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What are some degenerate cases of vector, multivariable, or just higher calculus theorems that can be applied to elementary (2-dimensional/on a plane) calculus and can solve problems easily that would be difficult to solve with elementary calculus alone? I know that many higher theorems are only generalizations of lower theorems so I would like to exclude these. Thank you for your responses. 72.128.95.0 (talk) 01:33, 18 March 2011 (UTC)[reply]

There is a nice application of Green's theorem to calculate area. Is this the sort of thing you have in mind? Sławomir Biały (talk) 02:40, 18 March 2011 (UTC)[reply]

Hi. Please see:

http://img145.imageshack.us/img145/5386/formula.gif

This is part of a formula generated by the WolframAlpha website. I don't understand these elements:

• The diagonal line of three dots at the end of line 1.
• What looks like a superscripted numeral 1 at the start of line 2.
• What look like fraction bars but with no numerators or denominators in lines 4 and 5.

Any ideas? 86.177.106.14 (talk) 03:57, 18 March 2011 (UTC)[reply]

I think they're all just bad formatting resulting from trying to split a long expression into several lines. Can you give us the query that generated this result? -- Meni Rosenfeld (talk) 07:24, 18 March 2011 (UTC)[reply]

See this. 86.183.3.118 (talk) 12:52, 18 March 2011 (UTC)[reply]

On the Wolfram Alpha webpage, if you hover the mouse to the bottom left corner of the box that contains all those formulas, a link to see the formulas in plain text will appear. Choose that link, get the formulas in plain text, and try to figure out the meaning from that. – b_jonas 10:21, 20 March 2011 (UTC)[reply]

And if you surround each of parentheses with curly braces, so that (....)^(1/3) renders as (....)^(1/3) instead of (....)⁽/3), and if you precede every ( with \left and every ) with \right, so that nested parens make a visible hierarchy, and if you finally put all the stuff inside wiki <math>....</math> tags, you'll get this:
${\displaystyle {\left(4F{\left(\sin ^{\left(-1\right)}{\left({\sqrt {\left({\left(-x{\left(t\right)}-1/4{\left(1+i{\sqrt {\left(3\right)}}\right)}{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}-{\left(1-i{\sqrt {\left(3\right)}}\right)}/{\left(4{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}\right)}-1/2\right)}/{\left(-1/4{\left(1+i{\sqrt {\left(3\right)}}\right)}{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}+1/4{\left(1-i{\sqrt {\left(3\right)}}\right)}{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}+{\left(1+i{\sqrt {\left(3\right)}}\right)}/{\left(4{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}\right)}-{\left(1-i{\sqrt {\left(3\right)}}\right)}/{\left(4{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}\right)}\right)}\right)}}\right)}|{\left(1/4{\left(1+i{\sqrt {\left(3\right)}}\right)}{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}-1/4{\left(1-i{\sqrt {\left(3\right)}}\right)}{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}-{\left(1+i{\sqrt {\left(3\right)}}\right)}/{\left(4{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}\right)}+{\left(1-i{\sqrt {\left(3\right)}}\right)}/{\left(4{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}\right)}\right)}/{\left(1/4{\left(1+i{\sqrt {\left(3\right)}}\right)}{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}+1/2{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}+{\left(1-i{\sqrt {\left(3\right)}}\right)}/{\left(4{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}\right)}+1/{\left(2{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}\right)}\right)}\right)}^{2}{\left(1/4{\left(1+i{\sqrt {\left(3\right)}}\right)}{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}-1/4{\left(1-i{\sqrt {\left(3\right)}}\right)}{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}-{\left(1+i{\sqrt {\left(3\right)}}\right)}/{\left(4{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}\right)}+{\left(1-i{\sqrt {\left(3\right)}}\right)}/{\left(4{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}\right)}\right)}{\left(x{\left(t\right)}-1/2{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}-1/{\left(2{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}\right)}+1/2\right)}{\left(x{\left(t\right)}+1/4{\left(1-i{\sqrt {\left(3\right)}}\right)}{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}+{\left(1+i{\sqrt {\left(3\right)}}\right)}/{\left(4{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}\right)}+1/2\right)}{\left(x{\left(t\right)}+1/4{\left(1+i{\sqrt {\left(3\right)}}\right)}{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}+{\left(1-i{\sqrt {\left(3\right)}}\right)}/{\left(4{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}\right)}+1/2\right)}\right)}/{\left({\left(-1/4{\left(1+i{\sqrt {\left(3\right)}}\right)}{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}-1/2{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}-{\left(1-i{\sqrt {\left(3\right)}}\right)}/{\left(4{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}\right)}-1/{\left(2{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}\right)}\right)}{\left(-1/4{\left(1+i{\sqrt {\left(3\right)}}\right)}{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}+1/4{\left(1-i{\sqrt {\left(3\right)}}\right)}{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}+{\left(1+i{\sqrt {\left(3\right)}}\right)}/{\left(4{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}\right)}-{\left(1-i{\sqrt {\left(3\right)}}\right)}/{\left(4{\left(-6c_{1}+2{\sqrt {\left(3\right)}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}\right)}\right)}{\left({\left(2x{\left(t\right)}^{3}\right)}/3+x{\left(t\right)}^{2}+c_{1}\right)}\right)}={\left(c_{2}+t\right)}^{2}}$
I have no idea, however, what the vertical bar at the end of arcsin means... --CiaPan (talk) 17:40, 24 March 2011 (UTC)[reply]
PS. Of course you would have also to precede every 'sqrt' with a backslash, and same with 'sin'. CiaPan (talk)

