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June 24

Divergent Series

Studying some analysis, I came across these two questions in my book. Give an example of a series that does not diverge to infinity but whose partial sums are increasing. One easy answer is just $\sum _{n=0}^{\infty }\left({\frac {1}{2}}\right)^{n}$ which converges to 1.5. The second question, the converse of the first one, asks for a series which diverges to positive infinity but whose partial sums do not form an increasing sequence. What would such a series look like?76.79.202.34 (talk) 00:44, 24 June 2008 (UTC)

The first series you mention actually converges to 2 (see geometric series). You should be able to come up with the example you're looking for by using positive terms and negative terms (so that the partial sums are not an increasing sequence), but the positive terms need to be bigger than the negative ones. Can you come up with one like that? 134.173.70.1 (talk) 00:56, 24 June 2008 (UTC)

(E/C) Your 1/2 sequence converges to 2, not 1.5, and is an example of Zeno's Paradox. -mattbuck (Talk) 00:58, 24 June 2008 (UTC)

Of course, I meant that the sum is two. Anyway, for the series in question, technically, my desired series can be something like 10-1+10-1+10-1+10... or some such series. The partial sums are both increasing and decreasing but the series does diverge. Is this good?76.79.202.34 (talk) 01:24, 24 June 2008 (UTC)

Yep, that does it. 134.173.70.1 (talk) 01:36, 24 June 2008 (UTC)

Well, the partial sums are neither increasing nor decreasing. You could also choose something much simpler, such as (7,1,2,3,4,5,6,7,8…) which clearly diverges to infinity, but is not increasing because from the first to the second partial sum it decreases. Now if you want something whose tail is not increasing nor decreasing, you need something slightly more interesting such as already given. GromXXVII (talk) 10:35, 24 June 2008 (UTC)

Huh? 7 + 1 > 7 last time I checked... If all the terms are positive, the partial sums will be strictly increasing. The OP's sequence is a good one. --Tango (talk) 15:01, 24 June 2008 (UTC)

Or did you mean those to be the partial sums? So the series is 7-6+1+1+1+1+1+...? --Tango (talk) 15:02, 24 June 2008 (UTC)

Interestingly enough, the harmonic series forms a decreasing sequence but diverges to infinity. shoy 14:47, 24 June 2008 (UTC)

In the harmonic series the terms are a decreasing sequence, but the sequence of partial sums is increasing. However, you can still have a summation where no tail of the sequence of partial sums is increasing (as Grom mentioned) and also make the terms of the series decay to 0. One way is to interleave a positive harmonic series and a negative geometric series: 1 - 1 + 1/2 - 1/2 + 1/3 - 1/4 + 1/4 - 1/8 + 1/5 - 1/16 + ... For bonus points, can you come up with one with the added restriction that the absolute values converge monotonically to 0? (Hint: make sure it doesn't satisfy the alternating series test!) For some reason this brings to mind the Riemann rearrangement theorem, which is a cool result. 69.111.164.70 (talk) 17:47, 24 June 2008 (UTC)

As long as we are talking about the harmonic series, I recall reading something about eliminating integers that contain the digit 9 will keep the resulting series bounded. Is it true that if we eliminate all the integers that have a 9 in their decimal expansion will keep the sum of the reciprocals bounded? Is it true for other digits? What if we eliminate the digit one instead? Can this be proven? How will one find the bound or the limit of the series?76.79.202.34 (talk) 21:35, 24 June 2008 (UTC)

It's true, it doesn't matter which digit you use (or which number base), it can be proven, the limit is about 23. See Kempner series. Algebraist 22:06, 24 June 2008 (UTC)

That’s curious, any idea how many integers have a 9 in their decimal expansion relative to the total number? I’d want to sum 1/10 + 19/100 +280/1000 +... whose terms seem to be converging to 1 GromXXVII (talk) 23:17, 24 June 2008 (UTC)

Sure, that is pretty easy. It is just a basic counting problem. For example, let us define a function from the set of natural numbers to the set of natural numbers. This function, for a given n, returns the number of natural numbers between 1 and n (inclusive) that have the digit 9 in their decimal expansion. So, for example, f(1)=0=f(2)=...=f(8) but then f(9)=1=f(10)=f(11)=...=f(18). And then f(19)=2. The answer to your question would be f(n) compared with n. It is pretty easy, I think, to see a pattern as numbers get large. For example, f(100)=19 because you have ten numbers which have a 9 in their ones place (9,19,29,39,49,59,69,79,89,99) and then you have 10 more numbers with a 9 in their tens place (90,91,92,93,94,95,96,97,98,99) so the total is 20 but 99 has been counted twice so f(100)=19. Now we know that each group of a hundred has 19 such numbers. So f(100)=19, f(200)=38, f(300)=57, f(400)=76, f(500)=95, f(600)=114, f(700)=133, f(800)=152, f(899)=171, and for the next 100 numbers, they all will have the digit 9 so f(1000)=f(899)+100=271. Now we know that each batch of a thousand numbers will have 271 numbers with a 9 digit in them. Therefore f(2000)=542, f(3000)=813, and so on. For an added bonus, does this function have a fixed point? In other words, does there exist an n (might be very large) such that f(n)=n? A similar problem was on one of the Google employment exams.A Real Kaiser (talk) 23:41, 24 June 2008 (UTC)

I think to really answer Grom's question, you need to work out the limit of f(n)/n as n tends to infinity. My guess would be that it tends to 1, since if you consider longer and longer numbers there are more numbers to consider and a greater proportion of them contain a 9. That would mean that a randomly chosen integer will almost surely have a 9 in it. --Tango (talk) 00:46, 25 June 2008 (UTC)

Wow, you seem to be correct. Originally, after reading your post, I disagreed with you. I thought that the ratio f(n)/n might be quite small (less than a half) as n gets large. So, I wrote a mini MATLAB program and ran it. And guess what, your conjecture seems to be true. First we observe that for a natural number k, our f(n) satisfies the recursive relationship,
$f(10^{1})=1,\,f(10^{k+1})=9(f(10^{k}))+10^{k}$.
This is exactly what we have above. So this way, we can quickly find out f(n) for large n. Let us also define g(n)=f(n)/n and now we have all the tools for a nice 10 liner code which can be run in MATLAB. MATLAB can only go up to k=308 but by the time $n=10^{308}$, our g(n) is indeed 0.9999999999999999. Now, after seeing this result, it is no wonder that the modified harmonic series converges. I mean you are eliminating almost all (pun intended) of the numbers. This would also mean that almost all the numbers have a digit 9 in their decimal expansion. But the same thing can be done for any digit like 1. So, after the same exact argument for each of the ten digits, can we say that almost all of the numbers contain all ten digits? So, from the harmonic series, if we eliminate all the terms with any one specific digit, it will converge. Awesome! We also know that the harmonic series with just the prime denominators also diverges. So, doing a proof by contradiction, we have that almost all of the prime numbers contain all 10 digits, because if they didn't, then we would have a harmonic series with denominators missing at least one digit which must converge which is a contradiction. Totally awesome!76.79.202.34 (talk) 01:53, 25 June 2008 (UTC)

I'm not sure your recursive relation is correct, it looks to me like you're counting some numbers twice. I suspect the 10^k at the end should be 10^k-f(10^k), since you don't want to count numbers with a 9 in the first place and somewhere else twice. Or, equivalently, change the 9 to an 8, since you can add any digit that isn't a 0 or a 9 to the beginning of a number with a 9 to get another number with a 9, or you can add a 9 to the beginning of any number to get a number with a 9. The rest of your comment may well be correct, but I think it would need a lot of work to make it rigorous. --Tango (talk) 15:11, 25 June 2008 (UTC)

This relationship is correct because it gives us the same exact numbers as above. And the numbers above came from counting carefully so that something like 99 or 991 will not be counted twice. I mean you can easily verify this yourself. No need to take my word for it. As for the rest of the comment, of course, this is not rigorous in any manner and will take some effort to be written up nicely but it does seem a very cool idea.A Real Kaiser (talk) 01:11, 26 June 2008 (UTC)

