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March 3

Conditional convergence of complex series

Resolved

The harmonic series $\Sigma 1^{n}/n$ diverges. I'm guessing that replacing the 1 with any other "unit circle" complex number in the numerator $e^{i\theta }$ would lead to convergence, a special case being the alternating harmonic series $\Sigma (-1)^{n}/n$. In this case the Taylor series for the complex Log function makes it easy to solve.

Question is, does this work in general? Given a real alternating sequence which is conditionally convergent but not absolutely convergent, is it always the case that replacing the (-1) with $e^{i\theta }$ results in a complex sequence that is conditionally convergent? Zunaid 08:59, 3 March 2009 (UTC)

No, you can see this by altering your example slightly. just put in two zeros ( or something that converges rapidly if you dont like 0) after every second term as in

1 - 1/2 + 0 - 0 + 1/3 - 1/4 + 0 - 0 + 1/5 - 1/6 + 0 - 0

Then put in e^iπ/2=i

You'll see you get both the real and imaginary part going infinite. You're quite right about the original sequence though, anything within on on the unit circle except 1 will make it converge. Dmcq (talk) 10:43, 3 March 2009 (UTC)

Ah but wait! Let's force the sequence of terms to be "absolutely decreasing" i.e. abs(a_n+1) <= abs(a_n) (perhaps the inequality needs to be strict?) so that you can't "cheat" by inserting throw-away rapidly-converging terms as in your counter-example. Where does that leave us? Zunaid 13:44, 3 March 2009 (UTC)

That's a much better question, and in fact it is a quite important theorem, see Abel's test. Dmcq (talk) 15:27, 3 March 2009 (UTC)

Thanks! That was perfect. Zunaid 16:47, 3 March 2009 (UTC)

why is financial charts plotted on logarithmic scale

why is financial charts plotted on logarithmic scale pls —Preceding unsigned comment added by 92.230.67.192 (talk) 16:31, 3 March 2009 (UTC)

Charts are generally plotted on a logarithmic scale when there is a large range of values to be plotted so a linear scale would have to be very small. This means small variations would be invisible. The other reason, often related, is when the values being plotted high an exponential trend - plotting such a trend logarithmically results in a straight line, which is easy to recognise. Financial charts aren't always plotted logarithmically, but those that are are done so for one of these reasons. --Tango (talk) 18:18, 3 March 2009 (UTC)

What makes a log scale appropriate is not just that the range of values is large, but that the relative size of variations is important and not the absolute size. For example, say that in 1990, something went from 2 to 5; but in 2000, it went from 10 to 15. If the 1990 increase is considered the larger one (because the relative increase is 150% rather than 50%), a log scale is appropriate. This is typical with things like prices, although they are often shown on a linear scale for other reasons. But if the 2000 increase is considered larger (because 5 points is more than 3 points), then you want a linear scale. This is typical with things like interest rates or unemployment rates. --Anonymous, 03:55 UTC, March 4, 2009.

To be perfectly honest, what matters if what you are trying to prove with your statistics. You choose a scale that makes the graph look like it shows what you are saying it should show. --Tango (talk) 13:30, 4 March 2009 (UTC)

Suppose you want to show the exchange rates between (say) the Italian lira and the Swiss franc from 1861 to 1999, during which time I believe the lira dropped by two orders of magnitude while the franc was relatively firm. If the number of lire to a franc is displayed linearly, it appears that the rate was nearly constant in the early period and much more volatile at the end, but this impression is (most likely) false. If the value of a thousand lire in francs is shown linearly, it gives an equally false impression the other way around. — Stock prices are usefully shown on a log scale because what interests the investor is the ratio between prices at different dates: you want to know what you'd have if you bought $1000 worth of FooCorp in 2001 and sold it in 2002, regardless of the nominal price. —Tamfang (talk) 18:35, 4 March 2009 (UTC)

I seem to remember reading that a log scale approximates to the rate of increase. 89.240.206.60 (talk) 22:27, 7 March 2009 (UTC)

A graph on a log scale shows the same slope for the same rate of proportional increase, regardless of the absolute amount of the variable in question. —Tamfang (talk) 04:43, 8 March 2009 (UTC)

ramanujan nested radicals problem

His article says that for this problem

${\sqrt {1+2{\sqrt {1+3{\sqrt {1+\cdots }}}}}}.$

He offered the solution

$x+n+a={\sqrt {ax+(n+a)^{2}+x{\sqrt {a(x+n)+(n+a)^{2}+(x+n){\sqrt {\mathrm {\cdots } }}}}}}$

But how do you find those constants x,n,a? Equating the parts of the problem to the solution, it looks like there are no solutions. Particularly:

$ax+(n+a)^{2}=1$

$a(x+n)+(n+a)^{2}=1$

n is nonzero because equating the scalar multiples from the problem to those of the solution

$x=2$

$(x+n)=3$

So what are x,n,a for the problem he offered? The article says that the answer is x+n+a=3. .froth. (talk) 18:28, 3 March 2009 (UTC)

Not sure what you're saying as you've already done most of the work, am I missing something? You'd have x=2, n=1 and a=0. They're nice these nested radical problems I think. Dmcq (talk) 19:22, 3 March 2009 (UTC)

Oh a=0 wowwwww .froth. (talk) 20:59, 3 March 2009 (UTC)

Yes, see zero. It's an interesting new idea from India so I guess that's where Ramanujan heard about it ;-) Dmcq (talk) 15:05, 4 March 2009 (UTC)

Comparing MATLAB and C++

I have written up a few simulations in MATLAB but they take forever to run. Each simulation takes about 30 hours on a dual processor 14GB RAM linux station. My questions is that if I programmed and ran the simulation in C++, would it really be faster? Does anyone have an idea of by what factor will this increase? Someone told me it will speed up by a factor of 10 but that sounds like an exaggeration.

On another note, can anybody recommend a good random number generator for C++? The built in srand and rand are crap. I need something with a uniform distribution for some serious SERIOUS number crunching. Even in the simplest case, I will call the random generator more than 10,000 times so I don't want the numbers to be biased or start repeating. Thanks!-Looking for Wisdom and Insight! (talk) 19:49, 3 March 2009 (UTC)

I've used a pseudorandom number generator (PRNG) based on the one in Numerical Recipes in C++ a lot, but I won't recommend it. Not because it isn't good (it is), but because (1) it is non-free software, and (2) NR in C++ is an excellent example of Fortran in any language - I had to spend quite a while to de-uglify it. Our article on the Numerical Recipes books states that the GNU Scientific Library provides many of the same functions, I'd be surprised if a good PRNG wasn't one of them. The Art of Computer Programming has an excellent chapter on pseudorandom number generators if you want to write your own. If you do, I'd also like to recommend the program ENT for testing it (and for testing other PRNG's as well). --NorwegianBlue ^talk 21:21, 3 March 2009 (UTC)

Note that the best choice of PRNG depends heavily on your requirements, and what sort of "bias" you're trying to avoid. Any halfway decent one will not have any detectable deviation from uniformity in the long run, so it's going to be more subtle than that. The known flaw, for example, in the linear congruential RNGs (besides their modest period) is that successive tuples taken from them "fall mainly in the planes" — there are some discrete hyperplanes that will capture most such tuples.

If your top two requirements are, in either order, speed and long period, then an excellent RNG is the Tausworthe one (hmm, that seems to be a redlink, but Google should help you out). It also gives excellent results in terms of autocorrelation. It begins to get into trouble when you start looking at third-moment measures — if you have a run of high numbers, then numbers soon after that are more likely to be anti-correlated, and if you have a run of low numbers, then numbers soon after that are more likely to be positively correlated.

If you want real unassailable lack of any detectable nonrandomness, you need to use something with cryptographic strength; say, MD5. These RNGs have excellent properties, but are slow as molasses compared to Tausworthe and even to linear congruential. --Trovatore (talk) 21:47, 3 March 2009 (UTC)

Depends on the application. I've moved code from MATLAB to C++ and seen as much as a factor of 30 improvement. Most of that probably came from the way C++ allowed me to more intelligently manage the allocation and deallocation of memory (MATLAB can be fairly dumb about this some times, and it can get to be a major source of overhead). For virtually any application compiled C++ will be faster than MATLAB, but whether it is a little faster or lot faster will often depend on how clever and thoughtful you are as a programmer. If your MATLAB code is already highly optimised (e.g. using matrix operations rather than for loops whenever possible, limiting the creation and destruction of large intermediate variables, etc.), then you will probably see smaller gains, i.e. a factor of 2 or 3. Dragons flight (talk) 21:53, 3 March 2009 (UTC)

Mersenne twister is a popular high-speed PRNG for numerical simulations, that has a very long period and is proven free of a bunch of statistical biases that plague some of the awful generators of olden times. It has a C implementation and may be callable from Matlab. The BSD random(3) function is not too bad either. None of these are built to withstand adversarial analysis though (i.e. they are not supposed to be cryptographically secure). The gold, er, standard if you require resistance to malicious attack is the Advanced Encryption Standard but you probably want to study some crypto theory (in addition to traditional numerical methods) if you are faced with that type of problem. 207.241.239.70 (talk) 05:00, 4 March 2009 (UTC)

March 4

Proof in baby Rudin Theorem 3.42

I just have a question on one step so hopefully I can just write that and it will make enough sense.