That can be further simplified by removing some parentheses and replacing some subexpresson with a single symbol:
${\displaystyle K:={\left(-6c_{1}+2{\sqrt {3}}{\sqrt {\left(3c_{1}^{2}+c_{1}\right)}}-1\right)}^{\left(1/3\right)}}$
${\displaystyle {\left(4F{\left(\sin ^{\left(-1\right)}{\left({\sqrt {\left({\left(-x{\left(t\right)}-1/4{\left(1+i{\sqrt {3}}\right)}K-{\left(1-i{\sqrt {3}}\right)}/{\left(4K\right)}-1/2\right)}/{\left(-1/4{\left(1+i{\sqrt {3}}\right)}K+1/4{\left(1-i{\sqrt {3}}\right)}K+{\left(1+i{\sqrt {3}}\right)}/{\left(4K\right)}-{\left(1-i{\sqrt {3}}\right)}/{\left(4K\right)}\right)}\right)}}\right)}|{\left(1/4{\left(1+i{\sqrt {3}}\right)}K-1/4{\left(1-i{\sqrt {3}}\right)}K-{\left(1+i{\sqrt {3}}\right)}/{\left(4K\right)}+{\left(1-i{\sqrt {3}}\right)}/{\left(4K\right)}\right)}/{\left(1/4{\left(1+i{\sqrt {3}}\right)}K+1/2K+{\left(1-i{\sqrt {3}}\right)}/{\left(4K\right)}+1/{\left(2K\right)}\right)}\right)}^{2}{\left(1/4{\left(1+i{\sqrt {3}}\right)}K-1/4{\left(1-i{\sqrt {3}}\right)}K-{\left(1+i{\sqrt {3}}\right)}/{\left(4K\right)}+{\left(1-i{\sqrt {3}}\right)}/{\left(4K\right)}\right)}{\left(x{\left(t\right)}-1/2K-1/{\left(2K\right)}+1/2\right)}{\left(x{\left(t\right)}+1/4{\left(1-i{\sqrt {3}}\right)}K+{\left(1+i{\sqrt {3}}\right)}/{\left(4K\right)}+1/2\right)}{\left(x{\left(t\right)}+1/4{\left(1+i{\sqrt {3}}\right)}K+{\left(1-i{\sqrt {3}}\right)}/{\left(4K\right)}+1/2\right)}\right)}/{\left({\left(-1/4{\left(1+i{\sqrt {3}}\right)}K-1/2K-{\left(1-i{\sqrt {3}}\right)}/{\left(4K\right)}-1/{\left(2K\right)}\right)}{\left(-1/4{\left(1+i{\sqrt {3}}\right)}K+1/4{\left(1-i{\sqrt {3}}\right)}K+{\left(1+i{\sqrt {3}}\right)}/{\left(4K\right)}-{\left(1-i{\sqrt {3}}\right)}/{\left(4K\right)}\right)}{\left({\left(2x{\left(t\right)}^{3}\right)}/3+x{\left(t\right)}^{2}+c_{1}\right)}\right)}={\left(c_{2}+t\right)}^{2}}$

