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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]