Hi. I'm looking for the unique positive real root of a certain 3rd degree polynomial with coefficients in ${\displaystyle \mathbb {Q} \left[{\sqrt {13}}\right]}$. This should be expressible in radicals, per history, if I'm not mistaken.

I've solved it, following Cardano's method in our article, cubic function, but I'm having trouble expressing the solution in terms of radicals of real numbers. I can get it in terms of radicals of complex numbers, or in purely real terms with a cosine.

The function is ${\displaystyle f(x)=x^{3}+3{\sqrt {13}}\cdot x^{2}+12x-16{\sqrt {13}}}$, and the solution I've got is:

• ${\displaystyle x={\frac {1}{81}}\left[{\overline {z}}z^{\frac {2}{3}}+81z^{\frac {1}{3}}-81{\sqrt {13}}\right]}$, with ${\displaystyle z={\sqrt {13}}+2i{\sqrt {179}}}$.

Or, equivalently,

• ${\displaystyle x=6\cos \left({\frac {\alpha }{3}}\right)-{\sqrt {13}}}$, where ${\displaystyle \alpha }$ is an angle in the first quadrant with ${\displaystyle \tan(\alpha )={\frac {2{\sqrt {179}}}{\sqrt {13}}}}$.

Is this the best I can do, or is there some way to massage the first expression into one involving only reals and radicals? I could look at the equation expressing ${\displaystyle \cos \left(\alpha \right)}$ as a cubic function of ${\displaystyle \cos \left({\frac {\alpha }{3}}\right)}$, but I think that'll take me in a circle.

What I really want to know is which field extension of ${\displaystyle \mathbb {Q} }$ the solution is in. I mean, I'm pretty sure it's in ${\displaystyle \mathbb {Q} \left[{\sqrt {13}}\right]\left[z^{\frac {1}{3}}\right]}$, but is there any way to find a real extension that it's in? Will ${\displaystyle {\sqrt {179}}}$ do it? Any ideas are appreciated; thanks in advance. -GTBacchus^(talk) 03:29, 25 February 2010 (UTC)[reply]

It should be a 6th degree extension so I'm not sure how sqrt(179) is supposed to help. I vaguely remember figuring out a messy way to do a hw problem like this with a bunch of linear algebra and then being told I had done a lot more calculation than was really needed. The article Minimal polynomial (field theory) doesn't look too helpful, unfortunately. The Mathworld reference at that article does mention that Mathematica can figure it out for you. 75.62.109.146 (talk) 03:49, 25 February 2010 (UTC)[reply]

I'm at least as interested in the journey as in the destination in this case, so simply getting the answer is only a partial solution. I realize that the splitting field of the polynomial is degree 6 over Q, but that doesn't mean that our particular x is a degree 6 number, right? I mean, the cube root of 2 satisfies x^6-4=0, but its degree as an algebraic number is only 3. -GTBacchus^(talk) 04:30, 25 February 2010 (UTC)[reply]

x^6-4 = (x^3+2)(x^3-2) and the splitting fields of those two factors are the same, so the overall splitting field still has degree 3. This is what you would expect--the degree of the splitting field is the degree of the minimal polynomial, which for the cube root of 2 is x^3-2. (edited) 75.62.109.146 (talk) 06:26, 25 February 2010 (UTC)[reply]

The degree of the splitting field of x^3-2 over Q is 6, not 3 - see splitting field. The field obtained by adjoining the real cube root of 2 to Q does have degree 3 over Q, but it is not a splitting field - you have to adjoin a primitive cube root of 1 as well to get a splitting field. In general, the splitting field of an irreducible polynomial of degree n may have a degree as large as n factorial.Gandalf61 (talk) 09:54, 25 February 2010 (UTC)[reply]

Thanks, I must have been confused. I had a diagram in mind involving adjoining one of the complex cube roots of 2 and getting the rest from there, but it was wrong. 75.62.109.146 (talk) 17:52, 25 February 2010 (UTC)[reply]

Well, you can plug the coefficients into the first of these formulae. There is undoubtedly a better way, though! (If you just want the answer and don't care about how you get it, any decent maths computer package will be able to get it for you.) --Tango (talk) 03:54, 25 February 2010 (UTC)[reply]

No, I'm interested in how to get there. As for those formulas you linked to, I tried that. The quantity under the square root comes out negative, and we end up with the first expression for x that I've got above. -GTBacchus^(talk) 04:29, 25 February 2010 (UTC)[reply]

Try one of the other formulae, then. For one of them, the i's should cancel out since any cubic with real coefficients will have a real root. --Tango (talk) 18:13, 25 February 2010 (UTC)[reply]

You see, that's exactly what my problem is. I know that the function has 3 distinct real roots, and those formulas all have imaginary numbers in them. I know that the imaginary parts must therefore cancel, but I can't seem to make it happen, unless I use trig functions. If I use the "evalf" command on Maple, I get a real number, but no indication of how Maple got there. (For "evalf", it's probably numerical methods anyway.)

My question is: can I algebraically eliminate the imaginary numbers in my first expression above, and if so, how? If expressing the answer in cosines is the best I can do, I accept that, but I'd like to know why. If, on the other hand, there is a way to actually eliminate the imaginary parts, then I'm asking for help to find that way. Is this number expressible in terms of radicals of real numbers? Does that make sense? -GTBacchus^(talk) 18:46, 26 February 2010 (UTC)[reply]

OK, so the base field ${\displaystyle \mathbb {Q} ({\sqrt {13}})}$ is formally real, the cubic is irreducible, and it has three real roots. Unless I'm missing something, this is exactly the situation of casus irreducibilis, hence the answer is no, it's not expressible using only real radicals.—Emil J. 19:03, 26 February 2010 (UTC)[reply]

Aha. In that case, I don't feel dumb for being unable to eliminate the i's. Thank you; that's an interesting link. -GTBacchus^(talk) 01:11, 27 February 2010 (UTC)[reply]

The mathworld link points to this which has a few examples of doing it by hand. It looks plausible that you can do your example with the same methods. 75.62.109.146 (talk) 03:55, 25 February 2010 (UTC)[reply]

I'll try this and report back whether it works. :) -GTBacchus^(talk) 04:29, 25 February 2010 (UTC)[reply]

Related previous question: [1]. If you're able to express the solution in radicals involving reals only, I (User_talk:NorwegianBlue) would love to hear about it! --85.200.133.168 (talk) 20:46, 26 February 2010 (UTC)[reply]

The casus irreducibilis of expressing the real solutions of the cubic equation using reals only, gave historically rise to the invention of the complex numbers. Bo Jacoby (talk) 20:28, 28 February 2010 (UTC).[reply]

I'm reading: "Assume GCD(r,s) = 1. Then the Euclidean algorithm gives polynomials B(t) and C(t) such that: ${\displaystyle B(t)(1+t+\dots +t^{r-1})+C(t)(1+t+\dots +t^{s-1})=1}$." This is outside my usual area, and I'm not sure how to interpret "the Euclidean algorithm gives". Do they mean that it's obvious for some reason that ${\displaystyle 1+\dots +t^{r-1}}$ and ${\displaystyle 1+\dots +t^{s-1}}$ have GCD = 1, and so the Euclidean algorithm (specifically Bezout's identity) gives the linear combination to 1? Or do they mean that you could (omitting the details) actually do the Euclidean algorithm by repeated long-divisions and you'd discover thereby that the GCD is 1?

