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February 22

Radians and calculus

Is there an intuitive/easy way to see why measuring angles in radians leads to nicer formulas in calculus? 74.15.139.132 (talk) 01:30, 22 February 2012 (UTC)[reply]

If you plot a curve of sine as a function of degrees with equal scaling on the x & y axes, then it is a long, long, stretched-out graph, with the maximum slope going up the curve of only π/180 ≈ 0.0175, while if you plot it as a function of radians, then it is a more "natural" looking graph, with a maximum slope going up the curve of 1. In fact, the slope of sin(x) at each point is exactly cos(x). Thus the first derivative of the sine is the cosine, or in formula d/dx sin(x) = cos(x). If we were to define a function sin_d(x) which was the sine as a function of degrees, then the derivative would be d/dx sin_d(x) = π/180 * cos_d(x). That is messy enough, but in radians we know that the fourth derivative of the sine function is the sine function itself, where in degrees, the fourth derivative of sin_d(x) is (π/180)^4 * sin_d(x). What a mess!. Radians give these functions the right horizontal scale, somewhat in the same sense that the e, the natural log base, is just the right choice for the exponential function so that it's derivative is itself. -- 203.82.91.152 (talk) 04:25, 22 February 2012 (UTC)[reply]

I know that, I'm wondering if there's an intuitive way of seeing why this is true for radians, defined as 1/(2*pi) of a circle, w/o explicit computation. 74.15.139.132 (talk) 18:35, 22 February 2012 (UTC)[reply]

OK. Our trig functions are easily represented on a unit circle because that sets the hypotenuse to 1 which make the ratios which define the trig functions trivial. And an angle is not just the central wedge shape that you measure with a protractor, but it can also be thought of as the length of the arc of the unit circle that is cut out by the wedge, and that is the definition of radians. A 1 radian angle cuts out an arc of the unit circle with length 1. Why is this important to calculus? The coordinates of a point on the unit circle, θ up (CCW) from the x-axis, are (x, y) = (cos θ, sin θ), for whatever measure of angle you are using (assuming that your sine and cosine function are intended to take that measure of angle). When you choose radians you get the advantage that for the initial, very small movement up from θ = 0, you find ∆θ ≈ ∆y, so that lim ∆θ -> 0 of ∆y/∆θ = 1. Since y = sin θ, we just showed that d/dθ sin θ = 1 for θ = 0, and in general we find that d/dθ sin θ = cos θ without some arbitrary and annoying scaling factor, such as π/180. So the intuitive feature of radians you are seeking is linked to their property of representing distance around the unit circle. -- ToE 02:36, 23 February 2012 (UTC)[reply]

One nice thing is looking at the area element

rdrd\theta

when integrating in polar coordinates. This its quite easy to see working in radians as the length of an arc is the radius times the angle.--Salix (talk): 21:45, 22 February 2012 (UTC)[reply]

As alluded to above, if we only used degrees, the derivative of sin at 0 would be π/180. In radians, it's exactly 1 instead, which is nice. For this to happen, we need the limit as θ tends to 0 of (sin θ)/θ to be 1. (This is because if you look at the derivative of sin at 0 through its definition as a limit, this is precisely the limit you obtain.) So what it boils down to is we want to know why sin θ and θ are roughly equal when θ is very small and measured in radians. Now have a look at this illustration. sin θ is the length of the dotted vertical segment. θ is the length of the circular sector to the right of the vertical dotted segment (but only if θ is measured in radians). Now imagine that θ becomes very small, so that the point on the circle slides towards (1,0). You can see, at least intuitively, that these two lengths are going to be very close in relation to one another (that is, their ratio is going to become close to 1). All of this only works if the angle is measured in radians. 96.46.204.126 (talk) 22:11, 22 February 2012 (UTC)[reply]

In addition, the series expansion for sin and cos are in radians. In all of these cases, using radians (the natural unit of angles) eliminates constants (which are akin to unit conversions). And because of the series, sin(x) is approximately x when x is small, and x is in radians. And the inverse trig functions naturally give their result in radians. Bubba73 ^{You talkin' to me?} 03:03, 23 February 2012 (UTC)[reply]

Matrix Eigenvalues

I need to find the eigenvalues and eigenvectors of the matrix

\mathbf {A} ={\begin{bmatrix}\mathbf {O} &\mathbf {I} &\mathbf {O} &\mathbf {O} \\-\mathbf {B} &\mathbf {O} &\mathbf {S} &\mathbf {O} \\\mathbf {O} &\mathbf {O} &\mathbf {O} &\mathbf {I} \\{\mathbf {DB} }&\mathbf {O} &-{\mathbf {D(C-S)} }&\mathbf {O} \\\end{bmatrix}}

where O is the zero matrix, I is identity matrix, B is a circulant tridiagonal matrix with elements (-1,2,-1) and C and D are diagonal matrices with constant diagonal terms (in other words, a scalar times the identity matrix.) S is a circulant matrix (or a diagonal matrix, if that helps.) I am hoping that the presence of large number of O's , I's and simple matrices would lead to a closed form solution for the eigenvalues and the eigenvectors. Any help will be sincerely appreciated. deeptrivia (talk) 03:35, 22 February 2012 (UTC)[reply]

you can at least evaluate the determinant more easily: Swap rows and columns around in groups. You can easily make the two I's appear on the top left and lower right by swapping rows 1 and 2, and rows 3 and 4. Then you have

\pm \left|B\right|\times \left|D(C-S)\right|\pm \left|S\right|\left|DB\right|

. The signs are given by the parity of the size of the two I's. HTH, Robinh (talk) 20:38, 22 February 2012 (UTC)[reply]

New method for long division

Please excuse my math ignorance, but I have no way of asking this question without seeming like a complete dunce when it comes to math (true): My son is starting long division in grade school. I tried helping him with his homework, but I can't make heads or tails of what is going on here. When I was in school (back in the 80's) we did long division by dividing into the first set of numbers, subtracting out, dropping down the next set of numbers, and continuing the process until completed with a remainder. It still works, but it definitely not the way they are teaching it now. And I am stumped. I know you can't help "do my son's homework for him," so I won't include the actual problem, but is there someone out there who is privy to this "new way" of teaching long division to elementary aged kids that can point me to some helpful resources? The problems are basic long division, no variables or square roots or anything like that yet. Quinn ^{░ RAIN} 05:34, 22 February 2012 (UTC)[reply]