$\left|\sum _{n=p}^{q-1}A_{n}(b_{n}-b_{n+1})+A_{q}b_{q}-A_{p-1}b_{p}\right|\leq M\left|\sum _{n=p}^{q-1}(b_{n}-b_{n+1})+b_{q}+b_{p}\right|$

The only info about $M$ given anywhere is that it is some $|A_{n}|\leq M$ for all $n$. It seems to make sense but I guess I'm not seeing exactly why this works. We're talking complex numbers here so this is modulus, not just absolute value but maybe it'd work the same either way.

Thanks for any help. StatisticsMan (talk) 01:46, 4 March 2009 (UTC)

Also, why is it that some math stuff looks so crappy if it's not on a separate line or whatever? StatisticsMan (talk) 04:13, 4 March 2009 (UTC)

Well, this seems to be the proof of Dirichlet's test for series. If so, besides the assumption you stated on the complex sequence (an) (i.e., it has bounded partial sums An), there is also that (bn) is a positive and decreasing sequence of real numbers (converging to 0). If it is positive and decreasing, the inequality you wrote is very clear (and if it is not, the inequality in general doesn't hold. Notice that your RHS is just 2M|bp|). --pma (talk) 09:00, 4 March 2009 (UTC)

Sorry, yes, this is what it is and I thought of those extra assumptions last night and how I should have put them here. I can tell what the right hand side is. My question was how exactly do I know the LHS is less than or equal to the RHS.

I guess we know b_n - b_n+1, b_q, b_p are all positive. So, the way to make LHS the biggest is to have A_n, A_q > 0 and A_p-1 < 0 or the opposite of this. Otherwise, they sort of cancel each other out a bit. So, I guess I can see that. The largest possible would be to make A_n = A_q = M and A_p-1 = -M and in that case I can pull out the M. But, now I'm thinking in terms of complex numbers. Of course the b_i are all real numbers but the A_i are not. However, since the b_i are all real, I guess I can see that to make the LHS the largest I'd want A_i real and just as described before. So, I guess I get it now.

The point is, like you say the inequality does not hold in general so it takes a bit of thinking and reasoning, though not very hard I guess. Yet, the proof is in a textbook so they do not explain the reasoning at all. Thanks StatisticsMan (talk) 13:1

9, 4 March 2009 (UTC)

OK; it was just this anyway: I write in separate lines as you like :)

$\left|\sum _{n=p}^{q-1}A_{n}(b_{n}-b_{n+1})+A_{q}b_{q}-A_{p-1}b_{p}\right|\leq$

$\leq \sum _{n=p}^{q-1}|A_{n}(b_{n}-b_{n+1})|+|A_{q}b_{q}|+|A_{p-1}b_{p}|=$

$=\sum _{n=p}^{q-1}|A_{n}|(b_{n}-b_{n+1})+|A_{q}|b_{q}+|A_{p-1}|b_{p}\leq$

$\leq \sum _{n=p}^{q-1}M(b_{n}-b_{n+1})+Mb_{q}+Mb_{p}=$

$=M\left(\sum _{n=p}^{q-1}(b_{n}-b_{n+1})+b_{q}+b_{p}\right)=$

$\textstyle =2Mb_{p}$

--pma (talk) 13:47, 4 March 2009 (UTC)

Probability: Normal & Exponential Expectations

Hi there refdesk!

Was wondering if anyone could give me a hand getting started on this probability question - just a hand in the right direction to get me started would be majorly appreciated! Thankyou very much in advance :)

Let k be a positive integer and let $X$ be distributed $N(0,1)$. Find a formula for ${\textbf {E}}X^{k}$. Find also a formula for ${\textbf {E}}e^{\lambda x}$.

How do I begin to tackle these problems? I'm not hugely familiar with them so any detailed(or not!)assistance would be incredibly helpful - thankyou very much again,

Spamalert101 (talk) 04:23, 4 March 2009 (UTC)Spamalert101

Have you looked at the article on the normal distribution? I just did, and saw some relevant info there. Baccyak4H (Yak!) 04:35, 4 March 2009 (UTC)

I suggest also looking at the article about Expected values and seeing how they are computed. Then just blast out the calculation directly. 207.241.239.70 (talk) 05:05, 4 March 2009 (UTC)

You might actually find the second part easier using completing the square and then get out the appropriate power of λ for the first part. Dmcq (talk) 15:15, 4 March 2009 (UTC)

If k is odd, the answer is obvious. Suppose k is even.

${\begin{aligned}&{\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }x^{2n}e^{-x^{2}/2}\,dx={\frac {2}{\sqrt {2\pi }}}\int _{0}^{\infty }x^{2n}e^{-x^{2}/2}\,dx={\frac {2}{\sqrt {2\pi }}}\int _{0}^{\infty }(2u)^{n}e^{-u}\,{\frac {du}{\sqrt {2u}}}\\&={\frac {2^{n}}{\sqrt {\pi }}}\Gamma \left(n+{\frac {1}{2}}\right).\end{aligned}}$

Then remember that Γ(1/2) = √π and do some algebra, including the fact that Γ(a + 1) = aΓ(a), repeated a sufficient number of times. You should get a bunch of cancelations along the way.

I've typed hastily, so check the details in what I wrote. Michael Hardy (talk) 17:27, 4 March 2009 (UTC)

what is the opposite of zero?

what is the opposite of zero? —Preceding unsigned comment added by 86.144.124.55 (talk) 17:31, 4 March 2009 (UTC)

It doesn't really have an opposite. In certain contexts, infinity can be viewed as the opposite of zero. In other contexts, one is the opposite of zero, and there are probably contexts with various other answers. --Tango (talk) 17:39, 4 March 2009 (UTC)

For one, zero is the opposite of zero. — Emil J. 17:49, 4 March 2009 (UTC)

The answer is obviously -0. -mattbuck (Talk) 18:04, 4 March 2009 (UTC)

Emil is right if "opposite" is meant in a mathematical sense. I don't know why Mattbuck is pulling our chain; shouldn't he still be writing down the exact value of pi as I demanded a short while ago? In simple language the opposite of zero is something. Cuddlyable3 (talk) 19:09, 4 March 2009 (UTC)

Oh, that. He finished. There's not nearly as much of it as you might imagine. --Trovatore (talk) 19:44, 4 March 2009 (UTC)

I may be old but I assure you there is nothing wrong with my imagination nor is there anything wrong with my imagination. Remind me please, what was the last digit of pi ? Cuddlyable3 (talk) 23:07, 4 March 2009 (UTC)

The last digit of pi is 3. Of course, if you go the other way it's the first digit. -mattbuck (Talk) 00:16, 5 March 2009 (UTC)

Who mentioned digits? --Trovatore (talk) 23:29, 4 March 2009 (UTC)

Someone called JacobJakoff[1] Cuddlyable3 (talk) 00:00, 5 March 2009 (U5TC)

Presumably Mattbuck is referring to -0. Algebraist 19:14, 4 March 2009 (UTC)

If you equate "zero" with "nothing" (which I'm assuming is what you mean by "simple language") then the opposite is either "something" or "everything", I'm not sure which... --Tango (talk) 19:18, 4 March 2009 (UTC)

It might be anything. Other than 0 or -0. Kittybrewster ☎ 19:39, 4 March 2009 (UTC)

(Unindented!) Then again, which opposite? The additive opposite of zero IS zero, whereas the multiplicative opposite is undefined, as everyone know. the boolean opposite of 0 is 1. if you want to be really pedantic, the color opposite of 0 would be a black field with a white circle, i guess. P.S. Algebraist: he's obviously just on one of the infinite zeros IN pi. --Evan ¤ Seeds 20:15, 4 March 2009 (UTC)

What's an infinite zero? --Trovatore (talk) 21:38, 4 March 2009 (UTC)

I think he means "infinitely many zeros", although I'm not sure it's been proven that pi contains infinitely many zeros (in any base other than binary, at least). --Tango (talk) 00:03, 5 March 2009 (UTC)

Normal number suggests this is unknown. Algebraist 01:37, 5 March 2009 (UTC)

"Unknown" in a technical sense specific to mathematics (not yet proved from any generally accepted set of axioms). Not really unknown; the right answer is completely clear. --Trovatore (talk) 02:01, 5 March 2009 (UTC)

The "obvious" answer has been proven incorrect before, it could happen again (unlikely, but possible). --Tango (talk) 00:19, 6 March 2009 (UTC)

Yes. It's also possible that Peano arithmetic will turn out to be inconsistent.