This is a general problem: How does one go about investigating the effects of rotating coordinate axes regarding tensors? For example, what does a 2-fold symmetry about all 3 orthogonal axes translate to in tensors and suffix notation? Thanks —Preceding unsigned comment added by 131.111.222.12 (talk) 10:47, 18 March 2011 (UTC)[reply]

The whole point of tensors is to abstract away from a particular coordinate system. So to represent things like that you have to use the representation of a tensor in a particular coordinate system in which the symmetry you are interested in holds. Dmcq (talk) 11:54, 18 March 2011 (UTC)[reply]

Thanks Dmcq, but how does one represent a tensor in a particular coordinate system? (sorry, I am new to this...) —Preceding unsigned comment added by 131.111.222.12 (talk) 15:57, 18 March 2011 (UTC)[reply]

An order-k tensor in a space with n dimensions can be represented as an array of n^k numbers. For example, a 2nd-order tensor in a 3-dimensional space is represented as a 3x3 matrix of 9 numbers. Gandalf61 (talk) 16:26, 18 March 2011 (UTC)[reply]

Take a look at the Covariance and contravariance of vectors article, and in particular the Use in tensor analysis subsection. — Fly by Night (talk) 19:01, 18 March 2011 (UTC)[reply]

Thanks, Gandalf and Fly by Night. —Preceding unsigned comment added by 131.111.222.12 (talk) 15:59, 19 March 2011 (UTC)[reply]

If you have two GPS co-ordinates, x & y is there an easy way to convert (x - y) into a distance in kilometers? Or does the curvature of the Earth make this a difficult problem? --CGPGrey (talk) 20:18, 18 March 2011 (UTC)[reply]

See Great-circle_distance. I don't exactly think this qualifies as "easy", but at least it explains how to compute the distance. Gscshoyru (talk) 20:23, 18 March 2011 (UTC)[reply]

Thank you. That is perfect. --CGPGrey (talk) 12:09, 20 March 2011 (UTC)[reply]

It depends a lot on how far apart they are. If you're talking about locations within, I don't know, a few hundred miles of each other, and you don't need extreme accuracy, you can probably do it in your head. Just remember that one nautical mile is one minute of arc. So to get the north-south distance, look at the difference in latitudes in degrees, multiply by 60, and that's the answer in nautical miles. Add another ten percent to get regular miles, or maybe 70 percent to get kilometers.

For east-west distance it's a little more complicated, but only a little — you have to multiply by the cosine of your latitude.

Now you have two edges of a right triangle, and you just use the Pythagorean theorem. --Trovatore (talk) 20:29, 18 March 2011 (UTC)[reply]

This is spherical geometry. The Pythagorean theorem is valid in plane geometry. Dolphin (t) 07:58, 20 March 2011 (UTC)[reply]

Yes, I understand that. I specifically said, if the two points are close, and you don't need extreme accuracy. In that case the Pythagorean theorem works just fine. (Well, at reasonable latitudes anyway.) --Trovatore (talk) 08:09, 20 March 2011 (UTC)[reply]

Ah, I was a little low on the conversion factor. Add fifteen percent to nautical miles to get miles, or 85 percent to get kilometers. --Trovatore (talk) 20:31, 18 March 2011 (UTC)[reply]

I think I'm using the right terms. What I want is the permutation group on N elements that has the highest minimum k where N^k = I. For example, if N=52, I'm looking for the defined shuffle that if that shuffle were to be applied repeatedly would take the longest to get back to the deck that we started with. For example, if N=5, the group would be (1 2 3)(4 5) since that would take 6 iterations/shuffles to get back to start. For N=6, it is (1 2 3) (4 5) (6). For N=8, it is (1 2 3 4 5)(6 7 8) since that would take 15 iterations/shuffles. Any idea where I can find a list, or any software that would help calculate it?Naraht (talk) 20:49, 18 March 2011 (UTC)[reply]

I think what you're looking for is an element of maximal order in the permutation group ${\displaystyle S_{N}}$. By considering the cycle decomposition, this is equivalent to seeking a partition of N that solves a certain maximization problem. This seems like a hard problem in general. Sławomir Biały (talk) 21:48, 18 March 2011 (UTC)[reply]

