could be mapped to anything.  Can I get a hint?  Thanks StatisticsMan (talk) 19:20, 7 May 2009 (UTC)


 * (1,0,0) can indeed be mapped to any element. In general, $${\rm Hom}(\mathbb Z,G)\simeq G$$ for any abelian group G, since $$\mathbb Z$$ is the free abelian group over one generator. On the other hand, the group of (not necessarily continuous) homomorphisms from $$\mathbb Z_p$$ to itself is going to be a horrible beast, I doubt that you can describe it in a reasonable way. (You can map (0,1,0) to anything, but that alone does not tell you much. $$\mathbb Z_p$$ is not cyclic, so there may be many different homomorphisms from $$\mathbb Z_p$$ mapping 1 to the same element.) Since you originally said that you are trying to count the number of homomorphisms: there is a trivial upper bound $$2^{2^{\aleph_0}}$$, and I bet that it is tight. — Emil J. 10:27, 8 May 2009 (UTC)


 * Yea, the problem says count, but I think it means to describe instead. Thanks for the information.  Hopefully my teacher is around today and I can ask her.  But, it is the last day of finals week so she might not be.  Thanks again. StatisticsMan (talk) 13:13, 8 May 2009 (UTC)


 * Let A be the additive group of the ring of p-adic integers. A is a large group (uncountable), but it has some amazing properties that make it behave more like a finitely generated group.  A is the inverse limit of Z/p^nZ, and Hom(A,Z/p^nZ) = Z/p^nZ, so Hom(A,A) is the inverse limit of Z/p^nZ = A (and indeed, as an endomorphism ring, it is the ring of p-adic integers). JackSchmidt (talk) 14:56, 8 May 2009 (UTC)


 * Wow, that's nice. So all homomorphisms from $$\mathbb Z_p$$ to itself are continuous after all. I got sidetracked by the fact that a "slight" change of the target group to $$\mathbb Q_p$$ results in the huge group $${\rm Hom}(\mathbb Z_p,\mathbb Q_p)\simeq(\mathbb Q_p)^{2^{\aleph_0}}$$ (that's Cartesian power). — Emil J. 12:20, 11 May 2009 (UTC)

for Buffon's needle does the length of the needle not matter
for Buffon's needle does the lenght of the needle not matter for the result, ie compared with the width of the strips or any other value (such as the width and length of the area it is being dropped over) ... the article doesn't mention any length in the introduction, but mentions a result of pi... but doesnt it depend on the length (for example an infinitely long needle will almost certainly cross boards) 79.122.21.123 (talk) 20:15, 6 May 2009 (UTC)
 * It looks like the beginning of the "solution" section requires that t≥l, where t is the width of the boards, and l is the length of the needle. So an infinitely long needle wouldn't be allowed, unless the boards were infinitely wide... but then the whole thing becomes meaningless. Staecker (talk) 20:33, 6 May 2009 (UTC)


 * Hmmm ? If the length of the needle l is less than the width of the strips/boards t then the probability that the needle will cross a line is $$\frac{2l}{\pi t}$$ - so it does depend on l (or, more precisely, on the ratio l / t). In the section Lazzarini's estimate it says "Lazzarini chose needles whose length was 5/6 of the width of the strips of wood. In this case, the probability that the needles will cross the lines is 5/3π". Gandalf61 (talk) 09:58, 7 May 2009 (UTC)