In either case I'm missing part of the argument. I don't know why those two polynomials have GCD 1 (though maybe it's an easy fact that I've forgotten), and I also don't know how to use the Euclidean algorithm to show it- I can do a step or two but things get too messy.

I'd also like to know exactly what the polynomials B and C are. They should be expressible in terms of s and r, right? Like I said I get lost in the details when I try to construct them by the Euclidean algorithm. Thanks- Staecker (talk) 14:38, 25 February 2010 (UTC)[reply]

Think of for instance with r 3 and s 7, then we can get numbers so 3r-7s=1 and in fact 3.5-2.7 works. now think of those polynomials as strings of 1s: 111 111 111 111 111 - 1111111 1111111 0 = 1. You can translate these directly into B and C by the groupings, B is ${\displaystyle t^{12}+t^{9}+t^{6}+t^{3}+1}$ and C is ${\displaystyle -(t^{8}+t)}$. Don't know if that's what they mean but it shows how you can get them easily when the GCD is 1. Dmcq (talk) 15:16, 25 February 2010 (UTC)[reply]

(edit conflict) Since ${\displaystyle t^{n}-1=(t-1)(1+t+\dots +t^{n-1})}$, the statement that your two polynomials have GCD 1 reduces to the statement that ${\displaystyle t^{s}-1}$ and ${\displaystyle t^{r}-1}$ have no root save 1 in common, i.e. that a number that is both an rth root of unity and an sth root of unity is a 1st root of unity, and this follows easily from r and s being coprime. Algebraist 15:24, 25 February 2010 (UTC)[reply]

Thanks a lot- two great answers. Just what I needed (I hope). Staecker (talk) 16:31, 25 February 2010 (UTC)[reply]

Are there any instances of a 3-4 dimensional representation of the Mandelbrot Set actually appearing in reality, and if so, what are the implications of such a representation as regards to its 2-dimensional counterpart, described here? In short, could there exist a fractal like the Mandelbrot set, but in a 3 dimensional format, (analogous to an actual carboard box shaped into a cube along with its square counterparts on its sides)? —Preceding unsigned comment added by 71.108.11.25 (talk) 08:53, 26 February 2010 (UTC)[reply]

See 3D Mandelbulb. 75.62.109.146 (talk) 10:37, 26 February 2010 (UTC)[reply]

Thank you. I really appreciate the information and the speedy reply that you gave me. But I also see how you might be missing my point. I actually wanted to know whether a real-life Mandelbulb, as you termed it, could be constructed naturally rather than artificially, and lying outside the realm of computerization and mathematics; take, for example, this 3d representation of a fractal File:Sa-fern.jpg. 71.118.39.219 (talk) 04:47, 27 February 2010 (UTC)[reply]

Romanesco broccoli? 75.62.109.146 (talk) 08:47, 27 February 2010 (UTC)[reply]

Thank you. I greatly appreciate the new input. :| TelCoNaSpVe :| (talk) 08:31, 28 February 2010 (UTC)[reply]

An editor asked for feedback on an article covering an aspect in differential algebra here. My algebra is quite rusty, as is my DifE, and I never studied differential algebra. I made a couple small suggestions, but I thought maybe some of the visitors here would be in a better position to opine in a useful way.

I also be curious to hear some math gurus weigh in on my observation that so many math articles do not follow WP referencing guidelines—is this considered acceptable going forward, or should it be discouraged?--SPhilbrickT 02:05, 27 February 2010 (UTC)[reply]

I have not read the article yet (P-derivations), but I will do so if possible (and perhaps comment on the article as well). With regards to your second point—namely, the lack of references in specific mathematics articles—it can occassionally be a problem. However, if the article in question is "advanced", I think that it is a safe assumption that whoever wrote it is correct in that which he/she has written (in most cases, one would not learn about Bott periodicity (for instance), unless one has a good amount of experience with algebraic topology and the like; therefore, if someone does write a significant chunk of the article on Bott periodicity (in good faith), we can trust that the chunk in question is accurate). On the other hand, of course, even experienced mathematicians make mistakes, and that is why it is useful to reference non-trivial claims in articles. Referencing can also be useful to the person reading an article, because he/she may want to delve deeper into the topics in question. But as a general rule of thumb, it is most important to write a decent article, after which the first priority is to reference non-trivial claims (another alternative is to add references as you are writing the article). PST 03:56, 27 February 2010 (UTC)[reply]

I have to disagree here. I've seen plenty of math articles with questionable information so the theory that math articles don't need to be verifiable doesn't hold water.--RDBury (talk) 04:42, 27 February 2010 (UTC)[reply]

I never said that mathematics articles do not need to be verifiable; all I said was that the problem of "verifiability" is not that significant for "advanced mathematics articles". PST 05:01, 27 February 2010 (UTC)[reply]

One thing that does come to mind (and which I am sure will come to Michael Hardy's mind as well), is that the title of the article is "P-derivations" and not "P-derivation" which contradicts WP:MOS. Perhaps someone else can fix this. PST 03:58, 27 February 2010 (UTC)[reply]

I moved the article.--RDBury (talk) 04:46, 27 February 2010 (UTC)[reply]

(ec)On the first question, the subject as defined in the article seems to specific to Buium's work, at the least it's not something that many people are going to be familiar with.