You're right. After staring at it for a few minutes, I realise I was thinking about how many k-digit numbers have 9's in, rather than have many k-or-less-digit numbers do... Ignore me! --Tango (talk) 14:07, 26 June 2008 (UTC)

More on Diverging Series

Working with harmonic series, the next question I have is that
"Find a simple function of n in terms of ln(n), call it w(n) so that
$\lim _{n\rightarrow \infty }\left(1+{\frac {1}{3}}+{\frac {1}{5}}+\cdots +{\frac {1}{2n-1}}-w(n)\right)=0$."
Now, the book doesn't give any hints on how to go about this. In fact, I think that the author just wants the reader to guess and check. So, I took the easiest approach of graphing the partial sums of the odd terms and my w(n), and then see how their graphs coincide. I started with w(n)=ln(n) and ended with $w(n)=0.5(\log(n)+1)+0.5$. I have also calculated and graphed the partial sums and my w(n) up to n=50,000 and the graphs are virtually indistinguishable. In addition, for n=50,000, the maximum vertical difference (the L1 norm of the difference) between the two graphs is 0.0182 which is better than any other linear modification of log(n) I could dome up with. My questions is, with my given w(n), is the above limit really zero? How can one analytically come up with an expression so that the above limit is zero rather than just guessing and graphing to "eyeball" the two graphs getting closer and closer together?76.79.202.34 (talk) 21:49, 24 June 2008 (UTC)

Here's a hint: think of what happens when you take the series and subtract it from an appropriate partial sum of the harmonic series. Compare with the Euler-Mascheroni constant. Confusing Manifestation(Say hi!) 23:25, 24 June 2008 (UTC)

If I'm not mistaken, the Euler-Mascheroni constant is a part of the formula for w(n). Here's a big hint: find a way to make use of the harmonic series to approximate ln(n) for large n. By the way, the formula you came up with by trial and error is actually very close to correct answer. 76.224.119.217 (talk) 00:21, 25 June 2008 (UTC)

You are both, of course, right. The exact expression that I got is
$w(n)={\frac {\ln(n)+\gamma }{2}}+\ln(2)$.
This is the correct one, right? Because I derived it analytically and then when I graphed it to check, the graphs are exactly the same for n=50,000. Thanks everyone!76.79.202.34 (talk) 02:44, 25 June 2008 (UTC)

That's correct. 76.224.119.217 (talk) 04:00, 25 June 2008 (UTC)

June 25

Frey curves and Integer Equations

Could somewhat explain the connection between Frey curves and equations of integers in a somewhat less jargon filled manner than: Frey curve#Application of the epsilon conjecture to Fermat.27s Last Theorem.

I'd like to clarify where the discriminant comes in and how the Modularity theorem allows one to map such equations onto other forms.

Feel free to improve the noted article in the process. Dragons flight (talk) 03:37, 25 June 2008 (UTC)

The article is on the epsilon conjecture, and that section does a pretty good job of explaining the history of the epsilon conjecture. The point of that section of the article is just that Frey curves are so weird that (now read next section) they do not exist. JackSchmidt (talk) 16:36, 25 June 2008 (UTC)

A Frey curve is simply the defined form y^2 = x(x-a^n)(x+b^n), which of course exists. One then somehow applies the algebra of curves to arrive at the conclusion that a^n + b^n can not equal c^n for any rational c. I'd like to see the process of reaching that conclusion explained in a way that doesn't require graduate study in number theory. In particular, it would be nice to define terms like conductor and discriminant in this context. Dragons flight (talk) 17:31, 25 June 2008 (UTC)

I think you misunderstood. A Frey curve is only defined given a solution of a^p + b^p = c^p, for on odd prime p (not for general n, and not for general a,b). The properties discovered by Frey depend essentially on this. I think if you want to understand the proof of Fermat's last theorem, you probably do require some graduate study in number theory. If you only want to see how algebraic curves can be used to answer elementary questions in number theory, then I recommend Murty and Ram's Problems in Algebraic Number Theory, which has a few examples of how projective curves relate to integer equations. This would also be a reasonable beginning at graduate study of number theory. If you are just interested in the development of modern number theory, then Edwards's book on a historical introduction to modern number theory is quite good, but very detailed. You might like [1][2] and Discriminant. In the number field case, the conductor measures a departure from being the maximal order, and the discriminant measures which ideals can ramify (who you've taken n'th roots of). Generically, they just measure which primes are interesting for your particular domain, and the structure of those two sections says that for a Frey curve, no prime is interesting, which means it must be the nicest kind of curve, in particular, it has genus 0. JackSchmidt (talk) 03:59, 26 June 2008 (UTC)

Logarithm of a sum?

Can I somehow take the logarithm of a sum?

For example, suppose I have $e^{ax+b}=y$, and I know a, b and y, and want to figure out x. I do this with $ax+b=\ln y$, making $x={\frac {\ln y-b}{a}}$.

But suppose I have $e^{ax}+e^{bx}=y$, and I again know a, b and y, and want to figure out x. Can this be done? JIP | Talk 17:05, 25 June 2008 (UTC)

Do you have use of a computer? otherwise stuck?87.102.86.73 (talk) 19:42, 25 June 2008 (UTC)

I don't think that there is an exact form for the solution to that equation, at least not in terms of elementary functions. As far as I know, there is no simple formula for taking the logarithm of a sum. ln(a + b) = ln(a) + ln(1 + b/a) for all real numbers a and b whenever the equation is defined, but this doesn't solve your problem. 76.238.8.4 (talk) 19:48, 25 June 2008 (UTC)

Should have found one solution x I could certainly talk about all the other solutions that satisfy the same equation, should you wish to go down that route.. Still no solution though.87.102.86.73 (talk) 19:54, 25 June 2008 (UTC)

Quick answer - Not easily.
Hard answer - According to Maple (a powerful maths software package) your answer will be of the form ${\frac {\gamma }{b}}$, where ${\gamma }$ is a root of $y-e^{z}-e^{\frac {az}{b}}=0$, which doesn't help much. Basically it could be anything, possibly complex, and will be very difficult to find unless a=b, in which case the answer is fairly trivial. Rambo's Revenge (talk) 20:06, 25 June 2008 (UTC)

Incidentally, you can get an analytic form if you happen to also know $e^{ax}-e^{bx}$, which I suppose you might depending on the context. In general though, there is no simple analytic form except in special cases (e.g. a = 2*b). Dragons flight (talk) 20:22, 25 June 2008 (UTC)

Question: Would it not be possible to treat the answer as a solution to $f(e^{x})={e^{x}}^{a}+{e^{x}}^{b}-y=0$? Then find the derivative with respect to $e^{x}$ and use the Newton-Raphson iteration to find $e^{x}$ (not an analytic solution but the OP didn't specify)? Presumably the OP is allowed a calculator since logs have to be calculated. Zain Ebrahim (talk) 20:43, 25 June 2008 (UTC)

Yes, you could use Newton's method to approximately determine e^x, from which you could find x. Anyway, if a and b are integers, make the substitution u = e^x and solve the polynomial (or rational expression) u^a + u^b = y (however, this equation can't always be solved exactly, either). Then ln u = x. As others have said, though, there is no general analytic solution to the original equation. As an afterthought, if a is an integral multiple of b, or vise versa, then you could make a substitution like u = e^bx instead. 76.238.8.4 (talk) 20:54, 25 June 2008 (UTC)

Given the equation $e^{ax}+e^{bx}=c$ and wanting to solve for x given a,b,c. Expand the exponentials into power series: $2-c+\sum _{k=1}^{\infty }{\frac {a^{k}+b^{k}}{k!}}x^{k}=0.$ Truncate at some degree n and solve the polynomial equation $2-c+\sum _{k=1}^{n}{\frac {a^{k}+b^{k}}{k!}}x^{k}=0$ numerically using the Durand-Kerner method. Bo Jacoby (talk) 21:56, 27 June 2008 (UTC).x=y+p+i i=29

