Mathematics desk
< July 8	<< Jun \| July \| Aug >>	July 10 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

July 9

Differentiability

Let us consider the function $f(x)=x^{2}\sin \left({\frac {1}{x^{2}}}\right)$ where x is not zero and let f(0)=0. Now this function is continuous at x=0 and the problem is to "explore" the differentiability of this function. For x nonzero, we can use standard techniques and obtain that, $f'(x)=2x\sin \left({\frac {1}{x^{2}}}\right)-{\frac {2}{x}}\cos \left({\frac {1}{x^{2}}}\right)$ . Now when take $\lim _{x\rightarrow 0}f'(x)$ , this limit does not exist because the function oscillates wildly between positive and negative infinity. But if we use the epsilon-definition, we can show that the derivative does indeed exist at x=0 and f'(0)=0 by simply letting epsilon=delta. I thought that whenever you could not simply take the derivative and evaluate, you could take the derivative and then take the limit and that would be the value of the derivative at that point (if it existed). But here, if I take the derivative and then take the limit of the derivative, I do NOT get the value of the derivative. The function itself is continuous and I know that this does not automatically imply that the derivative must be continuous as well. But how can this continuous function have a discontinuous derivative which is not equal to its rate of change at x=0? Any insight?--76.79.202.34 (talk) 00:04, 9 July 2008 (UTC)[reply]

Your observations are essentially all correct and indeed this is a function which has a derivative at every point and yet the derivative is not continuous. If you find this surprising, you are not alone. Perhaps if you plot the function (if you have not done so yet), this may help you understand this phenomenon better. There is just one point in your comment which requires some clarification. When you say "rate of change" I interpret this to be "slang" for "derivative". If this is synonimous with derivative, then the rate of change is equal to the derivative. Oded (talk) 00:37, 9 July 2008 (UTC)[reply]

So basically, we already know a theorem, that for a continuous function f(x), if $\lim _{x\rightarrow a}f'(x)$ exists then f'(a) must also exist and $\lim _{x\rightarrow a}f'(x)=f'(a)$ . In my above example, $\lim _{x\rightarrow 0}f'(x)$ does not exist so obviously, it does not equal f'(0) which is zero. I understand that now. Here is the next question. In this theorem I just stated, when we say that the limit "exists", are we including the case when $\lim _{x\rightarrow a}f'(x)=\pm \infty$ in which case we can deduce that $\lim _{x\rightarrow a}f'(x)=f'(a)=\pm \infty$ . I mean $f(x)=x^{1/3}$ is such an example at x=0. Is this always the case?--76.79.202.34 (talk) 00:57, 9 July 2008 (UTC)[reply]

Yes, that's correct. If

\lim _{x\to 0}f'(x)=\infty

, then

f'\,(0)=\infty

, and similarly with

-\infty

. Oded (talk) 01:02, 9 July 2008 (UTC)[reply]

Makes much more sense now. Thanks!--A Real Kaiser (talk) 04:18, 9 July 2008 (UTC)[reply]

This and many similar functions are more readily understood by considering their behaviour on the complex plane. Your function has an essential singularity at the origin, where it oscillates very violently; but on the real axis, all the ghastliness cancels out and one is left (by chance, so to speak) with a nicely behaving function that just happens to be continuous and differentiable at the origin. HTH, Robinh (talk) 07:45, 9 July 2008 (UTC)[reply]

Now here are two follow up questions. The first question is, if I am given an arbitrary function f(x) and I want to find f'(a) at some point a and $\lim _{x\rightarrow a}f'(x)=DNE$ , then how can I tell between the cases when f is simply not differentiable at a (like f(x)=abs(x) at x=0) or f is differentiable at a but the derivative cannot be found using ordinary methods (like $f(x)=x^{2}\sin \left({\frac {1}{x^{2}}}\right)$ at x=0? Is there any way to tell that the function is just not differentiable at a or that it is differentiable but you cannot just take the limit of the derivative? The second question is, assuming that I CAN tell that a function is differentiable at x=a somehow, how to deduce/calculate its value? For example, in my original question up above, I was already told that f'(0)=0 and then one can prove it is so using the definition of the derivative. But what if I don't know a value to conjecture with? How to come up with a number? Is it just an educated guess? What if up above $f\,'(0)=\pi /167$ or some other number?--69.224.229.251 (talk) 23:20, 9 July 2008 (UTC)[reply]