On the second question, you should probably raise it at Wikipedia talk:WikiProject Mathematics. I will say that about 10% of math articles don't list any references at all, which is about the same as Wikipedia over all.--RDBury (talk) 04:28, 27 February 2010 (UTC)[reply]

Suppose a normally distributed random variable x has a mean μ and standard deviation σ. If the variable is multiplied by n, what is the mean and standard deviation of the new variable in terms of n, μ and σ?--220.253.101.232 (talk) 09:40, 27 February 2010 (UTC)[reply]

Come to think of it, the new mean would be nμ. Not sure about the standard deviation though.--220.253.101.232 (talk) 09:42, 27 February 2010 (UTC)[reply]

Presuming this is homework, you could try working through an example, or seeing how the variance is affected (that will tell you what happens to the s.d.). Our article on the normal distribution is pretty good, by the way. 75.62.109.146 (talk) 09:44, 27 February 2010 (UTC)[reply]

This is not actually homework, I'm just interested. Specifically I was reading Modern portfolio theory, and I was wondering what the expected return and standard deviation would be of n shares given the expected return and standard deviation of one of those shares. —Preceding unsigned comment added by 220.253.101.232 (talk) 09:54, 27 February 2010 (UTC)[reply]

Is it (nσ^2)^0.5 ? --220.253.101.232 (talk) 10:02, 27 February 2010 (UTC)[reply]

If n is constant, the mean is nμ and the standard deviation is |n|σ, and it's easy to see why if you look at the definitions of mean and standard deviation. Michael Hardy (talk) 16:31, 27 February 2010 (UTC)[reply]

...and expressions involving √n occur in sums of n independent random variables. If you multiply a random variable by n, that's a sum of n copies of the same random variable—as far from independence as you can get. Michael Hardy (talk) 16:33, 27 February 2010 (UTC)[reply]

The quick way to see what happens to mean and SD is to realise that multiplying (or dividing) all values by a constant is equivalent to a change of units - a mean of 5cm and an SD of 1cm, for example, represent physical measurements which could be expressed alternatively as 50mm and 10mm respectively. If all values are expressed in mm rather than cm, they are still the same size as before, and so must be their mean and SD - everything will be ten times greater in the different units.→86.152.79.95 (talk) 19:18, 27 February 2010 (UTC)[reply]

For a random variable x having mean μ and standard deviation σ, the notation x ≈ μ±σ has the benefit that the formulas for adding a constant or multiplying by a constant take the familiar forms of associative and distributive rules: a+x ≈ a+(μ±σ) = (a+μ)±σ and ax ≈ a(μ±σ) = aμ±aσ. Bo Jacoby (talk) 22:18, 1 March 2010 (UTC).[reply]

I have an interest in biostatistics and bioinformatics, but my maths background is mostly limited to a few semesters of college calculus and whatever (very) basic statistics I have been able to pick up since. Recently, this paper (which has a corresponding PPT presentation) caught my eye as something that could help me with a side project I've been considering on and off for a few years, but I'm not familiar with some of the notation that's used. What do I need to learn to manage a basic understanding of the algorithms presented in that paper? – ClockworkSoul 18:49, 27 February 2010 (UTC)[reply]

The notation (and all of the algorithms) in that paper are standard for a computer science algorithms class. If you don't want to take a class, but want the standard textbook that most universities use, get Introduction to Algorithms (commonly referred to as CLRS). -- kainaw™ 21:18, 27 February 2010 (UTC)[reply]

Also note, bioinformatics is heavy on database algorithms, not really mathematics. My job (in bioinformatics) is mostly taking algorithms other have written that will take a year or two to complete and reworking them to run in a day or two. It requires extensive knowledge of data organization, parallel processing algorithms, and statistical approximations. As a bonus, you should study a lot of medical jargon since the data is all medical data. It makes it much easier to know how to identify TIA in the data instead of being held up waiting for a doctor to explain it to you. -- kainaw™ 21:22, 27 February 2010 (UTC)[reply]

Thank you for your help, Kainaw! Fortunately, that's where my formal training is, so at least I've made some progress on that front! – ClockworkSoul 22:06, 27 February 2010 (UTC)[reply]

That article is about combinatorial graph algorithms and also some linear algebra (the part where it discusses least squares) that you'd probably find out about in either a numerical analysis or a statistics class. 75.62.109.146 (talk) 09:42, 28 February 2010 (UTC)[reply]

Hello everyone, reviewing for a numerics exam, I came across the following statement in my book. Consider the initial value problem ${\displaystyle y'={\sqrt {y}}}$ with y(0)=0. This ODE can be solved by separating the variables and we get the nontrivial solution ${\displaystyle y(t)=t^{2}/4}$. But when we apply the explicit Euler method we only get zeros as the computed solution. Can someone please explain this conundrum? What in the world is going on? Thanks! 174.29.98.151 (talk) 00:37, 28 February 2010 (UTC)[reply]

That differential equation does not satisfy the Lipschitz condition, and the solution to your IVP is not unique. See the Picard–Lindelöf theorem for more.(Igny (talk) 01:38, 28 February 2010 (UTC))[reply]

The article Envelope (mathematics) has some information on this type of situation.--RDBury (talk) 06:37, 28 February 2010 (UTC)[reply]

You've encountered a singular solution. By the way, neither Mathematica (DSolve) nor Maxima (ode2) can find it and even don't warn! ${\displaystyle {\ddot {\frown }}}$ The general solution given by separation of variables is a family of parabolas, all tangent to the line y=0. Whenever a solution family has a common envelope, it may also be a solution by itself. Namely, separating the variables, you've divided by y while implicitly assuming that y≠0. That's where the information is lost. — Pt _(T) 12:38, 28 February 2010 (UTC)[reply]

Obviously, ${\displaystyle \log(x^{n})}$ when n is even has a different domain (all values except 0). So when stating the equivalence ${\displaystyle \log(x^{n})=n\log x}$, is the standard practice to add that qualification "when n is odd" or is ${\displaystyle \log(x^{n})}$ often assumed to only refer to the part of the domain where x is positive unless otherwise specified? 71.161.56.189 (talk) 00:34, 28 February 2010 (UTC)[reply]

As functions of x the two expressions have different domains and are therefore not technically the same function when n is even. If you want to be completely correct then you might add "(x>0)" so the even/odd issue doesn't even come up. But this kind of thing is a minor issue since most people know not to try to use negative values in a log function without specifically being told.--RDBury (talk) 06:32, 28 February 2010 (UTC)[reply]

Would another way of handling this be to say conventionally that "f(x)=g(x)" means that they are equal when x is in the domain of both functions? (in this case, ${\displaystyle f(x)=\log(x^{n})}$ and ${\displaystyle g(x)=n\log x}$) --COVIZAPIBETEFOKY (talk) 13:22, 28 February 2010 (UTC)[reply]

Yes, in that case, for even n, g is the restriction of f to the positive real line, or equivalently, f is an extension of g to the set of all non-zero reals. -GTBacchus^(talk) 18:34, 28 February 2010 (UTC)[reply]