Well, basically you want a partition of 52 such that the least common multiple of the pieces is as large as possible. I think that's right. I don't know whether this really gives the largest values, but a good strategy for getting a large one seems to me to just see how many different primes you can add before you get there. 2+3+5+7+11+13 adds up to 41; I think that's the best you can do. So take a permutation that has disjoint cycles of all those lengths. What happens on the remaining cards, I think, doesn't matter, and the order of the permutation is 2*3*5*7*11*13=30030. Is this the largest? At a brief glance, I think so. --Trovatore (talk) 21:55, 18 March 2011 (UTC)[reply]

Whoops, no, you can do a little better than that. Take eight of the remaining eleven cards and put them in a cycle. That bumps up the length by a factor of four, to 120120. Don't know if this is the best yet. --Trovatore (talk) 22:01, 18 March 2011 (UTC)[reply]

Ah, but then you don't need the 2-cycle anymore, so there are still five cards left. Splice four of them into the 13-cycle to get 17 instead, and we have 1*8*3*5*7*11*17=157080. –Henning Makholm (talk) 00:21, 19 March 2011 (UTC)[reply]

OEIS (see below) says that the best possible for 52 elements is 180180, which factors as 4*5*7*9*11*13. –Henning Makholm (talk) 00:35, 19 March 2011 (UTC)[reply]

My understanding is it's supposed to work for every N, but this is one possible strategy. Another strategy is to try to maximize the product of mutually coprime integers that sum to N. Sławomir Biały (talk) 22:01, 18 March 2011 (UTC)[reply]

OK, we can work it out ourselves for 52, but it is trial and error. Is there actually any mechanical method and is actually solving this for a specific N an NP problem?Naraht (talk) 00:29, 19 March 2011 (UTC)[reply]

Compute the first half-dozen entries by hand and stick the list into OEIS. Out comes OEIS: A000793, with references, links, and a name: Landau's function. –Henning Makholm (talk) 00:35, 19 March 2011 (UTC)[reply]

Nice find! --Trovatore (talk) 00:41, 19 March 2011 (UTC)[reply]

Agreed nice find. I should have tried OEIS in the first place. I know there is an infobox for integer sequences, but Landau's function doesn't have it. (not sure if it should or not.Naraht (talk) 01:14, 19 March 2011 (UTC)[reply]

A terminological aside: If the problem is (in) "NP", it means that it is easier (in a certain technical sense) than all problems that are not NP. You're looking for NP-complete, which means that it is NP but among the hardest (i.a c.t.s) NP problems. But it is not obvious to me that this problem is even in NP. –Henning Makholm (talk) 00:53, 19 March 2011 (UTC)[reply]

Sorry screwed up the concept. I looked at one of the links from oeis, one of the papers http://arxiv.org/abs/0803.2160 is on calculating them up to one million billions. It indicates that the running time for calculating g(n) is $\co(N^{3/2}/\sqrt{\log N}). Not sure what the $co function corresponds to though.Naraht (talk) 01:14, 19 March 2011 (UTC)[reply]

The \co seems to be "curly O" or some such. So this is just big O notation. You can download the full paper from your link. --Trovatore (talk) 01:22, 19 March 2011 (UTC)[reply]

Thought that I'd have to pay for the full paper, but I don't. And definitely not exponential time.Naraht (talk) 01:32, 19 March 2011 (UTC)[reply]

Well, it is at least pseudo-polynomial then. Without the pseudo it's less clear, but probably not even NP. –Henning Makholm (talk) 12:55, 19 March 2011 (UTC)[reply]

Ah, thanks, now a piece falls into place. I hadn't realized that N was the actual number, rather than the number of digits. That explains the bit in the abstract about there being an oh-of-whatever algorithm, that works up to 10^6. I was really scratching my head about that one, wondering if maybe this was one of those arXiv papers that was some distance from being ready for journal submission. --Trovatore (talk) 08:21, 20 March 2011 (UTC)[reply]

They write "... a basic algorithm to compute g(n) for 1≤n≤N; its running time is O(..N..)", that is, compute all of the N initial values of the sequence in such-and-such time. Sounds like some variation on bottom-up dynamic programming. –Henning Makholm (talk) 12:42, 20 March 2011 (UTC)[reply]

Consider the function ${\displaystyle 3x^{\frac {1}{3}}(e^{x}-1)}$. Is this function differentiable at zero? Drawing a graph on a computer seems to show that the derivative is zero at zero but the derivative ${\displaystyle x^{\frac {-2}{3}}(e^{x}-1)+3x^{\frac {1}{3}}e^{x}}$ is not defined at zero since ${\displaystyle x^{\frac {-2}{3}}}$ is not defined at zero.