You weren't given the value of f'(0), you were given the value of f(0). This is part of the definition of the function. You then had to prove that f was differentiable at 0 with derivative 0. The usual way to do this for nasty functions (when you can't use the chain rule and such) is to go right to the definition of differentiability. This works here. Taking the limit is a valid alternate method as long as the limit exists and you know already that the function is differentiable, but I can't think of an example for which it's not easier just to use the definition. Algebraist 14:37, 10 July 2008 (UTC)[reply]

Okay, you are right (I used equivocal language and didn't make myself clear). What I meant to say that, the author said to prove that the derivative of f at x=0 is zero. My question was, in this case, the author told us which value to prove it for, but what if I was given another function and I had no idea what the derivative at a point, how would I know? It seems to me that the prescience of the derivative at zero was required and then we proved that the derivative is indeed zero. If the book had said here is f(x) and find f'(0) then how would I have done it? This is the same problem as trying to prove by induction that $\sum _{k=1}^{n}k={\frac {n(n+1)}{2}}$ . The proof requires this knowledge in advance.--69.106.206.37 (talk) 20:53, 10 July 2008 (UTC)[reply]

By definition, the derivative of f at x is

\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}

, if the limit exists. So for the above function, the derivative at zero is the limit as x tends to zero of

x\sin \left({\frac {1}{x^{2}}}\right)

, which is easy to calculate. Algebraist 21:02, 10 July 2008 (UTC)[reply]

Wow, Algebraist, that was awesome. I can't believe I didn't see that. It makes perfect sense now. Thanks--76.79.202.34 (talk) 21:45, 10 July 2008 (UTC)[reply]

iNFINITE dISTANCE

i just can't get my mind around this...can anybody help?... assuming space is infinite, is an infinite distance possible? i.e. if you start at point A and I start at point B, and we move in a straight line towards each other, we NEVER meet....if infinite distance is possible, is it possible to have an infite space WITHOUT it?....thanx....---stolf —Preceding unsigned comment added by 64.19.74.59 (talk) 14:05, 9 July 2008 (UTC)[reply]

Yes if you have infinite space then infinite distance is possible.

Note if you never meet the other person whilst travelling in the same direction then you are travelling in parallel.. (see Parallel (geometry). It's worth noting that parallel lines never meet even at infinity, though some people believe they do.87.102.86.73 (talk) 14:21, 9 July 2008 (UTC)[reply]

I disagree with both points. Infinite space means your co-ordinates are unbounded, but any given point will always have finite co-ordinates, so the distance between two points is always finite. That's for standard Euclidean space and anything similar to it, at least. In projective space, things get a little more interesting. You have "points at infinity", which, in a rather imprecise sense, have infinite co-ordinates. In that sense, parallel lines *do* meet at infinity. I'm not sure if it's meaningful to talk about infinite distance in a projective space, though, since I'm not sure there is a sense of distance at all, I've certainly never worked with one - not globally, anyway. There may well be other kinds of space in which you can have infinite distance, but I don't know of any. --Tango (talk) 15:36, 9 July 2008 (UTC)[reply]

Mmmh lines y=1, y=0 at infinty (∞,1) and (∞,0) ? I mis-read that didn't I.87.102.86.73 (talk) 17:05, 9 July 2008 (UTC)[reply]

Like I said, thinking of points at infinity as having infinite co-ordinates is very imprecise. You need to actually use projective co-ordinates (or Homogeneous coordinates). The projective equivalent of the affine y=1 and y=0 are Y=Z and Y=0, which are both satisfied by the point [1:0:0], which is the "point at infinity" where they meet (Z=0 corresponds to infinity, roughly speaking). --Tango (talk) 18:16, 9 July 2008 (UTC)[reply]

According to Metric (mathematics)#Extending the range, you can considered extended metrics which allow infinite distance, but they are always equivalent topologically to a standard metric. --Tango (talk) 15:39, 9 July 2008 (UTC)[reply]