Far too many mathematicians are still stuck on Euclidean foundationalism, on the idea that there's a difference in kind between mathematics and the empirical sciences. There isn't. There's only a difference in degree, and in subject matter. We do know that there are infinitely many zeros in the decimal representation of pi. It's not a perfect, apodeictically certain sort of knowledge, no. But there's no such thing. --Trovatore (talk) 00:29, 6 March 2009 (UTC)

I'm surprised the real projective line hasn't been mentioned yet... — Charles Stewart (talk) 20:23, 4 March 2009 (UTC)

The opposite of "zero" is clearly "orez", duh.  :-) Dragons flight (talk) 23:10, 4 March 2009 (UTC)

It's not true that zero is nothing because a zero comprises an amount of ink that is spread around some space. The opposite to that is everything that is not ink and not occupying any space. A black hole fits that profile nicely. (We need not wikiquibble about the ink because this is an inkless encyclopedia.)Cuddlyable3 (talk) 23:21, 4 March 2009 (UTC)

No, the number zero is not comprised of ink. The written word 'zero' does in some cases, but we aren't discussing that. Algebraist 01:39, 5 March 2009 (UTC)

But the number zero is most definitely not "nothing" either, it's no less something than any other mathematical object. — Emil J. 10:43, 5 March 2009 (UTC)

Actually the opposite of 0 is 0, except that in handwriting it might be narrower on the bottom than on the top. This is just like how the opposite of 6 is 9. – b_jonas 19:46, 5 March 2009 (UTC)

The empty set is comprised of any other set. —Tamfang (talk) 06:00, 8 March 2009 (UTC)

I'm really quite amazed at all these wrong answers. The question makes no sense if "zero" is the number, but it makes perfect sense if "zero" is a predicate (as in "x is zero"). So that is the only possible meaning of the question and the answer is of course "non-zero". McKay (talk) 09:39, 6 March 2009 (UTC)

Zero is a placeholder. Therefore its opposite is a place that nothing is holding. Ooops, nothing equals zero. So the opposite of zero is a double negative "full of sound and fury, and signifying.......nothing." Cuddlyable3 (talk) 13:03, 8 March 2009 (UTC)

March 5

Length of a curve

Can I get some tips on integrating equations such as ${\begin{aligned}&\int {\sqrt {a+x^{2}}}\,dx\end{aligned}}$ , which come up when finding the length of a curve? 72.200.101.17 (talk) 02:20, 5 March 2009 (UTC)

These are usually the differentials of things like sin^-1x, iirc. If you've got a table of standard integrals, it's likely to be on that. -mattbuck (Talk) 03:12, 5 March 2009 (UTC)

This is a classic case of where the solution is found by thinking and working outside the box. See Trigonometric substitution. (And be sure to note that the $a$ in that article's examples is not quite the same as the $a$ in your problem.) -- Tcncv (talk) 04:31, 5 March 2009 (UTC)

The trigonometric substitution x = (√a) tan; θ will transform that one into the integral of secant cubed. Michael Hardy (talk) 00:46, 6 March 2009 (UTC)

Why use such a substitution when you have hyperbolic trigonometry? Put $x={\sqrt {a}}\sinh y$ (a monotonic function on the whole real line) and use $\cosh ^{2}\alpha -\sinh ^{2}\alpha =1$ together with $\sinh '\alpha =\cosh \alpha$, followed by double-"angle" formula for $\cosh ^{2}\alpha$. — Pt _(T) 11:10, 12 March 2009 (UTC)

Continued - Random Generator

This is a continuation of the questions I asked above about MATLAB, C++, and a good random number generator. How about a generator that will give me a decimal between zero and one? I know I can always use any generator and then divide the output by the maximum output possible but I really don't want to be dividing by really large numbers 20,000 times.-Looking for Wisdom and Insight! (talk) 05:54, 5 March 2009 (UTC)

Use J (programming language). Get 20000 random numbers between 0 and 1 by typing

? 20000 $ 0

Bo Jacoby (talk) 06:23, 5 March 2009 (UTC).

If you're worried about computation overhead, keep in mind that the computation time spent on calculating the next random integer likely far dwarfs a single float division computation. Also, if you are dividing by a constant power of two, the computation time is negligible (depending on how smart your compiler is): a constant just needs to be subtracted from the exponent.

Although I didn't comment on your discussion above, I've used the GNU Scientific Library's random number generators before, and found them easy to use and flexible. (But I did not have specific requirements for a random number generator, any one would do for me.) The documentation can be found here. Click "Random Number Generator Algorithms" to get the list of high quality RNGs implemented in the GSL. Eric. 131.215.158.184 (talk) 08:49, 5 March 2009 (UTC)

(ec):Did you check out the Gnu Scientific Library? I assume that what you want is sampling from a uniform distribution in the interval [0..1]. The function gsl_rng_uniform does almost that, it returns numbers in the interval [0..1), i.e. it may return 0, but never 1. Internally, it performs the division that you're reluctant to do. Btw, why are you worried about 20,000 divisions in C++? If you're interested in other languages, the command in R (programming language) is runif(20000), where runif is short for random uniform. --NorwegianBlue ^talk 08:57, 5 March 2009 (UTC)

20000 doesn't matter nowdays on a computer, but if you're really worried by divides then simply precompute the inverse and multiply by that. Dmcq (talk) 13:09, 5 March 2009 (UTC)

Uniform distribution and independence

Could anyone possibly explain briefly to me how one would go about showing that if X and Y are independent, identically distributed random variables, each uniformly distributed on [0,1], $U=X+Y$ and $V={\frac {X}{Y}}$, $U$ and $V$ aren't independent?

I think it's something to do with the fact that if U is large within the given range, the ratio of X:Y is approximately 1, but I'm not sure... help! Thanks :) —Preceding unsigned comment added by 131.111.8.98 (talk) 15:15, 5 March 2009 (UTC)

Yes, follow your idea. Consider the event U>3/2 and the event V>2, for instance, and their intersection --pma (talk) 16:09, 5 March 2009 (UTC)

(slight topic hijack, hope it's ok) What is a good book to read about this stuff? One that would let me write a sentence like the one you (pma) just wrote. Something that explains what random variables and events are in a practical enough way to show how to do such calculations, but also mathematically rigorous enough (let's say, starting from basic real analysis) to explain what the words really mean. Thanks. 207.241.239.70 (talk) 04:45, 7 March 2009 (UTC)

Differential Equations for Chemical Engineering

I'm currently reading a little into reactor design for a project I am doing (in particular, this collection of lectures. It's going well, except for some problems with differential equations (which I only started investigating 10 minutes ago, so apologies if I'm asking obvious questions). For example, the design equation of a batch reactor is:
${\frac {dm_{a}}{dt}}=R_{a}(t)=-km_{a}(t)$
I am reliably informed that this is an ordinary differential equation and can be solved by seperating the variables. Hence:
$dm_{a}=-km_{a}(t)dt$
${\frac {dy}{m_{a}(t)}}=-kdt$
So far, I'm happy with that. It then says "integrating gives":
$\ln m_{a}(t)=-kt+{\text{constant}}$
This is where I get stuck. First of all, integrating with respect to what? As I don't understand this, I don't understand why dt becomes t and dy becomes 1 (i.e. so $1/m_{a}$ integrates to give $\ln {m_{a}}$). I have a feeling I've made a complete hash of this, can anyone set me straight? Thanks. --80.229.152.246 (talk) 21:30, 5 March 2009 (UTC)

That should be

${\frac {dm_{a}}{m_{a}(t)}}=-kdt$

Separation of variables is a short-cut, but somewhat of an abuse of notation when expressed like this, as it trades on the (intentional) resemblance of Leibniz's notation to a fraction. A more rigorous approach is:

${\frac {1}{m_{a}}}{\frac {dm_{a}}{dt}}=-k$

then you integrate both sides with respect to t to get

$\int {\frac {1}{m_{a}}}{\frac {dm_{a}}{dt}}\,dt=\int -k\,dt$

$\Rightarrow \int {\frac {1}{m_{a}}}\,dm_{a}=\int -k\,dt$

$\Rightarrow ln(m_{a})=-kt+c$ Gandalf61 (talk) 21:58, 5 March 2009 (UTC)

Thanks very much. It makes a lot more sense now. --80.229.152.246 (talk) 23:34, 5 March 2009 (UTC)

If you are familiar with group theory (which has many applications in chemistry, by the way), you will know that manipulations with the quotient operator are also an "abuse of notation" analagous to manipulating the differentials. --PST 04:17, 6 March 2009 (UTC)

Calculation of Pi

Suppose I randomly pick a point inside a 1x1 square, determine whether its distance from a certain vertex is less than 1, and repeat the process n times. I then calculate pi using pi=number of hits/number of misses*4. How large does n have to be if I want pi to be accurate to k digits?

Out of curiosity, I wrote a Java program to calculate pi in this way. The first 10 million repetitions gave me 3.142, but the next 250 billion only managed to determine 1 extra digit. --Bowlhover (talk) 22:13, 5 March 2009 (UTC)

With n trials, you get a random variable with mean π and standard deviation sqrt((4π-π²)/n)≈1.6/sqrt(n). To get k reliable digits, you want the s.d. to be well under 10^−k, say under 10^−k/3, so you want over 25*10^2k trials. So your 250 billion trials should be good for 4 or 5 digits, as you discovered. Algebraist 22:23, 5 March 2009 (UTC)

This reminds me the story of the famous Buffon's needle. Various people enjoied the experimental measure of $\pi$ by counting intersections of needles thrown on the parquet. Results:

• Wolf (1850), 5000 needles, $\pi$=3.15..
• Smith (1855), 3204 needles, $\pi$=3.15..
• De Morgan (1860), 600 needles, $\pi$=3.13..
• Fox (1864), 1030 needles, $\pi$=3.15..
• Lazzerini (1901), 3408 needles, $\pi$=3.141592..
• Reina (1925), 2520 needles, $\pi$=3.17..