I understand the concepts behind both the Needleman–Wunsch algorithm and the Smith–Waterman algorithm. However, they look pretty much the same to me. I see very superficial differences (NW adds gap costs and SW subtracts gap costs). I am left wondering what is the real difference? Is it merely that Smith-Waterman went the extra step of backtracking through the Wagner-Fischer algorithm to find a "best fit" substring? -- kainaw™ 04:25, 28 February 2010 (UTC)[reply]

The NW algorithm is used for global alignment, and the SW for local alignment. They are used in different contexts. For example, if one is looking for a small motif in a large sequence, SW is a good algorithm, since SW will find all occurrences of the motif in the large sequence. If one applies NW in this case, the motif may be stretched in a weird way (by inserting gaps) and the purpose will be lost. On the other hand, NW is a good algorithm when one is sure that two sequences are very similar and one just wants to examine the dissimilarities which are important. --[[User:MoneyFromAir|MoneyFromAir] —Preceding undated comment added 02:34, 2 March 2010 (UTC).

If you like, Smith-Waterman has no negative weights so "starting a sequence in the wrong position" isn't penalised, hence the local alignment. x42bn6 Talk Mess 02:05, 3 March 2010 (UTC)[reply]

What I never really noticed until just now is that the NW algorithm sets the top/left roes to 0 1 2 3... while the SW algorithm sets them all to 0. So, that is a huge difference. I noticed it when looking why you get local alignment with SW. Thanks. -- kainaw™ 03:48, 3 March 2010 (UTC)[reply]

I am trying to write a proof that given a Hermitian matrix A,

${\displaystyle ||(A-\lambda I)^{-1}||_{2}={\frac {1}{\min _{\lambda _{i}\in spec(A)}|\lambda -\lambda _{i}|}}.}$

This is what I have so far. I know that for a Hermitian matrix B,

${\displaystyle ||B||_{2}=\max _{\mu _{i}\in spec(b)}|\mu _{i}|.}$

So if I define the matrix ${\displaystyle B=A-\lambda I}$ which is basically a perturbation of matrix A, then the eigenvalues of A also get perturbed by the same amount. The eigenvalues of B are ${\displaystyle \mu _{i}=\lambda _{i}-\lambda }$. Then I say that because A is Hermitian, B is also Hermitian (which is not true if ${\displaystyle \lambda }$ is a arbitrary complex constant). So ${\displaystyle ||B||_{2}=\max |\mu _{i}|}$ and then by the inverse eigenvalue theorem we have

${\displaystyle ||(A-\lambda I)^{-1}||_{2}=||B^{-1}||_{2}=\max {\frac {1}{|\lambda -\lambda _{i}|}}={\frac {1}{\min |\lambda -\lambda _{i}|}}.}$
This is just a basic outline with details missing. But my problem is that I need to show this for any arbitrary constant lambda (including complex values) but that one step I take (where I say that B is also Hermitian because A is Hermitian) is wrong. So any help in patching this proof up would be appreciated. Thanks! -Looking for Wisdom and Insight! (talk) 10:37, 28 February 2010 (UTC)[reply]

Hint: B might not be Hermitian, but it is normal. -- Meni Rosenfeld (talk) 13:37, 28 February 2010 (UTC)[reply]

To avoid original research, I'm searching for papers/books that describe methods for converting data to vectors. There is the simple method of making a set of boolean vectors. For example, gender would become gender_m and gender_f with each set to 1 or -1. What is it called when you turn a vector that has 2 values into something like 1 for 'm' and -1 for 'f' (and possibly 0 for unknown)? Further, what if multiple values are set in a circle where the degree of rotation selects a value? The best example I can think of off the top of my head is a data field that contains a person's movie genre choice with four options: comedy, action, drama, romance. In this case, comedy=0, action=90, drama=180, romance=270 - which places comedy opposite drama and action opposite romance. If I had some "official" names of these methods, I could find papers and add cited information to the appropriate articles. -- kainaw™ 23:15, 28 February 2010 (UTC)[reply]

The closest thing I can think of at the moment is the idea of an enumerated type in computer programming. To me it would seem more natural, in your example, to say comedy=0, action=1, drama=2, and romance=3, and this is often how enumerated types are implemented. —Bkell (talk) 00:43, 1 March 2010 (UTC)[reply]

I agree, but that doesn't allow for vector comparison. In vector comparison, you want similar topics to be similar vectors and opposite topics to be opposite vectors. Almost everyone uses boolean-style vectors, but I'm interested in ones that use multiple values. -- kainaw™ 03:40, 1 March 2010 (UTC)[reply]

I'd call it something like "decomposition into independent axes", but that's a description rather than a name. You need to find a certain number of axes such that any data point can be described by its position on each of those axes (ideally with no redundancy - that's the "independent" bit). There are different ways of doing that, depending on what you are starting with. In most real life cases that can't be done in any useful way. For example, you will be able to find films that are funny and dramatic or action filled and romantic (there aren't many, but they are out there if you look for them). You can definitely find people that like both action and romance or both comedy and drama. You may find Opponent-process theory interesting, it's not quite what you are looking for, but closely related. --Tango (talk) 03:58, 1 March 2010 (UTC)[reply]

I agree. I do not feel that vector comparison of data is an optimal way to compare data. However, it is very common. So, in an attempt to learn as much about it as possible (and to cite as much as possible in related Wikipedia articles), I have to study stuff that I do not agree with. -- kainaw™ 04:37, 1 March 2010 (UTC)[reply]

Map each of the four values into a complex fourth root of unity: comedy=1, action=i, drama=−1, romance=−i. This smoothly generalizes male=1, female=−1. Bo Jacoby (talk) 08:00, 1 March 2010 (UTC).[reply]

As an added bonus, you get an automorphism that will turn any romance movie into an action movie, and vice versa. :-) —Bkell (talk) 09:54, 1 March 2010 (UTC)[reply]

Every remote control needs a complex conjugate button! --Tango (talk) 05:10, 2 March 2010 (UTC)[reply]

It sounds as though you want to rank objects in some linearly-ordered way for each of several attributes. How about a "multidimensional spectrum" (a plausible-sounding phrase I just made up but perhaps has been used before), like a political spectrum, for example? —Bkell (talk) 10:01, 1 March 2010 (UTC)[reply]

I was asked this question at Arabic Wikipedia: Does the constant function have infinite number of critical points? I thought it shouldn't have at all but the definition of the critical point allows this since its derivative is always zero at any point.--Email4mobile (talk) 02:59, 1 March 2010 (UTC)[reply]

Technically, yes, every point is critical. --Tango (talk) 03:01, 1 March 2010 (UTC)[reply]

... or, more precisely, every point is a stationary point (since the whole function is stationary). Dbfirs 12:42, 1 March 2010 (UTC)[reply]

But for a critical point or a stationary point, I think there should be some description as: ... where the function "stops" increasing or decreasing. This is not the same in case of a constant function.--Email4mobile (talk) 03:19, 2 March 2010 (UTC)[reply]

That's the motivation behind the definition, certainly, but definitions often behave oddly in corner cases. One could explicitly exclude points around which the function is stationary from the definition, but there really isn't much to be gained by doing so. --Tango (talk) 03:53, 2 March 2010 (UTC)[reply]

I have a process that takes two random numbers A and B - each in the range 0..1 with an even distribution (ie every value is equally probable). What is the distribution of (A+B)/2, (AxB) and the fractional part of (A+B)?

