Cantor's theorem proves that there are different sized infinities. By analogy, are there different sized infinitesimals as well? such that d(dx) < dx? --Yanwen (talk) 01:10, 10 February 2009 (UTC)[reply]

Firstly, you should be aware that in conventional analysis, notations such as dx do not refer to infinitesimals; in fact infinitesimals are not involved in conventional real analysis at all. There are, however, various settings other than conventional real analysis in which infinitesimals (or things like them) are allowed. In all of these that I am aware of, such as the hyperreal numbers and surreal numbers, there are infinitely many different sizes of infinitesimal. By the way, if you do (like Leibniz) interpret dx as an infinitesimal, then it makes sense to say d(dx) < dx, since d(dx) is an infinitesimal increment in dx. Algebraist 01:19, 10 February 2009 (UTC)[reply]

This makes me think, you can work with derivatives and infinitesimals quite nicely in ${\displaystyle \mathbb {R} [\varepsilon ]/(\varepsilon ^{2})}$, for example, as the article on dual numbers explains, if we have ${\displaystyle p(x)=a_{0}+a_{1}x+\cdots +a_{n}x^{n}}$, then ${\displaystyle p(x+a\varepsilon )=p(x)+ap'(x)\varepsilon }$. This allows us to reverse the process and consider derivatives ${\displaystyle {\frac {dy}{dx}}}$ as a quotient of infinitesimals. But this then fails when trying to do second derivatives as we just get 0 out, and working in ${\displaystyle \mathbb {R} [\varepsilon ]/(\varepsilon ^{3})}$ loses many of the properties we wanted. Is there a nice way to get this working? --Xedi^Talk 02:04, 10 February 2009 (UTC)[reply]

Why do you just get zero out? The derivative of a polynomial is a polynomial, so the same definition works just fine. --Tango (talk) 11:35, 10 February 2009 (UTC)[reply]

Yeah sorry, I didn't really explain, the polynomial thing was just a motivational example, what I meant was that you use ε to define derivatives instead, so that ${\displaystyle f'(x)={\frac {f(x+\varepsilon )-f(x)}{\varepsilon }}}$. You then run into trouble if you try to do second derivatives. You have a valid point though, I guess you can consider ${\displaystyle f'(x)}$ as another function and repeat the process, but you can't write things like ${\displaystyle f''(x)={\frac {f(x+2\varepsilon )-2f(x+\varepsilon )+f(x)}{\varepsilon ^{2}}}}$ --Xedi^Talk 17:54, 10 February 2009 (UTC)[reply]

Is that how you usually define 2nd derivatives? I just define them as the derivative of a derivative... Using that definition, once you have a definition of a derivative you're set. --Tango (talk) 17:56, 10 February 2009 (UTC)[reply]

Of course, with the standard definition of a derivative, the two are equivalent - assuming you replace epsilon with h and some limits and those limits all behave themselves nicely. --Tango (talk) 17:58, 10 February 2009 (UTC) [reply]

If e squares to zero, it's what's called a nilpotent element and you can't divide by it in the ordinary way. The standard way to get around that is to rewrite all rules in terms of multiplication. In other words, ${\displaystyle f'(x)={\frac {f(x+\varepsilon )-f(x)}{\varepsilon }}}$ becomes ${\displaystyle \varepsilon f'(x)=f(x+\varepsilon )-f(x)}$, where f' might be any function that satisfies it, if any such function exists. Repeating this gives definitions for higher-order derivatives, but doesn't give a formula for them, because the next step would be ${\displaystyle \varepsilon ^{2}f''(x)=0=f(x+2\varepsilon )-2f(x+\varepsilon )+f(x)}$ Black Carrot (talk) 23:35, 10 February 2009 (UTC)[reply]

Although, come to think of it, you could use the Im() notation from complex numbers to take the infinitesimal part of this, making it f'(x) = Im(f(x+e)-f(x)) and f"(x) = Im(Im(f(x+2e)-f(x+e))-Im(f(x+e)-f(x))). Black Carrot (talk) 14:16, 12 February 2009 (UTC)[reply]

From [1], "There are, however, some rather counterintuitive properties of coin tossing. For example, it is twice as likely that the triple TTH (tails, tails, heads) will be encountered before THT than after it, and three times as likely that THH will precede HHT. Furthermore, it is six times as likely that HTT will be the first of HTT, TTH, and TTT to occur than either of the others (Honsberger 1979). "

and

"More amazingly still, spinning a penny instead of tossing it results in heads only about 30% of the time (Paulos 1995)."