June 26

what is the solution to this

x^x = y^y

Where x < y

202.147.44.80 (talk) 03:09, 26 June 2008 (UTC)

There are an infinite number of solutions. If you have a graphing calculator, zoom in to see that the function x^x is initially decreasing near x = 0. As a hint for an exact solution, look at the number 0.5 (think about it). 76.238.8.4 (talk) 04:13, 26 June 2008 (UTC)

I guess the question should be read as 'What function f it is, such that for any x from properly chosen domain its output y=f(x) satisfies x^x=y^y?' Of course f not being Identity function. --CiaPan (talk) 10:03, 26 June 2008 (UTC)

For 0 < x < e^-1

$y=f(x)={\frac {x\ln(x)}{W(x\ln(x))}}$

where W is the Lambert W function. Note that for x in the given range, we have −e^-1 < x ln(x) < 0 and W(x ln(x)) has two values, one of which is ln(x). Take the other value of W(x ln(x)), for which |W(x ln(x))| < |ln(x)|, and you have y > x such that y^y = x^x. Gandalf61 (talk) 13:01, 26 June 2008 (UTC)

Fractional Coloring

Is the fractional chromatic number of a graph necessarily close to the regular chromatic number, or can they be extremely different? Black Carrot (talk) 04:49, 26 June 2008 (UTC)

They are within a logarithmic factor of each other according to [[3]]. 84.239.133.47 (talk) 11:23, 26 June 2008 (UTC)

Logic

I have a (subsequent, follow-up) question, but must first begin with a preliminary clarifying question ... it's about logic. I have heard / read / studied that ... you can never prove a negative statement ... you can only prove a positive statement. Thus, for example, one cannot prove "I did not visit Hawaii" ... but one can prove "I did visit Hawaii." That is, it is logically impossible to prove the negative statement. In fact, in criminal trials (in the USA), this is why the government can prove that the defendant is "guilty" (he did commit the crime) ... but the defense can never prove that the defendant is "innocent" (he did not commit the crime) ... at the very best, the defense can merely conclude that the defendant is "not guilty" (meaning, there is not enough evidence to prove that he did commit the crime). So, is my understanding correct? That one can logically prove that an event happened, but one cannot logically prove that an event did not happen? That it is logically impossible to prove a negative statement? Thanks. (Joseph A. Spadaro (talk) 10:55, 26 June 2008 (UTC))

Do you mean one can prove that, say, 'two is less than three', but can't prove that 'two is not greater than three'...? --CiaPan (talk) 11:10, 26 June 2008 (UTC)

No, I don't think that's what I mean. "Two is less than three" is a factual statement ... it is a statement of being or existence, no? It is either true or false, period. I am referring to proving events ... like I did not go to Hawaii ... I did not commit the murder ... I did not eat an apple. Etc. I think "events occuring" are different than "states of being" or factual statements like "2 is less than three" or "Bush is the current President". I think? (Joseph A. Spadaro (talk) 12:37, 26 June 2008 (UTC))

This isn't really a matter of logic, at least as the term is used in mathematics. As CiaPan points out, any statement is logically equivalent to a negative statement (can you prove that you didn't not visit Hawaii?). In any case, in the sense of 'prove' used in mathematics, nothing about the real world can be proved. It's more about the amount of evidence needed to establish certain claims about the real world. Thus if you had lots of pictures of yourself in Hawaii, eyewitness reports, etc, we would probably be forced to admit that you'd been to Hawaii. On the other hand, far more evidence would be required to prove that you'd never been there: a reliable account of all your movements from birth, a totally reliable list of everyone who's ever made it to Hawaii, including (somehow) people using false names or illegal channels, or somesuch thing. It is in this sense that the existence of (for example) Russell's teapot cannot be disproved. Algebraist 11:19, 26 June 2008 (UTC)

Forgot to say: I believe this sort of thing is part of what philosophers call 'logic', so you might want to try the humanities desk. And I don't think your second example is really applicable: one could establish beyond reasonable doubt that someone had not committed a crime, for example by showing that they were on another continent at the time. The point in criminal trials is that the burden of proof rests with the prosecution, so the defence doesn't have to prove innocence. Algebraist 11:23, 26 June 2008 (UTC)

Algebraist ... you are convoluting the (real) issue by adding in standards such as "beyond reasonable doubt". That is merely an artificial legal concept that man has injected into legal proceedings. Of course, we can prove "beyond reasonable doubt" that Johnny did not kill Jimmy. But can it be proven absolutely (that is, wthout the limiting qualification or reduced standard of "beyond reasonable doubt"?) I thought that I had heard / read / studied that the negative can never be proved ( or is it, "proven" )? (Joseph A. Spadaro (talk) 12:45, 26 June 2008 (UTC))

I agree the humanities desk might be better suited for this type of logic problem. However I agree mostly with your first point..

It is easy to proove A (ie A=I've been to paris) because all I need is a single example of proof (and a level of trust) and there you go.

However to proove (not A) (ie I've not been to paris) is more difficult because it requires proof by exhaustion - however it's not logically impossible - but it is much more difficult to proove.87.102.86.73 (talk) 12:25, 26 June 2008 (UTC)

So, "User 87" ... you are saying that to prove "I have never been to Paris" would be the logical equivalent of taking every other moment in my life (all 8 zillion of them) and "proving that I was in Spain / USA / Japan / etc. (anywhere that is non-Paris)" ...? And adding up all those (hypothetical) 8 zillion proofs logicallys amounts to the equivalent proof that "I have never been to Paris" ... yes? (Joseph A. Spadaro (talk) 12:51, 26 June 2008 (UTC))

Yes, that's what proof by exhaustion involves, hence it's name. In a purely mathematical sense it's possible, obviously it's easier to do this for the last week rather than you whole life..87.102.86.73 (talk) 12:59, 26 June 2008 (UTC)

I think you are getting to the heart of the issue here, which is dealing with an issue of quantification. While universal quantification is possible is mathematics where the domain of quantification can be appropriately axiomaticly controlled, it isn't feasible (and certainly isn't practical) in the physical world. The issue of not being able to prove a negative boils down to the equivalence $\neg \exists x.\phi (x)\leftrightarrow \forall x.\neg \phi (x)$. What you really can't do is prove universally quantified propositions unless you have some suitable controls on the universality. -- Leland McInnes (talk) 16:20, 26 June 2008 (UTC)

To the OP:I used 'beyond reasonable doubt', because that's the highest standard we can expect in 'proofs' of statements about the real world. If we admit unreasonable doubts (as we do in mathematics), then you can't prove either positive or negative statements: you say you've been to Hawaii, and I say neither you nor Hawaii exist, and I'm just a brain in a vat orbiting the third planet of Betelgeuse. Algebraist 13:31, 26 June 2008 (UTC)

Generally, when we want to "prove" (in the legal sense) a negative, we do so by proving a contradictory positive. For example, I can't direct prove "I did not shoot the deputy", but I can directly prove "I was busy shooting the sheriff at the time the deputy was shot", which means I can't have shot the deputy. In other words, I have an alibi. --Tango (talk) 14:14, 26 June 2008 (UTC)

Yes, of course you can prove a negative statement ! If I prove that this horse is dead, then I have also proved that this horse is not alive. If I prove that this cow has exactly four legs, then I have also proved that this cow does not have exactly three legs; does not have exactly five legs etc. etc. Even simpler, if I prove that any statement X is true, then I have also proved that X is not false - and vice versa. See this essay for a readable analysis by a professional philosopher. (Perhaps you were thinking of the negative proof fallacy ??). Gandalf61 (talk) 14:19, 26 June 2008 (UTC)

Proving a negative statement (X does not exist, I never went to Hawaii) needs the whole domain of potential instances to be proven empty (X exists nowhere, my name is not on any hotel register in Hawaii), which may be difficult but not impossible. It is easy if a contradictory proved positive statement preempts the whole domain (Y exists everywhere, Hawaiians would certainly remember my face if I had ever shown it there but they don't). Proving a positive statement (an X exists, I visited Hawaii) needs no consideration of domain (look at this X here!, look at this pic of me with a hula girl!). In legal trials the defence does not have to prove the defendent never committed a crime, only that the positive assertion by the prosecutor is insufficiently proven. Cuddlyable3 (talk) 14:53, 26 June 2008 (UTC)

From a mathematical perspective, negation is a relationship between two logical statements, not a property of a single logical statement. For example, the statement "This jar is open." is the negation of the statement "This jar is closed.", and vice versa. However, it's not the case that one of these statements is "the negative" and the other is "the positive". So from a mathematical perspective, there is no (meaningful) way to categorize statements into two categories, "positive statements" and "negative statements". The question, "Can I prove a negative statement?" doesn't have any mathematical meaning.