I should be able to figure this out - but somehow I have a brain blockage this morning! (Be gentle - I'm not a math geek!)

TIA SteveBaker (talk) 14:38, 1 March 2010 (UTC)[reply]

If A and B are independent and uniformly distributed on (0,1), then (A+B)/2 has a triangular distribution, while the fractional part of A+B is again uniformly distributed on (0,1). My cursory search hasn't found a standard name for the distribution of the product, but it's given by Prob(A*B<x)=x-ln(x) for x in (0,1). Algebraist 15:07, 1 March 2010 (UTC)[reply]

Many thanks! That's exactly what I needed. SteveBaker (talk) 15:11, 1 March 2010 (UTC)[reply]

The last one can't be right, as ${\displaystyle \lim _{x\to 0+}x-\ln x=+\infty }$.—Emil J. 15:13, 1 March 2010 (UTC)[reply]

The correct formula is Pr(A × B < x) = x(1 − ln x) for x in (0,1].—Emil J. 15:25, 1 March 2010 (UTC)[reply]

Yea it looks like it was a typo. --pma 17:13, 1 March 2010 (UTC)[reply]

And maybe it's worth adding the distribution of max(A,B) which is just Pr(max(A,B)<x)=x² -just because max(A,B)<x is the same as A<x AND B<x, and by the assumed independence the probability of the latter event is Pr(A<x)Pr(B<x)=x²--pma 17:13, 1 March 2010 (UTC)[reply]

Mathworld has Uniform Product Distribution for the product of a number of uniform distributions. Dmcq (talk) 18:49, 1 March 2010 (UTC)[reply]

Consider the map f : Z --> S^1, from the integers to the unit circle, given by f(n) = (cos(n),sin(n)). I'm interested in the image of this map. How will the images of the integers fill up the circle? For example, is it like the irrationals fill up the reals, or like the transcendentals fill up the reals? (This is an open answer, and not a dichotomy between my two examples) •• Fly by Night (talk) 18:21, 1 March 2010 (UTC)[reply]

The transcendentals are uncountable, so it definitely won't be like them. --Tango (talk) 18:29, 1 March 2010 (UTC)[reply]

And so are the irrationals. In fact, transcendentals and irrationals are homeomorphic. The image of the map is a countable dense subset, so in that respect it is similar to the situation of rationals within reals. However, it is hard to answer the question better without a hint of what kind of "being like" you have in mind.—Emil J. 18:37, 1 March 2010 (UTC)[reply]

I was hoping for a tag or a name for the way that the image is distributed about the circle. I mean topologically. •• Fly by Night (talk) 18:47, 1 March 2010 (UTC)[reply]

Well, it's dense and countable, as I wrote. These properties determine it uniquely, as far as topology is concerned: for every two countable dense subsets of S¹ there is a homeomorphism of S¹ to itself which transforms one of the sets into the other.—Emil J. 19:27, 1 March 2010 (UTC)[reply]

So it's like the rational numbers? •• Fly by Night (talk) 20:27, 1 March 2010 (UTC)[reply]

Yes, it's analogous to the rationals in the reals - a countable set that is dense within an uncountable set. --Tango (talk) 20:34, 1 March 2010 (UTC)[reply]

pma's reply should have been here. -- Meni Rosenfeld (talk) 08:04, 2 March 2010 (UTC)[reply]

Sorry, I misread that as "rationals". Yes, the transcendentals and irrationals are essentially the same (they differ by a countable, and thus null, set, the algebraic irrationals). --Tango (talk) 20:34, 1 March 2010 (UTC)[reply]

Very nice. You can make this homeomorpism as close to the identity as you wish, right? I'm quite sure one can even make it an analytic diffeomorphism. --pma 21:21, 1 March 2010 (UTC)[reply]

I beg your pardon, but I didn't mention anything about making the identity close to anything (whatever that might mean). What's an analytic diffeomorphism? •• Fly by Night (talk) 22:00, 1 March 2010 (UTC)[reply]

I think pma's comment was a (misplaced) reply to Emil J.. An analytic diffeomorphism is a homeomorphism that is analytic (can be expressed as a power series) with respect to some differential structure (see those articles for a more precise/rigorous definition). For S¹, there is an obvious differential structure inherited from the plane. --Tango (talk) 22:13, 1 March 2010 (UTC)[reply]

Btw, to be precise you need much more than a differentiable structure to define analytic maps, that is, an analytic structure. And you don't need to derive the (real) analytic structure of S¹ from the plane. It has analytic charts of the form exp(it). Or, think to the quotient R/2πZ.--pma 09:20, 2 March 2010 (UTC)[reply]

Was it misplaced? Indentation does have a meaning, and if possible posts should follow the chronological order in the thread.--pma 22:45, 1 March 2010 (UTC)[reply]

Your comment was immediately below mine and indented more than it, which would usually mean it was a reply to mine, which it wasn't. So, yes, it was misplaced. --Tango (talk) 03:55, 2 March 2010 (UTC)[reply]

Is what Tango says above the common rule? I'm a bit curious to know --pma 07:38, 2 March 2010 (UTC)[reply]

Yes, keeping replies adjacent to posts trumps chronological order. Think of it as a tree structure. I've marked the place where your reply should have been. See also Wikipedia:Indentation. -- Meni Rosenfeld (talk) 08:04, 2 March 2010 (UTC)[reply]

thank you both! --pma 09:21, 2 March 2010 (UTC)[reply]

You're welcome. Wikipedia:Talk page#Indentation is the official guideline on the subject (although it isn't actually tagged as a guideline, for some reason). --Tango (talk) 19:31, 2 March 2010 (UTC)[reply]

I've recently been faced with quite a surprising result, and am having difficulty explaining it - I'll try to pose the problem as succinctly as possible.

