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July 6

name of formula

What is the name of this formula by Leonhard Euler: e^it = cos t + i sin t ? Ugly bag of mostly water 03:59, 6 July 2007 (UTC)[reply]

I think it's one of many that are named Euler's formula. I haven't heard it named anything else. Confusing Manifestation 04:07, 6 July 2007 (UTC)[reply]

Suggested interpretations of images

There is a collection of images created by Hamedog that should maybe be converted to SVG and moved to the Commons (perhaps in the opposite order), if only we can figure out what they are meant to illustrate. One such image (Image:1 d xinter.png), which illustrates the x-intercepts of a cubic polynomial, is already on the Commons, but the other ones, shown below, are somewhat more of a mystery.

B
C

C
D

D
E

E

It seems that image C is illustrating the effects of scaling a polynomial by a constant factor. Any ideas about what the others represent, and whether they could be of any use if uploaded to the Commons? —Bkell (talk) 04:25, 6 July 2007 (UTC)[reply]

E is the shifting of (x-1)x(x+1) through the transformation x -> x - 1. Let's see what else we can pull out. Confusing Manifestation 06:44, 6 July 2007 (UTC)[reply]

It's a pity he failed at drawing arrows - Capuchin 07:12, 6 July 2007 (UTC)[reply]

It's hard to say for sure what he was going for, but he does seem in general to be displaying characteristics of various cubics. In A, the two curves are very symmetric to each other. The cryptic remark about "two, one" is unhelpful, but it is the case that they intersect at (1,0) and (0,2), that the blue curve is the purple curve translated by (-5/3,-50/27), and that the blue curve is the purple curve rotated 180 degrees about the point (1/2,-2). Similar points apply to B, and it's possible he was trying to show something in connection with that. I agree that C appears to be pointing out both the growth and the decline of a cubic on scaling, and would like to add that D appears to show that a curve that starts out below can end up above, for the same reason. I agree with CM about E, and with Capuchin regarding his artistic skill. Black Carrot 11:36, 11 July 2007 (UTC)[reply]

Bring Radicals

I have two questions. One's on the talk page of that article, if anyone wants to have a shot at it. The other is, where online can I find out more about them? I searched Google of course, but almost every single site that mentioned them had a verbatim copy of our current article instead of original content. Black Carrot 05:08, 6 July 2007 (UTC)[reply]

Additional Mathematics Project Work 2007

Question: 1 (C) A, B and C decides to open an account of RM 10000 for three years without withdrawal. A keeps the amount at an interest rate of 2.5% per annum for a duration of 1 month renewable at the end of each month.

I have completed the question this far: 102.5/100*10000=10250 But I do not understand the phrase "for a duration of 1 month renewable at the end of each month".

Can anyone please help me? —Preceding unsigned comment added by 60.50.173.111 (talk • contribs)

I think the point of the question is that the interest is paid and compounded monthly. At the end of one month the amount in the account is

10,000 + 10,000 x 0.025 / 12 = 10,020.83

This sum now earns interest during the second month, so the amount in the account at the end of the second month is

10,020.83 + 10,020.83 x 0.025 / 12 = 10,041.71

... and at the end of month 3 the amount in the account is

10,041,71 + 10,041.71 x 0.025 / 12 = 10,062.63

You need to find the amount in the account at the end of 3 years. You can carry on working it out month by month for 36 months - or you can look for a shortcut. Gandalf61 09:12, 6 July 2007 (UTC)[reply]

Multidimensional limit

Can somebody help me with this limit? (x^2+y^2)/(x+y) when x+y=0 and (x^2+y^2)/(x+y) when (x,y)->(0,0). Is the second case already analyzed with the first one (x+y=0)? Thanks. --Taraborn 13:36, 6 July 2007 (UTC)[reply]

When x+y = 0, the value of (x²+y²)/(x+y) is undefined. Therefore it is not a limit, and it is unclear what it means to ask for its limit (if that is the question of he first part). The second part is more clear, but should properly be stated as follows: "What is the limit of (x²+y²)/(x+y) as (x,y) → (0,0)?" I assume that you know the "multidimensional" definition of limit of a function. If the limit of f(x,y) as (x,y) → (0,0) exists, the value of f(x,y) should be everywhere defined in a neighbourhood of (0,0) with the possible exception of the point (0,0) itself. However, arbitrarily close to (0,0) there are points (x,y) such that x+y = 0, so there is a little problem . --Lambiam ^Talk 16:27, 6 July 2007 (UTC)[reply]