PS:It has something to do with doubling the divisor, if that helps. Quinn ^{░ RAIN} 05:39, 22 February 2012 (UTC)[reply]

I'm not familiar with a different way to do division, but the doubling you mention sounds like it might be a square-root algorithm. Could it be this: http://www.homeschoolmath.net/teaching/square-root-algorithm.php (scroll down a bit)?--121.74.109.179 (talk) 05:51, 22 February 2012 (UTC)[reply]

Yes it seems that Long Division Teaching Aid, "Double Division" is basically what is going on here. Can anyone "dumbdown" the method for me so I can grasp what they're wanting? We have to show-the-work and I'm an English major, so I "need" to understand the concept before I can wrap my brain around it to try to explain to my son. 06:32, 22 February 2012 (UTC)

Or maybe an example with explanation? I don't have to understand how A=B, I just need to know how A-gets-to-B (if that' makes sense) so I have a broad understanding of what I'm trying to do here. After that, I can trial and error the problems, and hopefully come up with a way to assist my son with his homework. Quinn ^{░ RAIN} 06:37, 22 February 2012 (UTC)[reply]

It's not really all that different from long division. They explicitly fill in the zeros to the right, where long division leave potentially confusing blanks. They also precompute multiples of the divisor, but only the 1x, 2x, 4x, and 8x multiples. Doing this involves the three doublings you mentioned. Doubling is cheap, but the price they pay is that where you would get a 7 in the answer via long division, they get a 4, a 2, and a 1 in three different steps, and have to sum them for a final answer. If this is used commonly enough, and we can find references in educational literature, it would certainly deserve mention in long division, if not its own article. -- ToE 06:41, 22 February 2012 (UTC)[reply]

I just noticed that the link I gave above has a nice applet on the right which will step you through example problems of your choice, with explanation for each step taken. Why don't you give that a try and ask here if something doesn't make sense? I was about to write up an example along the lines of those in Long division#Method, but that applet will probably do a better job of explaining things. Cheers! -- ToE 06:50, 22 February 2012 (UTC)[reply]

I don't think they should change methods like this unless the new method is much better than the old one. After all, people need to be able to check each other's work, which we can't do if we all use different systems. I've sometimes experienced this when getting change back, when instead of counting up the change, they count down from the amount tendered until they get to the amount of the bill. I find that very confusing, and insist on recounting it myself. So, this sure doesn't save any time. StuRat (talk) 07:11, 22 February 2012 (UTC) [reply]

The method sounds related to Ancient Egyptian multiplication or Egyptian multiplication and division. Bubba73 ^{You talkin' to me?} 17:35, 22 February 2012 (UTC)[reply]

Well I've got the hang of it now (the applet, which I didn't notice at first, was the key, so thanks for that), but I have to say in the long run it is not a time saver vs. the "old" version, but oh well. Concur with StuRat above. Thanks all! Quinn ^{░ RAIN} 01:16, 23 February 2012 (UTC)[reply]

Great, I will mark this resolved. StuRat (talk) 04:41, 24 February 2012 (UTC)[reply]

Resolved

Quinn, do you know if your son's school is teaching this method instead of long division, or as a precursor to long division? -- ToE 15:20, 24 February 2012 (UTC)[reply]

I do not know. That's an interesting question. I realize this thread is marked resolved, but I am extremely curious why you ask this. Do you think this is something I should inquire into (beyond simply being interested in his schooling) and why? Please reply here or on my talk. Thanks, Quinn ^{░ RAIN} 03:10, 25 February 2012 (UTC)[reply]

I think the point is, that if they only teach this method, and not the normal way, those kids will be at a disadvantage when trying to work with others who use the normal method. StuRat (talk) 03:17, 25 February 2012 (UTC)[reply]

I was only curious. Were I the parent of a child who was being taught this method to the exclusions of traditional long division, I would feel compelled to at least demonstrate the old method while pointing out the similarities and differences, and I'd have to consider if it was worthwhile fully teaching the old method, but I did not intend to imply any parental advice in my question. -- ToE 08:26, 25 February 2012 (UTC)[reply]

... and I suspect that the denizens of this ref desk are sufficiently interested in developments in mathematics education that, even if you don't meet with your son's teacher for a couple of months (they still do semi-regular parent-teacher get-togethers, right?), we'd enjoy seeing a followup section posted here. -- ToE 10:52, 25 February 2012 (UTC)[reply]

Quadratic Eigenvalue Problem with Circulant Matrices

I have solve the following quadratic eigenvalue problem:

det(s^{2}\mathbf {I} +s\mathbf {B} +\mathbf {C} )=0

where B and C are circulant matrices. Since the eigenvalues of circulant matrices are known in closed form, I'm hoping there would be a closed form solution in terms of eigenvalues of B and C. deeptrivia (talk) 11:17, 22 February 2012 (UTC)[reply]

This is just a stab in the dark, so check it thoroughly. Since all three of the summands are circulant, isn't the matrix inside your determinant circulant itself? After that, it looks like its eigenvalues are the sum of the eigenvalues of the three matrices. That sounds to me like a connection you're looking for. I'm doing this hastily, so apologies if I made a mistake and I mislead you. Rschwieb (talk) 18:59, 23 February 2012 (UTC)[reply]

I removed "Matrix Eigenvalues" as the title, since that is also a title of a previous section, and section names must be unique, both for proper human indexing and wiki indexing. StuRat (talk) 19:16, 24 February 2012 (UTC) [reply]

Finding quadratic formula

long ago I found a wikipedia article about how to "get" the quadratic formula from $ax^{2}+bx+c=0$ ?? But I cannot find it now? someone may help? Thanks! 190.158.184.192 (talk) 17:53, 22 February 2012 (UTC) (PS: The process of getting the formula itself.)[reply]