From a non-mathematical perspective, we have an intuitive notion that some statements (e.g., "I have been to Hawaii.") are vaguely "positive" and that other statements (e.g., "I have not been to Hawaii.") are vaguely negative. It frequently is easier to find convincing evidence supporting a true positive-feeling statement than supporting a true negative-feeling statement; but this general tendency has nothing to do with mathematics. Eric. 86.153.207.223 (talk) 01:45, 27 June 2008 (UTC)

I think this definition of "negative" means "prove that a specific event didn't occur" (prove you were never in Hawaii, didn't commit the murder, etc.). This definition doesn't apply to states, like "open" and "closed", where, as just stated, neither is any more negative than the other. StuRat (talk) 04:46, 3 July 2008 (UTC)

Straightness

What is the best way to measure the straightness of a line? For example, if I wanted to objectively compare 2 roads in terms of their relative straightness, how would I do this? 12.147.18.2 (talk) 18:48, 26 June 2008 (UTC)

You could take a sum of the a function of the deviation (at right angles) from a central straight line, measured at regular intervals, and then divide by the distance.. A suitable function would be the absolute value, |x| or x squared amongst others.

In general some sort of statistical analysis of the deviation from straightness would be required.87.102.86.73 (talk) 19:03, 26 June 2008 (UTC)

"Best" is a little vague. For a somewhat mathematically rigorous idea, try measuring the mathematical curvature of the centerline at various points, and proceed as the anon suggests (sum them). For a practical method, try comparing the mileage on a car driving the road versus the distance from start to finish as the crow flies. The shortest path between two points is a straight line, and if the roads are approximately straight this should work, though it obviously fails for two loops around a city. JackSchmidt (talk) 19:20, 26 June 2008 (UTC)

"Straight" is a vague term. 87.102's definition is "proximity to a straight line", which isn't bad, but leads to some counterintuitive results. For example: say Trip A involves driving 10 miles north, then 10 miles east. Trip B involves driving 1 mile north, then 1 mile east, then repeating that 10 times. By this definition, Trip B would be much straighter than Trip A. If I were driving it, I'd certainly consider Trip A straighter.

Instead, I'd suggest a measure of the ratio between a straight line and a non-straight line, with the same endpoints. If it's 8 miles as the crow flies between points R and S, and your route takes 10 miles, you have a straightness of 0.8. jeffjon (talk) 19:20, 26 June 2008 (UTC)

Your metric wouldn't support the conclusion that Trip A in your example is straighter. It would rate Trip A and Trip B the same ($1/{\sqrt {2}}$). --Prestidigitator (talk) 19:51, 26 June 2008 (UTC)

Perhaps some angle turned per unit length metric might be what you wish?87.102.86.73 (talk) 19:58, 26 June 2008 (UTC)

You may even calculate the sum (or integral) of absolute values of all turns your road takes. However this may be misleading, too. Say you go straight from point X to Y, but you take a little loop (like the lowercase phi letter) at the halfway. The loop does not make a big change in your trip, however it adds from 2π up to about 3π radians to the 'total direction change'. Another counterexample are piecewise straight lines with a single right angle turn on the picture below:

They all have the same 'total direction change' equal 90°, however those for B close to A or to C seem more 'straight' than those wih B far apart from AC line. CiaPan (talk) 09:04, 27 June 2008 (UTC)

Being more specific might help here - did you mean the straightness of relatively straight lines, or the straightness of obviously curved lines.(see radius of curvature)87.102.86.73 (talk) 19:58, 26 June 2008 (UTC)

How about just dividing the length of the line by the distance of its endpoints to get a measure of straigtness? – b_jonas 12:05, 27 June 2008 (UTC)

Yeah, that's a much clearer way of stating what I suggested above. Prestidigitator pointed out one shortcoming, but it seems to me like the closest to workable, pending any clarification from the OP. jeffjon (talk) 12:32, 27 June 2008 (UTC)

One might also consider the methods described in the article Tortuosity, although some of them have already been mentioned here. --Martynas Patasius (talk) 11:39, 28 June 2008 (UTC)

How about calculating the aspect ratio of the smallest rectangle that contains the line ? Take the previously mentioned examples (A) (10 units north, 10 units east) and (B) (1 unit north, 1 unit east, repeated 10 times). The smallest rectangle containing A is $5{\sqrt {2}}$ by $10{\sqrt {2}}$ and has an aspect ratio of $1/2$. The smallest rectangle containing B is $0.5{\sqrt {2}}$ by $10{\sqrt {2}}$ and has an aspect ratio of $1/20$. B is thus 10 times straighter than A. Turiacus (talk) 22:33, 29 June 2008 (UTC)

Matrix definition

Hi, where does the term "matrix" come from? --88.104.167.212 (talk) 19:17, 26 June 2008 (UTC)

Matrix_(mathematics)#History states that "The term "matrix" was coined in 1848 by J. J. Sylvester" - the word is much older, see http://www.etymonline.com/index.php?term=matrix

87.102.86.73 (talk) 20:04, 26 June 2008 (UTC)

Thank you - --88.104.167.212 (talk) 06:10, 27 June 2008 (UTC)

And Sylvester may have had this type of matrix in mind when he coined the term (although I can't immediately find a source for this). Gandalf61 (talk) 08:41, 27 June 2008 (UTC)

June 27

Optimization

Hello. A factory wants to make the cheapest 250 cm³ right cone. Plastic at 0.03¢/cm² must cover the base. Waffle at 0.05¢/cm² must cover the lateral area. I read the article titled optimization (mathematics). I recently learned linear programming. A system of equations would have several points of intersection forming a polygon. These vertices are the optimal values. One of those is the minimum value. The volume formula is $h={\frac {750}{\pi r^{2}}}$ where r is the radius in cm (r > 0) and h is the height in cm perpendicular to the base (h > 0). The limits are that r and h near but never equal 0. The volume equation on a Cartesian coordinate plane exists within the first quadrant. The r- and h-axes will not form any points of intersection with the volume formula. Do I need at least three equations and/or inequalities, other than the cost equation and the two inequalities aforementioned, intersecting at three distinct points to substitute the x- and y-values of the optimal values into the cost equation: C = [$\pi$r(0.03s + 0.05r)]¢ where C is the cost of the cone in cents and s is the slant height in cm? By trial-and-error, the radius is about 3.93 cm and the cost 8.333 cents. Thanks in advance. --Mayfare (talk) 01:08, 27 June 2008 (UTC)

This is not a linear programming problem because the volume of a cone is a nonlinear function of its radius, so the constraint that the cone have a volume of 250 cm³ is nonlinear. Here's how I would approach it: The family of all cones is parameterized by two real numbers (for example, radius and height). But the constraint that the volume be 250 cm³ removes one degree of freedom, so you can parameterize the family of "all cones with volume 250 cm³" by a single real number. Now simply express the cost in terms of this one parameter, and optimize it with standard single-variable calculus. Good luck! —Keenan Pepper 01:59, 27 June 2008 (UTC)

Note however, that you can take the base surface as a variable in place of the base radius. Then the base cost becomes linear to the variable. --CiaPan (talk) 05:20, 27 June 2008 (UTC)

I was thinking the same thing (volume as bh/3) until I realized that cost is also a function of the lateral area, which is still a function of the radius. --Prestidigitator (talk) 18:49, 27 June 2008 (UTC)

Oops. You're right. --CiaPan (talk) 05:24, 30 June 2008 (UTC)

Hektar (ha) conversions?