I'm thinking that there's an equals sign missing from the first part of the problem. That's the only way the problem might make sense. Donald Hosek 17:29, 6 July 2007 (UTC)[reply]

I don't see a spot where inserting an equals sign makes the first part more plausible, but perhaps you are thinking of changing x+y=0 to x=y=0. With the same mod applied to my answer, it would equally stand. --Lambiam ^Talk 19:31, 6 July 2007 (UTC)[reply]

I was thinking of doing something along the lines of defining

f(x,y)=0

when

x+y=0

then finding

\lim _{(x,y)\to (0,0)}f(x,y)=(x^{2}+y^{2})/(x+y)

. I think that's sufficient for the limit to be defined, but it's all idle speculation without the original poster's clarifying the problem. Donald Hosek 20:13, 6 July 2007 (UTC)[reply]

Is that so? Take L to be any value other than zero, and let (x,y) approach zero along the parabolic curve y = –x + 2x²/L. --Lambiam ^Talk 20:46, 6 July 2007 (UTC)[reply]

Errrrr... Well, I have a function which is defined this way: (x^2+y^2)/(x+y) when x+y!=0 and 0 when x+y=0. I am asked to discuss the continuity of the function, and, for this purpose, I'll have to calculate a limit, isn't it? --Taraborn 21:06, 6 July 2007 (UTC)[reply]

Lambian, I was just speculating about the problem that was offered (although my surmise was correct about the missing part, at least), I didn't actually do any of the math to see what happens with that condition. Taraborn, discussing continuity includes the possibility that there is discontinuity. Donald Hosek 22:45, 6 July 2007 (UTC)[reply]

Taraborn, discussing continuity includes the possibility that there is discontinuity.... was that an insult? I'm not an idiot, and I'm quite aware of that. By saying "calculating a limit" I considered either giving its value or proving its nonexistence. I don't think the problem is that hard to understand, seriously. I tried doing the limit when x->(-y) (remember the condition x+y=0) and got infinity, but I'm by no means sure of that, that's why I'm asking for help. Thanks. --Taraborn 08:17, 7 July 2007 (UTC)[reply]

I don't have any reason to think you're more of an idiot than the rest of us; I also don't know what you are aware of. So forgive me if I state familiar facts.

Without regard to dimension, we can take a continuous function and replace the value at one point with something arbitary. Thus we should not assume that a value and a limit coincide.
A principal feature in the multidimensional case is the additional freedom in how we approach a point. In the one-dimensional case, we can only approach from above or below, though these can already give different answers. In your plane example, we can approach along many different paths. What we hope for is the limit does not depend on the approach.
Another important difference can be illustrated with polynomials. A polynomial in x alone, such as x²−4, has zeros at isolated points. But a polynomial in x and y, such as x²+y²−4, has zeros of a more interesting character (here, a circle). Thus a ratio of polynomials, such as y/(x²+y²−4), can have a geometrically interesting locus of points where it blows up.

The likely point of the exercise is for you to discover and explore these facts in a specific case. --KSmrq^T 10:58, 7 July 2007 (UTC)[reply]

GODDAMN. I think I'm wasting my time here, well, forget about my question. --Taraborn 16:35, 7 July 2007 (UTC)[reply]

It did not help that you managed to completely mangle up the question as you initially posed it. If you study my reply above to Donald Hosek, you'll find it contains the answer. But the issue is: DO YOUR OWN HOMEWORK. If you are too lazy for that, do not waste our time by posting your homework questions here. --Lambiam ^Talk 08:09, 8 July 2007 (UTC)[reply]

I guess calling people lazy, stupid and wasting MY time with your incompetence goes against WP:CIVIL and that assume good faith thing, I guess. It's not a homework question, you Einstein, since I rarely have homework on holidays. If you could read, because at this point I'm really doubting so, you'll see that mr. Donald has only made conjectures about the nature of the problem, not about the solution. You all were just wasting my time, because it's clear that you all have no idea about how to solve this problem. You were pointing obviousnesses such as "discussing continuity includes the possibility that there is discontinuity." or and trying to figure out what I was asking like here "When x+y = 0, the value of (x²+y²)/(x+y) is undefined. Therefore it is not a limit, and it is unclear what it means to ask for its limit (if that is the question of he first part)." (then sin(x)/x when x->0 isn't a limit either by your ""logic""), which was pretty clear from the beginning, if you have an idea of the topic, of course. Next time, try to answer to questions that, at the very least, you can understand, thank you. It's not that hard to understand that function has problems when x+y=0, and studying what happens when x+y APPROACHES 0 isn't that hard to understand either. I guess you need a course in basic vector calculus urgently. --Taraborn 08:28, 8 July 2007 (UTC)[reply]