Try Quadratic equation#Derivations of the quadratic formula. Qwfp (talk) 18:12, 22 February 2012 (UTC)[reply]

Curl

In maxwell's equations the curl of electric and magnetic fields are used, but curl only operates on functions of 3 variables and E and B generally are 4 variables because they depend on time as well. Can someone explain? Money is tight (talk) 18:20, 22 February 2012 (UTC)[reply]

The curl acts only on the spatial variables. Sławomir Biały (talk) 22:59, 22 February 2012 (UTC)[reply]

The E and B fields are generally functions of four variables (coordinates x, y, z and t), but the curl only involves partial derivatives with respect to three of these variables. Each partial derivative treats the other three variables as constant. The partial derivative with respect to the fourth variable t generally occurs in the same equations alongside the curl (or div). This 3+1 separation is an artifact of the choice of opertors. In a treatment such as in geometric algebra that puts all four the these coordinates on an equal footing replaces the curl, div and time derivative with one that combines all four partial derivatives into one operator, and in the process simplifies the statement of Maxwell's equations. — Quondum ^☏✎ 04:52, 23 February 2012 (UTC)[reply]

Ok, so does this mean say we express the electric field as (F(x,y,z,t),G(x,y,z,t),H(x,y,z,t)), the curl of E is calculated using the usual formula with partial derivatives only involving x,y,z? Money is tight (talk) 22:53, 25 February 2012 (UTC)[reply]

And then you would have the curl vector field with t as a parameter (I think). Rschwieb (talk) 01:57, 29 February 2012 (UTC)[reply]

February 23

Math Problem

5≤ -s/12 -3 — Preceding unsigned comment added by 68.80.157.180 (talk) 02:17, 23 February 2012 (UTC)[reply]

Did you have a question? RudolfRed (talk) 02:48, 23 February 2012 (UTC)[reply]

I added a title. I think we can assume you want us to help solve the inequality for s. It works pretty much like solving an equality, with one big exception: "When you multiply or divide both sides of an inequality by a negative number, you also flip the sign of the inequality". Thus, you will flip ≤ to ≥. (There are other differences in solving inequalities, such as dealing with square roots, but those don't come into play here.) If you show us your work, we would be glad to check it for you. StuRat (talk) 03:09, 23 February 2012 (UTC)[reply]

Try www.wolframalpha.com : [1] Bo Jacoby (talk) 08:36, 23 February 2012 (UTC).[reply]

Trig function help

Hello, how would I accomplish this trig problem?

cos (arcsin (5/13) - arctan (3/4))

thanks.--Prowress (talk) 03:43, 23 February 2012 (UTC)[reply]

Hint: determine the missing lengths of the triangles involved, then use the identity $\cos {(A-B)}=\cos {A}\cos {B}+\sin {A}\sin {B}$ . --Kinu ^t/_c 04:19, 23 February 2012 (UTC)[reply]

Difference between a series and a progression

Could you please explain the difference between a Series and a progression with examples? Kasiraoj (talk) 08:09, 23 February 2012 (UTC)[reply]

You could start by working out the difference between a sequence and a series, and then seeing how the listed examples of a progression fit into the picture. — Quondum ^☏✎ 09:04, 23 February 2012 (UTC)[reply]

Uniqueness of injective/flat resolutions

The article resolution (algebra) article mentions projective resolutions are unique up to chain homotopy. I would guess the same can be said for injective and flat resolutions, but I'm not familiar enough with the subject matter to verify. Is this the case? Thanks! Rschwieb (talk) 18:27, 23 February 2012 (UTC)[reply]

Yes injective resolutions are unique up to homotopy, however I don't know about flat ones, although I think they're unique up to quasi-isomorphism. Money is tight (talk) 22:48, 25 February 2012 (UTC)[reply]

If you have the time to point me to a reference that I can look at, I can take care of the footwork to put in the info :) Thanks! Rschwieb (talk) 14:16, 27 February 2012 (UTC)[reply]

Heart-shaped function

I found it quite interesting to find out that the function:

$(\cos(100x){\sqrt {\cos(x)}}+{\sqrt {|x|}}-0.7)(4-x^{2})^{0.01}$

gives a good approximation of a heart shape when viewed graphically (e.g. here).

How would this fairly complicated equation have been derived? Trial and error, or chance? Is there software for generating a function that will deliver a certain shape? --Iae (talk) 20:01, 23 February 2012 (UTC)[reply]

I don't know, but a Limaçon can be heart-shaped. Bubba73 ^{You talkin' to me?} 20:20, 23 February 2012 (UTC)[reply]

You might also be interested in [2]. There was probably some trial and error involved in finding the function but there is some cleverness involved as well. For example the cos 100x produces the up and down squiggles that fill in the shape. The sqrt(cos x) gives the rounded bits on either side and the sqrt(|x|) produces the V shape in the middle. You might try playing with the formula to see what effects small changes have, for example to find out what happens if you change the .7 to a .8 or a .6.--RDBury (talk) 22:45, 23 February 2012 (UTC)[reply]

MATH GESSUING

I am a two digit number over 50.When you put me in groups of 7,2 are left over.The sum of my digits are 11. What number am I ? — Preceding unsigned comment added by 24.163.1.170 (talk) 22:01, 23 February 2012 (UTC)[reply]

You are not a number! You are a free man. --Trovatore (talk) 22:17, 23 February 2012 (UTC)[reply]

The answer is 65. The title of this question is misleading, though. --COVIZAPIBETEFOKY (talk) 22:50, 23 February 2012 (UTC)[reply]

I'm still trying to figure out why the "over 50" caveat is given. Number is of the form 7n+2 from the one criteria and a subset of those of the form 9n+2 from the other so the answers are a subset of the numbers of the form 63n+2, for n=0 the answer is 2 (which doesn't fit the second criteria exactly, n=1 gives 65 and n=2 gives 128 which is too big. Without the requirement that it be a two digit number, the numbers start with (marked * sum to another 9n+2 other than 11) are 2*,65, 128, 191, 254, 317, 380, 443, 506, 569*, 632, 695*, 758*, 821, 884*, 947*, 1010*, 1073, etc. I don't know if there is a clean way to find out what the last number of the form 63n+2 whose digits sum to 11, but we know it can't be more than 11111111111.Naraht (talk) 03:52, 24 February 2012 (UTC)[reply]