--pma (talk) 00:10, 6 March 2009 (UTC)

What did Lazzerini do differently to the others? That can't just be good luck... --Tango (talk) 00:15, 6 March 2009 (UTC) I clicked the link... surprising how informative that can be! --Tango (talk) 00:17, 6 March 2009 (UTC)

That's some interesting cheating. If Lazzerini had done his experiment the proper way, it would probably have taken him millions of years to get 3.141592. Even my program, which has finished 360 billion trials, is still reporting around 3.14159004, with no indication that the first "0" will become "2" anytime soon.

Algebraist: how did you get the standard deviation expression sqrt((4π-π²)/n)? I know very little about statistics, so please explain. Thanks! --Bowlhover (talk) 06:07, 6 March 2009 (UTC)

A single trial gives rise to a Bernoulli random variable with parameter p=π/4. It thus has mean π/4 and variance π/4-π²/16. Multiplying by 4 gives a r.v. with mean π and variance 4π-π². Averaging out n of these (independently) gives mean π and variance (4π-π²)/n. Standard deviation is the square root of variance. Algebraist 09:03, 6 March 2009 (UTC)

It will take me a while to learn those concepts, but thanks! --Bowlhover (talk) 08:15, 7 March 2009 (UTC)

Are you using something faster than Math.random() to generate random numbers, or do you have a really fast computer? After some vague amount of time (15 minutes?) I've only got 2 billion trials... though I already have 3.1415 stably.

I am vaguely contemplating the difficulty of writing a program that cherry-picks results: given a fixed amount of time (measured in trials), it would calculate every few million trials how close the current approximation to pi is, and calculate whether continuing running the program or restarting the program gives a lower expected value for your final error. Or maybe, a-la Lazzerini, it chooses a not-too-ambitious continued fraction approximant for pi and attempts to hit it exactly, aborting when that happens. Eric. 131.215.158.184 (talk) 07:01, 6 March 2009 (UTC)

I'm using java.util.Random, the same class that Math.random uses, so I don't think that has anything to with it. My computer's processing speed is probably not relevant either because it is only 2 GHz, probably slower than your computer's. Maybe you're outputting the value of pi every iteration? I used "if (num_trials%10000000==0)" so that printing to screen doesn't limit the program's speed.

You might be interested to know that I modified the program to only output a calculated value of pi if it is the most accurate one made so far. Here are the results (multiply all the fractions by 4):

Data dump

4.0 1/1 3.5 7/8 3.272727272727273 9/11 3.142857142857143 11/14 3.140534262485482 1352/1722 3.1410330818340104 1353/1723 3.1415313225058004 1354/1724 3.1416202844774275 2540/3234 3.1415743789284645 2624/3341 3.1416072990876143 6284/8001 3.1416015625 6434/8192 3.141587606581002 6540/8327 3.141589737441554 6551/8341 3.1415929203539825 6745/8588 3.1415924255132652 519694/661695 3.1415928668580926 519698/661700 3.1415928268290365 520214/662357 3.1415924928442895 520254/662408 3.1415925871823793 521535/664039 3.1415926140510604 521901/664505 3.1415926673317953 521923/664533 3.14159265069769 2757939/3511517 3.141592653304311 2758832/3512654 3.1415926537518883 2789534/3551745 3.1415926534962155 8230357/10479216 3.1415926535954957 8415806/10715337 3.1415926535873497 8493712/10814530 3.1415926535881242 31499778/40106763 3.141592653590579 41739544/53144438 3.1415926535901835 260989009/332301527 3.141592653589953 261093002/332433935 3.1415926535896586 409395251/521258223 3.1415926535898775 484979918/617495610 3.141592653589793 39509540665/50305109569

It's neat that the last, 3.141592653589793, is accurate to the last printed digit. --Bowlhover (talk) 08:16, 7 March 2009 (UTC)

My computer's new, but it's a low-end laptop. Hmmmm. No, I print out every 2^24 iterations. I don't do a square root, either, which is an easy mistake. Maybe I'm just impatient with not running the program long enough. Good idea with printing only the best approximation so far... it sort of "cheats" because as your current estimate drifts from being too low to too high it must pass very near pi in between. Eric. 131.215.158.184 (talk) 21:27, 7 March 2009 (UTC)

I suspect that Lazzarini's point was to show how a delicate matter is a statistical experiment... --pma (talk) 09:05, 6 March 2009 (UTC)

March 6

About Lagrange's Theorem

Lagrange's four-square theorem states that every positive integer can be written as the sum of four squares of integers.

If S is the set of all non-negative integers except 7, is it still the case that every positive integer is the sum of the squares of four elements of S?

I feel sure that this sort of question has been considered before. Does anyone know where to find it in the literature?Partitioin (talk) 00:43, 6 March 2009 (UTC)TC)

I fooled around with a computer program for a few minutes and didn't find any counterexamples in a few thousand trials, but that means nothing. I wikilinked the theorem for you, hope that's ok. 207.241.239.70 (talk) 05:37, 7 March 2009 (UTC)

I've confirmed solutions (sans 7s) for integers up through 300,000,000. The average number of solutions per value appears to have a slightly less than linear relationship with numbers. In the 1,000,000 range the average number of solutions per integer is on the order of 25,000 or a ratio of about 0.025. However, I've found some numbers having significantly fewer solutions, such as 10,485,760 with only 2 solutions – that's quite a statistical bump. Of course, none of this has anything to do with a proof. -- Tcncv (talk) 05:47, 8 March 2009 (UTC)

I am not surprised that the number of solutions does not behave smoothly - there are an infinite number of integers which only have one representation (up to order) as the sum of four squares - see OEIS A006431. If you omit 5 from S instead of 7 then empirical evidence suggests that every integer has a representation as a sum of four squares of integers in S apart from 79. And, interestingly, 79 has 3 distinct representations as the sum of four squares:

79 = 7² + 5² +2² + 1²

79 = 5² + 5² + 5² + 2²

79 = 6² + 5² + 3² + 3²

... but they all happen to include 5². However, I don't see an approach to a proof. Gandalf61 (talk) 13:17, 8 March 2009 (UTC)

Physics question

Please see Wikipedia:Reference_desk/Science#Force_meter_question, (somebody sensible), for some reason the usually sentient editor "Gandalf61", appears to have lost their marbles, or I'm going blind...

ie Third opinion213.249.232.187 (talk) 14:34, 6 March 2009 (UTC)

Nope, he's right. "Tango" proposes a good analogy near the end of the thread. yandman 16:26, 6 March 2009 (UTC)

Finding $y(x)$ from ${\frac {d}{dx}}$ Finding ${\frac {dy}{dx}}$ from ${\frac {d}{dx}}$

How do you find $y(x)$ from ${\frac {d}{dx}}$?The Successor of Physics 15:19, 6 March 2009 (UTC) How do you find ${\frac {dy}{dx}}$ from ${\frac {d}{dx}}$?The Successor of Physics 13:15, 8 March 2009 (UTC)

It's a little difficult to understand what you mean, but if you mean that knowing ${\frac {dy}{dx}}$ as a function of $x$, you want to express $y$ as a function of $x$, then Antiderivative is your article. —JAO • T • C 17:29, 6 March 2009 (UTC)

Perhaps I should restate my question as "How do you find ${\frac {dy}{dx}}$ from ${\frac {d}{dx}}$?". Thanks anyway.The Successor of Physics 13:15, 8 March 2009 (UTC)

${\frac {d}{dx}}$ is an operator. Applying it to $y$ is how you find ${\frac {dy}{dx}}$. Your question sounds like "How do you find $x+y$ from $+$?" and has the same answer: knowing what operator to apply (${\frac {d}{dx}}$ or $+$, for instance) is not useful if you don't know what to apply it to. —JAO • T • C 17:37, 8 March 2009 (UTC)

I want to know it so in for example ${\frac {dy}{dx}}={\frac {d}{dx}}y=ty,t={\frac {d}{dx}}$ because you can express derivatives more easily, and if ${\frac {dy}{dx}}=ty,{\frac {d}{dx}}=t$, and sometimes I want know y and knows t but not y so that is why I asked the question.The Successor of Physics 10:04, 9 March 2009 (UTC)

Your expression $t={\frac {d}{dx}}$ is complete nonsense. Algebraist 10:58, 9 March 2009 (UTC)

If you do know that ${\frac {d}{dx}}y=t(x)y$ for some known function $t$ which you can integrate, then you can use separation of variables to find $y$. But as Algebraist says, don't try to divide both sides by $y$, because the left-hand side isn't a multiplication to begin with. —JAO • T • C 14:21, 9 March 2009 (UTC)

I'm reading a book on vector calculus, and suddenly remembered that ${\frac {d}{dx}}y=\nabla y$! does this help? —Preceding unsigned comment added by Superwj5 (talk • contribs) 14:53, 9 March 2009 (UTC)

Not really. An actual example of a case where you "know $t$" would help more. (Separation of variables was not helpful?) —JAO • T • C 18:01, 9 March 2009 (UTC)