A couple questions. First, WTF is going on? Second, is this (if it's even true) due more to a counterintuitive property of mathematics or a counterintuitive property of physics? Recury (talk) 20:33, 10 February 2009 (UTC)[reply]

The second, if true (I make no claim either way) would be due to physical asymmetries of the coin. I would believe it is not exactly 50% but 30% or thereabouts sounds dubious to me. Baccyak4H (Yak!) 20:41, 10 February 2009 (UTC)[reply]

The first set of claims can be checked by probability calculus (assuming 50%, ignoring the second claim for now). For example, consider TTH vs. THT. As we start the series, any Hs don't contribute to either pattern...yet. As soon as we get a T, then let's see what happens. If the next flip is another T (50%), then TTH will happen first...as soon as the first H is flipped. If it is an H, (50%), then a T following (50%, or 25% total) gives the second. If another H (50% or 25% total), then neither triple can now be made and the flips continue until another T appears, and the same calculus applies, except we have now conditioned on the event that we did not see either triple after the first T, which is an event with 25% probability. One can use either induction (2:1 ratio of probabilities after the first T, 2:1 ratio if not after the first but after the second, etc.) or geometric series (TTH = 50% + (25% × 50%) + (25% × 25% × 50%) + ...) to realize that TTH does appear first twice as often. Similar arguments could be used to check the other claims (which I have not done). Baccyak4H (Yak!) 20:54, 10 February 2009 (UTC)[reply]

I have verified the other claims: there is a 3/4 probability of THH being the first of (THH, HHT), and there is a 3/4 probability of HTT being the first of (HTT, TTH, TTT), with 1/8 probability for each of the other two. The reasoning I used is analogous to the above. Eric. 131.215.158.184 (talk) 21:22, 10 February 2009 (UTC)[reply]

The last one is especially easy to see. TTT and TTH can only occur first if they're the first three tosses. (Otherwise the first occurrence would follow an H or a T, but if it was an H then HTT already happened, and if it was a T then TTT already happened.) The chance of that is 1/8 each, which leaves 3/4 for HTT. -- BenRG (talk) 00:58, 11 February 2009 (UTC)[reply]

Thanks, I admit I don't totally have my head around all of that, but I can at least see how it would be possible. The 30% on the penny on the other hand... Recury (talk) 21:35, 10 February 2009 (UTC)[reply]

It is easy (and fun) to check for yourself what happens when you spin a coin on its edge. The ratio of heads to tails varies from coin to coin. It is seldom close to 50% and can be quite far from 50%. McKay (talk) 21:37, 10 February 2009 (UTC)[reply]

It's not easy - getting a coin to spin properly is hard, I just tried... I spun a UK 1p coin 5 times and got 4 heads (so 80%), but it took far more than 5 attempts! --Tango (talk) 21:41, 10 February 2009 (UTC)[reply]

If it were 30%, the odds of 4 or more heads would be a tiny bit over 3%, so that's pretty statistically significant evidence that 30% is not true for my coin. --Tango (talk) 22:04, 10 February 2009 (UTC)[reply]

There might also be a psychological reason for the coin-spinning example. People might tend to always hit the coin on either the face or tail to start it spinning, which may make it more likely to land a particular way. StuRat (talk) 23:00, 10 February 2009 (UTC)[reply]

There was something within the past three years or so in the Monthly about this. I'll see if I can find it.

This is NOT about physical asymmetries. Michael Hardy (talk) 21:17, 11 February 2009 (UTC)[reply]

This dude seems to think it is about mass distribution or the nature of the edge of the coin (for American pennies). [2] 68.144.30.20 (talk) 23:13, 11 February 2009 (UTC)[reply]

The claim that spinning a penny gives heads 30% of the time is about physical asymmetries, if it's true at all. Algebraist 23:43, 11 February 2009 (UTC)[reply]

Note the fact 1 has nothing to do with coins, it is just quite easy combinatorics. You can find all about statistics on the occurrences of a given substring within a random words, e.g. in Knuth's concrete mathematics. The second claim is most likely exaggerated, but in any case is clearly related with the physics of coin tossing. Coupling these two statements with no comment about their explanation is somehow misleading: a person with no scientific background may think that the first claim also refer to a mysterious behaviour of coins. After checking it experimentally, this person may think that the second fact also is true. So, either, yes, it is a kind of joke, or it is a case of quite bad scientific divulgation (maybe the source was a joke, and contained a further explanation, like it was in the style of great Martin Gardner; and then the explanation has been missed by the writer of the online article. Very bad!) pma (talk) 08:39, 12 February 2009 (UTC)[reply]

The asymetry of probabilities of heads or tails when spinning a coin is because the decision is made by the unavoidable imbalance at the very start of the spin. Thus if you always hold the coin a particular way around to start the spin, the average result will be biased away from the ideal 50/50. By how much is an empirical result and the alleged 30% heads is only anecdotal. Cuddlyable3 (talk) 15:25, 12 February 2009 (UTC)[reply]