What about 10000001111111111, 10000000000001111111111, 10000000000000000001111111111, ... ?--RDBury (talk) 12:01, 24 February 2012 (UTC)[reply]

There isn't a largest. 6 + 3 + 2 = 11, so setting n to any (positive integer) power of 10 gives you a number that fits the form. 632, 6302, 63002, 630002... Smurrayinchester 11:29, 27 February 2012 (UTC)[reply]

86.174.199.35 (talk) 16:40, 24 February 2012 (UTC)[reply]

Area enclosed

Hello. Suppose I have an arbitrary function defined by a parametric which on some interval of the parameter is closed and simple. How would I find the area enclosed by the curve? (without doing something ugly like splitting it into two regular functions and integrating each, then subtracting of course... *shudder*...) Thanks. 24.92.85.35 (talk) 23:50, 23 February 2012 (UTC)[reply]

See Green's theorem#Area_Calculation – b_jonas 10:08, 24 February 2012 (UTC) (Update: deleted the formula. It was wrong. – b_jonas 10:09, 24 February 2012 (UTC))[reply]

This means that if you have the point of the curve parametrized as (x(t), y(t)) where ~~(x runs)~~ t runs from a to b, and this is a closed curve so x(a) = x(b) and y(a) = y(b), then you can get the area as the absolute value of

\int _{a}^{b}y{\frac {\partial x(t)}{\partial t}}\,dt

– b_jonas 10:13, 24 February 2012 (UTC)[reply]

t runs from a to b.86.174.199.35 (talk) 16:40, 24 February 2012 (UTC)[reply]

Right. Fixed it. – b_jonas 17:08, 25 February 2012 (UTC)[reply]

February 24

Solids of revolution by cylindrical shells

Resolved

– I got

20π/3

which matches the computer answer, and it's due in an hour. Feel free to check my work. --Ks1stm (talk) ^{[alternative account of Ks0stm]} 17:24, 24 February 2012 (UTC)[reply]

My calculus textbook says "the volume of the solid obtained by rotating about the y-axis the region under the curve $y = f(x)$ from $a$ to $b$ is $V = \int b a 2 π x f(x) dx$ where $0 \leq a < b$ ." That's fine and all except one of my homework problems is to find the volume of the solid obtained by rotating the region between $y = x 2$ and $y = 2 - x 2$ about $x = 1$ , and this region would produce an $a$ less than 0. Will it still get me the correct answer if I attempt to find the volume by doing $V = \int 1 -1 2 π (1- x) (2-2 x 2) dx$ or will it be thrown off by $a$ being less than 0? Ks0stm ^{(T•C•G•E)} 00:55, 24 February 2012 (UTC)[reply]

Just impose a new coordinate system such that x=1 in the old one is x'=0 (y axis) in the new one and you won't have that problem. 24.92.85.35 (talk) 03:13, 24 February 2012 (UTC)[reply]

If I'm not mistaken, the definition is worded in a manner to ensure that the radii are all positive (and only included in the integration once, of course). In this case, because r=(1-x) and your limits are x=-1..1, these conditions still hold. --Kinu ^t/_c 03:26, 24 February 2012 (UTC)[reply]

Length of the curve of any polynomial

How is the length of any polynomial between two points calculated? --Melab±1 ☎ 04:24, 24 February 2012 (UTC)[reply]

See Arc_length#Finding_arc_lengths_by_integrating. StuRat (talk) 04:38, 24 February 2012 (UTC)[reply]

Would all polynomials substituted into the equation be integrable though? --Melab±1 ☎ 17:50, 24 February 2012 (UTC)[reply]

No, this problem is one of the motivations behind Elliptic integrals.--Salix (talk): 18:23, 24 February 2012 (UTC)[reply]

Clarification: all the integrands at arc length are integrable functions, in the sense that the definite integrals exist as finite real numbers (they also satisfy the conditions given by Riemann_integral#Integrability whenever f is a poynomial). However, there is no guarantee that the indefinite integral has an analytic expression, which leads us to Salix's comment above. (I only bring this up because, as I understand it, Melab is using 'integrable' to mean 'has an analytic antiderivative', which is not standard usage. (also feel free to chastise/correct me if I've missed something here;)) SemanticMantis (talk) 21:29, 24 February 2012 (UTC)[reply]

That is quite alright. But, I am confused if you mean that not all polynomials will produce an analytic antiderivative. I can understand that not all non-polynomial functions would have an analytical arc length. --Melab±1 ☎ 15:11, 25 February 2012 (UTC)[reply]

Yep, you have it. The antiderivative of a polynomial is a polynomial. But the integrands for arc length of a polynomial are not always polynomials, as Salix mentions. SemanticMantis (talk) 15:28, 25 February 2012 (UTC)[reply]

Question: limits of partially defined functions on the reals

I was alerted to a quirk about limits which I do not recall hearing before. Suppose f(x)=1 on the irrationals, and is undefined on the rationals. Topologically, when considered as a function from the irrationals into the reals, it has limit 1 as x approaches 0. It is appealing then to say that this function has a limit as a partially defined function on R. However the current version of the wiki article on function limits requires that f be defined at least on a (connected) open interval with endpoint 0. This stricter requirement would exclude the function described above from having a limit, due to the dense set of removable discontinuities.

Question(s): Is there anything foundationally wrong with taking the wiki definition and "excusing" countably many removable discontinuities? (I.e. we would say "if x is less than delta from c AND f(x) is defined, then f(x) is less than epsilon from L") Does anybody know if any authors address this?

