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October 5

Sine x tending towards x without L'Hôpital's rule.

Hi,

Was wondering if anybody had a method of proving that sine x tends towards x as x becomes smaller and smaller. Using this for a proof that the derivative of sine x is cosine x, therefore I cannot use L'Hopitals rule.

Thanks Jack — Preceding unsigned comment added by 2.27.38.121 (talk) 19:08, 5 October 2011 (UTC)[reply]

Rephrase as $\lim _{x\to 0}{\frac {\sin x}{x}}$ . Using the Taylor series for sin (I'm not sure if you're allowed to use that though), you get $\lim _{x\to 0}{\frac {1}{x}}\left({\frac {x^{1}}{1!}}-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+{\frac {x^{9}}{9!}}-\cdots \right)$ This gives ${\frac {1}{1!}}-{\frac {x^{2}}{3!}}+{\frac {x^{4}}{5!}}-{\frac {x^{6}}{7!}}+{\frac {x^{8}}{9!}}-\cdots$ . As x tends to 0, every term but the first tends to 0. 1/1! is 1. $\lim _{x\to 0}{\frac {\sin x}{x}}=1$ , so $\lim _{x\to 0}\sin x=x$ KyuubiSeal (talk) 20:16, 5 October 2011 (UTC)[reply]

There's a purely geometrical argument to show that

\lim _{x\to 0}{\frac {\sin x}{x}}=1\,.

You can find it here. — Fly by Night (talk) 21:28, 5 October 2011 (UTC)[reply]

Dang, we just learned that a few hours ago in class. Beaten to the punch. KyuubiSeal (talk) 00:33, 7 October 2011 (UTC)[reply]

Significance testing with subgroups

I'm looking to do some significance testing, but the way the problem is set up, I'm unsure as to the best way to proceed. I'm looking at the difference between two "treatments" (actually there's multiple, but I'm currently only looking at pairwise comparisons), where each treatment is applied across a (fixed) number of subgroups (subgroups can be matched across treatments). Within each of the subgroups there is a (potentially variable) number of replicates, each of which is considered to be equivalent/indistinguishable from any other replicate in the same subgroup (and there is no matching of replicates between subgroups, or between treatments). Each replicate in each subgroup has a readout of #successes/#trials, where the total number of trials is fixed (but different) within each subgroup. Some simplifying factors, if they help: the number of replicates for each subgroup is the same (it's also the same between treatments). Also, for the purposes of comparison, each success is considered equivalent. That is if treatment 1 resulted in 5/10 for group A and 4/6 for group B, that would be considered identical performance to a treatment where there's 6/10 for group A and 3/6 for group B.

I considered doing a straight sum of #successes across the groups, but as the replicates aren't matched across groups, I was unsure of how to do the confidence interval/variance estimations. What's the best way to do significance testing in this situation? If it comes with an estimate on how the error bars vary with respect to number of replicates (like the square-root-of-n rule for a typical normal distribution) so much the better. Thanks. -- 140.142.20.229 (talk) 19:34, 5 October 2011 (UTC)[reply]

See Bayesian_inference#Posterior distribution of the binomial parameter although this is not the orthodox frequentist approach. Bo Jacoby (talk) 23:31, 5 October 2011 (UTC).[reply]

Hmm ... for some reason I hadn't considered treating them as independent Bernoulli trials, though it makes sense in retrospect. So instead of having n replicates, each with up to t successes, each subgroup would have nt pass/fail replicates. As I have equal n for each subgroup, and each pass/fail trial is considered equivalent between subgroups, I can combine all of the pass/fail trials into

\sum _{i}\;nt_{i}

Bernoulli trials for each treatment, and then use either the Bayesian approach you pointed out, or perhaps some other binomial-based significance test (c.f. binomial proportion confidence interval). -- 140.142.20.229 (talk) 02:27, 6 October 2011 (UTC)[reply]

Quintic Equations Solutions

Is there a way to solve all quintic equations and above by algebraic means? — Preceding unsigned comment added by 196.201.51.21 (talk) 21:07, 5 October 2011 (UTC)[reply]

No, not generally. See algebraic equation. But there are ways to compute approximations to the roots. See root-finding method. Bo Jacoby (talk) 23:20, 5 October 2011 (UTC).[reply]

A young Ghanaian Researcher Known as Emmanuel Nii Afrah Sackey Has got the Solution for all Quintic equations and above. I met him in Accra-Ghana and his Email address is [redacted]. Anyone can contact him for any further issues about his discovery. I really mean what am writing because i have met him before. — Preceding unsigned comment added by 196.201.51.17 (talk) 21:39, 6 October 2011 (UTC)[reply]

You may know him, but we do not know you. Bo Jacoby (talk) 22:43, 6 October 2011 (UTC).[reply]

.. and publishing his email address here is not appropriate. I've removed it. AndrewWTaylor (talk) 08:47, 7 October 2011 (UTC)[reply]

... and if you believe that he has found a flaw in Niels Henrik Abel's 1824 proof that equations of degree 5 (and higher) are not always solvable using radicals, then perhaps he will win a Nobel prize? There are algebraic methods for many quintics, of course, but not all. Dbfirs 12:21, 7 October 2011 (UTC)[reply]

In that improbable scenario, it would be more likely to be a Fields Medal. AndrewWTaylor (talk) 13:00, 7 October 2011 (UTC)[reply]

You should say what is meant by "by algebraic means". If you mean "in terms of rational expressions with radicals", then the answer is no. If you allow some other kinds of expressions then the answer could be yes. The question is similar to (and in fact deeply related to) questions about constructing certain geometric figures using straight-edge and compass. The answer to "is it possible" depends strongly on which tools you're allowed to use. Staecker (talk) 12:58, 7 October 2011 (UTC)[reply]

To AndrewWTaylor: what i am saying is very true i even e-mailed him about what i posted on this site and he gave me his number if anyone wants to ask questions about it. He also said he is trying to publish it in a journal. so what do you say?

I say he's wrong if he thinks he has a solution to the quintic using only addition, subtraction, multiplication, division, and root extraction. I found an article about a person with that name; that article claimed a 'new' solution for cubic equations rather than quintics. These are solvable, unlike quintics, although the article also says that the formula avoids the use of complex numbers (which means it's incomplete or wrong). So on the whole I don't expect much.

CRGreathouse (t | c) 17:54, 10 October 2011 (UTC)[reply]