The usual metric given to projective spaces is the Fubini-Study metric. There is a good explanation of its natural geometric meaning, only hinted at on that article's talk page, in one of the appendixes to V. I. Arnold's book on classical mechanics.John Z (talk) 17:52, 9 July 2008 (UTC)[reply]

Clarify: infite space/distance is possible, but to actually be at infinty is not so possible..87.102.86.73 (talk) 17:05, 9 July 2008 (UTC)[reply]

You can have infinite distance in a finite space, it is worth noting, just not in straight lines. -mattbuck (Talk) 11:39, 10 July 2008 (UTC)[reply]

That's more infinite length than infinite distance - the OP is clearly talking about the distance between two points, which will always be measured in a straight line (or at least, a geodesic). --Tango (talk) 14:11, 10 July 2008 (UTC)[reply]

i'm still not getting it...i can see that in A's frame of ref, B (at an infinite distance from A) couldn't exist, and vice versa...but couldn't both A and B exist in a third frame of ref?...they would split space into 2 mutually exclusive sections: anyplace you could get to from A, you could not get to from B, and vice versa....but why couldn't both these spaces exist, "meeting" at infinity, which is to say, the infinite space extending from A is the same infinite space extending from B...make any sense?....---stolf —Preceding unsigned comment added by 64.19.74.59 (talk) 15:48, 10 July 2008 (UTC)[reply]

You probably could construct something like that, but it wouldn't have very nice global behaviour. For example, you could take two copies of the complex projective line (which is topologically a 2-sphere) and identify ("stick together") their points at infinity. There wouldn't be one coordinate system that describes the whole space, and the shared point at infinity would be a singularity, so the space wouldn't be a manifold, but it would just about satisfy your description. --Tango (talk) 16:06, 10 July 2008 (UTC)[reply]

It strikes me that you might be able to make some fairly natural definition of distances on the long line (or at least the "open long ray" in the terminology of that article) that could take infinite values (in fact, values from the long line itself). I haven't really tried to work this through. --Trovatore (talk) 16:18, 10 July 2008 (UTC)[reply]

You might be on to something there. The natural distance between

(\omega _{0},0)

and

(2\omega _{0},0)

is presumably infinite. You have (countably) infinitely many unit intervals between the points, so the length ought to be infinity times one, which is infinity. I'm not sure how rigorously you can define that definition of distance, but intuitively it seems to work. --Tango (talk) 16:31, 10 July 2008 (UTC)[reply]

2\omega _{0}=\omega _{0}

. You want

\omega _{0}2

. Algebraist 20:57, 10 July 2008 (UTC)[reply]

True, thank you! --Tango (talk) 00:05, 11 July 2008 (UTC)[reply]

Better to typeset it as

\omega \cdot 2

; much easier to read that way. Interesting history about this notation: When Cantor first published on ordinal arithmetic, he indeed notated this ordinal as

2\omega \,

, which to me at least seems more intuitive for one copy of

\omega \,

followed by another than for

\omega \,

copies of 2 (the latter being what it modernly means). Apparently the reason he reversed it, in his second major publication, is that with the modern notation you get

\alpha ^{\beta \cdot \gamma }={\left(\alpha ^{\beta }\right)}^{\gamma }

whereas with the older notation it would have to be written as

\alpha ^{\beta \cdot \gamma }={\left(\alpha ^{\gamma }\right)}^{\beta }

--Trovatore (talk) 02:46, 11 July 2008 (UTC)[reply]

Yes, there are spaces where two points can be at infinite distance from each other, and that does tend to split the space into equivalence classes. You may be interested in nonstandard analysis, where exactly that idea is used to define the halo and galaxy of a nonstandard real number. Start with the real numbers. Add a number that you'll call "infinity", and assume that most of standard arithmetic works. (The details there are a bit tricky, but don't worry about it.) Therefore, infinity plus one must not equal infinity, because then (subtracting them out) one would equal zero. Same for infinity plus or minus any finite number. So, there's an area around infinity where you can move around comfortably. None of these numbers can be finite, because then (by a similar argument as before) infinity would be finite. This gives you two completely separate sections of the number line - the area containing numbers a finite distance from zero, and the area containing numbers a finite distance from infinity. Similarly, twice infinity has its own area, as does negative infinity, infinity squared, half of infinity, and so on. By taking ordered pairs or triples of coordinates in the nonstandard reals, you can produce a two- or three-dimensional space with similar properties. This space, apart from being inherently awesome, can be useful in calculus and real analysis. Black Carrot (talk) 15:57, 11 July 2008 (UTC)[reply]