I'm not interested in altering the wiki article, this is just curiosity. Rschwieb (talk) 16:35, 24 February 2012 (UTC)[reply]

You could say that the property holds almost everywhere, if nothing else. Looie496 (talk) 18:18, 24 February 2012 (UTC)[reply]

That doesn't work. Consider the function that is 1 on the irrationals and 0 on the rationals instead of undefined. The limit of this function at 0 doesn't exist. The function you describe is actually continuous (the pre-image of open sets is open), but its domain is the irrational numbers and so we're dealing with the topology of the irrationals. There's no sense in which a function can be undefined at a point in its domain. Rckrone (talk) 18:29, 24 February 2012 (UTC)[reply]

Limit of a function#Functions on metric spaces appears to comfortably accommodate this (including sparse/countable domains); it involves only defining the set (domain) over which the limit is taken (as opposed to allowing a domain on which a function is undefined(!)), and that there must exist elements of the domain within any nonzero distance from the limiting point not counting the limiting point itself. — Quondum ^☏✎ 18:35, 24 February 2012 (UTC)[reply]

(edit conflict) What you're talking about makes perfect sense. It's not usually discussed in elementary calculus texts because it's not necessary for talking about derivatives.

If f is a real-valued function defined on a subset E of R, then it makes sense to speak of the limit of f at a number a whenever a is a limit point of E. This means that every open interval containing a, no matter how small, contains a point of E other than a itself. (In fact, limits will still make sense if you drop the "other than a itself" part, in which case the notion is that of an adherent point.) It is not necessary that the set of points where f is undefined be countable, though this is a sufficient condition to ensure that every point is a limit point.

These kinds of limits will be addressed, at least implicitly, in any mathematical analysis textbook that discusses topology (namely, metric spaces or topological spaces). But this subsumes ordinary limits within a much larger theory and requires a degree of abstraction that goes far beyond what is needed for our purposes. The mathematical analysis textbook by Zorich discusses the more general concept of a limit we've been talking about well before it reaches topology. I can't give you a page number because I don't have the English version.96.46.204.126 (talk) 18:41, 24 February 2012 (UTC)[reply]

Haha, well if it's not in English then don't worry about it. I'm glad to hear my intuition was OK. Historical context: this all arose when I discovered a user "improved" l'Hopital's Rule by showing the g'(x) nonzero hypothesis was superfluous. All the analysis texts on my shelf used that hypothesis, so I was doubtful this person had outsmarted 20 generations of mathematicians. Just recently I found one reference which uses the same logic as that user. (First line of proof: "since lim f'/g' exists, g' does not vanish on a connected interval with endpoint a"). Is this not just a misconception? Maybe it's not so much of a misconception as an overstrong definition of limit. If one demands that in the definition of a limit that the function has to be defined on an interval adjacent to the limit point, then yes, the g'(x) nonzero condition is superfluous. Rschwieb (talk) 19:20, 24 February 2012 (UTC) Update: Checked 10 analysis books on a shelf. One of the authors (Kenneth Rogers) uses the proof I find questionable. S.G. Krantz assumes the two functions to be differentiable on an entire neighborhood including c, which I think probably does imply the g'(x) nonzero hypothesis. The other eight (including Rudin and Stewart) use the g'(x) nonzero hypothesis. Rschwieb (talk) 19:53, 24 February 2012 (UTC)[reply]

No, sorry, I meant that there is an English translation of it, but I don't have the translation, so I can't give you a page number in the English version. In Theorem 5.5.1 of the book you linked to, most mathematicians would include the hypothesis g′(x) ≠ 0 over some interval rather than relying on the technicalities of their particular definitions. It's likely that these technicalities are made clear wherever he defines limits. By the way, I can see no reason that the set of points at which g'(x) = 0 should be countable. 96.46.204.126 (talk) 19:59, 24 February 2012 (UTC)[reply]

I think that if you allow g′(x) to be zero for some x and choose to say nothing about what f′(x) does at those points, then L'Hopital's rule becomes false. For example, let f′ and g′ both be continuous, and let g′ be identically zero on infinitely many intervals approaching a, but nonzero elsewhere. Make f′ identical to g′ except on the intervals where g′ is zero, where you let f′ go crazy (while still tending to zero at a). You'll be able to pick f′ in such a way that L'Hopital's rule fails. (f and g will be the corresponding antiderivatives taking the value 0 at a. For example, f(x) will be the integral of f′ from a to x.) 96.46.204.126 (talk) 20:28, 24 February 2012 (UTC) [Correction: This may only work if you let g′ spend enough time being 0. 96.46.204.126 (talk) 20:37, 24 February 2012 (UTC)][reply]

I think some careful thought is needed for this one. Just because it is possible to define a limit over a subset of R doesn't mean that the limit over R is defined, and in l'Hôpital's rule we are working over R. If we allowed such an "excusing" (even on a countable subset), we would be able to find limits in contexts where they do not exist. I also think it is possible that the apparent leap in the proof (of theorem 5.5.2) from the existence of a limit to a the nonzero statement may be tricky enough and uninteresting enough that it simply has not tackled adequately in the literature. Surely if the nonzero requirement was both interesting and necessary, there should be a reference that shows that it is necessary? It is not at all clear to me that the leap is false. I would guess that it may be possible to show that the existence of a limit implies the existence of a neighborhood on which the function (of which the limit is being taken) is defined. — Quondum ^☏✎ 05:30, 25 February 2012 (UTC)[reply]

It pivots on what you define as "limit". If one insists that limited functions are defined on a connected interval approaching c, then by definition when you say lim f'/g' exists, you can infer g'(x) nonzero on said interval. This enhanced definition of limit implies the g'(x) nonzero hypothesis. If one uses the more general definition of existence of limit, then the example I gave on the l'Hopital's talk page shows that there exist functions f, g such that f/g has a limit 0 as x approaches 0, but f/g is undefined at many points in every interval (0,b). Thus in comparison, the g'(x) nonzero hypothesis is the more general of the two. Strengthening the definition of limit is only useful for simplicity. I do not recall ever hearing this enhanced definition in my career, so that's why I'm now wary of it. Rschwieb (talk) 19:08, 27 February 2012 (UTC)[reply]

And I agree with you in terms of what is to be presented; changes are new and non-standard. On a tangent, it is a little discomforting that there is no obviously robust rule for when l'Hôpital's rule applies; all of them seem to have twiddles to exclude pathological cases, or to be rather sensitive to details. But this is by the by; this kind of thing is also not too unusual I guess. — Quondum ^☏✎ 19:28, 27 February 2012 (UTC)[reply]

Products

A product can be expressed in terms of factorials;

\prod _{k=0}^{n-1}(N-k)={\frac {N!}{(N-n)!}}

Can a product of every other number be similarly expressed?