Look at it this way: if you know that ${\sqrt {}}y=ty$, does that mean that $t={\sqrt {}}$? Of course not. Just because it looks like a fraction instead of a funny symbol doesn't mean that ${\frac {d}{dx}}$ isn't an operator, and to deal with an equation with a differential operator you can't deal with it like it was a normal fraction or took a single value. Confusing Manifestation(Say hi!) 21:23, 9 March 2009 (UTC)

Determinants as Tensor Fields or Matrix Functions

Are Determinants Tensor Fields or Matrix Functions? If yes which and please state the equation and state its inverse.(Please don't be to harsh on me. I'm only new in this field.)The Successor of Physics 15:27, 6 March 2009 (UTC)

The most common use of the word "determinant" is as in "the determinant of a matrix", which is of course just a scalar, not a function. But there's also the determinant function, $\det$, which maps certain matrices to their respective determinants, so that's a matrix function ("matrix function" usually means a function from matrices to matrices, but scalars can be identified with 1×1 matrices so that makes little difference). I don't know enough about tensor fields to answer your first question, and I have no idea what equation you refer to. —JAO • T • C 17:38, 6 March 2009 (UTC)

The determinant is a (rank 0, i.e. somewhat trivial) tensor on the rows (or also the columns) of the matrix, if that's what you're asking. 207.241.239.70 (talk) 05:57, 7 March 2009 (UTC)

Cesaro mean

I am looking into the Cesaro mean right now. My question is, if I take any sequence that is bounded and I do the Cesaro mean over and over, do I eventually get a sequence that converges after a finite number of turns? Thanks for any help. StatisticsMan (talk) 18:10, 6 March 2009 (UTC)

I don't know, but looking at Cesàro summation's iterated section, "The existence of a (C, α) summation implies every higher order summation, and also that a_n=o(nα) if α>-1.", does seem to suggest bounded is a very strong hypothesis. (I think (C,n) summation means do Cesaro mean n times). JackSchmidt (talk) 20:46, 6 March 2009 (UTC)

Ah, unfortunately, it looks like things similar to your question are well studied (so beyond my limited skills). Peyerimhoff's Lectures on Summability doi:10.1007/BFb0060951 MR0463744 appears to have some reasonable discussion of this sort of thing, and in particular describes who and when people studied iterated Cesàro summation (Hölder, Schur) and how to more easily discuss it. I didn't find any specific counterexamples, and couldn't tell if any of the theorems applied to your case, but you might have an easier time. JackSchmidt (talk) 21:00, 6 March 2009 (UTC)

Oooh, you want a Tauberian theorem for iterated Cesàro means, and there is one in that book, Theorem III.2, and I think it says that a bounded sequence is Cesàro summable if and only if it is (C,n) summable for some n ≥ 1. There is an example in the Cesàro means chapter of a bounded sequence that is not Cesàro summable, so i think that means the answer to your question is no. JackSchmidt (talk) 21:07, 6 March 2009 (UTC)

Of course the answer is no. The counterexample has the form of a sequence $c_{k}$ alternating two values on consecutive intervals of larger and larger length. Instead of doing a quantitative computation for you, I'll just try to convince you by this example. You are in your sofa watching TV, say zapping from channel 1 to 9. This way you generate a sequence in {1,2...,9}, each second k, switching from channel $c_{k}$ to channel $c_{k+1}$. The trivial switch is also allowed, so at a certain moment, if you wish, you may start pressing channel 1 button compulsively for 100 times or more, one after the other (as actually happens to addicted TV watchers). The sequence of the Cesaro means, $\scriptstyle {\frac {1}{n}}\sum _{k=1}^{n}c_{k}$, somehow follows the sequence $c_{k}$ (in fact, with a regularization-delay effect). Anyway, after you have pressed channel 1 consecutively for 100 times or more, the means also are quite close to 1, say less than 2, and also the sequence of the means of the means starts being close to 1. Now, you switch to channel 9, pressing it each second, for 10,000 times or more, untill the mean, and the mean of the mean, are both close to 9 (say >8). At that point you go back to channel 1, pressing it till the first 4 iterated Cesaro means are all less than 2 again, even 100,000,000 times if needed. As you see, this way you can make a two-value sequence with no converging iterated Cesaro mean. With a bit of patience you can make the construction quantitative. I think $\scriptstyle c_{k}=0$ for $\scriptstyle (2n-1)!\leq k<(2n)!$ and $\scriptstyle c_{k}=1$ for $\scriptstyle (2n)!\leq k<(2n+1)!$ suffices.--pma (talk) 00:05, 7 March 2009 (UTC)

Thanks all. I can not read that book online. As far as pma's example, I have thought of this. I just alternated 1 and -1. I could say I want the average to go to at least 1/2 and then at least -1/2 each time to make sure it's not converging to something. So, 1, -1, -1, 1, 1, 1, 1, 1, ... . But, if I do the averages, I get 1, 0, -1/3, 0, 1/5, 1/3, 3/7, 1/2, and if I do averages again this time on the 2nd sequence, I get 1, 1/2, 2/9, 1/6, 13/75, 1/5, 57/245, 149/560, ...

The point is, I am barely getting above 1/4 now (I realize I did very few terms) and all values are positive. So, I am closer to convergence in only one step. No matter how weird and wild you make a sequence, if I iterate this process, I will at least get closer and closer to convergence. In your example with the TV, you also get closer to convergence after only one iteration. No matter what sequence you construct, even with factorial number of terms, each iteration you get closer and closer to convergence. So, your example may be correct, but I am not convinced yet.

I originally asked this question because of a homework problem that we would need this in, involving Banach Limits. But, we decided to use the shift operator instead of this operator in order to invoke the Hahn-Banach theorem (actually a slight generalization of it). Then, the problem became easy. But, I thank you all for answer. I am interested in this just in general and may look into it more sometime when I have more time. StatisticsMan (talk) 14:16, 7 March 2009 (UTC)

No problem. You would probably like the book (or other books with Tauberian theorems or Divergent series in the title). It works out PMajer's example, as well as the theorem that says that if one Cesàro mean is not enough, no finite number of iterations will be enough. It should be available in any reasonably sized university library (over 350 libraries according to OCLC 47540). JackSchmidt (talk) 15:42, 7 March 2009 (UTC)

OK, let's see if I convince you, sorry for the TV example. It's very elementary. Start defining $N_{1}=1$ and $c_{1}=-1$; then define $c_{2}=...=c_{N_{2}}$ to be 1, choosing $N_{2}$ so large that the first 2 Cesaro means at $N_{2}$ differ from 1 less than 1/2. Then define $c_{N_{2}+1}=...=c_{N_{3}}$ to be -1 choosing $N_{3}$ so large that the corresponding first 3 iterated Cesaro means at $N_{3}$ differ from -1 less than 1/3. And so on: by induction on k you define $c_{j}$ to be constantly equal to $(-1)^{k}$ for all $N_{k-1}<j\leq N_{k}$, choosing $N_{k}$ so large that the first k iterated Cesaro means differ from $(-1)^{k}$ less than 1/k. As you see, you do not get any closer to convergence; all the iterated Cesaro means have liminf=-1 and limsup=1.--pma (talk) 19:28, 7 March 2009 (UTC)

Alright, now what you are saying seems to make sense. Within the one sequence there are chunks that guarantee the kth Cesaro mean will not converge, for each positive integer k. Easy: that is, now that I have seen it, it is easy. As far as that book, I looked online and my library does not have it. But, that's okay. I have so much to do that I am required to do, I don't have much time to read some other book, even if it would be interesting. Thanks for the suggestion! StatisticsMan (talk) 20:45, 7 March 2009 (UTC)

You are welcome. In fact, the best solution is the interpolation result quoted by JackSchmidt, that is, as I understand, "a bounded sequence with convergent second Cesaro mean has the first Cesaro mean convergent too". --pma (talk) 23:26, 7 March 2009 (UTC)

Yea, that is a very bizarre result. It seems as if each iteration should get you closer to convergence. But, if one does not work, then 1 billion will not either. StatisticsMan (talk) 01:11, 8 March 2009 (UTC)

We can restate it this way, taking the quotient on the space c of converging sequences: the Cesaro mean defines a bounded linear operator M on ${\frac {\ell _{\infty }}{c}}$ (it's a Banach space, in which the norm of [x] is $\scriptstyle 1/2(\limsup x_{i}-\liminf x_{i})$ for all $x\in \ell _{\infty }$). Then we have $\scriptstyle \ker M=\ker M^{2}$, that is 0 is a semisimple eigenvalue of M (of infinite multiplicity). --pma (talk) 10:01, 8 March 2009 (UTC)

March 7

derivative of a series

Suppose we are given the series: $u=\sum _{l=-\infty }^{\infty }\sum _{n=1}^{\infty }\epsilon ^{n}U_{n}^{(l)}e^{il(kx-\omega t)}$. Here $u=u(U,x,t)$ where $U=U(\zeta ,\tau )$.