Call H the standard epimorphism from the unit quaternions, ${\displaystyle S^{3}}$, to SO(3) (described here). Let ${\displaystyle q=(a,b,c,d)\in S^{3}}$ and define a homomorphism ${\displaystyle \alpha }$ from SO(3) to PSL(2,C) via

${\displaystyle \alpha (H(q))={\begin{pmatrix}a+id&-c+ib\\c+ib&a-id\end{pmatrix}}}$. Now, let ${\displaystyle \sigma }$ be the spinor map from PSL(2,C) to the orthochronous, proper Lorentz transformations, L.

It turns out that the top left 3x3 block matrix of ${\displaystyle \sigma (\alpha (H(q)))}$ is ${\displaystyle H(q)}$ -- so, in some sense, ${\displaystyle \sigma \circ \alpha }$ is the identity on SO(3), since there is a monomorphism embedding SO(3) into L.

I was quite surprised by this -- should I be? I just composed ${\displaystyle \sigma }$ with ${\displaystyle \alpha }$ because I could, I wasn't expecting this to fall out so neatly, and can't explain why it happens. Is there something trivial that I'm missing? Obviously ${\displaystyle \alpha }$ maps onto PSU(2), and the spinor map takes this (or, certainly SU(2)) onto the embedding of SO(3) in L, but that doesn't really demystify anything. Any help is appreciated - thanks, Icthyos (talk) 22:28, 1 March 2010 (UTC)[reply]

I don't really believe your map ${\displaystyle \alpha }$ is well-defined. H(q)=H(-q), but unless I'm missing something ${\displaystyle \alpha }$ does not respect this Tinfoilcat (talk) 11:11, 2 March 2010 (UTC)[reply]

Well, ${\displaystyle \alpha }$ maps into PSL(2,C), so ${\displaystyle \alpha (H(q))={\begin{pmatrix}a+id&-c+ib\\c+ib&a-id\end{pmatrix}}=-{\begin{pmatrix}a+id&-c+ib\\c+ib&a-id\end{pmatrix}}={\begin{pmatrix}-a-id&c-ib\\-c-ib&-a+id\end{pmatrix}}=\alpha (H(-q))}$ . Icthyos (talk) 12:12, 2 March 2010 (UTC)[reply]

I knew I was missing something...Tinfoilcat (talk) 12:38, 2 March 2010 (UTC)[reply]

Think of the quaternions as (certain) 2x2 matrices over the complex numbers in the usual way: then the surjection from the group of unit quaternions to SO(3) composed with your map alpha is simply the natural map that sends a matrix in SL(2,C) to its image in PSL(2,C). Remember that the action of quaternions on 3-space by rotations is as follows: the quaternion q acts on a vector v by sending v to ${\displaystyle qvq^{-1}}$, where v is identified with a quaternion with zero real part. So the image of q in SO(3,R) is this "conjugation rotation". But the spinor map composed with the natural map SU(2) to PSL(2,C) also sends q to the map it induces by conjugation, this time on a certain set of matrices. The 3+1ness is because these matrices are of the form tI + (something independent of t), so conjugation doesn't move tI, but the action on the x,y,z part will be the same as the action on 3-space.Tinfoilcat (talk) 12:57, 2 March 2010 (UTC)[reply]

Thanks so much - I can't believe I overlooked the conjugation connection. I've had pretty much every fact you used written out on a side of A4, but just couldn't thread it together. Thanks again, Icthyos (talk) 15:24, 2 March 2010 (UTC)[reply]

Hi all,

How would I find a diffeomorphism which preserves area on the square |u|<1, |v|<1 with the 2 Riemannian metrics ${\displaystyle {\frac {du^{2}}{(1-v^{2})^{2}}}+{\frac {dv^{2}}{(1-v^{2})^{2}}}}$ and ${\displaystyle {\frac {dv^{2}}{(1-v^{2})^{2}}}+{\frac {du^{2}}{(1-v^{2})^{2}}}}$ (i.e. the diffeomorphism maps from the square with one Riemannian metric to the square with the second Riemannian metric). I really have no idea how to approach this one, since I have basically no experience with diffeomorphisms, and the problem just says 'show one exists', without any hint as to what it might be!

In a similar geometrical vein, how would I show that ${\displaystyle g(z)={\frac {z-a}{z-{\bar {a}}}}}$ defines an isometry from the upper half plane H to the disc D, where ${\displaystyle a\in H}$? I know that the distance in H is given by ${\displaystyle d(x,y)=2tanh^{-1}(|{\frac {x-y}{1-{\bar {x}}y}}|)}$, and I tried substituting ${\displaystyle x={\frac {z_{1}-a}{z_{1}-{\bar {a}}}},y={\frac {z_{2}-a}{z_{2}-{\bar {a}}}}}$ into the distance formula in H, and I ended up with ${\displaystyle 2tanh^{-1}(|{\frac {z_{1}-z_{2}}{(z_{1}-Re(a))+(z_{2}-Re(a))}}}$, and setting ${\displaystyle z_{1}=Re(a)+re^{i{\theta }_{1}},z_{2}=Re(a)+re^{i{\theta }_{2}}}$, we get ${\displaystyle d=2tanh^{-1}(|tanh({\frac {{\theta }_{1}-{\theta }_{2}}{2}})|)=|{\theta }_{1}-{\theta }_{2}|}$, i.e. the argument between the 2 points z1 and z2 when taken around the circle of centre Re(a); but where do I go from here? I have a feeling I'm very close - but is that really the correct distance between z1 and z2 in the Disc model?