\prod _{k=0}^{n-1}(N-2k)

Thanks, Spinning Spark 19:00, 24 February 2012 (UTC)[reply]

If N is even, then we have something like

\prod _{k=0}^{n-1}(N-2k)={\frac {N!2^{N/2-n}(N/2-n)!}{(N-2n)!2^{N/2}(N/2)!}}.

A similar expression is available if N is odd. Sławomir Biały (talk) 19:33, 24 February 2012 (UTC)[reply]

Beautiful! <3 --COVIZAPIBETEFOKY (talk) 20:09, 24 February 2012 (UTC)[reply]

Thanks for that. That's what I was afraid of, it has to be divided into odd and even expressions. Spinning Spark 21:22, 24 February 2012 (UTC)[reply]

Actually, that does not work even for even N. For instance,

8 × 6 × 4 = 192

has N = 8 and n = 3. Feeding those values into the expression I make 105 which is incorrect. Spinning Spark 12:23, 25 February 2012 (UTC)[reply]

Sorry, I was giving the product of the odds rather than the evens. The correct expression is

\prod _{k=0}^{n-1}(N-2k)={\frac {2^{N/2}(N/2)!}{2^{N/2-n}(N/2-n)!}}

for N even. For N odd it's

\prod _{k=0}^{n-1}(N-2k)={\frac {N!2^{(N-1)/2-n}((N-1)/2-n)!}{(N-2n)!2^{(N-1)/2}((N-1)/2)!}}.

Sławomir Biały (talk) 12:43, 25 February 2012 (UTC)[reply]

Still not working for the even case. The test example above is not even returning an integer now. The odd case seems ok though, at least for N=9. Spinning Spark 14:47, 25 February 2012 (UTC)[reply]

misprint fixed. It should work now. Sławomir Biały (talk) 15:09, 25 February 2012 (UTC)[reply]

Simplified, the formula for even N is

\prod _{k=0}^{n-1}(N-2k)={\frac {2^{n}(N/2)!}{(N/2-n)!}}

.

If you want the same expression to work for both even and odd N, then you can use the Gamma function:

\prod _{k=0}^{n-1}(N-2k)={\frac {2^{n}\Gamma (N/2+1)}{\Gamma (N/2-n+1)}}

. 98.248.42.252 (talk) 15:27, 25 February 2012 (UTC)[reply]

Ah...gamma function. Thanks. Spinning Spark 16:14, 25 February 2012 (UTC)[reply]

You've got the formulas now, but this being the reference desk, let me point to a reference. Factorial#Double_factorial tells about the product of every other number starting from 1 or 2, and how to write this in close form using the factorial function. Dividing two of these gives you the product you want. – b_jonas 17:16, 25 February 2012 (UTC)[reply]

February 25

Integrating with the natural logarithm

In my textbook this example is given.
$\int {1 \over 4x-1}\,dx={1 \over 4}\int ({1 \over 4x-1})4\,dx={1 \over 4}\int {1 \over u}\,du={1 \over 4}\ln(\left\vert u\right\vert )+C={1 \over 4}\ln(\left\vert 4x-1\right\vert )+C$
Where does the 4 in the denominator of the fraction to the left of the integrand come from? If it had been $\int {1 \over 4x^{2}-1}\,dx$ would the same denominator be 8x on account of differentiation or something? --Melab±1 ☎ 21:33, 25 February 2012 (UTC)[reply]

The just multiplied and divided by 4, then used the linearity property of

\int

. The answer to your second question is "no". Widener (talk) 21:42, 25 February 2012 (UTC)[reply]

But, I don't understand why I multiply and divide by 4. I don't know where the 4 comes from. If the denominator of the fraction inside the integral was 5x^2+7x-1, what would I multiply and divide by? --Melab±1 ☎ 22:00, 25 February 2012 (UTC)[reply]

In rewriting the integrand in the form

\int {\dfrac {1}{u}}\,du

, your

du=4\,dx

, so

{\dfrac {1}{4}}\,du=dx

. --Kinu ^t/_c 23:24, 25 February 2012 (UTC)[reply]

... and to follow up on the second part of your question, the use of the log rule only works because the derivative of

u=4x-1

is expressible as a multiple of the

dx

present in the original integrand. In the second example, unless the integrand contains a multiple of

(10x+7)dx

(i.e., the derivative of

u=5x^{2}+7x-1

), this method wouldn't work. Chances are this isn't something that would be given in the section of the textbook you're looking at right now, of course. --Kinu ^t/_c 23:30, 25 February 2012 (UTC)[reply]

So if

du=2x\,dx

then I would multiply by

1 \over 2x

? --Melab±1 ☎ 00:41, 26 February 2012 (UTC)[reply]

No, because

\int {\frac {f(x^{2})}{2x}}2xdx\neq {\frac {1}{2x}}\int f(x^{2})2xdx

. To integrate

{\frac {1}{4x^{2}-1}}

, notice that

{\frac {1}{4x^{2}-1}}={\frac {1}{4x-2}}-{\frac {1}{4x+2}}

Widener (talk) 01:45, 26 February 2012 (UTC)[reply]

Still does not answer the question of how I get the denominator. --Melab±1 ☎ 01:47, 26 February 2012 (UTC)[reply]

Make the substitution

u=4x-1

. It then follows that

du=4dx

. Therefore, we need

4dx

in the integrand. We achieve this by multiplying and dividing the integrand by 4. That's where the 4 comes from. Widener (talk) 01:55, 26 February 2012 (UTC)[reply]

To avoid the appearance of the 4 without explanation, an alternative derivation is

\int {1 \over 4x-1}\,dx=\int {1 \over u}\,dx=\int {1 \over u}\,{du \over 4}={1 \over 4}\int {1 \over u}\,du\dots

where we use the fact that

du=4dx

to go from the second to the third expression. Gandalf61 (talk) 09:01, 26 February 2012 (UTC)[reply]

I am still confused, so I'll pose it more generally.