The newly created Tiergarten Nürnberg article uses what I'm guessing is a German method of measurement, Hektar (ha). I have no idea how to convert this to square miles/meters, me being an American and an English major (i.e. horrible at math) to boot, so could someone enlighten me? María (habla conmigo) 13:36, 27 June 2008 (UTC)

I'm guessing it's a German spelling of Hectare. shoy 13:56, 27 June 2008 (UTC)

Great, good to know. Is there a wiki conversion template that automatically converts ha to mi/km within the text? María (habla conmigo) 14:11, 27 June 2008 (UTC)

You can use Google to help with the conversion. AndrewWTaylor (talk) 14:49, 27 June 2008 (UTC)

There is a wiki conversion template, called Template:Convert. For hectare to square miles, you could use {{convert|x|ha|sqmi}} where x is whatever number in hectares. The template is very versatile, you can convert many different units with it. Jkasd 16:29, 27 June 2008 (UTC)

Thank you so much, that's just what I needed. :) María (habla conmigo) 18:48, 27 June 2008 (UTC)

Estimating Median in finite space

Is there a method for producing a good estimate of the median in finite space?

For example I have 10,000 numbers, would it be possible to iteratively take these numbers in 1,000 chunks and estimate the median?

These chunks can be sorted and computed, but on the next chunk the previous values are not available, but any amount of running state values can be kept.

For background these restrictions of a software platform, which has storage capacity of 8000 bytes (about 2000 integers) and need to calculate the median of large data sets of more than 10,000 integers. The additional restriction is that there is only a single pass of the data.

Oh, and distribution characteristics of the data are unknown. Thanks 62.24.129.68 (talk) 15:13, 27 June 2008 (UTC)

Some information and ideas can be found at Selection algorithm; it doesn't address your question, though. Suppose that you want to find the median of n integers, and you can store m integers in memory. Here are some suggestions:

• You could keep track of the m-middlemost elements seen so far. However, if a lot of larger numbers appeared all at once, for example, then you begin to lose information: some of the m-middlemost elements seen so far become "unknown". After all n numbers have been read, you either know the value of the median for sure, or you have no idea what the median is. If you know that the input integers are statistically independent (or if you can choose which order you read the integers, in which case you would want to read them in random order), then we could explicitly calculate the probability that you successfully calculate the median. Although I don't know how to calculate that, I would guess that at n = 2000 and m = 10000, the probability of failure would be negligible.

• Fix any real number k > 1; you could keep track of the km-middlemost elements seen so far, except space them out by k at a time. For example, if m = 4 and k = 3, then the km-middlemost elements might at some point in time be:

4, unknown, unknown, 4, unknown, unknown, 8, unknown, unknown, 9, unknown, unknown.

Then proceed as with the previous method. This allows you to "fake" having a larger memory of km integers, even though you only have a true memory of m elements, at the cost of some inaccuracy of the result: you would only know that the median lies in some range, e.g., above we know that the median lies in the range 4 to 8. Thus, compared to the previous technique, you get a much higher probability of success (since you have a much larger "memory") but with possibly less accuracy in the answer. Note that if your distribution has a nice peak near the median (e.g., is Gaussian), then the range which you known the median lies in would likely be very narrow.

• If you know the value of n ahead of time, you could set k = n / (2m) and follow the second method suggested above. Then your fake memory of km elements is large enough to guarantee that you will successfully find the (range containing the) median regardless of the order the elements come in.

Hopefully some of these ideas prove useful. Eric. 144.32.89.104 (talk) 16:28, 27 June 2008 (UTC)

Thanks,

The number of elements is unknown to the function (it could be between 2 and a million), just to restate a bit more clearly. I was really hoping for a function, such that

$median(S)=M(m0(S_{1}),m0(S_{2}),m0(S_{3})\dots m0(S_{n}))$

Where S1…Sn can be in any order.

Here is my interpretation inspired by your suggestions (I think) without knowing $n$.

• We have an array $S$ with a maximum size of 1000 and start adding numbers to the array until I have reached 1000.
• Once we have reached 1000, we shrink the array to 500, by sorting, then taking 2 elements at a turn and calculating the median. i.e. $P(j)=(S(i)+S(i+1))/2$. The median of P should be approximately equal to that of S
• We can then start adding another 500 numbers.
• The problem with this “adaptive” version is that the overall shrinking will have to be maintained. So in 1…500 there are really 1,000 (1:2) elements in 501..1000 there are 500 (1:1). To make sure the array always ends up smaller, we will have to shrink 1.500 to 250 (1:4 ) and 501…1000 to 125 (1:4). The final array becomes 1:4.

I shall attempt this and see what the results are like. Trying to find some test data with a variety of statistical distributions will be the hard part though :) … 78.148.177.85 (talk) 18:25, 27 June 2008 (UTC)

One quick questio though, suppose I had the array shrunk to 1:2, then there 3 elements in the final batch (8,9,10), what two values would I put in? (8+9)/2 and ((8+9)+10)/2 maybe.. ? 78.148.177.85 (talk) 18:36, 27 June 2008 (UTC)

I don’t know if this addresses what you want to actually do, but if you allow yourself to access the numbers multiple times, you could find the median in I think O(n^2 log n) time, only ever storing m at a time: by finding upper and lower bounds on the median m/2 numbers at a time, a total of n/m times. GromXXVII (talk) 20:11, 27 June 2008 (UTC)

Sounds similar to this[4] which I found while searching.

These are the full constraints btw, (real world problem. not just made up for a quiz or something) :

1. You are handed the numbers one at a time
2. You may store as much state information as you like while processing rows. BUT…
3. At any time (in reality about every 4000 rows) you can and will be asked to save all the state information in at most 8000 bytes.
4. You will then be asked to restore from the 8000 bytes, and continue processing rows.
5. Finally you will be asked to provide the result using the 8000 bytes of state.
6. (Additionally you may be asked to merge your state with that of another state information and finally provide a result as if you had seen the same numbers)

I have performed some tests on the "shrinking array" version that was inspired by the previous posters response, and it gives reasonable approximations, except for the case of an even numbered set as I mentioned 78.148.177.85 (talk) 20:56, 27 June 2008 (UTC)

Here is an example of what I mean by the algorithm I loosely described in the first bullet. Suppose m = 3 and n = 9, and the numbers are:

1, 7, 6, 3, 4, 9, 8, 2, 5

At each point in time, we keep track of 3 numbers, the middle-most numbers that we have seen so far. This is the numbers held in memory after each successive number is read in:

1. Read the number 1: 1, none, none
2. Read the number 7: 1, 7, none
3. Read the number 6: 1, 6, 7
4. Read the number 3: 3, 6, 7
5. Read the number 4: 3, 4, 6
6. Read the number 9: 4, 6, unknown
7. Read the number 8: 4, 6, unknown
8. Read the number 2: 4, 6, unknown
9. Read the number 5: 4, 5, 6

This is an example of a successful run: the median is known to be exactly 5, with no error.

I'll explain where the "unknown" comes in, above. When the number 9 is read, the 3 middle-most numbers are 4, 6, 7. However, the program cannot know this: the number 7 had been read in earlier, and was "pushed" off the end of the array, and thus forgotten, since the program does not have enough memory to remember all of the numbers seen so far. The best it can do it mark that spot as "unknown".

