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March 1

LaTeX question

Hello, I type lots of things with LaTeX, mostly homework and notes. I recently made a separate file with many of my own commands/shortcuts. But, since I have a separate folder for each class, I have to put my shortcut file in every folder. Then, if I want to make a change, I have to change one and then copy and paste to a few other folders. So, is it possible to do this a better way? Is it possible, for example, to put my shortcut file in the same directory as things such as the amsmath file I can invoke with \usepackage, so that I can just do \usepackage{MyShortcuts} or something like that?

Thanks, any help would be much appreciated. Also, of course this is not a math question but I thought this would be the best place to put this as mathematicians/scientists are the most likely to use LaTeX and be proficient. StatisticsMan (talk) 01:25, 1 March 2009 (UTC)

Where tex looks for files varies based on the actual software you are using. On unix like systems (linux, Mac OS X) you can put the files in a directory "texmf" in your home directory, but you need to run a command like "texhash" or "mktexlsr" to let tex know about them. For miktex, there is something similar, but you would have to read the documentation. Just putting your file with the others won't be sufficient, you have to have it "rehash" or look again at what files it has. I think you have to do this whenever you install a new package in miktex, say from CTAN. JackSchmidt (talk) 02:25, 1 March 2009 (UTC)

I found using the texmf directory stucture rather confusing, so I have a bash script which takes a given file, copies it to a fixed temporary directory, copies my personal style file to that temporary directory, compiles the file, and copies the result back to the starting location. Although this isn't a very nice solution (and causes a bit of a mess when you include images in your document) it's less of a learning curve and easier to do then the "right" way. As an added bonus I don't have a ton of ".log" and ".aux" files scattered everywhere -- those stay in my temporary directory. Eric. 131.215.158.184 (talk) 08:45, 1 March 2009 (UTC)

Instead of maintaining separate copies of your macro file, you can keep one copy and use symbolic links in the other locations. This works great on unix-like systems (including Mac OS X). Not sure about Windows. Staecker (talk) 13:35, 1 March 2009 (UTC)

Use the TEXINPUTS environment variable to specify a path to search. In modern implementations of latex it understands a trailing "//" to mean recursive search of that directory. — Charles Stewart (talk) 14:04, 1 March 2009 (UTC)

Equality of some differentiated inner products

I'm trying to work through the details of lemma 16.1 in this book, and have gotten stuck on the following:

Finally, taking the inner product in (16.15) with $t^{2}A^{2}u(t)$, we obtain

${\frac {1}{2}}{\frac {d}{dt}}(t^{2}\|Au(t)\|^{2})+t^{2}(A^{2}u(t),Au(t))=t\|Au(t)\|^{2},$ (...)

Here (16.15) is a generalized version of the heat equation, ${\dot {u}}(t)+Au(t)=0$. A is a linear transformation satisfying $(Aw,v)=(w,Av)$ and $(Av,v)\geq 0$ for any v, w. (The inner products are done over the space domain; the dependence of u on a space variable is implicit.)

Following the hint,

$(t^{2}A^{2}u(t),{\dot {u}}(t))+(t^{2}A^{2}u(t),Au(t))=0$

$t^{2}(A^{2}u(t),{\dot {u}}(t))+t^{2}(A^{2}u(t),Au(t))=0$

and so to reach the desired result it must be that

$t^{2}(A^{2}u(t),{\dot {u}}(t))={\frac {1}{2}}{\frac {d}{dt}}(t^{2}\|Au(t)\|^{2})-t\|Au(t)\|^{2}.$

Why is this equation true / how do I derive this? (Or am I making an error somewhere?) Fredrik Johansson 22:47, 1 March 2009 (UTC)

Just compute

${\frac {d}{dt}}(t^{2}\|Au(t)\|^{2})=2t\|Au(t)\|^{2}+t^{2}{\frac {d}{dt}}\|Au(t)\|^{2}$

$=2t\|Au(t)\|^{2}+t^{2}{\frac {d}{dt}}(Au(t),Au(t))=2t\|Au(t)\|^{2}+2t^{2}(Au(t),A{\dot {u}}(t))=2t\|Au(t)\|^{2}+2t^{2}(A^{2}u(t),{\dot {u}}(t)).$--pma (talk) 00:03, 2 March 2009 (UTC)

Thank you! I had the idea, but got stuck somewhere... Fredrik Johansson 00:19, 2 March 2009 (UTC)

Infinity Norm of A Fourier Series

Hi, I was wondering how to compute the infinity norm of a finite Fourier series with the function its approximating.

My starting point is: $||f(x)-F_{20}(x)||=max_{x\in [-2\pi ,2\pi ]}|f(x)-F_{20}(x)|,$

From looking at the graph of the two functions together, it is pretty obvious the maximum lies at the ends of the interval but i don't know how to prove it.