Many many thanks! Delaypoems101 (talk) 02:35, 2 March 2010 (UTC)[reply]

Please Explain in detail HOW TO DRAW ANGLES MEASURING 70 degrees and 40 degrees WITHOUT USING PROTRACTOR. Also explain me whether all the angles can be drawn without using protractor. If not what type of angles can be drawn without using a protractor.Kasiraoj (talk) 04:53, 2 March 2010 (UTC)[reply]

If you don't have a protractor, what do you have? A straightedge and compasses? If so, see Straightedge and compass#Constructible angles. As for your specific examples, I don't think either can be constructed. I know 20 degrees can't (that article mentions that is it impossible to trisect ${\displaystyle {\frac {\pi }{3}}=60^{\circ }}$) and you can easily get from either 70 degrees or 40 degrees to 20 degrees (30 is easy to construct and 70-30=40, you can then bisect 40 to get 20). --Tango (talk) 05:08, 2 March 2010 (UTC)[reply]

Neither is possible by classical (precise) methods, but if your ruler has accurate markings you can use Angle trisection for an approximate method. (e.g. for 70 degrees, construct two 60 degree angles with a common arm, bisect one to get 30 degrees, then trisect this to give ten degrees next to the sixty degrees). This method will not be accepted by Pure Mathematicians, but may well be taught in Engineering as a good approximation for most practical purposes. There may be even more accurate methods, but mathematicians can prove that none of them can be exact in a super-precise mathematical sense. Surprisingly, it is possible to construct 69 degrees exactly (I'm working on the method). Dbfirs 08:21, 2 March 2010 (UTC)[reply]

Why do you say pure mathematicians wouldn't accept angle trisection by compass and marked ruler? Euclid might not have done, but pure maths has moved on quite a bit since him. Algebraist 13:23, 2 March 2010 (UTC)[reply]

Because, from a "perfect precision" point of view, all trisection methods involve measuring a length, and this inevitable relies on the accuracy of the ruler and the trisecter's estimating ability. There is a sense in which Euclidean constructions depend only on the perfection of ideal compasses and straight edge, and not on the user's ability to estimate. Dbfirs 14:36, 2 March 2010 (UTC)[reply]

You can trisect an angle without even the ruler and compass :) See [2] for instance. Dmcq (talk) 13:27, 2 March 2010 (UTC)[reply]

Yes, I like that method, but it still involves estimating a length doesn't it? Dbfirs 14:36, 2 March 2010 (UTC)[reply]

No estimation, but it is a Neusis construction like the various classical methods. Dmcq (talk) 14:47, 2 March 2010 (UTC)[reply]

Thanks for the link, but may I quote from it "Neusis became a kind of last resort that was invoked only when all other, more respectable, methods had failed." I still consider that it involves a form of visual estimation. Dbfirs 16:14, 2 March 2010 (UTC)[reply]

Depends on your standards. It is more difficult than extending a compass from one point to another though that also involves estimating. Dmcq (talk) 18:51, 2 March 2010 (UTC)[reply]

Yes, good point! Dbfirs 19:37, 2 March 2010 (UTC)[reply]

Am I correct in thinking that the only (non-Neusis) constructible integer angles are those that are multiples of three degrees? Dbfirs 19:53, 2 March 2010 (UTC)[reply]

Yes. Algebraist 23:42, 2 March 2010 (UTC)[reply]

Hello you all over the world. Please excuse my poor English, I'm a froggy. I'm puzzled by the folowing enigma that shows that ALL x=0! I can't find the mistake though I think it must be just a small trick. For your answers, please consider that I should be able to anderstand an advenced level in maths (4 years in university but it was 30 years ago).

For all x belonging to Z we have the identity :

e^ix = (e^ix)¹ = e^ix(2π/2π) = (e^i2π)^x/2π = 1^x/2π = 1

but knowing that e^ix=cos x + i sin x it brings

cos x = 1 and sin x = 0 , that is x=0 FOR ALL x ! Of course, in fact the correct conclusion should be x=2kπ, k in Z, but anyway the trouble is the same.

Thank you for your help. In a hurry to read your answers. Joël DESHAIES-Rheims-France---82.216.68.31 (talk) 13:01, 2 March 2010 (UTC)[reply]

Bonjour Joël. It's because ${\displaystyle a^{bc}=(a^{b})^{c}}$ is not always true when a,b,c are complex numbers. Have a look at the complex numbers secton of Exponentiation Tinfoilcat (talk) 13:13, 2 March 2010 (UTC)[reply]

Well, it's quite simple: raising to power is not commutative in complex numbers. Additionaly, and even more important, raising to power with non-natural exponent is NOT a function—it may be a multi-function (think eg. of (complex)√1), so it is important to carefully analyse which of possibly many values you take.
Compare (complex) (1^2)^(1/2) and (1^(1/2))^2. --CiaPan (talk) 13:20, 2 March 2010 (UTC)[reply]

Hello, I'm the asker. Thank you for clear explanations, I got it.Joël DESHAIES.Rheims-France---82.216.68.31 (talk) 15:50, 2 March 2010 (UTC)[reply]

That cos x = 1 and sin x = 0 does not imply that x = 0. Rather, it implies that x is an integer multiple of 2π, i.e. x ∈ {0, ±2π, ±4π, ...}. Michael Hardy (talk) 15:54, 2 March 2010 (UTC)[reply]

Hello Michael, please see my 9th line at the top, I wrote this remark. My title is x=0 because it's more "dramatic", more thrilling! Joël.--82.216.68.31 (talk) 16:17, 2 March 2010 (UTC)[reply]

Hi Joël. Note that the point here is not specific of complex numbers. You may do the same fallacy, and maybe see it better, with -1=(-1)^2/2=[(-1)²]^1/2=1. The point is that a left inverse to a map need not to be a right inverse to it; see left and right inverses. --pma 17:10, 2 March 2010 (UTC)[reply]

I was looking at the logistic function. This looks like cumulative distribution function of some sort. I decided to discover which distribution corresponded to the logistic function. If p(t) is the distribution that we want, then the logistic function and p(t) are related by

${\displaystyle \int _{-\infty }^{x}p(t)\ {\mbox{d}}t={\frac {1}{1+e^{-x}}}.}$

Differentiating both sides of the equation by x gives

${\displaystyle p(x)={\frac {e^{-x}}{(1+e^{-x})^{2}}},\ \ {\mbox{i.e.}}\ \ p(t)={\frac {e^{-t}}{(1+e^{-t})^{2}}}}$

We can show that this is right by integrating this last expression for p(t) with respect to t with t ≤ x. Doing this, we arrive back at the first equality. What kind of a probability distribution is this p(t)? It looks like a bell-shaped curve but the formula doesn't look like the expression for the Gaussian distribution. Any ideas? •• Fly by Night (talk) 16:30, 2 March 2010 (UTC)[reply]

See Logistic distribution. -- Meni Rosenfeld (talk) 16:44, 2 March 2010 (UTC)[reply]

Ahhh... D'OH! •• Fly by Night (talk) 16:52, 2 March 2010 (UTC)[reply]

Let S=1+2+4+8+... Then, 2S=2+4+8+16+... Subtracting produces

 2S=  2+4+8+...
- S=1+2+4+8+...
---------------
  S=-1

This is clearly absurd, but where's the error in this proof? --J4\/4 <talk> 16:37, 2 March 2010 (UTC)[reply]

The error is in the subtraction. Since S is infinite, you can't subtract ${\displaystyle 2S-S}$.