I wish to compute ${\frac {\partial ^{2}u^{2}}{\partial x^{2}}}$. How do I find a closed form expression for the above series so that I can compute the desired derivative. Thanks--122.160.195.98 (talk) 06:28, 7 March 2009 (UTC)

Because ${\frac {d(f(x)+g(x))}{dx}}={\frac {d(f(x))}{dx}}+{\frac {d(g(x))}{dx}}$, I suppose you just need to find the derivatives of the terms. Also, if you decompose the $e^{f(x)}$ which I wouldn't specify, I guess the series would turn into a sum of Fourier Series'.The Successor of Physics 07:20, 7 March 2009 (UTC)

How to compute u^2 is the main problem---OP —Preceding unsigned comment added by 122.160.195.98 (talk) 09:44, 7 March 2009 (UTC)

I suspect this is supposed to be an application of Parseval's theorem or the related Parseval's identity but I'm not up to working it out. 76.195.10.34 (talk) 12:53, 7 March 2009 (UTC)

You can't just blinding differentiate an infinite series term-by-term. That requires uniform convergence. --Tango (talk) 14:00, 7 March 2009 (UTC)

To OP: From my comment you should see that if differentiating once is ok, then just differentiate it again!The Successor of Physics 15:05, 7 March 2009 (UTC)

To Tango: I don't understand, Tango, why uniform convergence is required.The Successor of Physics 15:05, 7 March 2009 (UTC)

It just is. If the series doesn't converge uniformly then its derivative may well be different from the term-by-term derivative. See Uniform convergence#to Differentiability. You can't just assume that limits behave as you want them to behave, you have to actually prove it. --Tango (talk) 15:11, 7 March 2009 (UTC)

Uniform convergence of the series isn't enough, actually. You need uniform convergence of the term-by-term derivative, as our article states. Algebraist 15:18, 7 March 2009 (UTC)

Strictly speaking, I never said it was! ;) I couldn't remember the exact theorem (I tend to avoid analysis where possible), so was careful to speak vaguely. --Tango (talk) 15:25, 7 March 2009 (UTC)

I am the OP (from a different IP). This problem is from a paper I am reading and the author has assumed all necessary convergence conditions. The main problem is how to compute u^2. Everything else will come after that--118.94.73.74 (talk) 15:42, 7 March 2009 (UTC)

To compute u^2 you need to multiply two infinite series: see Cauchy product for how to do that. Your situation is a bit messy but a direct extrapolation from the description in the article. Eric. 131.215.158.184 (talk) 04:20, 8 March 2009 (UTC)

confusion about the proof of the splitting lemma in the wikipedia article

I have been reading the proof of the splitting lemma in the wikipedia article of that name and was wondering if anyone could help me to understand the very first part

at the very start of the proof to show that 3.(direct sum) implies 1.(left split) they take t as the natural projection of (A×C) onto A, ie. mapping (x,y) in B to x in A now why does this satisfy the condition that tq is the identity on A. similarily to show that 3. implies 2. they take u as the natural injection of C into the direct sum of A and C (A×C) ie. mapping y in C to (1,y) how does this satisfy the condition that ru is the identity on C.

It would apear to me that they mean something else by the "natural" projection and injection but i cant see what this would be??? thanks for your help and im sorry if this is badly worded —Preceding unsigned comment added by Jc235 (talk • contribs) 16:44, 7 March 2009 (UTC)

3. States not only that B is the direct sum of A and C, but also that the maps in the SES are the obvious ones: that q is the natural injection from A to A×C, i.e. q(a)=(a,0) and that r is the natural projection from A×C to C, i.e. r(a,c)=c. Then, setting t to be the natural projection of A×C onto A, we have that tq(a)=t(a,0)=a for all a in A. Similarly, if u is the natural injection of C into A×C, then ru(c)=r(0,c)=c for all c in C. (note: for the case of a general abelian category, talking about elements like this makes no sense, but this suffices for concrete examples) Algebraist 17:07, 7 March 2009 (UTC)

Confusion about differentiability & continuity

Hi there - I'm looking at the function $e^{-{\frac {1}{x^{2}}}}$ - the standard example for an infinitely differentiable non-analytic function - and I'm wondering exactly how you prove that the function has zeros at x=0 for all derivatives. In general, is it invalid to differentiate the function as you would normally would (assuming nice behaviour) to get, in this example, ${\frac {2e^{-{\frac {1}{x^{2}}}}}{x^{3}}}$, and then simply say it may (or may not) be differentiable at the 'nasty points' such as x=0? Or are there functions which have a derivative which is defined for well behaved points, but also has a well-defined value for non-differentiable points? Apologies if that made no sense. How would you progress then to show that all $f^{(k)}(0)$ are equal to 0? Do you need induction?

On continuity, I'm using the definition effectively provided by Heine as in [2] - i.e. that a function is continuous at c if $lim_{x\to c}f(x)=f(c)$: but what happens if both the function and the limit of the function are undefined at a single point and continuity holds everywhere else? Because it's not like the function has a -different- limit to its value at c explicitly, they're just both undefined - does that still make it discontinuous? Incidentally, I'm thinking of a function like ${\frac {sin(1/x)}{x^{2}}}$, where (if I'm not being stupid) both the function and its limit are undefined at 0, but wondering more about the general case - would that make it discontinuous?

Many thanks, Spamalert101 (talk) 21:23, 7 March 2009 (UTC)Spamalert101

Firstly, $e^{-{\frac {1}{x^{2}}}}$ is not defined at 0, any more than ${\frac {sin(1/x)}{x^{2}}}$ is. In the case of $e^{-{\frac {1}{x^{2}}}}$, however, there is a unique continuous function f on the real line extending the given function, and it's a standard abuse of notation to call this function (which takes the value $e^{-{\frac {1}{x^{2}}}}$ for x not zero and the value 0 at x=0) by the same name. Once you've done that, showing that f has derivative 0 at 0 is just a matter of applying the definition of differentiation: you have to show that ${\frac {e^{-{\frac {1}{x^{2}}}}}{x}}$ tends to zero as x tends to zero. At all other points, the chain rule (which only requires differentiability at the points involved; no nicer behaviour is needed) tells you that the derivative is ${\frac {2e^{-{\frac {1}{x^{2}}}}}{x^{3}}}$. Thus the derivative of f is the function g where g(x)=${\frac {2e^{-{\frac {1}{x^{2}}}}}{x^{3}}}$ for x not zero and g(0)=0. Then you can show that g is also differentiable at 0 with derivative 0. Showing that f is infinitely differentiable at 0 requires some sort of induction. Inducting on the statement 'the nth derivative of f is 0 at zero and is some rational function times $e^{-{\frac {1}{x^{2}}}}$ elsewhere' ought to work. Algebraist 21:32, 7 March 2009 (UTC)

On your second question, the function f from R\{0} to R given by f(x)=${\frac {sin(1/x)}{x^{2}}}$ is a continuous function. It can't be extended to a function continuous at 0, though, so by an abuse of terminology it might be thought of as being a function on R discontinuous at 0. Algebraist 21:35, 7 March 2009 (UTC)

Or, without an abuse of terminology, you can just define f(0)=0, or whatever, and then it simply is a function continuous everywhere except 0. --Tango (talk) 21:53, 7 March 2009 (UTC)

Let me add this. In order to prove that a $C^{\infty }$ function $f:\mathbb {R} \setminus \{0\}\to \mathbb {R}$ extends to a $C^{\infty }$ function on $\mathbb {R}$ it is necessary and sufficient that all derivatives of f have limit as $x\to 0$, which is easily seen to be the case of exp(-1/x²). One applies repeatedly the following well-known lemma of extension of differentiability, which is a simple consequence of the mean value theorem: "Let f be a continuous function on $\mathbb {R}$, differentiable in $\mathbb {R} \setminus \{0\}$; assume that there exists the limit $\lim _{x\to \ 0}f^{\prime }(x)=l$. Then f is differentiable at x=0 too, and $f^{\prime }(0)=l$" (going back to f(x):=exp(-1/x²), the thing works because the k-th derivative of f has the form mentioned above by Algebraist, that is f(x) times a rational function). --pma (talk) 23:54, 7 March 2009 (UTC)

March 8

Pascal Matrix

Is there a way to invert the Pascal matrix U_infinity? Black Carrot (talk) 02:16, 8 March 2009 (UTC)

The exponential definition should give a simple formula for the inverse. Upper triangular matrices behave reasonably well, even in infinite dimensions (even over non-discrete linear orders). If you work a few finite examples, the pattern for the full inverse should become clear too. JackSchmidt (talk) 03:10, 8 March 2009 (UTC)

And for the symmetrical one - since S_n = L_nU_n, this means that $S_{n}^{-1}=U_{n}^{-1}L_{n}^{-1}$. עוד מישהו Od Mishehu 10:20, 8 March 2009 (UTC)

Formula for nxn Singular Matrices

Is there a formula to calculate any arbitrary nxn sized singular matrices, and if yes, what?The Successor of Physics 03:56, 8 March 2009 (UTC)

I think you can create one by creating an arbitrary (n-1)×n matrix and then add any two rows to give you the final row. This should yield a matrix whose determinant is zero. More generally, you can calculate $Row_{n}=\sum _{i=1}^{n-1}a_{i}\cdot Row_{i}$ for arbitrary $a_{i}\,\!$ values. -- Tcncv (talk) 05:59, 8 March 2009 (UTC)

Thanks!The Successor of Physics 13:10, 8 March 2009 (UTC)

Hard to compute mathematical constant

Is there a prominent example of a mathematical constant which is so hard to compute, that (relatively) few decimal places are known? I'm aware that you easily can construct numbers to be hard to compute, but I'm asking for mathematical constant which is reasonanly important in its field and is hard to compute "by accident" and not by design. --Pjacobi (talk) 13:49, 8 March 2009 (UTC)

Off the top of my head, I'd suggest looking at Feigenbaum constants. Apparently, only a thousand digits have been worked out of δ. Pallida  Mors 14:32, 8 March 2009 (UTC)

Chaitin constant is hard for sure, even if not by accident... --pma (talk) 16:29, 8 March 2009 (UTC)

A hard to compute function is the zeta function. If you "compute" that, expect to get a fields medal.