The function i am approximating is $f(x)=x^{2}$, and my approximation is $F_{N}(x)=\Sigma _{k=1}^{N}{\frac {4}{k}}(-1)^{k}cos(kx)$

The Fourier series is 2$\pi$ periodic, so the graphs disperse at $\pi$ and $-\pi$, the Fourier series continuing periodicly and the $x^{2}$ graph increasing quickly. So in the [$-\pi ,\pi$] interval the difference is very small, but outside of this it grows so i'm guessing the greatest point is on the edge of the [$-2\pi ,2\pi$] interval. (The edge of the interval also happens to be very close to a minimum of the Fourier series, if not the minimum.)

Any help is greatly appreciated. Thanks —Preceding unsigned comment added by 82.32.212.250 (talk) 23:35, 1 March 2009 (UTC)

Anyone? (Sorry about not signing)

144.173.6.74 (talk) 15:04, 2 March 2009 (UTC)

So you are intersted in $\scriptstyle d_{N}:=\max _{x\in [-2\pi ,2\pi ]}|f(x)-F_{N}(x)|$, that is, the uniform distance on the interval [-2π,2π], where $\scriptstyle f(x)=x^{2}$, is it so? By the uniform convergence you have immediately that $\scriptstyle d_{N}=(2\pi )^{2}+o(1)$, and that $d_{N}$ is attained in a point $\scriptstyle x_{N}=2\pi +o(1)$, as $\scriptstyle N\to \infty$. The stronger statement that $\scriptstyle d_{N}$ is eventually attained at $\scriptstyle 2\pi$, shall follow proving for instance that $\scriptstyle F_{N}^{\prime }$ converges to $\scriptstyle f'(x)=2x$ uniformly in a nbd of $0$ (therefore, by periodicity, to $\scriptstyle 2(x-2\pi )$ uniformly in a nbd of $\scriptstyle 2\pi$). Indeed, this implies that there exists a left nbd $\scriptstyle J$ of $\scriptstyle 2\pi$ such that it eventually holds true that $\scriptstyle x^{2}-F_{N}(x)$ has positive derivative for $\scriptstyle x\in J$, so it is increasing there. This implies that $\scriptstyle x_{N}=2\pi$ as soon as $\scriptstyle x_{N}\in J$: and we do know that the latter eventually holds true. Is it clear? Then, should you need it, you can make everything quantitative and compute an $\scriptstyle N_{0}$ such that $\scriptstyle d_{N}$ is attained at $2\pi$ for all $\scriptstyle N\geq N_{0}$. Warning: your $F_{N}$ lacks the $\scriptstyle {\frac {a_{0}}{2}}$ term, and $\scriptstyle {\frac {1}{k}}$ should be squared.

Just for completeness: showing that $\scriptstyle F_{N}^{\prime }(x)$ converges uniformly for x bounded away from odd multiples of $\pi$. You can write

$\scriptstyle F_{N}^{\prime }(x)=-\sum _{k=1}^{N}{\frac {4}{k}}(-1)^{k}\sin(kx)=-4\Im \left(\sum _{k=1}^{N}{\frac {(-1)^{k}}{k}}e^{ikx}\right)$.

Now, putting $\scriptstyle z:=-e^{ix}$, the stated convergence is just the convergence of the logarithm series $\scriptstyle \sum _{k=1}^{\infty }{\frac {z^{k}}{k}}$, which is uniform on compacta of the unit circle non containing 1. --pma (talk) 19:10, 2 March 2009 (UTC)

March 2

A good book in Tensor Calculus?

What do you recommend as a reader-friendly "Tensor Calculus" Book? I've read some pages of this [1] book. What do you think about it? Re444 (talk) 11:05, 2 March 2009 (UTC)

"Tensor Analysis on Manifolds" by Bishop and Goldberg; it's friendly, inexpensive, light on prereqs, rigorus, and covers a decent ammount of topics for an introduction(in my opinion). If you look it up on Amazon the reviews can tell you more. I've also had people tell me that Spivaks Calculus on manifolds is good, I have a copy but didn't much like it myself(though you might :) Phoenix1177 (talk) 11:28, 2 March 2009 (UTC)

How about googling "Wikibook Tensor Calculus". This will give you a book on tensor calculus, made by the people here. :) --PST 13:41, 2 March 2009 (UTC)

Well, Wikipedia is a great and wide source of quick maths references, and is maybe the future; still I would prefer a real book by real authors. Moreover, as you know wery well, nothing is comparable to sniffing the glue of a maths book. --pma (talk) 22:37, 2 March 2009 (UTC) ;)

Spivak's Calculus on Manifolds is rather terse. Vol. 1 of his "Comprehensive Introduction to Differential Geometry" may be easier to deal with. It covers more subjects, but is more conversational. 76.195.10.34 (talk) 06:17, 3 March 2009 (UTC)

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[2] 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)

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)

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)

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)

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)

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 ${\frac {d}{dx}}$

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

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)

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)