Cool down a bit, will you? If you're not a native English speaker you may perhaps not realize how rude and offensive your use of the interjection Goddamn is. If the limit exists at the origin, you will get the same limit along any curve approaching it. I wrote above: "Take L to be any value other than zero, and let (x,y) approach zero along the parabolic curve y = –x + 2x²/L." This settles the question. --Lambiam ^Talk 09:36, 8 July 2007 (UTC)[reply]

I've repeated this a thousand times, so I don't think one thousand and one will make any difference. Your idea will prove the nonexistence of the limit at the origin, not along the straight line y=-x. --Taraborn 14:05, 8 July 2007 (UTC)[reply]

You wrote: "I am asked to discuss the continuity of the function". Obviously, the function is not continuous at those points (x,y) for which x+y = 0, except possibly at the origin. You did not state what the points of interest are, but it is so obvious that the function is not continuous at, for example, (1,–1), that the idea did not cross my mind that this might be an issue. For the origin, I have supplied a solution. Maybe I have miscounted, but the number of times I can find that you have stated, suggested or intimated before that you are interested in taking the limit along the straight line y=-x is exactly 0 (zero, zilch, niente, nada), which is a bit less than one thousand. But here goes. On that straight line (at least according to what you wrote, namely "I have a function which is ... 0 when x+y=0"), the function value is everywhere 0. It is a constant function. Usually, constant functions are rather continuous, and this one is no exception. For a continuous function, to find a limit, you can just take the function value. In this case, it is 0. I have used some algebra, in particular the fact that y = –x implies x+y = 0. --Lambiam ^Talk 19:41, 8 July 2007 (UTC)[reply]

Taraborn, mathematicians like it when the statements and notation are nice and tidy. You have used notation which, to my knowledge, is highly nonstandard (if at all sensible), which is the source of much conufsion. What I understand is that you ask not what happens when you move along the line

y=-x

(as Lambiam has now interpreted), but rather when

x+y

approaches 0, or more accurately, when we are approaching a point on the line

y=-x

. As Lambiam has also remarked, the function is clearly discontinuous at those points. Finally, it's not nice to lash out at people only because you have not succeeded in describing the problem correctly. -- Meni Rosenfeld (talk) 20:14, 8 July 2007 (UTC)[reply]

(re-tab and reply to everybody). Ok. Maybe I'm misreading this, but it seems both Taraborn and Lambiam are getting snippy and making assumptions. I call a small time-out.

Lambiam provided a solution to problem point at the origin. OK.
Of course the function f(x,y) = (x^2 + y^2)/(x+y) is discontinuous at x=-y. It seems the function Taraborn is asking about actually is. $g(x,y)={\begin{cases}{\frac {x^{2}+y^{2}}{x+y}},&{\mbox{if }}y\neq -x\\0,&{\mbox{if }}y=-x\end{cases}}$

the function

Yes, it is. --Lambiam ^Talk 20:55, 8 July 2007 (UTC)[reply]

Lets take 1,-1 then. evaluating points like g(1.001,-0.999)=1000.001 and g(0.999,-1.001)=-1000.001 gives evidence that all along y=-x, (except x=y=0) the function is divergent or asymptotic (if either term makes sense for functions on planes), which means that setting g(x,y) = 0 for y=-x does not glue any surfaces together. The key is proving this with absolute certainty.

Root⁴(one) 20:34, 8 July 2007 (UTC)[reply]

Okay, this means I got the problem right. My apologies to everyone that might have been offended by my tone, especially Lambiam. Sorry for not making the problem clear enough, I thought it definitely was but, as Meni Rosenfeld pointed out, I'm not a mathematician and I can't speak "in your language" :). Thank you for your help and your time. --Taraborn 21:57, 8 July 2007 (UTC)[reply]

I moved the tabs over for these next two messages as they are not related to my above post. Root⁴(one) 20:39, 8 July 2007 (UTC)[reply]