What's so hard about computing the (presumably Riemann) zeta function? People do it all the time. Algebraist 12:07, 9 March 2009 (UTC)

The inverse to the Riemann zeta function is difficult to compute (which is, presumably, what the person without a signature meant), but I don't think that's really relevant to this discussion. --Tango (talk) 13:39, 9 March 2009 (UTC)

If you can compute n digits of a monotonic function in O(f(n)), shouldn't you be able to compute n digits of the inverse in O(n(f(n))? (Assuming you have finite bounds on the value of the inverse.) Eric. 131.215.158.184 (talk) 18:15, 9 March 2009 (UTC)

Of course, you can define constants that are currently impossible to compute at all, such as C=0 if Goldbach's conjecture is true, otherwise C=the least even number that is not the sum of two primes; or the lowest counterexample to the Collatz conjecture; or the position of the first sequence of a googol consecutive zeros in the decimal expansion of pi. AndrewWTaylor (talk) 12:36, 9 March 2009 (UTC)

Chaitin's constant beats them by being uncomputable period, not just currently uncomputable. It's less contrived, too. Algebraist 12:45, 9 March 2009 (UTC)

Percentage-wise I'd imagine it's fairly hard to compute the digits of Graham's number. You'd probably run out of atoms to write the digits down with before you got any significant percentage of the overall number. Readro (talk) 13:18, 9 March 2009 (UTC)

Brun's constant is only known to a dozen or so decimal places. Here, "known" actually means more like "made an educated guess based on a massive computation and unproved conjectures", IIUIC there is not even an explicit upper bound on the constant which would be rigorously proved. There's also a discussion in [3]. — Emil J. 13:59, 9 March 2009 (UTC)

Good example. Eric. 131.215.158.184 (talk) 18:15, 9 March 2009 (UTC)

How about the area of the Mandelbrot set? Fredrik Johansson 15:15, 9 March 2009 (UTC)

conditions for a map to be a homomorphism

if a map between two groups preserves group structure on some non-trivial normal subgroup of the domain, is this enough to say that it is a homomorphism. If the answer is no are there any further conditions under which this is the case.

No, it's not sufficient. Consider the map $f:\mathbb {Z} _{2}\times \mathbb {Z} _{2}\rightarrow \mathbb {Z} _{4}$ that maps (0,0) to 0 and (1,0) to 2. That's a homomorphism on a normal subgroup of the domain. Then map (0,1) to 1 and (1,1) to 3. You how have f(2(0,1))=f(0,0)=0, but 2f(0,1)=2*1=2, so not a homomorphism everywhere. I don't know if there is a similar result that works, with just a few extra conditions, I doubt it, though. --Tango (talk) 15:23, 8 March 2009 (UTC)

what about if the map is surjective and the normal subgroup maps once over injectively onto the image

The counter-example I gave is bijective. --Tango (talk) 16:37, 8 March 2009 (UTC)

Note that if any automorphism of a group, G, fixes a subgroup H of G, H is said to be characteristic in G and we write H char G. If every inner automorphism of G fixes H, H is normal in G. In particular, every characteristic subgroup of a group is normal. Examples of characteristic subgroups are: the center of a group, the whole group, and the trivial group. A question you might like to investigate is whether any automorphism of a normal subgroup of a group G, extends to an automorphism over the whole group. If not, what additional conditions can you impose? --PST 11:44, 9 March 2009 (UTC)

Solving Exponentials... with a twist

Is it possible to solve exponential functions (mathematically) when the variable also appears in the base. For example: x^2x=3 ?

Thanks, 160.39.193.25 (talk) 18:37, 8 March 2009 (UTC)

I think the answer will end up being in terms of the Lambert W function. --Tango (talk) 18:55, 8 March 2009 (UTC)

In Lambert_W_function#Examples, Example 2 shows that the solution to $x^{x}=z$ is $x={\frac {\ln z}{W(\ln z)}}$. So in the case of $x^{2x}=3$, $x={\frac {\ln {\sqrt {3}}}{W(\ln {\sqrt {3}})}}$. Incidentally, does anyone know if there's an incantation to get Maxima to use that? By default, solve(x^(2*x) = 3, x); gives $x^{x}={\sqrt {3}}$ and doesn't solve further. --Delirium (talk) 02:48, 9 March 2009 (UTC)

March 9

sum of n terms of a series

Hello all. If $a_{1},a_{2}\cdots a_{n}$ are in AP and $a_{i}>0$ for each i, then what will be the sum of the first n terms of the series: ${\frac {1}{{\sqrt {a_{1}}}+{\sqrt {a_{2}}}}}+{\frac {1}{{\sqrt {a_{2}}}+{\sqrt {a_{3}}}}}\cdots$? Essentially I am looking for a method to find expressions for similar problems also.--Shahab (talk) 06:33, 9 March 2009 (UTC)

In your particular case, if you first rationalize, you get the same denominators, and you are left with alternate sums of square roots of the $a_{k}$ with nice cancellations. But in general, a simple closed formula like this one, is not available for partial sums of a series; look at Euler summation formula. --pma (talk) 08:28, 9 March 2009 (UTC)

Thanks. Your link is what I had been searching for since a long time.--Shahab (talk) 08:55, 9 March 2009 (UTC)

Fractional derivatives of exp(-1/x^2)

Just expanding on the question asked above, are all the fractional derivatives of $e^{-1/x^{2}}$ also equal to zero? Zunaid 09:25, 9 March 2009 (UTC)

Of course you mean: equal to zero at x=0. Yes, they are. They define the fractional integrals of f, or $\textstyle D^{-\alpha }f=J^{\alpha }f$, for $\scriptstyle \alpha >0$, as the convolution

$D^{-\alpha }f(x)=\gamma _{\alpha }*(f\chi _{\mathbb {R} _{+}})\,(x)=\int _{0}^{x}\gamma _{\alpha }(x-t)f(t)dt$,

where $\textstyle \gamma _{\alpha }(x):={\frac {x_{+}^{\alpha -1}}{\Gamma (\alpha )}}$ (check your link). Then in general, $\textstyle D^{s}f$ is defined via the identity $\textstyle D^{s}f=D^{m}D^{-\alpha }f$, with any positive integer $\textstyle m>s$ and $\textstyle \alpha :=m-s>0$. For the function $\textstyle f(x)=e^{-1/x^{2}}$ (and vanishing at 0), all $\textstyle D^{s}f$ vanish at x=0. Indeed, in this case $f\chi _{\mathbb {R} _{+}}$ is a smooth function, and $D^{m}(f\chi _{\mathbb {R} _{+}})=(D^{m}f)\chi _{\mathbb {R} _{+}}$, so we can switch derivation and convolution in the above definition of $\textstyle D^{s}f$:

$D^{s}f=D^{m}D^{-\alpha }f=D^{m}\left(\gamma _{\alpha }*(f\chi _{\mathbb {R} _{+}})\right)=\gamma _{\alpha }*D^{m}(f\chi _{\mathbb {R} _{+}})=D^{-\alpha }D^{m}f$,

showing that all fractional derivatives $\textstyle D^{s}f(x)$ vanish at 0, as all the $\textstyle D^{m}f(x)$ do. PS: I took the liberty of changing the header. --pma (talk) 16:35, 9 March 2009 (UTC)

Communicating meaning with distant space aliens - no pictures allowed

[This discussion began at the Language Reference Desk. -- Wavelength (talk) 14:56, 9 March 2009 (UTC)]

Imagine that the two-way communication of signals between us and some space-aliens orbiting a distant star has been established. They are blind and immobile and cannot use pictures or diagrams of any kind. There is no pre-established code or alphabet. While I can imagine that eventually the meaning of mathematical or logical symbols might eventually be established (for example tranmitting many messages such as "..+..=...." would give meaning to + and =), would it be possible to eventually build up enough meaning from a zero base so that in time they would understand what was meant by the message "Last thursday my Uncle Bill went to the supermarket"? Helen Keller springs to mind. 89.240.206.60 (talk) 02:01, 8 March 2009 (UTC)

I don't see how it's possible to go from 2+2=4 to any non-math concept. Remember, it was impossible to decipher hieroglyphics without help from the Rosetta Stone, even though they were written by human beings, and this would be n times worse (n >> 1). Clarityfiend (talk) 05:22, 8 March 2009 (UTC)