Also consider a function like ${\displaystyle |x|(e^{x}-1)}$. Again, |x| is not differentiable at zero so this function can't be either, yet it would appear that the slope at zero is zero for that function too. Widener (talk) 21:19, 18 March 2011 (UTC)[reply]

Just because you take a derivative symbolically and it's not defined at zero, it doesn't follow that the function itself has no derivative at zero. You need to go back to the epsilon-delta definition and try to figure out whether there is a value that satisfies that definition, or not. --Trovatore (talk) 21:26, 18 March 2011 (UTC)[reply]

Maybe you're exaggerating. How about just going back to the definition of the derivative as the limit of a difference quotient? Michael Hardy (talk) 21:46, 18 March 2011 (UTC)[reply]

If you get 0 in the denominator but some other number in the numerator, then you've got something that blows up. But here you get 0 in both the numerator and the denominator, and that means you need to be more careful. Michael Hardy (talk) 21:37, 18 March 2011 (UTC)[reply]

.....The methods taught in calculus generally say that if some formula holds, then a function is differentiable and the derivative is thus-and-so. For example, in the product rule, one says that if ƒ and g are differentiable then so is ƒg, etc. The conclusions of derivative-computing methods like those don't say that under specified circumstances a function is not differentiable. For that you sometimes need to go back to this:

${\displaystyle f'(x)=\lim _{w\to x}{\frac {f(w)-f(x)}{w-x}}.}$

If the limit doesn't exist, then the function is not differentiable. If the limit does exist then the function is differentiable.

Sometimes you can conclude non-differentiability by observing that the derivative has a vertical asymptote. I don't think that happens in this case. So apply the limit above with x = 0. Michael Hardy (talk) 21:45, 18 March 2011 (UTC)[reply]

....also, the product rule, and the like, doesn't say that if ƒ or g is not differentiable, then some conclusion holds.

Anyway, the bottom line in this case is that the derivative does exist at 0. You didn't correctly apply the product rule at the point x = 0. The rule says if ƒ and g are differentiable then a conclusion holds. But in this case, one of the two, ƒ or g, is not differentiable at x = 0, and therefore the product rule is simply not applicable at x = 0. So go back to the definition:

${\displaystyle f'(0)=\lim _{w\to 0}{\frac {f(w)-f(x)}{w-x}}=\lim _{w\to 0}{\frac {3w^{1/3}(e^{w}-1)}{w}}.}$

By various methods, among them L'Hopital's rule, you can sho that that limit exists and is 0. Michael Hardy (talk) 21:51, 18 March 2011 (UTC)[reply]

OK, to be more specific:

${\displaystyle \lim _{w\to 0}{\frac {3w^{1/3}(e^{w}-1)}{w}}=\left(\lim _{w\to 0}3w^{1/3}\right)\left(\lim _{w\to 0}{\frac {e^{w}-1}{w}}\right).}$

L'Hopital's rule gives you the second limit, and the first is easy. Michael Hardy (talk) 21:53, 18 March 2011 (UTC)[reply]

Easier: ${\displaystyle e^{x}-1}$ is just x plus terms of smaller order. So the function in question might as well be ${\displaystyle 3x^{\frac {4}{3}}}$. Just differentiate that. This sort of approximation is usually easier to work with than l'Hôpital in all but very simple cases. --Trovatore (talk) 22:51, 18 March 2011 (UTC)[reply]

Your "easier" method is pretty much the same thing as L'Hopital's rule, except that many students in a first-semester calculus course know L'Hopital's rule but not power series. Michael Hardy (talk) 23:06, 18 March 2011 (UTC)[reply]

The difference is that you can manipulate the power series of the component pieces, and you never have to take derivatives of the numerator or denominator as a whole. --Trovatore (talk) 23:56, 18 March 2011 (UTC)[reply]

c'mon folks. It's just the definition of the derivative of the exponential function at the origin. 23:21, 18 March 2011 (UTC)

True. If I told that to a class of students sitting in front of me, it would confuse them. Michael Hardy (talk) 23:29, 18 March 2011 (UTC)[reply]

Two points start at a point on a circle of circumference 1 and travel in the same direction with different speeds, x and y (distance covered in 1 second). How to find out when will they meet again at the starting point? 93.136.161.26 (talk) 22:10, 18 March 2011 (UTC)[reply]