\int {1 \over 4x-1}\,dx={1 \over a}\int ({1 \over 4x-1})4\,dx={1 \over a}\int ({1 \over u})\,du

How do I get

a

? The problem I am working on is

\int {x \over x^{2}+1}\,dx

.

u=x^{2}+1

and

du=2x\,dx

so I have:

\int {x \over x^{2}+1}\,dx={1 \over 1}\int {(x \over x^{2}+1})2x\,dx=\int {x \over u}\,du

Now the book emphasizes

\int {u' \over u}\,dx=ln(\left\vert u\right\vert )+C

, so I how do I use it in the above. --Melab±1 ☎ 23:47, 26 February 2012 (UTC)[reply]

The equality

\int {1 \over 4x-1}\,dx={1 \over a}\int ({1 \over 4x-1})4\,dx

obviously holds only if

a=4

.
How did you figure out

\int {x \over x^{2}+1}\,dx={1 \over 1}\int {(x \over x^{2}+1})2xdx

? That's obviously false. Widener (talk) 01:47, 27 February 2012 (UTC)[reply]

\int {x \over x^{2}+1}\,dx={\frac {1}{2}}\int {2x \over x^{2}+1}dx

Notice that in the integrand the numerator is now the derivative of the denominator and you can use the theorem emphasized in your book. Widener (talk) 01:52, 27 February 2012 (UTC)[reply]

I think I get it now. It has to be the numerator in the integrand must be manipulated to make it the derivative of its denominator. --Melab±1 ☎ 22:42, 27 February 2012 (UTC)[reply]

You need to be careful to distinguish between which transformations are valid (because the two sides of the equality are indeed equal), and which of the valid transformations will make progress in solving a problem.

You can multiply and divide a quantity by the same number and it will not change. So

3={\frac {3}{7}}\cdot 7

and

3x^{2}-2x={\frac {3x^{2}-2x}{5}}\cdot 5

and

\int {\frac {1}{4x-1}}\ dx=\int {\frac {1}{4}}{\frac {1}{4x-1}}\cdot 4\ dx

.

It is also true in general that for a constant number a,

\int af(x)\ dx=a\int f(x)\ dx

, and so

\int {\frac {1}{4}}{\frac {1}{4x-1}}\cdot 4\ dx={\frac {1}{4}}\int {\frac {1}{4x-1}}\cdot 4\ dx

.

It is just as true to say that

\int {\frac {1}{4x-1}}\ dx={\frac {1}{3}}\int {\frac {1}{4x-1}}\cdot 3\ dx

, the difference is that the latter will not help you solve the problem. The former will, because if you substitute

u=4x-1

(the denominator), you'll have

4dx=du

.

So for similar problems with a linear denominator, you'll want to multiply and divide by the coefficient of x. But you cannot expect this to work for any kind of denominator - integration is a hard problem and there's no general solution, you need to apply transformations that are both true and bring you closer to your goal. In your

\int {x \over x^{2}+1}\,dx

example, both

\int {x \over x^{2}+1}\,dx={1 \over 1}\int {x \over x^{2}+1}2x\,dx

and

\int {x \over x^{2}+1}\,dx={1 \over 2x}\int {x \over x^{2}+1}2x\,dx

are of course completely false. You can use for example

\int {x \over x^{2}+1}\,dx={\frac {1}{2}}\int {\frac {1}{x^{2}+1}}2x\ dx={\frac {1}{2}}\int {\frac {1}{u}}\ du

, or you can do a partial fraction decomposition as suggested by Widener (which in this case would involve complex numbers). -- Meni Rosenfeld (talk) 06:23, 27 February 2012 (UTC)[reply]

February 26

relating entropy ratchet to stock market fluctuations

suppose that the ones and tens digit of the closing dow jones industrial average each day was a stochastic process. could you explain why it is impossible to use an entropy ratchet type thing to keep getting money out of it? (like maxwell's demon). is it because of broker's fees or is there a fundamental mathematical reason? --80.99.254.208 (talk) 14:57, 26 February 2012 (UTC)[reply]

you may use this, which I don't understand: http://en.wikipedia.org/wiki/Parrondo%27s_paradox . This is not homework! Also please bear in mind that the dow jones industrial average is not hte only average: there are a couple that we could combine as with the paradox mentioned. I would like a general proof or explanation of why this is not possible. Or is it just broker's fees. --80.99.254.208 (talk) 14:59, 26 February 2012 (UTC)[reply]

Could you explain how you would make money out of it? If you explain your algorithm, I expect we can find the flaw in it. In general terms, the market tends to price assets so that there are no arbitrage possibilities (that is, ways of making risk-free profit). If an arbitrage did exist, lots of investors would take advantage of it and that would move the prices until the arbitrage was no longer possible. --Tango (talk) 15:42, 26 February 2012 (UTC)[reply]

I would like you to supply the winning algorithm, please. (Or else a proof that it could not exist). --80.99.254.208 (talk) 09:12, 27 February 2012 (UTC)[reply]

February 27

Logarithms

I'm having trouble with this problem. I am pretty sure logs are used to solve it. 40^(3x) = 5^(2x+1). Any help would be appreciated. 70.12.91.235 (talk) 04:09, 27 February 2012 (UTC)[reply]

Take the logarithm of each side

3x\log 40=(2x+1)\log 5

Widener (talk) 04:43, 27 February 2012 (UTC)[reply]

Or rewrite the equation as (40³/5²)^x = 5. 96.46.204.126 (talk) 09:20, 27 February 2012 (UTC)[reply]

How to reverse XORs

m(0..79) are 80 binary bits

Given m(0..15) we calculate m(16..79) using

  m(i) = m(i-16) xor m(i-14) xor m(i-8) xor m(i-13)

Alternatively if given m(64..79) please explain how to calculate m(0..15).