Alternatively, if your definition of infinite sums doesn't allow convergence to infinity, the first error is writing S=1+2+4+8+... to begin with.

Note, however, that this is only absurd in the (extended) reals. It is perfectly valid in, say, the 2-adics. -- Meni Rosenfeld (talk) 16:42, 2 March 2010 (UTC)[reply]

But since that's the same logic that's used to derive the formula S=a/(1-r) for an infinite series, doesn't that mean that the formula is invalid as well, since it's also an infinite series? --J4\/4 <talk> 16:46, 2 March 2010 (UTC)[reply]

What I meant is not that this is an infinite series but that the sum is infinite. If ${\displaystyle |r|<1}$ then ${\displaystyle \sum _{n=0}^{\infty }ar^{n}}$ converges to a real number, and the above manipulations are valid. -- Meni Rosenfeld (talk) 16:52, 2 March 2010 (UTC)[reply]

See 1 + 2 + 4 + 8 + · · ·. Gandalf61 (talk) 16:56, 2 March 2010 (UTC)[reply]

You can only use the standard arithmetic rules to manipulate infinite series if they converge. Yours doesn't, so the rules don't apply. (If you use a different topology, in which it does converge, then your answer is correct - the 2-adic numbers have one such topology.) --Tango (talk) 19:58, 2 March 2010 (UTC)[reply]

See also Divergent series#Theorems on methods for summing divergent series. Zunaid 09:22, 3 March 2010 (UTC)[reply]

This is homework. The assignment makes me consider the fourth order taylor polynomial p(x) of the function ${\displaystyle f(x)={\sqrt[{3}]{x+1}}}$, around x=0, and then (quoting literally): "with error ${\displaystyle E(x)=f(x)-p(x)}$, show that ${\displaystyle |E(x)|\leq {\frac {1}{3}}\cdot 10^{-6}}$ for ${\displaystyle |x|\leq 10^{-1}}$

I'm quite sure I didn't make any mistakes in the polynomial, derivatives or anything of that kind. I end up with the following equation using Taylor's theorem:

${\displaystyle |Error|\leq {\frac {|x|^{5}}{5!}}\cdot \max |{\frac {880}{243\cdot (x+1)^{14/3}}}|}$

The maximization on the right hand side being the fifth derivative. I'm quite sure about the above, too. Now comes the question: if I take ${\displaystyle x\in [-0.1,0.1]}$, the right hand side of the above equation results in approximately ${\displaystyle {\frac {1}{2}}\cdot 10^{-6}}$ at x=-0.1.

Assuming my professor isn't fallible and doesn't ask me to prove things he cannot, what did I do wrong here? Don't give the answer away please, though :) User:Krator (t c) 01:37, 3 March 2010 (UTC)[reply]

Not sure what your professor meant to have you do, but how about this: expand the series to five terms. The total error must be less than the sum of the fifth term plus the result of the error formula applied with 5 terms. For most functions seen in practice the actual error will be less than that given by the error formula.--RDBury (talk) 04:22, 3 March 2010 (UTC)[reply]

That's what I would suggest too. Note that the integral form of the remainder, evaluated with maple, gives R<0.255*10^-6. PS: sorry, I thought to 0≤x≤1/10, I didn't notice you wrote |x|≤1/10. --pma 08:09, 3 March 2010 (UTC)[reply]

According to my calculations, your calculations are incorrect, since the true error is >0.327*10^-6 at -0.1.

Also, five terms is not strong enough for the required estimate. Six terms work, though. -- Meni Rosenfeld (talk) 08:30, 3 March 2010 (UTC)[reply]

Thanks. So a possible explanation is that the original problem had 0≤x≤1/10 and not |x|≤1/10, in which case the professor's bound should be correct. --pma 10:04, 3 March 2010 (UTC)[reply]

Sorry, I wasn't clear. The professor's bound for the error of the fourth order expansion is correct for |x|≤1/10. To prove this you need to consider the sixth order expansion.

Anyway, your bound for R isn't correct for 0.1 either. -- Meni Rosenfeld (talk) 10:13, 3 March 2010 (UTC)[reply]

The assignment specifically tells me to consider the fourth order expansion, so I calculated the error using the fifth order expansion, for which the error bound is incorrect as I wrote earlier. Why do I need to consider the sixth? I'm afraid I don't understand that. User:Krator (t c) 11:42, 3 March 2010 (UTC)[reply]

The assignment asked you to calculate the error of the fourth order expansion. What you tried to do is bound the error using Taylor's theorem for the fourth order expansion, which involves the fifth derivative. As you have noticed, this doesn't give you the bound you want.

RDBury suggested using Taylor's theorem for the fifth order expansion (which involves the sixth derivative) to tighten the bound, and I corrected that fifth isn't enough - you need the theorem for the sixth order expansion, which uses the seventh derivative.

The proof involves the triangle inequality - if ${\displaystyle p_{n}(x)}$ is the nth order expansion of ${\displaystyle f(x)}$, then ${\displaystyle |f(x)-p_{4}(x)|\leq |f(x)-p_{6}(x)|+|p_{6}(x)-p_{4}(x)|}$. You need to bound each of these summands, and then the bound you want will follow.

Think of it this way - the more terms you use, the more accurate you are. Usually what we want to be accurate about is ${\displaystyle f(x)}$, but here we want to be accurate about ${\displaystyle f(x)-p_{4}(x)}$. -- Meni Rosenfeld (talk) 13:11, 3 March 2010 (UTC)[reply]

${\displaystyle .321419721*10^{-7}}$ ? User:Krator (t c) 14:13, 3 March 2010 (UTC)[reply]

Hello. Consider the equation tanθ=sinφ/(r+cosφ). If r<1, tanθ can equal any value, and therefore there is no limit to the value of θ. However, if r>1, then there should be a maximum value for θ. How can you find this value? I tried finding when the derivative equals 0, but I just ended up with cosφ=-r, which clearly isn't possible if r>1. —Preceding unsigned comment added by 173.179.59.66 (talk) 11:49, 3 March 2010 (UTC)[reply]

Check your calculations, you should have gotten ${\displaystyle \cos \varphi =-1/r}$ which means ${\displaystyle |\tan \theta |\leq {\frac {1}{\sqrt {r^{2}-1}}}}$. -- Meni Rosenfeld (talk) 13:18, 3 March 2010 (UTC)[reply]

Regardless of whether its domain is a sub-domain of the real domain, or of the complex domain, or of any other topological domain. HOOTmag (talk) 13:56, 3 March 2010 (UTC)[reply]