Suppose the one travelling at speed x winds around the circle n times while the one at speed y winds around m times, and then they're both at the starting point again for the first time. The distance covered in a fixed time is proportional to the speed, so the ratio of speeds, x/y, must equal the ratio of distances, n/m. This means that the ratio of speeds must be a rational number, i.e. a ratio of two integers (n and m). So they will never meet exactly at the starting point if the ratio of their speeds is irrational, although there will be future times when one of them passes the starting point and the other is very close. This is true no matter how small you want "very close" to be, although if you make it very small, you may have to wait a long time before it happens. See Diophantine approximation. The fraction n/m must be in lowest terms, since if n = ka and m = kb, where k,a, and b are integers, then they would have met earlier when the first particle had gone a times around the circle and the second b times. Michael Hardy (talk) 23:01, 18 March 2011 (UTC)[reply]

Ok so assuming x and y are both rational numbers, what would the formula be? 212.15.182.119 (talk) 23:41, 18 March 2011 (UTC)[reply]

${\displaystyle n/x}$, where ${\displaystyle n}$ is as described by Michael Hardy. –Henning Makholm (talk) 00:12, 19 March 2011 (UTC)[reply]

There's no need for x and y to be rational; rather it is the ratio x/y that needs to be rational. Michael Hardy (talk) 03:51, 19 March 2011 (UTC)[reply]

Question moved from "Finance Question about Bond Prices and Compounded Interest Rates" on the Humanities desk. -- OP, 68.75.28.230 (talk) 22:10, 18 March 2011 (UTC) [reply]

When calculating the present value of a bond that pays semi-annual interest, why do we use exactly half the annual market interest rate to calculate? Since the present value of bonds is based on compounded interest, I cannot understand the mathematical justification for this method of calculating present value.

As a simple example, suppose that we have a one-year bond that pays 10% annual interest in two semi-annual payments (i.e., each is 5% of the principal). Suppose that the market interest rate is 12% annually. The correct way to find the present value of this bond is as follows:

${\displaystyle 1000\times (1+0.06)^{-2}+1000\times 0.05\times {\dfrac {1-(1+0.06)^{-2}}{0.06}}=}$

${\displaystyle \$981.67\,}$

See Investopedia article for explanation. Notice that we’ve simply taken the annual market interest rate and divided it by two to get the semi-annual market interest rate. I don’t understand why, because a 6% compounded interest rate over two periods gives very different results than a 12% compounded interest rate over one period.

${\displaystyle 1.06^{2}=1.1236\,}$

${\displaystyle 1.12^{1}=1.12\,}$

To me it would make more sense if the above bond was calculated as follows:

${\displaystyle 1000\times (1+0.12)^{-1}+1000\times 0.05\times (1+0.12)^{-0.5}+1000\times 0.05\times (1+0.12)^{-1}=}$

${\displaystyle 892.86+47.25+44.64=\,}$

${\displaystyle \$984.75\,}$

I’d understand that because our interest actually works out to 12% compounded annually. However, this is NOT how bond present values are calculated, according to every source that I can find, in print or on the Web. (Check out Investopedia’s Bond Price Calculator for an example which gives the same result as the first calculation).

What am I missing here? Thanks!

P.S.: I struggled over whether to post this to math or humanities, but since the humanities description specifically says “finance,” I came here. Not anymore! 68.75.28.230 (talk) 21:19, 18 March 2011 (UTC)[reply]

Probably this desk is for the pecuniary or monetary or legal/political/social/cultural aspects of finance. The mathematical aspects are still best discussed on the Mathematics desk. -- Jack of Oz [your turn] 21:23, 18 March 2011 (UTC)[reply]

OP here. Thanks for the clarification. Should I just copy and paste this over to math desk, then? 68.75.28.230 (talk) 21:51, 18 March 2011 (UTC)[reply]

Yep. Best to also strike it out here so that readers know not to answer it here. -- Jack of Oz [your turn] 21:55, 18 March 2011 (UTC)[reply]

Presumably in this formula the "market interest rate" is expected to be quoted on the same basis as the bond interest rate. So, a "market interest rate" of 12% would actually mean two 6% payments per year, just as a 12% semi-annual bond would pay. 86.181.169.37 (talk) 23:48, 18 March 2011 (UTC)[reply]