Since the original asker has abandoned the question, I'll go ahead and ask my related question, only for the sake of my curiosity. What do you call the concept of finding the limit points of a multidimensional function as the function of points on a curve or a set of curves defined parametrically on the x,y plane that either cross or "ends up" at the point, like say y=t^3,x=t or the Golden spiral? Does this concept have a name? I know once you've defined the curve parametrically, it essentially becomes the problem of finding the limit of a function of a single variable.... but it still seems interesting and noteworthy enough to have its own name, or at least a less wordy description. Root⁴(one) 02:28, 8 July 2007 (UTC)[reply]

Well, okay, a multidimensional limit normally has to cover approach from all directions. However, you could take the limit along a particular plane, like x=0 or y=0 or x=y or something like that; it seems this would be a compact way to describe it. "Limit as (x,y) approaches (0,0) for f(x,y) where x=y"? Now, I don't really see the mathematical significance of a limit that is the same from both axis-aligned planes (i.e. x=0 and y=0), so I can't say that I think there's a particular name for it. It does, however, remind me of the gradient operation where you take the derivatives with respect to x and y separately in order to find the components of a vector representing that function's change (though again, if the function isn't actually continuous from all directions, the gradient doesn't really exist). - Rainwarrior 05:31, 8 July 2007 (UTC)[reply]

Integration question

Hi, all. I've got a bit stuck on an exercise in my A2 further maths book and I'm hoping someone can help.

The question is:

By expressing $\cos ^{2n}\theta$ in terms of cosines of multiple angles, prove that

$\int _{0}^{\pi }\cos ^{2n}\theta d\theta ={\frac {(2n)!\pi }{2^{2n}(n!)^{2}}}$

This is set in the context of de Moivre's theorem, and preceding questions have taken forms such as "Express $\cos ^{4}\theta$ in terms of multiple angles." The idea is to use $\cos ^{n}\theta =\left({\frac {e^{i\theta }+e^{-i\theta }}{2}}\right)^{n}$ , expanding the fraction according to the binomial theorem, collecting terms with powers of equal magnitude, then using $\cos n\theta =\left({\frac {e^{in\theta }+e^{-in\theta }}{2}}\right)$ to end up with terms like $\cos 2\theta$ in the answer.

However, I'm not sure how to proceed with the expansion in the case of a variable, 2n, rather that a constant.

I guess the 2n means that the fact it's even is significant, but I can't really see how. The factorials in the answer given seem to suggest the binomial coefficients come in to play there, but, again, I can't quite see how I can get an expression for $\cos ^{2n}\theta$ . I know that I can express the expansion of $\left({\frac {e^{i\theta }+e^{-i\theta }}{2}}\right)^{n}$ using sigma notation, but since there's no sigma in the RHS of the equation in the question, I'm a bit puzzled as to how to proceed. Thanks for any help you can give. Seth Bresnett 20:58, 6 July 2007 (UTC)[reply]

Remember that

\int \sum f_{n}(x)=\sum \int f_{n}(x)

I think that might help. Donald Hosek 22:40, 6 July 2007 (UTC)[reply]

One possible reason (I haven't worked it out) that there might be no "sigma" in the right side is the following: recall that

\sum _{i=1}^{n}C=nC

if C is a constant. So if every term of the sum works out to the same thing, you could collect all the terms together to remove the sigma. More generally, perhaps the sum can be simplified to remove the explicit sum (even if the terms are not all constant). Tesseran 06:57, 7 July 2007 (UTC)[reply]

Let's start with the case n=2:

\cos ^{4}\theta =\left({\frac {e^{i\theta }+e^{-i\theta }}{2}}\right)^{4}

=\left({\frac {1}{2^{4}}}\right)(e^{4i\theta }+4e^{2i\theta }+6+4e^{-2i\theta }+e^{-4i\theta })

=\left({\frac {1}{2^{3}}}\right)\left(\left({\frac {e^{4i\theta }+e^{-4i\theta }}{2}}\right)+4\left({\frac {e^{2i\theta }+e^{-2i\theta }}{2}}\right)+3\right)

=\left({\frac {1}{2^{3}}}\right)(\cos(4\theta )+4\cos(2\theta )+3)

Then integrate from 0 to π. The cos parts of the integral are all 0 (can you see why ?) and you are just left with 3π/8. Now generalise this to any value of n. When you have done that you can think about why 2n is used in the question - what do you get if you integrate an odd power of cos from 0 to π ? Gandalf61 08:31, 7 July 2007 (UTC)[reply]