If the numbers tend to come in increasing or decreasing order, or if they come in long runs of high numbers and long runs of low numbers, then this increases the chance of the algorithm failing. Here is an example:

1, 4, 3, 6, 2, 9, 8, 7, 5

1. Read the number 1: 1, none, none
2. Read the number 4: 1, 4, none
3. Read the number 3: 1, 3, 4
4. Read the number 6: 3, 4, 6
5. Read the number 2: unknown, 3, 4
6. Read the number 9: 3, 4, unknown
7. Read the number 8: 3, 4, unknown
8. Read the number 7: 4, unknown, unknown
9. Read the number 5: 4, unknown, unknown

This is an example of a failed run: the median is not known (although we know that it is at least 4, a small consolation).

I don't know how critical your application is, or whether occasional but unlikely failures are acceptable. This algorithm will always give the exact answer when it succeeds, but not always succeed. By using a large value for the k I described, you can increase the chance of success at the cost of approximation in the answer. If failure is unacceptable, then some further tweaking can produce something that will never fail, but simply give a rather large error in bad corner cases.

I'm glad now that my original explanation was so vague because I like the algorithm that you suggested in response. Eric. 86.153.207.223 (talk) 01:11, 28 June 2008 (UTC)

So basically, if you can read the numbers in sequence multiple times, it's possible to do an exact computation in few passes. If you can only read them once (one-pass), I think you could take a random sample of so many numbers that fit into your memory, and compute the median of those numbers. – b_jonas 13:24, 30 June 2008 (UTC)

Compact closed intervals in the order topology

There was a recent dispute about compactness in order topologies, and the following question was brought up: Is there a totally ordered space with a least and a greatest element, and with the least upper bound property, but that is not compact?

For finite unions of intervals in the reals it seems that such a space must be compact, since you can only omit the lower endpoint of an interval, but I think [0,1] union (2,3] is homeomorphic to [0,2] in the order topology, though not in the subspace topology, right? JackSchmidt (talk) 18:16, 27 June 2008 (UTC)

Yes, there's an obvious order-preserving bijection. For the original question, doesn't the usual least-upper-bound proof of (one-dimensional) Heine-Borel show that any complete bounded total order is compact in the order topology? Algebraist 18:23, 27 June 2008 (UTC)

Cool, thanks. I've actually had virtually no use for topology for years, and haven't looked at Heine–Borel for over 10 years, so I wanted to make sure I had everything in place. I mean, earlier today I didn't think [0] U (1,2] was compact (which it isn't in the subspace dangit!), so I don't want to depend on my recollection. If you want to state the compactness result clearly/citably (again), there is a {{fact}} tag at compact space that could use fixing. JackSchmidt (talk) 18:34, 27 June 2008 (UTC)

I've changed the wording a bit and cited it to Counterexamples. (my wording is equivalent since the order topology on an interval in an ordered space is the same as the subspace topology) Algebraist 18:51, 27 June 2008 (UTC)

Looks good, thanks again! I think leaving out the "closed interval" wording makes it much easier to understand. I managed to remove most of it by the time I posted here, but it was still lingering. JackSchmidt (talk) 19:21, 27 June 2008 (UTC)

Cartesian from Distance/Angle/Dihedral

Given 3 points in 3D-space (A, B & C, all with known Cartesian coordinates) is there a general formula for finding the Cartesian coordinates for a fourth point (D) given the distance CD, the angle BCD, and the dihedral angle for A->B->C->D (valued -π to π, as in dihedral angle#Alternative definitions)? I tried to use a computer algebra system to back-calculate it from the distance formula, the dot product formula, and the formula found under dihedral angle#Alternative definitions, but that results in two points. Unfortunately, for my purposes I'm looking at symbolic forms (e.g. I don't know if the dihedral is positive or negative), so procedural-type comparisons to narrow down which point is correct do not work too well. -- 128.104.112.147 (talk) 19:53, 27 June 2008 (UTC)

I get up to given the distance CD, the angle BCD - which gives me a circle of points - next I expect you have a dihedral angle to further define D.

I think two points is right.. they are diametrically opposed right?

To define it down to a single point I imagine you would need to specificy - the dist CD, angle BCD and define a direction of rotation about BC (eg with the vector BA being at angle 0) about BC - that way you should only get one point.87.102.86.73 (talk) 20:06, 27 June 2008 (UTC)

In other words if you take your line CD to be a vector Av_BC + Bv_{vector at right angles to BC} (A,B are scalars) then the two points you are getting will be due to +B and -B (I think) - this is because the dot product formula gives the same angle for cos(angleBCD) for both cases.

If you wan't to go through the maths involved in getting an equation for a single point I can certainly do that for you..87.102.86.73 (talk) 21:16, 27 June 2008 (UTC)

((One alternative is to let the dihedral angle only range 'pi' in valaue ie 0 to pi. —Preceding unsigned comment added by 87.102.86.73 (talk) 21:36, 27 June 2008 (UTC) ))

My understanding of the set-up is that the sign of the dihedral is what is specifying the direction of the dihedral rotation. The intersection of the sphere of distance and the cone of angle specify a circle of points in a plane perpendicular to the line BC. The dihedral angle then specifies where on that circle the point is - the full 2π range is needed to uniquely specify the complete circle. I think you are correct that I'm losing the distinction between the two halves because the tangent of the dihedral is not uniquely valued over the full -π to π range. I'm not sure how to fix that without resorting to an if-then-else construct, though. -- 128.104.112.147 (talk) 23:55, 27 June 2008 (UTC)

If you try an alternative method eg

1. find vector orthogonal to AB and BC call this v(y) (using cross product)

2. find vector orthogonal to v(y) and BC call this v(z)

The v(y),v(z), and BC form a set of right angled axis.

3. Now make a vector Q1 along CB of length |CD|

4. Rotate this vector Q1 about C using axis v(y) (ie in the plane given by v(z) and BC) by the angle BCD - call the result Q2

5. Now rotate Q2 about axis BC (in the plane given by v(z) and v(y) ) by the dihedral angle (here is where you must specify the correct direction of rotation)

6. The resultant vector is Q3

7. The point C+Q3 = D

that avoids any ifs - would that be ok/make sense ? (note it might be posssible to simplify this proceedure in terms of length - but I've split it into all the steps for simplicity)87.102.86.73 (talk) 09:48, 28 June 2008 (UTC)

Question on percentiles

According to the percentile rank article, being in the 85th percentile means a score above 85% of the other statistics in the data range. But, what does that mean on the other end? I'll give an example:

I am 6'6" tall, and a website claims that I am in the 99.8th percentile. Now, I know that means I'm taller than 99.8% of the sample group (I think it was American males), but how many am I shorter than? 0.2%? 0.1%? If I'm in a room with 1,000 people who reflect the results of the data, I am shorter than how mant of those people? Thanks. 70.105.164.43 (talk) 19:57, 27 June 2008 (UTC)

By convention, the percentile is always rounded down at the last digit shown, even if the next decimal place (which is not shown) is greater than 5. Thus, being in the 99.8th percentile means that you are taller than 99.8% to 99.9% of the population and shorter than 0.1% to 0.2% of the population. If you went to full precision on percentiles, adding up the percent above you and the percent below you would not equal 100% since, technically, you yourself are not counted as being above or below. 76.224.121.13 (talk) 08:36, 28 June 2008 (UTC)

Interesting; I was not aware of this. --Proficient (talk) 10:02, 30 June 2008 (UTC)

I wasn't aware of the convention that percentages are rounded down. It makes sense so as to avoid (for example) rounding 99.8% to 100% and saying that the person is taller than 100% of the sample (when clearly this is false). By extension, it seems reasonable that this rounding down would only apply to percentages greater than 50%, and that percentages less than 50% should be rounded *up* so as to avoid (for example) rounding 0.1% to 0%. Is this the case? Wikiant (talk) 11:32, 30 June 2008 (UTC)