84.209.89.214 (talk) 12:11, 27 February 2012 (UTC)[reply]

Try and get an equation with m(i-16) on the left instead of m(i) and work backwards. Dmcq (talk) 12:53, 27 February 2012 (UTC)[reply]

You may also need to know that bitwise xor is its own inverse. In other words, if a xor b = c then a = c xor b. Gandalf61 (talk) 13:29, 27 February 2012 (UTC)[reply]

U.S. income tax revenue

how much more revenue would be created if the income tax rate was increase back to 39.6 % this seems to be the rate people refer to when they talk about raising taxes ? — Preceding unsigned comment added by Poisonpaws (talk • contribs) 19:19, 27 February 2012 (UTC)[reply]

Diophantine equations

Given any Diophantine equation with no solutions, does there exist a proof that it has no solutions? Widener (talk) 21:46, 27 February 2012 (UTC)[reply]

We don't know for sure that a given Diophantine equation has no solutions unless we can prove it. If the question whether all Diophantine equations are known to have solutions or not, I would think the answer is "no". Consider that a rather famous family of Diophantine equations in Fermat's last theorem were only recently proved to have no solutions. I'm pretty sure you could find specific Diophantine equations or families that are conjectured to have no solutions, but as of yet have no proof. Rckrone (talk) 23:18, 27 February 2012 (UTC)[reply]

Is it possible that, given a Diophantine equation that has no solutions, there exists no proof that it has no solutions? A bit like the continuum hypothesis? Widener (talk) 23:32, 27 February 2012 (UTC)[reply]

If a Diophantine equation does have a solution, then there always exists a proof that it has a solution. Therefore, any proof that a Diophantine equation having solutions is undecidable would imply that there are no solutions, therefore it is not possible that the question of a Diophantine equation having solutions is provably undecidable. However, the question as to whether that problem is undecidable could itself be undecidable (and you could extend that back infinitely). Widener (talk) 23:35, 27 February 2012 (UTC)[reply]

This is mistaken. Unfortunately, you're confusing two different notions of "have solutions". One is "have solutions in the standard model", and one is "have solutions in any model". As noted below, Godel's incompleteness theorem can be used to explicitly construct such equations. — Preceding unsigned comment added by 121.74.97.164 (talk) 03:35, 28 February 2012 (UTC)[reply]

Our Diophantine set article discusses this question in some detail. Let me quote the last section of the article: "Matiyasevich's theorem has since been used to prove that many problems from calculus and differential equations are unsolvable. One can also derive the following stronger form of Gödel's first incompleteness theorem from Matiyasevich's result: Corresponding to any given consistent axiomatization of number theory,[5] one can explicitly construct a Diophantine equation which has no solutions, but such that this fact cannot be proved within the given axiomatization." Looie496 (talk) 00:11, 28 February 2012 (UTC)[reply]

Calculating the equation of a line tangent to a circle that passes through a particular point

What would be the generalized equations that can be used to find the $A$ , $B$ , and $C$ constants that describe the lines tangent to a circle that pass through a given point? The variables are as follows:

$P_{x}$ is the x coordinate of the point.
$P_{y}$ is the y coordinate of the point.
$C_{x}$ is the x coordinate of the circle.
$C_{y}$ is the y coordinate of the circle.
$r$ is the radius of the circle.

--Melab±1 ☎ 22:51, 27 February 2012 (UTC)[reply]

Let Q be the point of tangency, so Q-C is a radius. We have that |Q-C| = r², and that Q-C is perpendicular to the line, which goes in the direction of Q-P, so (Q-P).(Q-C) = 0. So then we can solve for Q = (Q_x,Q_y) using these constraints, namely (Q_x-C_x)² + (Q_y-C_y)² = r² and (Q_x-P_x)(Q_x-C_x) + (Q_y-P_y)(Q_y-C_y) = 0. Once you find Q, the equation for the line through P and Q can be found with Linear_equation#Two-point_form. Rckrone (talk) 23:33, 27 February 2012 (UTC)[reply]

I was thinking of expressing it like:

A=...

B=...

C=...

and then making

Ax+By+C=0

--Melab±1 ☎ 02:12, 28 February 2012 (UTC)[reply]

In the coordinate system where P=(0,0) and C=(1,0) the tangent equations are $y=\pm (r^{-2}-1)^{-{\frac {1}{2}}}x$ . Bo Jacoby (talk) 00:18, 28 February 2012 (UTC).[reply]

February 28

differentiation operator "d" and partial derivatives

I want to clarify something about the "operators" for total and partial derivatives.

Say I have the function

$Y=C(Y,i)+I(Y,i)$

and take the total derivative, "dY"

$dY={\frac {\partial C}{\partial Y}}dY+{\frac {\partial C}{\partial i}}di+{\frac {\partial I}{\partial Y}}dY+{\frac {\partial I}{\partial i}}di$

Not sure what I'm taking the derivative with respect to there, but anyway, this then gets rearranged

$dY(1-{\frac {\partial C}{\partial Y}}-{\frac {\partial I}{\partial Y}})=({\frac {\partial C}{\partial i}}+{\frac {\partial I}{\partial i}})di$

which results in

${\frac {\partial i}{\partial Y}}={\frac {1-\partial C/\partial Y-\partial I/\partial Y}{\partial C/\partial i+\partial I/\partial i}}$

I'm just not sure why they switch from $dY/di$ to using the partial derivative operator at the end. — Preceding unsigned comment added by Thorstein90 (talk • contribs) 03:39, 28 February 2012 (UTC)[reply]

Solutions to a transcendental equation

This might be a vague question, but can anything meaningful be said about the solutions to

A*sin(x)+B*cos(x)+C*sin(3*x)+D*cos(3*x)+E*sin(5*x)+F*cos(5*x) = 0 ?

Any conditions on A,B,C,D,E,F for real solution(s) to exist? Any clever ways to find a closed form solution? Thanks a ton! deeptrivia (talk) 04:23, 28 February 2012 (UTC)[reply]