CONVERSION OF GRAMS TO OUNCES

I have a table to convert grams to ounces but there are 2 different formulas 1 to (AVDP) ounces and 1 to (TROY) ounces what is the difference… —Preceding unsigned comment added by 72.237.213.42 (talk) 14:45, 9 July 2008 (UTC)[reply]

See ounce#definitions (AVDP is avoirdupois). Algebraist 14:49, 9 July 2008 (UTC)[reply]

(after EC) Avoirdupois and Troy are two different systems for measuring weight (or more properly mass). The linked articles give more details. --LarryMac | Talk 14:51, 9 July 2008 (UTC)[reply]

Once you understand this difference, you can tell the following joke:

"Which weighs more, a pound of gold or a pound of feathers?"

The obvious answer, "they weigh the same" is not correct. The pound of feathers weighs more than the pound of gold, since the gold is measured in Troy ounces; thus, despite being slightly larger ounces, there are only 12 of them in a troy pound, vs 16 for the pound of feathers.

Boy, does it lose a lot when you have to write out the explanation! --Danh, 63.231.161.34 (talk) 23:01, 11 July 2008 (UTC)[reply]

PPPs from the OECD

I'm trying to wrap my head around some statistics from the OECD regarding Purchasing power parity.

In their semi-annual report, Main Science and Technology Indicators (subscription required, unfortunately), the OECD has had a table since at least 2001 of Purchasing Power Parity values that they presumably used when calculating PPPs for data in the report itself.

My problem is that they don't line up year to year and that confuses me.

For example, in the 2001 report, they have a table that looks like this:

	1994	1995	1996	1997	1998	1999	2000
Australia	1.34	1.29	1.30	1.32	1.31	1.30	1.31
Austria	13.91	13.73	13.58	13.60	13.72	13.60	13.39
...
Turkey	12 096.22	22 334.21	39 274.65	70 999.93	123 168.96	185 566.63	272 836.30
United States	1.0	1.0	1.0	1.0	1.0	1.0	1.0

Whereas in the 2007 report, for 1995 you get:

	1995
Australia	1.32
Austria	0.95
...
Turkey	0.02
United States	1.0

Now why should historical PPPs vary this much? I thought the whole point of having a PPP was that it wouldn't change? How can the 1995 PPP of Turkey have changed by so much between the 2001 and 2007 reports? I'm totally confused. I'm trying to combine data from multiple reports to make a larger data set than any one report has, but the values can change dramatically for the same table in reports separated by a few years.

Anybody have any idea what's going on? I'm no economist. --98.217.8.46 (talk) 21:14, 9 July 2008 (UTC)[reply]

In 2005, Turkey knocked 6 zeros off their currency to create the Turkish new lira, so one table is measuring in old lira the other in new lira. --Tango (talk) 00:22, 10 July 2008 (UTC)[reply]

(edit conflict) I believe you are getting caught by countries that intentionally redefined their currency. For example, Turkey replaced the Turkish Lira with the Turkish New Lira at an exchange rate of 1 million to 1. The effect being that if you express PPP as New Lira / US Dollar you get a value that is 1 millionth the PPP expressed as Old Lira / US Dollar. Dragons flight (talk) 00:25, 10 July 2008 (UTC)[reply]

I thought the whole point of PPP was that you weren't currency-dependent? Man, that makes it impossible to do historical work with these values... sigh. --98.217.8.46 (talk) 16:17, 10 July 2008 (UTC)[reply]

Things are priced in the currency of that country. The point of PPP is that you can compare things priced in difference currencies based on the actual value of the money to a person in the country, rather than the value of the money on the foreign exchange markets. Once you've converted everything to the same currency, using the PPP rates, *then* you are currency independent. --Tango (talk) 17:02, 10 July 2008 (UTC)[reply]