June 28

Closed Form Formula

Working on a problem, I have come across this recursively defined sequence,
$0,2,{\frac {9}{2}},{\frac {44}{6}},{\frac {250}{24}},{\frac {1644}{120}},\cdots$.
The denominators are clearly the factorials but is there a closed form formula to produce the numerators. The recursive relation (if it helps) is
$a_{0}=0,\,a_{k}=(a_{k-1}+1){\frac {k+1}{k}}$
Thanks!--08:24, 28 June 2008 (UTC)

The numerators are sequence A052881 at Sloane's. The fractions themselves are $a_{n}=(n+1)\sum _{i=1}^{n}{\frac {1}{i}}.$ Algebraist 10:20, 28 June 2008 (UTC)

Wow, this is exactly what I was looking for. But how do you know what the explicit form is? How can one derive it (from the recursive relation or otherwise)? Thanks--A Real Kaiser (talk) 21:58, 28 June 2008 (UTC)

I derived it by looking at Sloane's (that's what it's there for!). As a check, I proved it by induction. Of course, there's still that summation in there, so it's not as closed a form as one might like. I suppose if you needed to approximate terms cheaply, you'd replace the sum by log(n) or log(n)+γ. Algebraist 22:18, 28 June 2008 (UTC)

Actually, believe it or not, this IS the form I needed. The summation is exactly what was needed. So this form is not bad at all for what I was doing, it actually made everything easier (even the (n+1) canceled out).--A Real Kaiser (talk) 02:14, 29 June 2008 (UTC)

Question about Floor and ceiling functions and INT(x) in computers etc

for reals, floor(x) returns int(x) except when x is already an integer - in which case I get x-0.5

ie floor (2.78)=2 floor (2.1)=2 floor(2)=1.5

But if I use f(x)=floor(floor(x)+n) where 0.5<n<1 I get f(x)=int(x) for all reals..

Floor function can be expressed as a fourier series or infinte polynomial, and so I imagine that with great difficulty I could express floor(floor(x)+n) as an infinite series too...

Is this series known? requesting links to look where to start, or explaining why it doesn't work.. Thanks.87.102.86.73 (talk) 10:53, 28 June 2008 (UTC)

Huh ? Which language or program are you using here ? I have never seen an implementation of floor which gives floor(2) = 1.5. There is sometimes a confusion between floor and int for negative reals, for which int may round up in some languages, but neither floor nor int should ever give a non-integer result. Gandalf61 (talk) 11:17, 28 June 2008 (UTC)

silly gandalf - for floor(x) I mean the mathematical version ie

$\lfloor x\rfloor =x-{\frac {1}{2}}+{\frac {1}{\pi }}\sum _{k=1}^{\infty }{\frac {\sin(2\pi kx)}{k}}.$ (copied from the wiki page)

and by 'INT(x)' I meant the computer implementations.. I was trying to get floor(x) (as a power or fourier series) to match INT(x).87.102.86.73 (talk) 11:37, 28 June 2008 (UTC)

Oh yes, silly me, apology.. I confused floor(x) and its power series...

What I meant was a power or fourier series etc that matches floor(x) or INT(x) for all real finite x including x=2,3,etc.. thanks.87.102.86.73 (talk) 11:52, 28 June 2008 (UTC)

Power series are far too well behaved to match a discontinuous function like floor. My memories of the ways of Fourier analysis are hazy, but I'm reasonably sure that the series you've got (with the value at discontinuities being the average of the left and right limits, rather than the correct value) is the best possible. Algebraist 12:25, 28 June 2008 (UTC)

And as a matter of interest, why do you want to do this? Algebraist 12:28, 28 June 2008 (UTC)

Question - I was wondering if I tried to constructed a power series from f(f(x)+n) (n=0.7?) what it would actually do? (ie f(x) is the power series or fourier series closest equivalent to int(x) )

I could easily get the first few terms of xⁿ.. there definately looks like there is a (set of) series...

I'm tentatively in agreement with you that I won't be able to get a discontinuity that goes from say 1 to 2 when x goes from 1.99... to 2.000...0001 , so what does happen? anyone know?

As to why I ask - pure curiosity - hoping I will become 'cleverer'...with your help

Could anyone explain why and if the fourier series above is the best possible..87.102.86.73 (talk) 13:01, 28 June 2008 (UTC)

I imagine this question boils down to "what happens/can go wrong with nested infinite power series" eg a similar expression would be e^(e^x) - which at first sight seems to be expressable as a power series.. does something 'go wrong' or not?87.102.86.73 (talk) 13:04, 28 June 2008 (UTC)

Power series are everywhere continuous, so can't work here. What do you mean when you say you can get the first few terms of xⁿ? Broadly speaking, if two functions are expressible as power series, then so is any combination of them. Thus $e^{e^{x}}$ is fine, for example. See analytic function for more info. Algebraist 13:13, 28 June 2008 (UTC)

Mmmh, you're right - I spoke before I actually checked I can't get ::$\lfloor x\rfloor =x-{\frac {1}{2}}+{\frac {1}{\pi }}\sum _{k=1}^{\infty }{\frac {\sin(2\pi kx)}{k}}.$ as a polynomial in x for reasons that are now obvious to me..87.102.86.73 (talk) 13:17, 28 June 2008 (UTC)

I think I made the assumption that because floor(x) can be expressed as a fourier series it would be an analytical function.. which it isn't.. is that right? (Thanks for your help I think I half understand now)87.102.86.73 (talk) 13:19, 28 June 2008 (UTC)

Yes. Fourier series are far more general than power series. For example, any function with finitely many discontinuities and local extrema has a Fourier series that converges to the right value (except maybe at discontinuities, as here), but almost all such functions are nowhere differentiable, and thus have no power series. Algebraist 13:57, 28 June 2008 (UTC)

Thanks. error in my reasoning now understood, a little wiser now.87.102.86.73 (talk) 15:01, 28 June 2008 (UTC)

Calculating infinite series fractions sum

Hi. This is not homework. I'm just curious as to the formulae used to calculate these, expecially with fractions, and if there is an article on this. I'm not talking about the more obvious ones like 1/3+1/30+1/300... or 1/10+1/200+1/1000... . I mean the other ones like 1/2+1/4+1/8+1/16... or 1/2+1/3+1/4+1/5... or 1/2+1/4+1/6+1/8+1/10... or 1/2+1/4+1/16+1/256... or 1/2+3/4+5/6+7/8... , etc. This is probably a silly or random or obvious question, but I'm just wondering how this is usually canculated in our base-10 number system. Thanks. ~AH1^(TCU) 14:38, 28 June 2008 (UTC)

If a series converges, then you can calculate the partial sums (i.e. sums of the first n numbers) to get (better and better) approximations to the limit. You can probably also calculate how good these approximations are (depending on the series). For especially nice series, you can calculate the limit explicitly, and not have to use approximations. Of course, if the series diverges, then there's no limit at all. There are a number of techniques for determining whether a series converges (and if so, if there's a nice form for the limit); fortunately, your examples are all dead easy. The first and fourth are geometric series, for which the limit is easy to find, and the other three can easily be shown to diverge, due to their relation to the harmonic series. For a nice example of a series whose limit is not obvious, see Goldbach–Euler theorem. Algebraist 14:53, 28 June 2008 (UTC)

Very powerful methods to calculate a number of different series are using the Taylor series or the Fourier series. (Igny (talk) 18:24, 28 June 2008 (UTC))

I disagree with Algebraist. The fourth is not geometric. I believe that the sum of the fourth series is unknown, perhaps it cannot be expressed in terms of elementary functions. It is approximately 0.8164. The big picture is that there are no general methods to determine the sum of an infinite series exactly. For a few simple series, this is easy. In other cases (such as the series $1+1/4+1/9+1/16+1/25+\cdots$) it requires some insight or tricks to find the sum. For most series that you would write they either diverge or the sum cannot be expressed in simpler terms except for the infinite series itself. In some of these cases, which are important for one reason or another, a name is given to to sum of the series. (For example, Apery's constant.) Oded (talk) 20:42, 28 June 2008 (UTC)

Sorry, I misread the fourth series. Algebraist 21:57, 28 June 2008 (UTC)

It looks like series four has been studied somewhat, but no nice form is known. The decimal expansion is A007404 at Sloane's. Plouffe's has 20000 digits. Algebraist 22:29, 28 June 2008 (UTC)

June 29

Very very hard problem

Prove that for every prime number greater than a googol, there exists a multiple of that prime number that is divisible by 7.