Earth has blind, immobile animals called barnacles, and some humans have done research on how to talk with animals (http://www.howtotalkwithanimals.com/), but I have never heard of anyone attempting to communicate with a terrestrial barnacle. Instead of contemplating communication with alien barnacle-like creatures, why not ponder how we humans can communicate better with each other? -- Wavelength (talk) 06:45, 8 March 2009 (UTC)

LINCOS was a whole elaborate language (developed at length in a book) based more or less around that premise (though I think there were some abstract mathematical images included)... AnonMoos (talk) 07:00, 8 March 2009 (UTC)

H. Beam Piper's much-reprinted story Omnilinual has terrans cracking the Martian language by finding a periodic table. Unfortunately, the idea in the story simply doesn't work: the English names for common elements only make sense in the context of the history of science, not modern science (eg oxygen = 'acid-maker' and hydrogen = 'water-maker]; these are Graeco-Latin rather than English, but German for example translates the roots and still perpetuates the errors), so why assume that the Martian names would be meaningful? --ColinFine (talk) 18:51, 8 March 2009 (UTC)

For what it's worth, I coincidentally ran into the following article today ---> Pioneer plaque ... in which NASA scientists are, in fact, trying to communicate with distant space aliens ... albeit with the use of pictures. (Joseph A. Spadaro (talk) 22:20, 8 March 2009 (UTC))

The essential bottleneck to get through may be that of naming geometric shapes, such as a triangle. A triangle could then be used to build up other shapes. The triangle could be named after being identified by its mathematical properties. If however they have no sense of the spatial, then you are stuffed. 89.243.72.122 (talk) 23:56, 8 March 2009 (UTC)

How can an organism distinguish between a random collection of perceptible stimuli and a purposeful collection of perceptible stimuli produced by intelligent design? How can it distinguish between a message and a non‑message?

-- Wavelength (talk) 02:09, 9 March 2009 (UTC)

Humans or even sheepdogs or bees seem to have no problems with doing that. And if we humans recieved a signal from a distant star in the form of the Fibonacci series or any other simple mathematical series, then that would indicate that the sender was an intelligent being. 89.242.94.128 (talk) 11:37, 9 March 2009 (UTC)

The series should not be too simple, as then we could not be sure it was not generated by some nonsentient physical process. The Fibonacci sequence in particular is a very bad example, as it is known to appear in nature without any involvement of intelligence, see Fibonacci number#Fibonacci numbers in nature. — Emil J. 13:42, 9 March 2009 (UTC)

This fellow's research into a generalization of information theory that assumes no prior common language might be of interest, for a formal take on a specific variation of the question, which he calls "Universal Semantic Communication". The general strategy is to frame it as goal-oriented communication, which allows us to conclude that we've successfully communicated something when we can achieve some goal as a result of the communication faster than we would've been able to do without it. --Delirium (talk) 02:55, 9 March 2009 (UTC)

It might be worth posting this question on the mathematics desk. I am sure that they would have ways of encoding mathematics that they would think recognisable (and going from simple operations to advanced formula). They might even have some insights in how to jump out of Mathematics. -- Q Chris (talk) 13:49, 9 March 2009 (UTC)

[The copied text ends here. -- Wavelength (talk) 22:46, 9 March 2009 (UTC)]

Commuting family of matrices

Hi, I am working on a representation theory problem but I need some linear algebra to do it. I have talked to the professor and he has given me much of the problem but I just still do not understand simply because I do not think we ever covered what he is saying in linear algebra. Here's a theorem I do have

If ${\mathcal {F}}\subseteq M_{n}$ is a commuting family, then there is a vector $x\in \mathbb {C} ^{n}$ that is an eigenvector of every $A\in {\mathcal {F}}$.

My professor is talking about something that is more powerful. He says, I think, if a commuting family leaves any subspace invariant, then that commuting family has a common eigenvector in that invariant subspace. Is this true?

Thanks for any help StatisticsMan (talk) 15:41, 9 March 2009 (UTC)

They are the same thing. If they have a common invariant subspace, then they act as operators on that subspace. Just think of the elements of F as functions rather than lists of numbers and it should be clear; you just restrict the functions to that invariant subspace. JackSchmidt (talk) 15:59, 9 March 2009 (UTC)

Okay, thanks. StatisticsMan (talk) 19:05, 9 March 2009 (UTC)

What do I have to be to get my profile on Wikipedia?

My name is Vo Duc Dien.

My question is what a person has to be to get his / her profile featured on your website?

Vo Duc Dien —Preceding unsigned comment added by 71.80.236.201 (talk) 17:44, 9 March 2009 (UTC)

If by "profile" you mean an article about you, then the answer is at Wikipedia:Notability (people). There are a lot of specialized subguidelines of that, like Wikipedia:Notability (academics) which would apply to mathematicians (I assume that would be the relation to mathematics here). If you have more questions about Wikipedia policy, Wikipedia:Village pump (policy) is a more suitable place. —JAO • T • C 17:50, 9 March 2009 (UTC)

Question on Perfect Squares

For which n does the following hold:

$1^{2}\pm 2^{2}\pm 3^{2}\pm \ldots \pm n^{2}=0$?

where it is possible to choose either + or - in any of the cases.

There are a lot of solutions for number 1 to 100, but how do I generalise this to the n case? I appreciate the contribution made already, albeit I typed the question into the incorrect section on Wikipedia.

--84.70.242.151 (talk) 17:48, 9 March 2009 (UTC)

I hope you don't mind I took the liberty of formatting the equation so it's clearer what the question is. I don't have much of an answer though. —JAO • T • C 17:55, 9 March 2009 (UTC)

Let's say that it is necessary that either n=0 mod 4 or n=-1 mod 4, otherwise the sum is odd. As you mentioned, here PrimeHunter shows that from n=7 to 100 it is also sufficient, while there are no solutions for 0<n<7. In general, observe that an algebraic sum of 8 consecutive squares with signs exactly (+ - - + - + + -) always vanishes. Therefore you can build solutions for any n=8m, n=7+8m, n=11+8m, n=12+8m, (with $\scriptstyle m\geq 0$) just taking the PrimeHunter's solutions resp for n=0, n=7, n=11, n=12, and attaching to them a null algebraic sum of the subsequent 8m consecutive squares, with signs choosen with the periodicity (+ - - + - + + -). This does all n>6 with n=0 mod 4 or n=-1 mod 4. A more challenging question would be, find an asymptotics for the number of solutions (it is exponential, at least $c2^{n/8}$, for you can replace any group of 8 signs with the opposite, if you wish) --pma (talk) 18:56, 9 March 2009 (UTC)

Ok, that does make sense. Thanks for the input. The 8 consecutive squares result is a new one to me...can use that again sometime no doubt! I wouln't have thought to use congruence here. I would have thought a series approch would have been the way to go about it. Thanks though --84.70.242.151 (talk) 23:47, 9 March 2009 (UTC)

An alternative to human mathematical concepts

Could a system of advanced mathematics develop without abstract concepts like vectors, imaginary numbers, or matricies, or are they somehow necessary? Could an alien species understand just as much science as humans do without, or with other, abstract ideas? As an example, humans use differential equations and complex numbers to solve simple harmonic motion. Is it possible to express simple harmonic motion using other ways that aren't used simply because they were discovered late in history? --99.237.96.33 (talk) 20:31, 9 March 2009 (UTC)

You just have to look at Newton's Principia to see for instance that a geometric way of expressing things could be used and an alien could possibly do that without a single line of equations. My own feeling is that people think about things in very different ways but still come to the same conclusions normally, so I've no difficulty thinking alien math might be quite unrecognizable and need quite a bit of reinterpretation - math is as much or more about how to go about it as the actual results. Dmcq (talk) 21:16, 9 March 2009 (UTC)

The basic concepts will probably need to be the same, but the way they are thought of could be completely different. For example, they will probably have complex numbers (they occur quite naturally when solving technological problems), but even with Earth maths we have two completely different ways of thinking about them (i=sqrt(-1) and i="90deg anticlockwise rotation") - an ETI could do things very differently. The answers, however, ought to be the same - maths is pretty absolute (we don't know if they would assume the continuum hypothesis or not, but we can be pretty sure they will have something close to Euclid's postulates, and if you have the same basic axioms, you'll draw the same basic conclusions, regardless of how you formulate them). --Tango (talk) 23:09, 9 March 2009 (UTC)

Topological Vector Spaces

Hi, I am working on a several part problem and I am stuck on this part. Specifically, if you have the book, it's Problem 10.30 part c from Royden. We have a base B at $\theta$ for a translation invariant topology, so the set of all basis elements for the whole set are of the form x + B. In this specific part, we are trying to show

   If multiplication by scalars is continuous (at $\langle 0,\theta \rangle$) from $\mathbb {R} \times X$ to X,
   then If $U\in B$ and $x\in X$, there is an $\alpha \in \mathbb {R}$ such that $x\in \alpha U$.

I, with friends, have been working on this part for a while and we do not know what to do. I am assuming it's pretty simple but we just can't see it. Hopefully I have provided enough information for you to be able to understand. Can any one help me out with a hint? Thanks StatisticsMan (talk) 23:09, 9 March 2009 (UTC)