This problem is darn so hard that it is fit only for a university student. Instead our maths teacher forced us mere school students to do it. Oh yes, the teacher tell us that googol is a very large number.

When I went on the web to google googol, I despair when I found out just how large a googol is. How on earth am I going to prove this? I dont even know where to start.

Please help. 122.107.149.37 (talk) 07:11, 29 June 2008 (UTC)

If the question is just as you state it, then it must be an illustration of misdirection. Forget the googol part. Forget the prime number part. Find a multiple of 1 that is divisible by 7. Find a multiple of 2 that is divisible by 7. Show that for any whole number at all (apart from 0) there is a multiple of that number that is divisible by 7. Gandalf61 (talk) 07:40, 29 June 2008 (UTC)

Surely with any sensible definition, 0 is itself divisible by 7? Algebraist 09:34, 29 June 2008 (UTC)

Yes, indeed - brain not working. Gandalf61 (talk) 13:02, 29 June 2008 (UTC)

They're probably only working in the positive integers, and with positive integer multiples. Black Carrot (talk) 11:31, 29 June 2008 (UTC)

Since 0 isn't prime, who cares? "Prime number" generally refers to positive integers. --Tango (talk) 12:27, 29 June 2008 (UTC)

About limits

Hi, I asked a question here. Would you be so kind and take a care of it? Mozó (talk) 11:08, 29 June 2008 (UTC)

Quick question.

Hey guys. I was wondering if any of you could give me a quick proof of the following statement before I head off to work: the real part of any non-trivial zero of the Riemann zeta function is 1/2.

Come on folks, looks like kindergarten stuff. I'm off to get my coffee. Answer pronto! TIA. —Preceding unsigned comment added by 124.191.113.253 (talk) 11:24, 29 June 2008 (UTC)

It's trivial. The Plouffe's inversion of a Jordan-Eisanhimmenheimann space produces a havamorphism from pi to the unit n-dimensional sphere. The result follows. Black Carrot (talk) 11:33, 29 June 2008 (UTC)

In case this is a serious question - see Riemann hypothesis. --Tango (talk) 12:25, 29 June 2008 (UTC)

A tricky one-to-one correspondence problem

Hi all:

I've been trying to establish a one-to-one correspondence between the intervals [0, 1] and (0, 1), but so far this problem has proven surprisingly difficult.

I got as far as constructing the following function as a bijective map from (-∞, +∞) to (0, 1):

$f(x)={\frac {1}{\pi }}\arctan x+{\frac {1}{2}}$

and realizes that I just need to find a way to map [0, 1] to (-∞, +∞) to solve the problem. But so far it's been eluding me.

Any good ideas?

Thanks,

76.65.15.166 (talk) 20:17, 29 June 2008 (UTC)

This can't be done with 'nice' functions, since [0, 1] and (0, 1) aren't homeomorphic. If I was asked this question, I would take the lazy option and use the Cantor–Bernstein–Schroeder theorem, but you might want something more direct. Algebraist 21:15, 29 June 2008 (UTC)

One direct method would be to deal with the rationals and irrationals separately. Algebraist 21:17, 29 June 2008 (UTC)

Consider the sequence $0,1,{\frac {1}{2}},{\frac {1}{3}},{\frac {1}{4}},\dots$. You take every element of this list to the element two slots further down. The other points in $(0,1)$ are mapped to themselves. This defines a simple bijection between $[0,1]$ and $(0,1)$. Oded (talk) 21:23, 29 June 2008 (UTC)

Yes, that was essentially my idea. Algebraist 21:32, 29 June 2008 (UTC)

Thank you both OdedSchramm and Algebraist. While OdedSchramm provided the final, definitive answer. I would still like to show my understanding of Algebraist's indirect argument with the following exposition:

---

Forward injection: $f(x)={\frac {1}{2}}x+{\frac {1}{4}}$

Backward injection: trivially $f(x)=x$

Thus, by the Cantor–Bernstein–Schroeder theorem, there must exist a bijective relationship between the sets [0, 1] and (0, 1).

---

Am I correct according to your expert opinions, Algebraist? also submitting to OdedSchramm's expert scrutiny my above self-composed answer ;)

76.65.15.166 (talk) 21:37, 29 June 2008 (UTC)

That's perfectly correct. Oded (talk) 21:41, 29 June 2008 (UTC)

For bonus marks, unravel the proof of Cantor–Bernstein–Schroeder to get an explicit bijection from those injections. Algebraist 21:43, 29 June 2008 (UTC)

Unfortunately the margin of this fine encyclopedia is too small to contain the bonus solution. So I will try to show this some other time. ;) Thanks again Oded and Algebraist, really appreciate it! 76.65.15.166 (talk) 21:46, 29 June 2008 (UTC)

June 30

point inside tetrahedron

Quick before its too late - a question.

Is there a very easy way to determine if a point lies inside an irregular tetrahedron defined by its four points?87.102.86.73 (talk) 23:58, 30 June 2008 (UTC)

Yes. Call your test point P. For each face, view it from the outside and label the points A, B, C in counter-clockwise (right-hand) order. ${\vec {v_{1}}}={\overline {AB}},{\vec {v_{2}}}={\overline {AC}},{\vec {v_{P}}}={\overline {AP}}$. Now if $({\vec {v_{1}}}\times {\vec {v_{2}}})\cdot {\vec {v_{P}}}>0$ then the point is outside the tetrahedron. If the point doesn't test outside for any of the faces, then it is inside or on the surface (I think you can figure out that edge case). --Prestidigitator (talk) 01:44, 1 July 2008 (UTC)

July 1

Analyticity on $\mathbb {R}$

We already know that all of the derivatives of $e^{x}$ at $x=0$ are equal to 1. Find another analytic function on $\mathbb {R}$, call it $f$, that is not equal to $e^{x}$ at any $x$ other than $x=0$, but for which the $n$th derivative of $f$ at 0, $f\,^{(n)}(0)$ is equal to 1 for all $n\geq 1$.

Now I thought that an analytic function has a power series expansion, the expansion is unique, and the coefficients are determined by the derivatives. In addition, the function would be equal to its expansion if the remainder term goes to zero. Now, if we find the answer to the above question, wouldn't that imply that the same power series represents two different functions? Because the coefficients of our f(x) and $e^{x}$ will be the same (if we expand around x=0). Is there something subtle that I am missing here or is the answer obvious but I am having one of my (too many) dumb moments? The question is verbatim from an analysis book. What would such a function look like? Thanks--76.79.202.34 (talk) 00:39, 1 July 2008 (UTC)

If you're allowed to have f(0) ≠ 1, then you can take f(x) = e^x + 1. If not, then I agree it's impossible. On the other hand, if what you're looking for is not an analytic function, but an infinitely differentiable one, then it is possible. I think e^x + e^-1/x2, suitably extended at 0, does the trick. The thing is, analytic and infinitely differentiable are synonymous for complex functions, but not real ones. 136.152.224.6 (talk) 03:52, 1 July 2008 (UTC)

FALSE POSITION

METHOD OF FALSE POSITION FOR FINDING A REAL ROOT OF A NONLINEAR EQUATION? —Preceding unsigned comment added by Wikifrommld (talk • contribs) 03:15, 1 July 2008 (UTC)

ALGORITHM

A CONCISE ALGORITHM OF THE FALSE POSITION METHOD —Preceding unsigned comment added by Wikifrommld (talk • contribs) 03:17, 1 July 2008 (UTC)